Class 11

Punjab State Board Syllabus PSEB 11th Class Environmental Education Book Solutions Guide Pdf in English Medium & Punjabi Medium & Hindi Medium are part of PSEB Solutions for Class 11.

PSEB 11th Class Environmental Education Guide | Environmental Education Guide for Class 11 PSEB in English Medium

PSEB 11th Class Environmental Education Book Solutions in Punjabi Medium

• Chapter 1 ਵਾਤਾਵਰਣ (Environment)
• Chapter 2 ਵਸੋਂ ਅਤੇ ਵਾਤਾਵਰਣ (Population and Environment)
• Chapter 3 ਮਨੁੱਖੀ ਗਤੀਵਿਧੀਆਂ ਦਾ ਵਾਤਾਵਰਣ ਉੱਪਰ ਪ੍ਰਭਾਵ (Impact of Human Activities on Environment)
• Chapter 4 ਆਰਥਿਕ ਅਤੇ ਸਮਾਜਿਕ ਵਿਕਾਸ (Economic and Social Development)
• Chapter 5 ਉਦਾਰੀਕਰਨ ਅਤੇ ਵਿਸ਼ਵੀਕਰਨ ਦਾ ਪ੍ਰਭਾਵ (Impact of Liberalization and Globalization)
• Chapter 6 ਵਿਕਾਸ ਅਤੇ ਵਾਤਾਵਰਣ ਵਿਚ ਸਮਾਜ ਦੀ ਭੂਮਿਕਾ (Role of Society in Development and Environment)
• Chapter 7 ਵਾਤਾਵਰਣਿਕ ਪ੍ਰਦੂਸ਼ਣ (Environmental Pollution)
• Chapter 8 ਪ੍ਰਦੂਸ਼ਣ ਅਤੇ ਰੋਗ (Pollution and Diseases)
• Chapter 9 ਵਿਸ਼ਵ ਵਿਆਪੀ ਮੁੱਦੇ ਅਤੇ ਵਾਤਾਵਰਣ ਦਾ ਸੁਧਾਰ (Global Issues and Improvement of Environment)
• Chapter 10 ਆਫ਼ਤਾਂ (Disaster)
• Chapter 11 ਊਰਜਾ ਦੀ ਖ਼ਪਤ (Energy Consumption)
• Chapter 12 ਉਰਜਾ ਦੇ ਰਵਾਇਤੀ ਸ੍ਰੋਤ (Conventional Sources of Energy)
• Chapter 13 ਊਰਜਾ ਦੇ ਗੈਰ ਰਵਾਇਤੀ ਸੋਤ (Non-Conventional Sources of Energy)
• Chapter 14 ਊਰਜਾ ਦਾ ਸੁਰੱਖਿਅਣ (Conservation of Energy)
• Chapter 15 ਸੁਰੱਖਿਅਤ ਕੰਮ ਕਾਜੀ ਵਾਤਾਵਰਣ (Safe Work Environment)
• Chapter 16 ਸੁਰੱਖਿਆ ਕਾਨੂੰਨ, ਦੁਰਘਟਨਾਵਾਂ ਅਤੇ ਮੁੱਢਲੀ ਸਹਾਇਤਾ (Safety Laws, Accidents and First Aid)
• Chapter 17 ਨਸ਼ਾ-ਮਾੜੇ ਪ੍ਰਭਾਵ (Drugs-Ill Effects-I)

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 12 Organic Chemistry: Some Basic Principles and Techniques Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

Very Short Answer Type Questions

Question 1.
How many σ bonds and π bonds are present in the second member of the alkene series?
Answer:
The second member of the alkene series is propene. The structual formula of the propene is

Question 2.
Show the polarisation of carbon-magnesium bond in the following structure.
CH₃ —CH₂ —CH₂ —CH₂ —Mg —X
Answer:
Carbon is more electronegative than magnesium. Therefore, Mg acquires a partial positive charge and carbon acquires a partial negative charge.

Question 3.
What are primary and secondary suffixes as applied to IUPAC nomenclature?
Answer:
The primary suffix indicates whether the carbon chain is saturated or unsaturated while the secondary suffix indicates the functional group present in the molecule.

Question 4.
Draw all position isomers of an alcohol with molecular formula, C3HgO.
Answer:

Question 5.
CH₂ = CH^– is more basic than HC = C^– . Explain why?
Answer:

Since, sp-carbon is more electronegative than sp²-carbon, therefore, CH ≡ C^– is less willing to donate a pair of electrons than H₂C = CH^–. In other words, H₂C = CH^– is more basic than HC ≡ C^–.

Question 6.
Why does SO₃ act as an electrophile? [NCERT Exemplar]
Answer:
In SO₃, three highly electronegative oxygen atoms are attached to sulphur atom. It makes sulphur atom electron deficient. Further, due to resonance, sulphur acquires a positive charge. Both these factors, make SO₃ an electrophile.

Question 7.
How will you separate a mixture of o-nitro phenol and p-nitrophenol?
Answer:
A mixture of o-nitrophenol and p-nitrophenol can be separated by steam distillation, o-nitrophenol being less volatile distils over along with water while p-nitrophenol being non-volatile remains in the flask.

Question 8.
In DNA and RNA, nitrogen atom is present in the ring system. Can Kjeldahl’s method be used for the estimation of nitrogen present in these? Give reason.
Answer:
DNA and RNA have nitrogen in the heterocyclic rings. Nitrogen present in rings, azo and nitro groups cannot be converted into (NH₄)₂SO₄. That’s why Kjeldahl’s method cannot be used for the estimation of nitrogen present in these.

Question 9.
Lassaigne’s test is not shown by diazonium salts, though they contain nitrogen. Why?
Answer:
Diazonium salts (C₆H₅N₂⁺X^–) readily lose N₂ on heating before reacting with fused sodium metal. Therefore, these salts do not give positive Lassaigne’s test for nitrogen.

Question 10.
Write three-dimensional wedge-dashed or wedge-line representations for the following:
(a) CH₃CH₂OH
(b) CH₂FCl
Answer:

Short Answer Type Questions

Question 1.
Draw all polygon formula for the molecular formula C₅H₁₀.
Answer:
The different polygon formula of the compound having molecular formula C₅H₁₀ are :

Question 2.
An alkane has a molecular mass of 72. Draw all its possible chain isomers and write their IUPAC names.
Answer:
First of all, we will derive the molecular formula.
Molecular formula of alkane is C_nH_2n+2
∴ Molecular mass = 72
∴ 12n + 2n + 2 = 72
n = 5
The alkane is C₅H₁₂. The possible chain isomers are

Question 3.
Arrange the following

Answer:

Question 4.
Suggest a method to purify
(i) camphor containing traces of common salt.
(ii) kerosene oil containing water.
(iii) a liquid which decomposes at its boiling point.
Answer:
(i) Sublimation-camphor sublimes while common salt remains as residue in the China dish.
(ii) Since the two liquids are immiscible, the technique of solvent extraction with a separating funnel is used. The mixture is thoroughly shaken and the separating funnel is allowed to stand. Kerosene being lighter than water forms the upper layer while water forms the lower layer.
The lower water layer is run off using the stop cork of the funnel and kerosene oil is obtained. It is dried over anhydrous CaCl₂ or MgSO₄ and then distilled to give pure kerosene oil.
(iii) Distillation under reduced pressure.

Question 5.
The structure of triphenylmethyl cation is given below. This is very stable and some of its salts can he stored for months. Explain the cause of high stability of this cation.

Answer:
In triphenylmethyl cation, due to resonance, the positive charge can move at both the o-and p-position of each benzene ring. This is illustrated below.

Since, there are three benzene rings, hence, there are, in all, nine resonance structures.
Thus, triphenylmethyl cation is highly stable due to these nine resonance structures.

Long Answer Type Questions

Question 1.
Consider structures I to VII and answer the following question (i) to (iv).


(i) Which of the above compounds form pairs of metamers?
(ii) Identify the pairs of compounds which are functional group isomers.
(iii) Identify the pairs of compounds that represent position isomerism.
(iv) Identify the pairs of compounds that represent chain isomerism.
Answer:
(i) V and VI or VI and VII form a pair of metamers since they differ in the number of carbon atoms on the either side of the functional group, i.e., O-atom.
(ii) I and V, I and VI, I and VII; II and V, II and VI, II and VII; III and V, III and VI; III and VII; IV and V; IV and VI and IV and VII are all functional group isomers.
(iii) I and II, III and IV and, VI and VII represent position isomerism.
(iv) I and III, I and IV, II and III and II and IV represent chain isomerism.

Question 2.
Write structural formulae for all the isomeric amines with molecular formula C^^N.
Answer:

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 11 The p-Block Elements Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 11 The p-Block Elements Important Questions

Very Short Answer Type Questions

Question 1.
What is the formula of kemite, an ore of boron?
Answer:
Formula of kernite, Na₂[B₄O₅(OH)₄] or Na₂B₄O₇.2H₂O

Question 2.
Complete the following chemical equations :
Z + 3LiAlH₄ → X + 3LiF + 3AlF₃
X + 6H₂O → Y + 6H₂
X + 3O₂ → B₂O₃ + 3H₂O
Answer:

Question 3.
Tl(NO₃)₃ acts as an oxidising agent. Explain.
Answer:
Due to inert pair effect, Tl in +1 oxidation state is more stable than that of +3 oxidation state. Therefore, Tl(NO₃)₃ acts as an oxidising agent and readily reduced to TlNO₃.

Question 4.
Why is B—X bond distance in BX₃ shorter than the theoretically expected value?
Answer:
This is due to pπ—pπ back bonding of the completely filled p-orbital of halogen X into the empty p-orbital of boron.

Question 5.
Are all the B—H bonds in diborane equivalent?
Answer:
No, there are two types of bonds in diborane two electron normal bonds and three centred two electron bonds.

Question 6.
Name the member of group 14 that forms the most acidic oxide.
Answer:
Among monoxides, CO is neutral and GeO is acidic while among dioxides, CO₂, SiO₂ are acidic, GeO₂ is also acidic but less acidic than Si02.

Question 7.
Silicones are used for making waterproof fabrics. Give reason.
Answer:
Silicones are synthetic polymers containing repeated units of R₂SiO where R is alkyl group.
Therefore, these are water repellants i.e., do not absorb water and are used for making waterproof fabrics.

Question 8.
Atomic radius of gallium (135 pm) is less than that of aluminium (143 pm).
Answer: It is due to poor shielding effect of 3d-electrons due to which effective nuclear charge increases in Ga, therefore, it is smaller than Al.

Question 9.
AlF₃ is high melting solid but AlCl₃ is low melting. Explain.
Answer: AlF3 is high melting solid because it is ionic in nature. On the other hand, A1C13 is covalent in nature and hence is a low melting solid.

Question 10.
How will you prepare an ahuninosilicate?
Answer:
Aluminosilicate is prepared by substituting some of the Si atoms in the three dimensional network of SiO₂ by Al atoms.

Short Answer Type Questions

Question 1.
Like CO, why its analog of SiO is not stable?
Answer:
CO is a resonance hybrid of the following two structures :

Thus, CO contains pπ-pπ multiple bonds. This is due to the reason that carbon has a strong tendency to form pπ—pπ multiple bonds due to its small size and high electronegativity. Silicon, on the other hand, because of its _ bigger size and lower electronegativity has no tendency to form pπ—pπ multiple bonds and hence, Si does not form SiO molecule analogous to CO molecule.

Question 2.
Account for the following:
(i) Graphite is used as lubricant.
(ii) Diamond is used as an abrasive.
Answer:
(i) Graphite has layered structure. Layers are held together by weak van der Waals’ forces and hence can be made to slip over one another. Therefore, graphite acts as a dry lubricant.
(ii) In diamond, each sp3 hybridised carbon atom is linked to four other carbon atoms. It has three-dimensional network of carbon atoms. It is very difficult to break extended covalent bonding and therefore diamond is a hardest substance on the earth. That’s why it is used as an abrasive.

Question 3.
Which one is more soluble in diethyl ether, anhydrous AlCl₃ or hydrated AlCl₃? Explain in terms of bonding.
Answer:
Anhydrous AlCl₃ is an electron-deficient compound while hydrated AlCl₃ is not. Therefore, anhydrous A1C13 is more soluble in diethyl ether because the oxygen atom of ether donates a pair of electrons to the vacant p-orbital on the Al atom in AlCl₃ forming a coordinate bond.

In case of hydrated AlCl₃, Al is not electron deficient since H₂O has already donated a pair of electrons to it.

Question 4.
BCl₃ is trigonal planar while AlCl₃ is tetrahedral in dimeric state. Explain.
Or
BCl₃ exists as monomer whereas AlCl₃ is dimerised through halogen bridging. Give reason. Explain the structure of the dimer of AlCl₃ also.
Answer:
Both BCl₃ and AlCl₃ are electron deficient molecules having six electrons in the valence shell of their respective central atoms. To complete their octets, the central atom in each case can accept a pair of electrons from the chlorine atom of another molecule forming dimeric structures. However, because of small size of B, it cannot accommodate four big sized Cl atoms around it. Therefore, BCl₃ prefers to exist as a monomeric planar molecule in which B atom is sp² -hybridised.

On the other hand, Al because of its bigger size can easily accommodate four Cl atoms around it. As a result, AlCl₃ exists as a dimer. In this dimer, since the covalency of Al has increased to 4.

Therefore, Al is sp³-hybridised and the four Cl atoms are held tetrahedrally around it.

Question 5.
Three pairs of compounds are given below. Identify that compound in each of the pairs which has group 13 element in more stable oxidation state. Give reason for your choice. State the nature of bonding also.
(i) TICl₃, TlCl
(ii) AlCl₃, AlCl
(iii) InCl₃, InCl
Answer:
(i) Due to strong inert pair effect, +1 oxidation state of T1 is more stable than +3. Since, compounds in lower oxidation state are ionic but covalent in higher oxidation state, therefore, TlCl₃ is less stable and covalent in nature but TlCl is more stable and is ionic in nature.

(ii) Due to absence of d-orbitals, Al does not show inert pair effect. Therefore, it is most stable than A1C1. Further, in the solid or the vapour state, AlCl₃ is covalent in nature but in aqueous solution, it ionises to form Al³⁺ (aq) and Cl^–(aq) ions.
(iii) Due to inert pair effect, indium exists in both +1 and +3 oxidation states, out of which +3 oxidation state is more stable than +1 oxidation
state. In other words, InCl₃ is more stable than InCl. Being unstable, InCl undergoes disproportionation reactions.
3InCl(aq) → 2In(s) + In³⁺(aq) + 3Cl^– (aq)

Long Answer Type Questions

Question 1.
(i) What are silicones? State the uses of silicones.
(ii) What are boranes? Give chemical equation for the preparation of diborane.
Answer:
(i) Silicones are a group of organosilicon polymers, which have (R2SiO) as a repeating unit. These may be linear silicones, cyclic silicones and cross-linked silicones. These are prepared by the hydrolysis of alkyl or aryl derivatives of SiCl₄ like RSiCl₃, R₂SiCl₂ and R₃SiCl and polymerisation of alkyl or aryl hydroxy derivatives obtained by hydrolysis.

Uses : These are used as sealant, greases, electrical insulators and for water proofing of fabrics. These are also used in surgical and cosmetic plants.
(ii) Boron forms a number of covalent hydrides with general formulae B_nH_n+4 and B_nH_n+6. These are called boranes. B₂H₆ and B₄H₁₀ are the representative compounds of the two series respectively.

Preparation of diborane : It is prepared by treating boron trifluoride with LiAlH4 in diethyl ether.
4BF₃ + 3LiAlH₄ → 2B₂H₆ + 3LiF + 3AlF₃
On industrial scale it is prepared by the reaction of BF₃ with sodium hydride.

Question 2.
Account for the following observations :
(i) AlCl₃ is a Lewis acid.
(ii) Though fluorine is more electronegative than chlorine yet BF₃ is weaker Lewis acid than BCl₃.
(iii) PbO₂ is stronger oxidising agent than SnO₂.
(iv) The +1 oxidation state of thallium is more stable than its +3 state.
(i) In AlCl₃, Al has only six electrons in its valence shell. It is an electron deficient species. Therefore, it acts as a Lewis acid (electron acceptor).
(ii) In BF₃, boron has a vacant 2p-orbital and fluorine has one 2p completely filled unutilized orbital. Both of these orbitals belong to same energy level therefore, they can overlap effectively and form pπ—pπ bond. This type of bond formation is known as back bonding. While back bonding is not possible in BCl₃ because there is no effective overlapping between the 2p-orbital of boron and 3p-orbital of chlorine. Therefore, electron deficiency of B is higher in BCl₃ than that of BF₃. That’s why BF₃ is a weaker Lewis acid than BCl₃.

(iii) Pb⁴⁺ is less stable thanPb²⁺, due to inert pair effect therefore, Pb⁴⁺ salts act as strong oxidising agents. Sn²⁺ is also less stable than Sn⁴⁺, thus Sn⁴⁺ can also act as an oxidising agent. But Pb⁴⁺ is a stronger oxidising agent than Sn⁴⁺ because inert pair effect increases down die group.
(iv) Tl⁺ is more stable than Tl³⁺ because of inert pair effect.

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 11 The p-Block Elements Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 11 The p-Block Elements

PSEB 11th Class Chemistry Guide The p-Block Elements InText Questions and Answers

Question 1.
Discuss the pattern of variation in the oxidation states of (i) B to T1 (ii) C to Pb.
Answer:

Boron and aluminium show an oxidation state of+3 only because they do not exhibit inert pair effect due to the absence of d- or f-electrons. Elements from Ga to T1 show two oxidation states, i.e., +1 and +3. The tendency to show +1 oxidation state increases down the group due to the inability of ns² electrons of valence shell to participate in bonding which is called inert pair effect. Therefore, Tl⁺ is more stable than Tl³⁺.

Carbon and silicon show an oxidation state of +4 only. In heavier members the tendency to show +2 oxidation state increases in the sequence Ge < Sn < Pb. It is due to the inability of ns² electrons of valence shell to participate in bonding (inert pair effect). Ge forms stable compounds in +4 state and only few compounds in +2 state. Sn forms compounds in both the oxidation states and lead compounds in +2 state is more stable than +4 oxidation state.

Question 2.
How can you explain higher stability of BC13 as compared to T1C13?
Answer:
Boron and thallium belong to group 13 of the periodic table. In this group, the +1 oxidation state becomes more stable on moving down the , group. BCl₃ is more stable than TlCl₃ because the +3 oxidation state of B is more stable than the +3 oxidation state of Tl. In Tl, the +3 state is highly oxidising and it reverts back to the more stable +1 state.

Question 3.
Why does boron trifluoride behave as a Lewis acid?
Answer:
BF₃ being electron deficient is a strong Lewis acid. It reacts with Lewis bases easily to complete the octet around boron.

Question 4.
Consider the compounds, BCl₃ and CCl₄. How will they behave with water? Justify.
Answer:
Being a Lewis acid, BCI3 readily undergoes hydrolysis. As a result, Boric ‘ acid is formed.
BCl₃ + 3H₂O → 3HCl + B(OH)₃
CCl₄ completely resists hydrolysis. Carbon does not have any vacant orbital. Hence, it cannot accept electrons from water to form an intermediate. When CCl₄ and water are mixed, they form separate layers.
CCl₄ + H₂O → No reaction.

Question 5.
Is boric acid a protic acid? Explain.
Answer:
Boric acid is not a protic acid. It is a weak monobasic acid, behaving as a Lewis acid.
B(OH)₃ + 2H₂O → [B (OH)₄]^– + H₃O₊
It behaves as an acid by accepting a pair of electrons from OH- ion.

Question 6.
Explain what happens when boric acid is heated.
Answer:
On heating boric acid (H₃BO₃) at 370 K or above, it changes to metaboric acid (HBO₂). On further heating, this yields boric oxide B₂O₃

Question 7.
Describe the shapes of BF₃ and \(\mathbf{B H}_{\mathbf{4}}^{-}\). Assign the hybridisation of boron in these species.
Answer:
(i) BF₃ : As a result of its small size and high electronegativity, boron tends to form monomeric covalent halides. These halides have a planar triangular geometry. This triangular shape is formed by the overlap of three sp² hybridised orbitals of boron with the sp-orbitals of three halogen atoms. Boron is sp² hybridised in BF₃.

(ii) \(\mathbf{B H}_{\mathbf{4}}^{-}\) : Boron-hydride ion (\(\mathbf{B H}_{\mathbf{4}}^{-}\)) is formed by the sp3 hybridisation of boron orbitals. Therefore, it is tetrahedral in structure.

Question 8.
Write reactions to justify amphoteric nature of aluminium.
Answer:
A substance is called amphoteric if it displays characteristics of both acids and bases. Aluminium dissolves in both acids and bases, showing amphoteric behaviour. ..
(i) 2Al(s) + 6HCl(aq) → 2Al³⁺(aq) + 6Cl^–(aq) + 3H₂(g)
(ii) 2Al(s) + 2NaOH(aq) + 6H₂O(Z) → 2Na⁺[Al(OH)₄]^–(aq) + 3H₂(g)

Question 9.
What are electron deficient compounds? Are BCI3 and SiCl4 electron deficient species? Explain.
Answer:
Electron deficient compounds are those in which the octet of all the atoms is not complete i.e., all the element present in the compound do not have 8e^– in their outer shell.
In trivalent state, the number of electrons around the central atom B in BCl₃ is six.

Such molecules have a tendency to accept a pair of electrons to achieve stability and hence, behave as Lewis acids.
In SiCl₄, the number of electrons around the central atom Si is eight so, it is electron precise molecule.

Question 10.
Write the resonance structures of \(\mathrm{CO}_{3}^{2-}\) and \(\mathrm{HCO}_{3}^{-}\).
Answer:

There are only two resonating structures for the bicarbonate ion.

Question 11.
What is the state of hybridisation of carbon in
(a) \(\mathrm{CO}_{3}^{2-}\)
(b) diamond
(c) graphite?
Answer:
(a) \(\mathrm{CO}_{3}^{2-}\) : C in \(\mathrm{CO}_{3}^{2-}\) is sp² hybridised and is bonded to three oxygen atoms.
(b) Diamond : Each carbon in diamond is sp³ hybridised and is bound to four other carbon atoms.
(c) Graphite : Each carbon atom in graphite is sp² hybridised and is bound to three other carbon atoms.

Question 12.
Explain the difference in properties of diamond and graphite on the basis of their structures.
Answer:

In diamond C is sp³ hybridised. Each C is tetrahedrally linked to four neighbouring carbon atoms through strong C – G sp³ – sp³ σ- bonds. The structure is highly rigid and this network extends to three dimensions. Whereas in graphite C is sp² hybridised. Each C is linked to three other C atoms forming hexagonal rings. Thus unlike diamond, graphite has a two-dimensional sheet like (layered) structure consisting of a number of benzene rings fused together. The various sheets or layers are held together by weak van der Waals forces. Based on the above structural differences between diamond and graphite, they can be summed up as follows:

Diamond:

1. It has a crystalline lattice.
2. In diamond, each carbon atom is sp³ hybridised and is bonded to four other carbon atoms through a o- bond.
3. It is made up of tetrahedral units.
4. The C—C bond length in diamond is 154 pm.
5. It has a rigid covalent bond network which is difficult to break.
6. It acts as an electrical insulator.

Graphite:

1. It has a layered structure.
2. In graphite, each carbon atom is sp² hybridised and is bonded to three other carbon atoms through a o- bond. The fourth electron forms a n bond.
3. It has a planar geometry.
4. The C—C bond length in graphite is 141.5 pm.
5. It is quite soft and its layers can be separated easily.
6. It is a good conductor of electricity.

Question 13.
Rationalise the given statements and give chemical reactions:
(a) Lead (II) chloride reacts with Cl₂ to give PbCl₄.
(b) Lead (IV) chloride is highly unstable towards heat.
(c) Lead is known not to form an iodide, Pbl₄.
Answer:
(a) Lead belongs to group 14 of the periodic table. The two oxidation states displayed by this group is +2 and +4. On moving down the group the +2 oxidation state’ becomes more stable and the +4 oxidation state becomes less stable.
This is because of the inert pair effect. Hence, PbCl4 is much less stable than PbCl₂. However, the formation of PbCl₄ takes place when chlorine gas is bubbled through a saturated solution of PbCl₂.
PbCl₂(s) + Cl₂(g) → PbCl₄(l)

(b) On moving down group IV, the higher oxidation state becomes unstable because of the inert pair effect. Pb(IV) is highly unstable and when heated it reduces to Pb(II)

(c) Lead is known not to form Pbl4. Pb (+4) is oxidising in nature and I^– is reducing in nature. A combination of Pb (IV) and iodide ion is not stable. Iodide ion is strongly reducing in nature. Pb(IV) oxidises I^– to I₂ and itself gets reduced to Pb (II)
PbI₄ → PbI₂ + I₂

Question 14.
Suggest reasons why the B-F bond lengths in \(\mathbf{B F}_{3}^{-}\) (130 pm) and \(\mathbf{B F}_{4}^{-}\) (143 pm) differ.
Answer:
The B—F bond length in \(\mathbf{B F}_{3}^{-}\) is shorter than the B—F bond length in \(\mathbf{B F}_{4}^{-}\). \(\mathbf{B F}_{3}^{-}\) is an electron-deficient species. With a vacant p-orbital on boron, the fluorine and boron atoms undergo pπ—pπ back bonding to remove this deficiency. This imparts a double-bond character to the B-F bond.

The double bond character causes the bond length to shorten in \(\mathbf{B F}_{3}^{-}\) (130 pm). However, when \(\mathbf{B F}_{3}^{-}\) coordinates with the fluoride ion, a change in hybridisation from sp² in (\(\mathbf{B F}_{3}^{-}\)) tp sp³ (in \(\mathbf{B F}_{4}^{-}\)) occurs.
Boron now forms 4σ-bonds and the double bond character is lost. This accounts for a B—F bond length of 143 pm in \(\mathbf{B F}_{4}^{-}\) ion.

Question 15.
If B—Cl bond has a dipole moment, explain why BC13 molecule has zero dipole moment.
Answer:
As a result of the difference in the electronegativities of B and Cl, the B—Cl bond is polar in nature. However the BC13 molecule is non-polar. This is because BCl₃ is trigonal planar in shape. It is a symmetrical molecule. Hence, the respective dipole-moments of the B-Cl bond cancel each other, thereby causing a zero-dipole moment.

Question 16.
Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous BF3 is bubbled through. Give reasons.
Answer:
Hydrogen fluoride (HF) is a covalent compound and has a very strong intermolecular hydrogen-bonding. Thus, it does not provide ions and aluminium fluoride (AlF) does not dissolve in it. Sodium fluoride (NaF) is an ionic compound and when it is added to the mixture, AlF dissolves. This is because of the availability of free F^–. The reaction involved in the process is:

When boron trifluoride (BF₃) is added to the solution, aluminium fluoride precipitates out of the solution. This happens because the tendency of boron to form complexes is much more than that of aluminium. Therefore, when BF₃ is added to the solution, B replaces Al from the complexes according to the following reaction :

Question 17.
Suggest a reason why CO is poisonous?
Answer:
Carbon monoxide is highly-poisonous because of its ability to form a complex with haemoglobin. The CO—Hb complex is more stable than the O₂—Hb complex. The former prevents Hb from binding with oxygen. Thus, a person dies because of suffocation on not receiving oxygen. It is found that the CO-Hb complex is about 300 times more stable than the O₂ – Hb complex.

Question 18.
How is excessive content of CO₂ responsible for global warming?
Answer:
Carbon dioxide is a very essential gas for our survival. However, an increased content of CO₂ in the atmosphere posses a serious threat. An increment in the combustion of fossil fuels, decomposition of limestone, and a decrease in the number of trees had led to greater levels of carbon dioxide. Carbon dioxide has the property of trapping the heat provided by sunrays. Higher the level of carbon dioxide, higher is the amount of heat trapped. This results in an increase in the atmospheric temperature, thereby causing global warming.

Question 19.
Explain structures of diborane and boric acid.
Answer:
(a) Diborane : B₂H₆ is an electron-deficient compound. B2H6 has only
12 electrons-6e~ from 6 H atoms and 3e^– each from 2 B atoms. Thus, after combining with 3 H atoms, none of the boron atoms has any electrons left. X-ray diffraction studies have shown the structure of diborane as:

H_t = terminal hydrogen
H_b = bridging hydrogen

2 boron and 4 terminal hydrogen atoms (H_t) lie in one plane, while the other two bridging hydrogen atoms (H_b) lie in a plane perpendicular to the plane of boron atoms. Again of the two bridging hydrogen atoms, one H atom lies above the plane and the other lies below the plane. The terminal bonds are regular two-centre two-electron (2c – 2e^–) bonds while the two bridging (B—H—B) bonds are three centre two electron (3c – 2e^–) bonds,

(b) Boric acid : Boric acid has a layered structure. Each planar BO₃ unit is linked to one another through H atoms. The H atoms form a covalent bond with a BO₃ unit, while a hydrogen bond is formed with another BO₃ unit. In the given figure, the dotted lines represent hydrogen bonds.

Question 20.
What happens when
(a) Borax is heated strongly,
(b) Boric acid is added to water,
(c) Aluminium is treated with dilute NaOH,
(d) BF3 is reacted with ammonia?
Answer:
(a) When heated, borax undergoes various transitions. It first loses water molecules and swells. Then, it turns into a transparent liquid, solidifying to form a glass-like material called borax bead.

(b) When boric acid is added to water, it accepts electrons from OH- ions.
B(OH)₃ +2HOH → [B(OH)₄]^– + H₃O⁺

(c) A1 reacts with dilute NaOH to form sodium tetrahydroxoaluminate (III). Hydrogen gas is liberated in the process.
Al(s) + 2NaOH(aq) + 6H₂O(I) → 2Na⁺Al(OH)₄(aq) + 3H₂(g)

(d) BF₃ (a Lewis acid) reacts with NH₃ (a Lewis base) to form an adduct. This results in a complete octet around B in BF₃.
F₃B + :NH₃ → F₃B ← NH₃

Question 21.
Explain the following reactions
(a) Silicon is heated with methyl chloride at high temperature in the presence of copper.
(h) Silicon dioxide is treated with hydrogen fluoride.
(c) CO is heated with ZnO.
(d) Hydrated alumina is treated with aqueous NaOH solution.
Answer:
(a) When silicon reacts with methyl chloride in the presence of copper (catalyst) and at a temperature of about 537 K, a class of organosilicon polymers called methyl-substituted chlorosilanes (MeSiCl3, Me₂SiCl₂, Me₃SiCl, andMe₄Si) are formed.

(b) When silicon dioxide (SiO₂) is heated with hydrogen fluoride (HF), it forms silicon tetrafluoride (SiF₄). Usually, the Si-0 bond is a strong bond and it resists any attack by halogens and most acids, even at a high temperature. However, it is attacked by HF.
SiO₂ + 4HF → SiF₄ + 2H₂O
The SiF₄ formed in this reaction can further react with HF to form hydrofluorosilicic acid.
SiF₄ + 2HF → H₂SiF₆

(c) When CO reacts with ZnO, it reduces ZnO to Zn. CO acts as a reducing agent.

(d) When hydrated alumina is added to sodium hydroxide, the former dissolves in the latter because of the formation of sodium meta-aluminate.
Al₂O₃ . 2H₂O + 2NaOH → 2NaAlO₂ + 3H₂O

Question 22.
Give reasons:
(i) Cone. HNO₃ can be transported in aluminium container,
(ii) A mixture of dilute NaOH and aluminium pieces is used to open drain.
(iii) Graphite is used as lubricant.
(iv) Diamond is used as an abrasive.
(v) Aluminium alloys are used to make aircraft body.
(vi) Aluminium utensils should not be kept in water overnight.
(vii) Aluminium wire is used to make transmission cables.
Answer:
(i) Concentrated HNO₃ can be stored and transported in aluminium containers as it reacts with aluminium to form a thin protective oxide layer on the aluminium surface. This oxide layer renders aluminium passive.
(ii) Sodium hydroxide and aluminium react to form sodium tetrahydroxoaluminate(III) and hydrogen gas. The pressure of the produced hydrogen gas is used to open blocked drains.
2A1 + 2NaOH + 6H₂O → 2Na⁺ [Al(OH)₄]^– + 3H₂
(iii) Graphite has a layered structure and different layers of graphite are bonded to each other by weak van der Waals’ forces. These layers can slide over each other. Graphite is soft and slippery. Therefore, graphite can be used as a lubricant.
(iv) In diamond, carbon is sp³ hybridised. Each carbon atom is bonded to four other carbon atoms with the help of strong covalent bonds. These covalent bonds are present throughout the surface, giving it a very rigid 3-D structure. It is very difficult to break this extended covalent bonding and for this reason, diamond is the hardest substance known. Thus, it is used as an abrasive and for cutting tools.
(v) Aluminium has a high tensile strength and is very light in weight. It can also be alloyed with various metals such as Cu, Mn, Mg, Si and Zn. It is very malleable and ductile. Therefore, it is used in making aircaft bodies.
(vi) The oxygen present in water reacts with aluminium to form a thin layer of aluminium oxide. This layer prevents aluminium from further reaction. However, when water is kept in an aluminium vessel for long period of time, some amount of aluminium oxide may dissolve in water. As aluminium ions are harmful, water should not be stored in aluminium vessels overnight.
(vii) Silver, copper, and aluminium are among the best conductors of – electricity. Silver is an expensive metal and silver wires are very expensive. Copper is quite expensive and is also very heavy. Aluminium is a very ductile metal. Thus, aluminium is used in making wires for electrical conduction.

Question 23.
Explain why is there a phenomenal decrease in ionisation enthalpy from carbon to silicon?
Answer: Electronic configuration of C is = 1s², 2s², 2p²
Electronic configuration of Si = 1s², 2s²2p⁶, 3s²3p²

As we proceed from C to Si in group 14, there is a sharp increase in covalent radius from 77 pm for C to 118 pm for silicon. Due to the increase in the size of the atom, there is a sharp fall in the value of ionisation enthalpy from carbon to silicon.

Question 24.
How would you explain the lower atomic radius of Ga as compared to Al?
Answer: Electronic configuration of Al and Ga are as ₁₃Al = 1s² 2s² 2p⁶ 3s² 3p¹ ;
₃₁Ga = 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p¹

The screening tendency of d-electrons is poor. Thus, On moving from Al to Ga, shielding effect of 10 d-electrons is unable to compensate increased nuclear charge. Therefore, atomic radius of Ga is smaller than that of aluminium due to effective nuclear charge.

Question 25.
What are allotropes? Sketch the structure of two allotropes of carbon namely diamond and graphite. What is the impact of structure on physical properties of two allotropes?
Answer:
The phenomenon of existence of an element in two or more forms which differ in physical properties but have almost same chemical nature is known as allotropy and the different forms of the element are known as allotropes.

Crystalline carbon occurs mainly in two allotropic forms (i) graphite and (ii) diamond. A third allotropic form of carbon called fullerene was discovered in 1985 by H.W. Kroto, E. Smalley and R.F. Curl.

In diamond, each carbon is sp³ hybridised and is linked to other four atoms tetrahedrally. There is three dimensional network of carbon atoms in diamond. In graphite, each carbon is sp² hybridised and makes three sigma bonds with three neighbouring carbon atoms. It has layered structure and the layers are held by weak van der Waals’ forces.

Impact of the structures of diamond and graphite on physical properties of the two allotropes are as follows :

1. Diamond because of its h. iness, is used as an abrasive and in making dyes while graphite is soft that it marks paper and it is used as a dry lubricant in machines.
2. Diamond does not conduct electricity whereas graphite is a good conductor of electricity because of the presence of one free electron in each carbon atom.
3. Diamond is transparent while graphite is opaque.

Question 26.
(a) Classify following oxides as neutral, acidic, basic or amphoteric:
CO, B₂O₃, SiO₂, CO₂, Al₂O₃, PbO₂, Tl₂O₃
(b) Write suitable chemical equations to show their nature. Answer: (a) CO is neutral oxide.
B₂O₃, SiO₂, CO₂ are acidic oxides.
PbO₂ Al₂O₃ are amphoteric oxides.
Tl₂O₃ is basic oxide.

(b) (i) Being acidic B₂O₃, SiO₂ and CO₂ react with alkalis to form salts.
B₂O₃ + CuO → CU(BO₂)₂
SiO₂ + 2NaOH → Na₂SiO₃ + H₂O
CO₂ + H₂O → H₂CO₃

(ii) Being amphoteric PbO₂ and Al₂O₃ react with both acids and basses.
PbO₂ + 4HCl → PbCl₄ + 2H₂O
PbO₂ + 2NaOH → Na₂PbO₃ + H₂O
Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O
Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O

(iii) Being basic Tl₂O₃ react with acid
Tl₂O₃ + 3H₂SO₄ → Tl₂(SO₄)₃ + 3H₂O

Question 27.
In some of the reactions thallium resembles aluminium, whereas in others it resembles with group I metals. Support this statement by giving some evidences.
Answer:
Thallium belongs to group 13 of the periodic table. The most common oxidation state for this group is +3. However, heavier members of this group also display the +1 oxidation state. This happens because of the inert pair effect. Aluminium displays the +3 oxidation state and alkali metals display the +1 oxidation state. Thallium displays both the oxidation states. Therefore, it resembles both aluminium and alkali metals.
Thallium, like aluminium, forms compounds such as TlCl₃ and Tl₂O₃. It resembles alkali metals in compounds Tl₂0 and TlCl.

Question 28.
When metal X is treated with sodium hydroxide, a white precipitate (A) is obtained, which is soluble in excess of NaOH to give soluble complex (B). Compound (A) is soluble in dilute HC1 to form compound (C). The compound (A) when heated strongly gives (D), which is used to extract metal. Identify (X), (A), (B), (C) and (D). Write suitable equations to support their identities.
Answer:
The given metal X gives a white precipitate with sodium hydroxide and the precipitate dissolves in excess of sodium hydroxide. Hence, X must be aluminium.

The white precipitate (compound A) obtained is aluminium hydroxide. When an excess of the base is added the compound B formed is sodium tetrahydroxoaluminate (III).

Now, when dilute hydrochloric acid is added to aluminium hydroxide, aluminium chloride (compound C) is obtained.

Also, when compound A is heated strongly, it gives compound D. This compound is used to extract metal X. Aluminium metal is extracted from alumina. Hence, compound D must be alumina.

Question 29.
What do you understand by (a) inert pair effect (b) allotropy (c) catenation?
Answer:
(a) Inert pair effect : As one moves down the group, the tendency of s-block electrons to participate in chemical bonding decreases. This effect is known as inert pair effect. In case of group 13 elements, the electronic configuration is ns² np¹ and their group valency is +3. However, on moving down the group, the +1 oxidation state becomes more stable. This happens because of the poor shielding of the ns² electrons by the d-and f-electrons. As a result of the poor shielding, the ns² electrons are held tightly by the nucleus and so, they cannot participate in chemical bonding.

(b) Allotropy : Allotropy is the existence of an element in more than one form, having the same chemical properties but different physical properties. The various forms of an element are called allotropes. For example, carbon exists in three allotropic forms: diamond, graphite, and fullerenes.

(c) Catenation : The atoms of some elements (such as carbon) can link with one another through strong covalent bonds to form long chains or branches. This property is known as catenation. It is most common in carbon and quite significant in Si and S. *

Question 30.
A certain salt X, gives the following results :
(i) Its aqueous solution is alkaline to litmus.
(ii) It swells up to a glassy material Y on strong heating.
(iii) When cone. H₂SO₄ is added to a hot solution of X, white crystal of an acid Z separates out.
Write equations for all the above reactions and identify X, Y and z.
Answer:
(i) Aqueous solution of salt X is alkaline. It indicates that ‘X’ is the salt of a strong base and a weak acid.
(ii) On strong heating, the salt ‘X’ swells up to a glassy material Y. It indicates that the salt ‘X’ is borax.
(iii) Hot aqueous solution of borax on reaction with cone. H₂SO₄ gives crystals of orthoboric acid.
The equations for the reactions involved in the question are as follows :

Question 31.
Write balanced equations for:
(i) BF₃ + LiH
(ii) B₂H₆ + H₂O
(iii) NaH + B₂H₆
(iv) im
(v) Al + NaOH
(vi) B₂H₆ + NH₃
Answer:

Question 32.
Give one method for industrial preparation and one for laboratory preparation of CO and CO₂ each.
Answer:
Carbon monoxide : In the laboratory, CO is prepared by the dehydration of formic acid with cone. H2S04 at 373 K. The reaction involved is as follows :

CO is commercially prepared by passing steam over hot coke. The reaction involved is as follows:

Carbon dioxide : In the laboratory, CO₂ can be prepared by the action of dilute hydrochloric acid on calcium carbonate. The reaction involved is as follows:
CaCO₃ + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)

CO₂ is commercially prepared by heating limestone. The reaction involved is as follows:

Question 33.
An aqueous solution of borax is
(a) neutral
(b) amphoteric
(c) basic
(d) acidic
Answer:
(c) Borax is a salt of a strong base (NaOH) and weak acid (H₃BO₃). It is, therefore, basic in nature.

Question 34.
Boric acid is polymeric due to
(a) its acidic nature
(b) the presence of hydrogen bonds
(c) its monobasic nature
(d) its geometry
Answer:
(b) Boric acid is polymeric because of the presence of hydrogen bonds, (as it has polar O—H bonds)

Question 35.
The type of hybridisation of boron in diborane is (a) sp (b) sp² (c) sp³ (d) dsp²
Answer:
(c) Boron in diborane is sp3 hybridised.

Question 36.
Thermodynamically the most stable form of carbon is
(a) diamond
(b) graphite
(c) fullerenes
(d) coal
Answer:
(b) Graphite is thermodynamically the most stable form of carbon.

Question 37.
Elements of group 14
(a) exhibit oxidation state of +4 only
(b) exhibit oxidation state of +2 and +4
(c) formM^2- and M⁴⁺ ions
(d) formM²⁺ and M⁴⁺ ions
Answer:
(b) The elements of group 14 have 4 valence electrons. Therefore, the oxidation state of the group is +4. However, as a result of the inert pair effect, the lower oxidation state becomes more and more stable and the higher oxidation state becomes less stable. Therefore, this group exhibits +4 and +2 oxidation states.

Question 38.
If the starting material for the manufacture of silicones is RSiCl3, write the structure of the product formed.
Answer:
Hydrolysis of alkyltrichlorosilanes followed by condensation polymerisation gives cross-linked silicones.

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 8 Redox Reactions Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 8 Redox Reactions

PSEB 11th Class Chemistry Guide Redox Reactions InText Questions and Answers

Question 1.
Assign oxidation numbers to the underlined elements in each of the following species:

Answer:
(a) Let the oxidation number of P be x.
Oxidation number of Na = +1
Oxidation number of H = +1
Oxidation number of O = – 2
In neutral compounds, the sum of the oxidation numbers of all the atoms is zero.
1 (+1) + 2 (+1) +1 00 + 4 (-2) = 0
1 + 2 + x – 8 = 0
3 + x + (-8) – 0
x = 8 – 3
⇒ x = + 5
Hence, the oxidation number of P is +5.

(b) Let the oxidation number of S be x.

1 (+1) +1 (+1) +1 (x) + 4 (-2) = 0
⇒ 1 + 1 + x – 8 = 0
⇒ x = + 6
Hence, the oxidation number of S is +6.

(c) Let the oxidation number of P be x.

4 (+1) + 2 (x) + 7 (-2) = 0
⇒ 4 + 2x – 14 = 0
⇒ 2x = +10
⇒ x = + 5
Hence, the oxidation number of P is + 5.

(d) Let the oxidation number of Mn is x.

2 (+1) + x + 4 (-2) = 0
⇒ 2 + x — 8 = 0
⇒ x = + 6
Hence, the oxidation number of Mn is + 6.

(e) Let the oxidation number of O be x.

1 (+ 2) + 2 (x) = 0
⇒ 2 + 2x = 0
⇒ x = – 1
Hence, the oxidation number of O is -1.

(f) Let the oxidation number of B be x.

1 (+1)+1 (x) + 4 (-1) = 0
⇒ 1 + x – 4 = 0
⇒ x = + 3
Hence, the oxidation number of B is + 3.

(g) Let the oxidation number of S is x.

2 (+1) + 2 (x) + 7 (-2) = 0
⇒ 2 + 2x -14 = 0
⇒ 2x = +12
x = +6
Hence, the oxidation number of S is + 6.

(h) Let the oxidation number of S be x.

1 (+1) +1 (+ 3) + 2 (x) + 8 (-2) +12 (2 x 1 + (-2)) = 0
⇒ 1 + 3 + 2x -16 + 24 – 24 = 0
⇒ 2x = 12
x = + 6
Hence, the oxidation number of S is + 6.

Question 2.
What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?


Answer:
(a) In

the oxidation number (O. N.) of K is +1. Hence, the average oxidation number of I is \(\frac{-1}{3}\). However, O.N. cannot be fractional.
Therefore, we will have to consider the structure of KI₃ to find the oxidation states.
In a KI₃ molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.

Hence, in a KI₃ molecule, the O.N. of the two I atoms forming the I₂ molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is -1.

(b) Let the oxidation number of S be x.

2 (+1) + 4 (x) + 6 (-2) = 0
=» 2 + 4x – 12 = 0
⇒ 4x = +10
⇒ x = + 2 \(\frac{1}{2}\)
However, O.N. cannot be fractional. Hence S must be present in different oxidation states in the molecule.

The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0.

(c) Let the oxidation number of Fe be x.

3(x) + 4(-2) = 0
3x – 8 = 0
x = \(+\frac{8}{3}\)
However O.N. cannot’be fractional. Here, one of the three Fe atoms exhibits the O.N. of +2 and the other two Fe atoms exhibit the O.N. of+3

(d) Let oxidation number of C be x.

2 (x) + 6 (+1) + 1 (-2) = 0
2x + 4 = 0
x = -2
Hence, the O.N. of C is – 2

(e) Let the oxidation number of C be x.

2 (x) + 4 (+1) + 2 (-2) = 0
2x = 0
x = 0
Therefore, the average oxidation number of C is zero.
Let us consider the structure of CH₃COOH

Oxidation number of atom = 1(+1) + x+1(-2) +1(-1) = 0
x = +2
Similarly, oxidation number of C₂ atom
3(+1) + x+ 1(-1) = 0
x = -2.

Question 3.
Justify that the following reactions are redox reactions:
(a) CuO (s) + H₂(g) → Cu(s) + H₂O(g)
(b) Fe₂O₃ (s) + 3CO(g) → 2Fe(s) + 3CO₂(g)
(c) 4BCl₃ (g) + 3LiAlH₄ (s) → 2B₂H₆(g) + 3LiCl (s) + 3 AlCl₃(S)
(d) 2K (s) + F₂ (g) → 2K⁺F^–(s)
(e) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O (g)
Answer:
(a) CuO (s) + H₂(g) → Cu(s) + H₂O(g)
Let us write the oxidation number of each element involved in the given reaction as :

Here, the oxidation number of Cu decreases from +2 in CuO to 0 in Cu i.e., CuO is reduced to Cu. Also, the oxidation number of H increases from 0 in H₂ to +1 in H₂O i.e., H₂ is oxidized to H₂O. Hence, this reaction is a redox reaction.

(b) Fe₂O₃ (s) + 3CO(g) → 2Fe(s) + 3CO₂(g)
Let us write the oxidation number of each element in the given reaction as:

Here, the oxidation number of Fe decreases from +3 in Fe₂O₃ to 0 in Fe i.e., Fe₂O₃ is reduced to Fe. On the other hand, the oxidation number of C increases from +2 in CO to +4 in CO₂ i.e., CO is oxidized to CO₂.
Hence, the given reaction is a redox reaction.

(c) 4BCl₃ (g) + 3LiAlH₄ (s) → 2B₂H₆(g) + 3LiCl (s) + 3 AlCl₃(S)
The oxidation number of each elements in the given reaction can be represented as:

In this reaction, the oxidation number of B decreases from +3 in BCl₃ to -3 in B₂H_6- i.e., BCl₃ is reduced to B₂H₆. Also, the oxidation number of H increases from -1 in LiAlH₄ to +1 in B₂H₆ i.e., LiAlH₄ is oxidized to B₂H₆. Hence, the given reaction is a redox reaction.

(d) 2K (s) + F₂ (g) → 2K⁺F^–(s)
The oxidation number of each element in the given reaction can be represented as:

In this reaction, the oxidation number of K increases from 0 in K to +1 in KF i.e., K is oxidized to KF. On the other hand, the oxidation number of F decreases from 0 in F₂ to -1 in KF i.e., F₂ is reduced to KF.
Hence, the above reaction is a redox reaction.

(e) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O (g)
The oxidation number of each elements in the given reaction can be represented as:

Here, the oxidation number of N increases from -3 in NH₃ to +2 in NO. On the other hand, the oxidation number of O₂ decreases from 0 in O₂ to -2 in NO and H₂O i.e.,O₂ is reduced. Hence, the given reaction is a redox reaction.

Question 4.
Fluorine reacts with ice and results in the change:
H₂O(S) + F₂(g) → HF(g) + HOF(g)
Justify that this reaction is a redox reaction.
Ans. Let us write the oxidation number of each atom involved in the given reaction above its symbol as:

Here, we have observed that the oxidation number of F increases from 0 in F₂ to +1 in HOF. Also, the oxidation number decreases from 0 in F₂ to -1 in HF. Thus, in the above reaction, F is both oxidized and reduced. Hence, the given reaction is a redox reaction.

Question 5.
Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) and \(\mathrm{NO}_{3}^{-}\). Suggest structure of these compounds. Count for the fallacy.
Answer:

2 (+1) +1 (x) + 5 (-2) = 0
⇒ 2 + x -10 = 0
⇒ x = + 8
However, the O.N. of S cannot be +8. S has six valence electrons. Therefore, the O.N. of S cannot be more than +6. The structure of H2S05 is shown as follows :

Now, 2 (+1) +1 (x) + 3 (-2) + 2 (-1) – 0
⇒ 2 + x – 6 – 2 = 0
⇒ x = + 6
Therefore, the O.N. of S is +6.

2 (x) + 7 (-2) = – 2 ⇒ 2x -14 = – 2
⇒ x = + 6
The structure of Cr2Oy is shown as follows:

Let the oxidation number of each Cr atom be
4(-2) + (-2) +1(-2) + 2x = 0
– 8 – 2 – 2 + 2x = 0
2x = +12
x = +6

Oxidation number of Cr in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) is same.
Hence, there is no fallacy.

1 (x) + 3(-2) = -1
= x – 6 = -1
x = +5
The structure of \(\mathrm{NO}_{3}^{-}\) is shown as follows:

Let the oxidation number of N be x.
1(-1) + x + 1(-2) + 1(-2) = 0
x = + 5
Oxidation number of N in \(\mathrm{NO}_{3}^{-}\) ion is same.
Hence, there is no fallacy.

Question 6.
Write formulas for the following compounds:
(a) Mercury (II) chloride
(b) Nickel (II) sulphate
(c) Tin (IV) oxide
(d) Thallium (I) sulphate
(e) Iron (III) sulphate
(f) Chromium (III) oxide
Answer:
(a) Mercury (II) chloride: Hg (II) Cl₂
(b) Nickel (II) sulphate: Ni (II) SO₄
(c) Tin (IV) oxide : Sn (IV) O₂
(d) Thallium (I) sulphate: Tl₂ (I) SO₄
(e) Iron (III) sulphate: Fe₂(III) (SO₄)₃
(f) Chromium (III) oxide: Cr₂(III)O₃

Question 7.
Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.
Answer:
The substances where carbon can exhibit oxidation states from -4 to +4 are listed in the following table :

Substance	Formula	Oxidation State of Carbon
Methane	CH4	-4
Ethane	C2H6	-3
Ethene	C2H4	-2
Ethyne	C2H2	-1
Dichloromethane	CH2Cl2	0
Hexachlorobenzone	C6Cl6	+1
Carbon monoxide	CO	+2
Oxalic acid	(COOH)2	+3
Carbon dioxide	CO2	+4

The substances where nitrogen can exhibit oxidation sates from -3 to +5 are listed in in the following table.

Substance	Formula	Oxidation State of Nitrogen
Ammonia	NH3	-3
Hydrazine	N2H4	-2
Hydride	N2H2	-1
Dinitrogen gas	N2	0
Nitrous oxide	N2O	+1
Nitric oxide	NO	+2
Dinitrogen trioxide	N2O3	+3
Nitrogen dioxide	NO2	+4
Nitrogen pentoxide	N2O5	+5

Question 8.
While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?
Answer:
(i) In sulphur dioxide (SO₂), the oxidation number (O.N.) of S is +4 and the range of the O.N. that S can have is from +6 to -2.
Therefore, SO₂ can act as an oxidising as well as reducing agent.
(ii) In hydrogen peroxide (H₂O₂), the O.N. of O is -1 and the range of the O.N. that O can have is from 0 to -2. O can sometimes also attain the oxidation numbers +1 and +2. Hence, H₂O₂ can act as an oxidising as well as reducing agent.
(iii) In ozone (O₃) the O.N. of O is zero and the range of the O.N. that O can have is from 0 to -2. Therefore, the O.N. of O can only decrease in this case. Hence,O₃ acts only as an oxidant.
(iv) In nitric acid (HNO₃) the O.N. of N is +5 and the range of the O.N. that N can have is from +5 to -3. Therefore, the O.N. of N can only decrease in this case. Hence, HNO₃ acts only as an oxidizing agent.

Question 9.
Consider the reactions:
(a) 6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆ (aq) + 6O₂(g)
(b) O₃(g) + H₂O₂(Z) → H₂O(1) + 2O₂(g)
Why it is more appropriate to write these reactions as:
(a) 6CO₂ (g) + 12H₂O(1) → C₆H₁₂O₆ (aq) + 6H₂O(l) + 6O₂ (g)
(b) O₃(g) + H₂O₂(Z) → + H₂O(Z) + O₂(g) + O₂(g)
Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.
Answer:
(a) The process of photosynthesis involves two steps :
Step 1: H₂O decomposes to give H₂ and O₂.
2H₂O(l) → 2H₂(g) + O₂(g)
Step 2: The H₂ produced in step 1 reduces CO₂ thereby producing glucose (C₆H₁₂O₆) and H₂O.
6CO₂(g) + 12H₂(g) → C₆H₁₂O₆(S) + 6H₂O(l)
Now, the net reaction of the process is given as:

It is more appropriate to write the reaction as given above because water molecules are also produced in the process of photosynthesis.
The path of this reaction can be investigated by using radioactive H₂O¹⁸ in place of H₂O

(b) O₂ is produced from each of the two reactants O₃ and H₂O₂. For this reason O₂ is written twice.
The given reaction involves two steps. First O₃ decomposes to form O₂ and O. In the second step H₂O₂ reacts with the O produced in the first step thereby producing H₂O and O₂.

The path of this reaction can be investigating by using \(\mathrm{H}_{2} \mathrm{O}_{2}^{18}\) or \(\mathrm{O}_{3}^{18}\).

Question 10.
The compound AgF₂ is unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why?
Answer:
AgF₂ → Ag + F₂
The oxidation state of Ag in Ag F₂ is +2. But +2 is an unstable oxidation state of Ag. Therefore, whenever Ag F₂ is formed, silver readily accepts an electron to form Ag⁺. This helps to bring the oxidation state of Ag down from +2 to a more stable state of +1. As a result, Ag F₂ acts as a very strong oxidizing agent.

Question 11.
Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving two illustrations.
Answer:
Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. This can be illustrated as follows:
(i) P₄ and F₂ are reducing and oxidising agents respectively.
If an excess of P₄ is treated with F₂ then PF₃ will be produced, where in the oxidation number (O.N.) of P is +3.

However, if P₄ is treated with an excess of F₂, then PF₅ will be produced, wherein the O.N. of P is +5.

(ii) K acts as a reducing agent, whereas O₂ is an oxidising agent.
If an excess of K reacts with O₂, then K₂O will be formed, wherein the O.N. of O is -2.

However, if K reacts with an excess of O₂, then K₂O₂ will be formed, wherein the O.N. of O is -1.

Question 12.
How do you account for the following observations?
(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction.
(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why?
Answer:
(a) Oxidation of toluene to benzoic acid in acidic medium.

Oxidation of toluene to benzoic acid in basic and neutral medium.

On industrial scale, alcoholic potassium permanganate is preferred to acidic or alkaline potassium permanganate because in the presence of alcohol, both the reactants KMnO₄ and C₆H₅CH₃ are mixed very well and form homogeneous solution and in homogeneous medium reaction takes place faster than in heterogeneous medium. Further more in neutral medium, OH^– ions are produced in the reaction itself.

(b)

HCl is a weak reducing agent. k cannot reduce H₂SO₄ to SO₂ that’s why pungent smelling gas HCl is obtained.

HBr is a strong reducing agent, it reduces H₂S0₄ to SO₂ and is itselfoxidised to Br₂. That’s why we get red vapours of bromine when conc.
H₂SO₄ reacts with inorganic mixture containing bromide salt.

Question 13.
Identify the substance oxidised, reduced, oxidising agent andreducing agent for each of the following reactions:
(a) 2AgBr(s) + C₆H₆O₂(aq) → 2Ag(s) + 2HBr(aq) + C₆H₄O₂(aq)
(b) HCHO(l) +2 [Ag (NH₃)₂]⁺ (aq) + 3OH^–(aq) → 2Ag(s) + HCOO^–(aq) + 4NH₃(aq) + 2H₂O (l)
(c) HCHO(l) + 2Cu²⁺ (aq) + 5 OH^– → Cu₂O (s) +HCOO^–(aq) + 3H₂O(l)
(d) N₂H₄ (Z) + 2H₂O₂ (1) → N₂ (g) + 4H₂O (l)
(e) Pb (s) +PbO₂ (s) + 2H₂SO₄ (ag) → 2PbSO₄(s) + 2H₂O(l)
Answer:
(a) Oxidised substance -C₆H₆O₂
Reduced substance – AgBr
Oxidising agent – AgBr
Reducing agent -C₆H₆O₂

(b) Oxidised substance – HCHO
Reduced substance – [Ag (NH₃)₂]⁺
Oxidising agent – [Ag (NH₃)₂]⁺
Reducing agent – HCHO

(c) Oxidised substance – HCHO
Reduced substance -Cu²⁺
Oxidising agent – Cu²⁺
Reducing agent – HCHO

(d) Oxidised substance – N₂H₄
Reduced substance -H₂O₂
Oxidising agent-H₂O₂
Reducing agent -N₂H₄

(e) Oxidised substance – Pb
Reduced substance – PbO₂
Oxidising agent – Pb O₂
Reducing agent – Pb

Question 14.
Consider the reactions:

Why does the same reductant, thiosulphate react differently with iodine and bromine?
Answer:
The average oxidation number (O.N.) of S in \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) is +2. Being a stronger oxidising agent than I₂, Br₂ oxidises \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-}\) in which the O.N. of S is +6. However I₂ is a weak oxidising agent. Therefore, it oxidises \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to \(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) in which the average O.N. of S is only +2.5. As a result, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) reacts differently with iodine and bromine. ,

Question 15.
Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.
Answer:
(i) F₂ can oxidize Cl^– to Cl₂, Br^– to Br₂ and I^– to I₂ as:
F₂(aq) + 2Cl^–(s) → 2F^–(aq) + Cl₂(g)
F₂(aq) + 2Br^–(aq) → 2F^–(aq) + Br₂(7)
F₂(aq) + 2I^–(aq) → 2F^–(aq) + I₂(s)

On the other hand, Cl₂, Br₂ and I₂ cannot oxidize F^– to F₂. The oxidizing power of halogens increases in the order of I₂ < Br₂ < Cl₂ < F₂. Hence fluorine is the best oxidant among halogens.

(ii) HI and HBr can reduce H₂SO₄ to SO₂, but HCl and HF cannot. Therefore HI and HBr are stronger reductants than HCl and HF.

2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O
2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O

Again, I- can reduce Cu²⁺ to Cu⁺ but Br^– cannot.

4I^–(aq) + 2Cu²⁺(aq) → Cu₂I₂(s) + I₂(aq)

Hence, hydroiodic acid is the best reductant among hydrohalic compounds.
Thus, the reducing power of hydrohalic acids increases in the order of
HF < HCl < HBr < HI.

Question 16.
Why does the following reaction occur?
\(\mathrm{XeO}_{6}^{4-}\)(aq) + 2F^– (aq) + 6H⁺(aq) → XeO₃(g) +F₂(g) + 3H₂O(i)
What conclusion about the compound Na₄XeO₆ (of which \(\mathrm{XeO}_{6}^{4-}\) is a part) can be drawn from the reaction.
Answer:
The given reaction occurs because \(\mathrm{XeO}_{6}^{4-}\) oxidizes being an oxidizing agent and F^– reduces \(\mathrm{XeO}_{6}^{4-}\)

In this reaction, the oxidation number (O.N.) of Xe decreases from +8 in \(\mathrm{XeO}_{6}^{4-}\) to +6 in XeO₃ and the O.N. of F increases from -1 in F^– to 0 in F₂.
Hence, we can conclude that Na₄XeO₆ is a strong oxidizing agent thanF^–.

Question17.
Consider the reactions:
(a) H₃PO₂ (aq) + 4 AgNO₃ (aq) + 2H₂O(l) → H₃PO₄ (aq) + 4Ag(s) + 4HNO₃ (aq)
(b) H₃PO₂ (oqr) + 2CuSO₄ (aq) + 2 H₂O(0 → H₃PO₄ (aq) + 2Cu(s) +H₂SO₄ (aq)
(c) C₆H₅CHO(l) + 2[Ag (NH₃)₂]+ (aq) + 3OH^– (aq) → C₆H₅COO^–(aq) + 2Ag(s) + 4NH₃ (aq) + 2H₂O(l)
(d) C₆H₅CHO(l) + 2Cu²⁺(aq) + 5OH^–(aqf) → No change observed.
What inference do you draw about the behaviour of Ag⁺ and Cu²⁺ from these reactions?
Answer:
Ag⁺ and Cu²⁺ act as an oxidizing agents in reactions (a) and (b) respectively.
In reaction (c), Ag⁺ oxidizes C₆H₅CHO to C₆H₅COO^–, but in reaction (d) Cu²⁺ cannot oxidize C₆H₅CHO. Hence, we can say that Ag⁺ is a stronger oxidising agent than Cu²⁺.

Question 18.
Balance the following redox reactions by ion-electron method:


Answer:
(a) Step 1 : The two half reactions involved in the given reaction are:
Oxidation half reaction :

Reduction half reaction:

Step 2 : Balancing I in the oxidation half reaction, we have
2I^– (aq) → I₂(s)
Now, to balance the charge, we add 2 e“ to the RHS of the reaction.
2I^–(aq) → I₂(s) + 2e^–

Step 3 : In the reduction half reaction, the oxidation state of Mn has reduced from +7 to +4. Thus, 3 electrons are added to the LHS of the reaction.
\(\mathrm{MnO}_{4}^{-}\) + 3e^– → MnO₂(aq)
Now, to balance the charge, we add 4 OH“ ions to the RHS of the reaction as the reaction is taking place in a basic medium.
\(\mathrm{MnO}_{4}^{-}(a q)^{-}\) + 3e^– → MnO₂(aq) + 4OH^–

Step 4 : In this equation, there are 6 O atoms on the RHS and 4 O atoms on the LHS. Therefore, two water molecules are added to the LHS.
\(\mathrm{MnO}_{4}^{-}\) (aq) + 2H₂O + 3e^– → MnO₂(aq) + 4OH^–

Step 5 : Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have:
6I^–(aq) → 3I₂(s) + 6e^–
2 \(\mathrm{MnO}_{4}^{-}\)(aq) + 4H₂O + 6e^– → 2MnO₂(s) + 8OH^–(aq)

Step 6 : Adding the two half reactions, we have the net balanced redox reaction as:
6I^–(aq) + 2\(\mathrm{MnO}_{4}^{-}\)(aq) + 4H₂O(l) → 3I₂(s) + 2MnO₂(s) + 8OH^–(aq)

(b) Following the steps as in part (a) we have the oxidation half reaction as:
SO₂(g) + 2H₂O(Z) → \(\mathrm{HSO}_{4}^{-}\)(aq) + 3H⁺ (aq) + 2e^– (aq)
And the reduction half reaction as:
\(\mathrm{MnO}_{4}^{-}\)(aq) + 8H⁺(aq) + 5e^– → Mn²⁺(aq) + 4H₂O(l)

Multiplying the oxidation half reaction by 5 and the reduction half reaction by 2, and then by adding them, we have the net balanced redox reaction as :

2Mno₄^–(aq) + 5SO₂(g) + 2H₂O(Z) + H⁺(aq) → 2Mn(aq) + 5HSO₄^–((aq)

(c) Following the steps as in part (a), we have the oxidation half reaction as:
Fe²⁺(aq) → Fe³⁺(aq) + e^–

And the reduction half reaction as:
H₂O₂(aq) + 2H⁺ (aq) + 2e^– → 2H₂O(l)

Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:
H₂O₂(aq) + 2Fe²⁺(aq) + 2H⁺(aq) → 2Fe³⁺ (aq) + 2H₂O(Z)

(d) Following the steps as in part (a), we have the oxidation half reaction as:

Question 19.
Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
(a) P₄(s) + OH^–(aq) → PH₃(g) + H₂PO₂^–(aq)
(b) N₂H₄(l) + CIO₃^–(aqr) → NO(g) + Cl^–(g)
(c) Cl₂O₇(g) + H₂O₂(aqr) → ClO^–₂(aq) + O₂(g) + H⁺
Answer:

O.N. increases by 1 per P atom.
P₄ acts both as an oxidising as well as a reducing agent.

Oxidation number method:
Total decrease in O.N. of P₄ in PH₃ = 3 x 4 = 12
Total increase in O.N. of P₄ in H₂PO₂ = 1 x 4 = 4
Therefore, to balance increase/decrease in O.N. multiply PH₃ by 1 and H₂PO₂^– by 3, we have,

P₄(s) + OH^–(aq) → PH₃(g) + 3H₂PO₂^–(aq)

To balance O atoms, multiply OH- by 6, we have,

P₄(s) + 6OH^–(aq) → PH₃(g) + 3H₂PO₂^–(aq)

To balance H atoms, add 3H₂O to L.H.S. and 3OH^– to the R.H.S. we have,

P₄ (s) + 6OH^–(aq) + 3H₂O(l) → PH₃(g) + 3H₂PO₂^–(aq) + 3OH^–(aq)
or P₄(s) + 3OH^–(aq) + 3H₂O(l) → PH₃(g) + 3H₂PO₂^–(aq) …(1)

Thus, Eq. (1) represents the correct balanced equation.

Ion-electron method : The two half reactions are:

Oxidation number method :
Total increase in O.N. of N = 2 x 4 = 8
Total decrease in O.N. of Cl = 1 x 6 = 6
Therefore, to balance increase/decrease in O.N. multiply N₂H₂ by 3 and \(\mathrm{ClO}_{3}^{-}\) by 4, we have,

3N₂H₄(l) + 4ClO₃^– (aq) → NO(g) + Cl^–(aq)

To balance N and Cl atoms, multiply NO by 6 and Cl^– by 4, we have,

3N₂H₄(l) + 4ClO₃^–(aq) → 6NO(g) + 4Cl^–(aq)

Balance O atoms by adding 6H₂O, in R.H.S.

3N₂H₄(l) + 4ClO₃^–(aq) → 6NO(g) + 4Cl^–(aq) + 6H₂O (l) …(1)

H atoms get automatically balanced and thus Eq. (1) represents the correct balanced equation.

Ion electron method :

Oxidation number method :
Total decrease in O.N. of Cl₂O₇ = 4 x 2 = 8
Total increase in O.N. of H₂O₂ = 2 x 1 = 2
∴ To balance increase/decrease in O.N. multiply H₂O₂ and O₂ by 4, we have,

Question 20.
What type of information can you draw from the following reaction?
(CN)₂(g) + 2OH^–(aq) → CN^–(aq) + CNO^–(aq) + H₂O(l)
Answer:
The oxidation number of carbon in (CN)₂ CN^– and CNO^– is +3, +2 and +4 respectively. These are obtained as shown below:
Let the oxidation number of C be x.
(i) (CN)₂
2 (x – 3) = 0
∴ x = +3

(ii) CN^–
x -3 = -1
∴ x = +2

(iii) CNO^–
x – 3 – 2 = -1
∴ x = + 4

The oxidation number of carbon in the various species is:

The following information we can drawn from the above reaction :
(i) Decomposition of cyanogen in the cyanide ion (CN^–) and cyanate ion (CNO^–) occurs in basic medium.
(ii) Cyanogen (CN)₂ acts as both reducing agent as well as oxidising agent.
(iii) The reaction is an example of disproportionation reaction.
(iv) Cyanogen (CN)₂ is called pseudohalogen while CN^–, CNO^– ions are called pseudohalide ions.

Question 21.
The Mn³⁺ ion is unstable in solution and undergoes disproportionation to give Mn²⁺, MnO₂ and H⁺ ion. Write a balanced ionic equation for the reaction.
Answer:
The given reaction can be represented as:

Mn³⁺ (aq) → Mn²⁺(aq) + MnO₂(s) + H⁺(aq)

The oxidation half equation is:

The oxidation number is balanced by adding one electron as:
Mn³⁺(aq) → MnO₂(s) + e^–

The charge is balanced by adding 4H+ ion as :
Mn³⁺(aq) → MnO₂(s) + 4H⁺ (aq) + e^–

The O atoms and H+ ions are balanced by adding 2H20 molecules as:
Mn³⁺(aq) + 2H₂O(l) → MnO₂(s) + 4H⁺ (aq) + e^– …(1)

The reduction half equation is:
Mn³⁺(aq) → Mn²⁺(aq)

The oxidation number is balanced by adding one electron as :
Mn³⁺ (aq) + e^– → Mn²⁺ (aq) … (2)

The balanced chemical equation can be obtained by adding equation (1) and (2) as :
2Mn³⁺(aq) + 2H₂O(l) → MnO₂(s) + Mn²⁺(aq) + 4H⁺(aq)

Question 22.
Consider the elements: Cs, Ne, I and F
(a) Identify the element that exhibits only negative oxidation state.
(b) Identify the element that exhibits only positive oxidation state.
(c) Identify the element that exhibits both positive and negative oxidation states.
(d) Identify the element which exhibits neither the negative nor the positive oxidation state.
Answer:
(a) F exhibits only negative oxidation state of-1.
(b) Cs exhibits only positive oxidation state of +1
(c) I exhibits both positive and negative oxidation states. It exhibits oxidation states of-1, 0, +1, + 3, + 5, and + 7.
(d) The oxidation state of Ne is zero. It exhibits neither negative nor positive oxidation states.

Question 23.
Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.
Answer:
The given redox reaction can be represented as:

Question 24.
Refer to the periodic table given in your book and now answer the following questions:
(a) Select the possible non-metals that can show disproportionation reaction.
(b) Select three metals that can show disproportionation reaction.
Answer:
In disproportionation reaction, one of the reacting substances always contains an element that can exist in at least three oxidation states.

Question 25.
In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with lO.OOg of ammonia and 20.00 g of oxygen?
Answer:
The balanced chemical equation for the given reaction is given as:

68 g of NH₃ reacts with 160 g of O₂
Therefore 10 g of NH₃ reacts with \(\frac{160 \times 10}{68}\). g of O₂ or 23.53 g of O₂
But the available amount of O₂ is 20 g which is less than the amount required to react with 10 g NH₃. So, O₂ is the limiting reagent and it limits the amount of NO produced. From the above balanced equation.
160 g of O₂; produces 120 g NO.
Therefore, 20 g of O₂; produces = \(\frac{120 \times 20}{160}\) = 15 g NO

Question 26.
Using the standard electrode potentials given in the table 8.1,
predict if the reaction between the following is feasible: *
(a) Fe³⁺ (aq) and I^– (aq)
(b) Ag⁺ (aq) and Cu (s)
(c) Fe³⁺ (aq) and Cu(s)
(d) Ag (s) and Fe³⁺ (aq)
(e) Br₂ (aq) and Fe²⁺(aqr)
Answer:

Question 27.
Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO₃ with silver electrodes.
(ii) An aqueous solution of AgNO₃ with platinum electrodes
(iii) A dilute solution of H₂SO₄ with platinum electrodes.
(iv) An aqueous solution of CuCl₂ with platinum electrodes,
Answer:
(i) AgNO₃ ionizes in aqueous solutions to form Ag⁺ and \(\mathrm{NO}_{3}^{-}\) ions.
On electrolysis, either Ag⁺ ions or H₂O molecules can be reduced at the cathode. But the reduction potential of Ag⁺ ions is higher than that of H₂O.

Hence, Ag+ ions are reduced at the cathode. Similarly, Ag metal or H20 molecules can be oxidized at the anode. But the oxidation potential of Ag is higher than that of H20 molecules.

Ag(s) → Ag⁺(aq) + e^– ;\(E^{\ominus}\) = -0.80V

2H₂O₂(g) → O₂(g) + 4H⁺(aq) + 4e^– ;\(E^{\ominus}\) = -1.23V
Therefore, Ag metal gets oxidized at the anode.

(ii) Pt cannot be oxidized easily. Hence, at the anode, oxidation of water occurs to liberate O₂. At the cathode, Ag⁺ ions are reduced and get deposited.

(iii) H₂SO₄ ionizes in aqueous solutions to give H⁺ and \(\mathrm{SO}_{4}^{2-}\) ions.

H₂SO₄ (aq) → 2H⁺(aq) + \(\mathrm{SO}_{4}^{2-}\) (aq)

On electrolysis, either of H⁺ ions or H₂O molecules can get reduced at the cathode. But the reduction potential of H⁺ ions is higher than that of H₂O molecules.

2H⁺(aq) + 2e^– → H₂(g); \(E^{\ominus}\) = 0.0 V
2H₂O(aq) + 2e^– → H₂(g) + 2OH^–(aq): \(E^{\ominus}\) = – 0.83V

Hence, at the cathode, H⁺ ions are reduced to liberate H₂ gas.
On the other hand, at the anode, either of \(\mathrm{SO}_{4}^{2-}\) ions or H₂O molecules can get oxidized. But the oxidation of \(\mathrm{SO}_{4}^{2-}\) involves breaking of more bonds than that of H₂O molecules. Hence, \(\mathrm{SO}_{4}^{2-}\) ions have a lower oxidation potential than H₂O. Thus, H₂O is oxidized at the anode to liberate O₂ molecules.

(iv) In aqueous solutions, CuCl₂ ionizes to give Cu²⁺ and Cl^– ions as
CuCl₂ (aq) → Cu²⁺ (aq) + 2Cl^– (aq)

On electrolysis either of Cu²⁺ ions or H₂O molecules can get reduced at the cathode. But the reduction potential of Cu²⁺ is more than that of H₂O molecules.

Cu²⁺(aq) + 2e^– → Cu(aq) ;\(E^{\ominus}\) = + 0.34V ;
H₂O(l) + 2e^– → H₂(g) + 2OH^– ;\(E^{\ominus}\) = – 0.83V

Hence, Cu²⁺ ions are reduced at the cathode and get deposited.
Similarly, at the anode, either of Cl^– or H₂O is oxidized. The oxidation , potential of H₂O is higher than that of Cl^–

2Cl^–(aq) → Cl₂(g) + 2e^– ;\(E^{\ominus}\) = -1.36V :K
2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e^– ;\(E^{\ominus}\) = -1.23V

But oxidation of H₂O molecules occurs at a lower electrode potential . than that of Cl^– ions because of over-voltage (extra voltage required to liberate gas). As a result, Cl^– ions are oxidized at the anode to liberate Cl₂ gas. :

Question 28.
Arrange the following metals in the order in which they displace each other from the solution of their salts.
Al, Cu, Fe, Mg and Zn. ‘
Answer:
A metal of stronger reducing power displaces another metal of weaker reducing power from its solution of salt.
The order of the increasing reducing power of the given metals is
Cu < Fe < Zn < Al < Mg. Hence, we can say that Mg can displace Al from its salt solution, but Al cannot displace Mg. Thus, the order in which the given metals displace each other from the solution of their salts is given below:
Mg > Al > Zn > Fe > Cu

Question 29.
Given the standard electrode potentials:
K⁺ / K = – 2.93V, Ag⁺ / Ag = 0.80 V, Hg²⁺ /Hg = 0.79 V
Mg²⁺/ Mg = -2.37 V, Cr³⁺/Cr = -0.74 V Arrange these metals in their increasing order of reducing power.
Answer:
The lower the electrode potential, the stronger is the reducing agent. Therefore, the increasing order of the reducing power of the given metal is
Ag < Hg < Cr < Mg < K.

Question 30.
Depict the galvanic cell in which the reaction Zn(s) + 2Ag⁺(aqr) → Zn²⁺(aqr) + 2Ag(s) takes place, further show:
(i) which of the electrode is negatively charged,
(ii) the carriers of the current in the cell, and
(iii) individual reaction at each electrode.
Answer:
The galvanic cell corresponding to the given redox reaction can be represented as:
Zn / Zn²⁺ (aq) | | Ag⁺(aq) / Ag
(i) Zn electrode is negatively charged because at this electrode, Zn oxidizes to Zn²⁺ and the leaving electrons accumulate on this electrode.
(ii) Ions are the carriers of current in the cell.
(iii) The reaction taking place at Zn electrode can be represented as:
Zn(s) → Zn²⁺ (aq) + 2e^–
and the reaction taking place at Ag electrode can be represented as:
Ag⁺(aq) + e^– → Ag(s)

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 9 Hydrogen Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 9 Hydrogen Important Questions

Very Short Answer Type Questions

Question 1.
Name the isotope of hydrogen which contains equal number of protons and neutrons.
Answer:
Deuterium (\({ }_{1}^{2} \mathrm{H}\))
Number of protons (p) = number of electrons
= atomic number = 1
Number of neutrons (n) = mass number – atomic number
= 2 – 1 = 1 .

Question 2.
Why is the ionisation enthalpy of hydrogen higher than that of sodium?
Answer:
Both H and Na contain one electron in the valence shell. But the size of H is much smaller as compare to that of Na and hence, the ionisation enthalpy of hydrogen is much higher (1312 kJ mol^-1) than that of Na (496 kJ mol^-1).

Question 3.
What do you mean by 15 volume H2O2 solution?
Answer:
‘15 volume H₂O₂’ means 1 mL of a 15 volume H₂O₂ solution gives 15 mL of O₂ at NTP.

Question 4.
Which isotope of hydrogen is radioactive?
Answer:
Tritium

Question 5.
Arrange H₂, D₂ and T₂ in the decreasing order of their
(i) boiling points
(ii) heat of fusion
Answer:
(i) T₂ > D₂ > H₂
(ii) T₂ > D₂ > H₂

Question 6.
Write the Lewis structure of hydrogen peroxide.
Answer:
The Lewis structure of hydrogen peroxide is :

Question 7.
Write one chemical reaction for the preparation of D₂O₂.
Answer:
D₂O₂ is prepared by distillation of potassium persulphate (K₂S₂O₈) with D₂O.

Question 8.
Suggest a method to show the electronegative nature of hydrogen.
Answer:
When sodium hydride is electrolysed, hydrogen is evolved at anode, which shows its electronegative nature.

Question 9.
What type of bonds are broken when water evaporates.
Answer:
Intermolecular hydrogen bonds are broken when water evaporates.

Short Answer Type Questions

Question 1.
Describe the industrial applications of hydrogen dependent on
(i) the heat liberated when its atoms are made to combine on the surface of a metal.
(ii) its effect on the unsaturated organic systems in the presence of a catalyst.
(iii) its ability to combine with nitrogen under specific conditions.
Answer:
(i) Due to this property, hydrogen is used in atomic hydrogen welding/cutting torch.
(ii) Due to this property hydrogen is used for the manufacture of vanaspati ghee from edible oils such as cotton-seed oil, soyabean oil, corn oil etc.

(iii) Due to this property dihydrogen is used for the manufacture of ammonia (Haber’s process).

Question 2.
Why does water show high boiling point as compared to hydrogen sulphide? Give reasons for your answer.
Answer:
Water show high boiling point as compared to hydrogen sulphide due to high electronegativity of oxygen (EN = 3.5), water undergoes extensive H-bonding as a result of which water exists as associated molecule.

For breaking these hydrogen bond, a large amount of energy is needed and hence the boiling point of H₂O is high. In other words, due to lower electronegativity of S (EN =2.5), hydrogen sulphide do not undergo H-bonding. Consequently, H₂S exists as discrete molecule and hence its boiling point is much lower than that of H₂O. That is why H₂S is a gas at room temperature.

Question 3.
If a given sample of water has degree of hardness equal to 46 ppm. If entire hardness is due to MgSO₄, how much MgSO₄ is present per kg of water?
Answer:
Given, degree of hardness = 46 ppm
Which means that 106 g of sample require 46 g of CaCO₃
∴ CaCO₃ present in 1000 g of water = \(\frac{46 \times 1000}{10^{6}}\) = 46 x 10^-3 g
1 mol (or 100 g) of CaC0₃ = 1 mol (or 120 g) of MgSO₄
∴ 46 x 10^-3 g of CaCO₃ = \(\frac{120 \times 46 \times 10^{-3}}{100}\)g = 0.055 g or 55 mg

Question 4.
What are the advantages in using hydrogen as a fuel?
Answer:
Hydrogen as a fuel has the following advantages :

1. It has high calorific value.
2. During combustion, it does not produce smoke or any unpleasant fumes.
3. It leaves no ash after burning. The only product of combustion is water.
4. It does not pollute the air because no pollutant is produced during its combustion.
5. It can be used in a fuel cell to generate electricity.
6. It can be used in the internal combustion engines with slight modifications.

Question 5.
Calculate the volume strength of a 3% solution of H₂O₂
Answer:
100 mL of H₂O₂ solution contains H₂O₂ = 3 g
∴ 1000 mL of H₂O₂ solution will contains
H₂O₂ = \(\frac{3}{100}\) x 1000 = 30 g
Consider the chemical equation,

Now 68 g of H₂O₂ gives O₂ at NTP = 22.7 L
∴ 30 g of H₂O₂ will give 02 at NTP = \(\frac{22.7}{68}\) x 30 = 10.014
But 30 g of H₂O₂ are present in 1000 mL of H₂O₂.
Hence, 1000 mL of H₂O₂ solution gives 02 at NTP = 1.0014 mL
∴ 1 mL of H₂O₂ solution will give O₂ at NTP = \(\frac{10014}{1000}\)= 10.01 mL
Hence, the volume strength of 3% H202 solution = 10.01

Long Answer Type Questions

Question 1.
(i) (a) How would you prepare dihydrogen from water by using a reducing agent?
(b) How would you prepare dihydrogen from a substance other than water?
(c) How would you prepare very pure dihydrogen in the laboratory?
(ii) Write a short note on hydrogenation of vegetable oils.
Answer:
(i) (a) Sodium metal is a good reducing agent. It reduces water to hydrogen (or dihydrogen).
2H₂O + 2Na → 2NaOH + H₂(g)
(b) Dihydrogen can be obtained by treating zinc with dilute HCl
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
(c) Highly pure dihydrogen (hydrogen gas) can be prepared by the following methods :
I. Fairly pure hydrogen can be obtained by treating pure magnesium or pure aluminium with chemically pure H₂SO₄ or HCl diluted with distilled water. The gas is passed over P₂O₅ and is collected by the displacement of mercury.
Mg(s) + H₂SO₄(aq) > MgSO₄(aq) + H₂(g)

II. Highly pure hydrogen gas can be obtained by electrolysing a warm solution of Ba(OH)2 in a U-tube using nickel electrodes. The gas is purified by passing it over heated platinum gauze when traces of oxygen combine with hydrogen forming water. The gas is then dried by passing it over caustic potash sticks and phosphorus pentoxide. Hydrogen is finally adsorbed in palladium and the impurities remain unadsorbed. On heating palladium under reduced pressure pure hydrogen is liberated.

(ii) When oils like groundnut oil or cotton seed oil (which are unsaturated compound i.e., have double bond) are treated with hydrogen in the presence of nickel as catalyst, they get converted into edible fats like margarine and vanaspati ghee (which are saturated compounds). This reaction is called hydrogenation of vegetable oils or hardening of oils.

Question 2.
Give ion electron equations for the following reactions :
(i) Oxidation of ferrous ions to ferric ions by hydrogen peroxide both in acidic and basic media.
(ii) Oxidation of iodide ion to iodine by hydrogen peroxide in acidic medium.
(iii) Reduction of acidified potassium dichromate solution.
(iv) Oxidation of sulphurous acid to sulphuric acid.
(v) Oxidation of ferrocyanide ions to ferricyanide ions in acidic medium.
Answer:

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 10 The s-Block Elements Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 10 The s-Block Elements Important Questions

Very Short Answer Type Questions

Question 1.
Complete the following reactions,
(i) \(\mathbf{O}_{2}^{2-}\) +H₂O →
(ii) O₂ +H₂O →
Answer:
(i) Peroxide ions react with water to form H202.
\(\mathbf{O}_{2}^{2-}\) + 2H₂O→ 20H’ + H₂O₂
(ii) Superoxides react with water to form H202 and 02.
\(2 \mathrm{O}_{2}^{-}\) + 2H₂O → 20H^– + H₂O₂ + O₂

Question 2.
(i) Predict giving reason, the outcome of the reaction

Answer:
(i) Large cation (K⁺) can stabilise large anion (I^–).
(ii) This is because the larger cation (K⁺) can stabilise larger anion (Cl^–).

Question 3.
Which colours are imparted to flame when the following elements are introduced in the flame one by one?
(i) Strontium (ii) Barium (iii) Calcium
Answer:
(i) Strontium — Brick red
(ii) Barium — Grassy green
(iii) Calcium — Crimson red

Question 4.
Sodium fire in the laboratory should not be extinguished by pouring water. Why?
Answer:
This is because sodium produces hydrogen gas with water which catches fire because of the exothermic nature of the reaction.

Question 5.
What is light soda ash? Wliy is it called so?
Answer:
Light soda ash is anhydrous Na₂CO₃. It is called so because it is fluffy solid with a low packing density of about 0.5 g cm^-3.

Question 6.
What is baryta water? Give its one use.
Answer:
Baryta is an aqueous solution of barium hydroxide. It can also be used for detection of CO₂.

Question 7.
Give the chemical formula of quick lime, slaked lime and lime water.
Answer:
Quick lime is CaO, slaked lime is Ca(OH)₂ and lime water is an aqueous solution of Ca(OH)₂.

Question 8.
Which magnesium compounds are the constituents of toothpaste?
Answer:
Mg(OH)₂ and MgCO₃ are the constituents of toothpaste.

Question 9.
What is the mixture of CaCN₂ and carbon known as?
Answer:
A mixture of calcium cyanamide (CaCN₂) and carbon is known as nitrolim. It is used as a fertiliser.

Question 10.
It is necessary to add gypsum in the final stages of preparation of cement. Explain why?
Answer:
Gypsum (CaSO₄ . 2H₂O) is added in the final stages of preparation of cement because it slows down the process of setting of cement so that it gets sufficiently hardened thereby imparting greater strength to it.

Short Answer Type Questions

Question 1.
How would you distinguish between
(i) Be(OH)₂ and Ba(OH)₂
(ii) BeSO₄ and BaSO₄
Answer:
(i) Be(OH)₂, beryllium hydroxide is soluble in aqueous sodium hydroxide solution whereas Ba(OH)₂, (barium hydroxide) does not, because Be(OH)₂ is amphoteric in nature and Ba(OH)₂ is basic in nature. Be(OH)₂ + 2NaOH → Na2[Be(OH)₄] (Sodium beryllate)
(ii) BeSO₄ is soluble in water whereas BaSO₄ is insoluble in water.

Question 2.
Element A bums in nitrogen to give an ionic compound B. Compound B reacts with water to give C and D. A solution of C becomes milky on bubbling carbon dioxide. Identify A, B, C and D.
Answer:
Element A is calcium

Compounds = Ca₃N₂ ; CompoundC = Ca(OH)₂ and Compound D = NH₃

Question 3.
What happens when
(i) chlorine gas is passed through a cold and dilute solution of NaOH?
(ii) yellow phosphorus is heated with NaOH solution?
(iii) carbon dioxide is passed through ammonical brine solution?
(iv) sodium hydrogen carbonate is heated?
Answer:
(i) Sodium hypochlorite and sodium chloride are obtained.
Cl2 + 2NaOH → NaCl + NaClO + H₂O
(ii) Phosphine gas is obtained.
P4 + 3NaOH + 3H₂O → 3NaH₂PO₂ + PH₃
(iii) Sodium hydrogen carbonate is precipitated.
NH₃ + H₂O + CO₂ + NaCl → NH₄Cl + NaHCO₃↓
(iv) Sodium ash is obtained.

Question 4.
Mention the various sources of sodium chloride and explain the preparation of sodium chloride from sea-water and salt mines.
Answer:
NaCl occurs abundantly in nature. Its major sources are (a) Sea water which contains 2.7 to 2.9 % NaCl.
(b) Water of inland lakes such as Sambhar Lake in Rajasthan.
(c) Salt-mines which contain rock salt are located in England, Australia, and Himachal Pradesh.
Preparation
(i) From sea water : Sea water is filled in big tanks where it slowly evaporates, leaving behind solid salt. In cold countries, where temperatures are very low, pure water get freeze. Ice formed is removed and concentration of NaCl in solution increases. The concentrated sodium can be separated and evaporated to get NaCl.
(ii) From salt-mines : Salt mines are located deep under the surface of the earth. Holes are made into these mines with the help of drillers and broken pieces of salt rocks are taken out by suitable means.

Question 5.
When water is added to compound (A) of calcium, solution of compound (B) is formed. When carhon dioxide is passed into the solution, it turns milky due to the formation of compound (C). If excess of carbon dioxide is passed into the solution, milkiness disappears due to the formation of compound (D). Identify the compound A, B, C and D. Explain why the milkiness disappears in the last step? [NCERT Exemplar]
Answer:
Appearance of milkiness on passing CO₂ in the solution of compound B indicates that compound B is lime water and compound C is CaCO₃. Since, compound B is obtained by adding H₂O to compound A, therefore compound A is quicklime, CaO. The reactions are as follows :

Long Answer Type Questions

Question 1.
Ions of an element of group 1 participate in the transmission of nerve signals and transport of sugars and amino acids into cells. This element imparts yellow colour to the flame in flame test and forms an oxide and a peroxide with oxygen. Identify the element and write chemical reaction to show the formation of
its peroxide. Why does the element impart colour to the flame?
Answer:
Yellow colour flame in flame test indicates that the alkali metal must be sodium. It reacts with O₂ to form a mixture of sodium peroxide, Na₂O₂ and sodium oxide, Na₂O.

Ionization enthalpy of sodium is low. When sodium metal or its salt is heated in Bunsen flame, the flame energy causes an excitation of the outermost electron which on reverting back to its initial position gives out the absorbed energy as visible light. That’s why sodium imparts yellow colour to the flame.

Question 2.
The stability of peroxide and superoxide of alkali metals increase as we go down the group. Explain giving reason.
Answer:
The stabilit-y of peroxides or superoxides increases as the size of metal ion increases, i.e., KO₂ < RbO₂ < CsO₂.
The reactivity of alkali metals towards oxygen to form different oxides is due to strong positive field around each alkali metal cation. Li⁺ is smallest, it does not allow 0^2- ion to react with O₂ further. Na⁺ is larger than Li, its positive field is weaker than Li⁺. It cannot prevent the conversion of O^2- into \(\mathrm{O}_{2}^{2-}\). The larger K⁺, Rb⁺ and Cs⁺ ions permit \(\mathrm{O}_{2}^{2-}\)ion to react with O₂ further forming superoxide ion (\(\mathrm{O}_{2}^{-}\)).

Further more, increased stability of the peroxide or superoxide with increase in the size of metal ion is due to the stabilisation of large anions by larger cations through lattice energy effect.

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 9 Hydrogen Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 9 Hydrogen

PSEB 11th Class Chemistry Guide Hydrogen InText Questions and Answers

Question 1.
Justify the position of hydrogen in the periodic table on the basis of its electronic configuration.
Answer:
Hydrogen is the first element in the periodic table. It has the electronic configuration 1s¹. It is similar to alkali metal (ns¹) of group I. It shows resemblance with alkali metals of group I of the periodic table. So it can be placed above the alkali metals in group I of the periodic table.
On the other hand, the electronic configuration of hydrogen shows that it is short of one electron to the nearest noble gas configuration (He) having the electronic configuration 1s². Like halogens it forms covalent bonds (H₂, Cl₂, Br₂, etc.) as well as ionic bonds (e.g. Na⁺ H^–). It forms H⁺ ion by giving one electron and hydride ion (H^–) by gaining one electron. On the basis of its electronic configuration (1s²) hydrogen is placed with other ns¹ elements namely alkali metals in the group I as well as in group 17 of the periodic table. Thus, the position of hydrogen in the periodic table is anomalous.
Hydrogen with so many unique characteristics is, therefore best placed separately in the periodic table of elements.

Question 2.
Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes?
Answer:
Hydrogen has following three isotopes :
1. Protium,\({ }_{1}^{1} \mathrm{H}\),
2. Deuterium, \({ }_{1}^{2} \mathrm{H}\) or D, and
3. Tritium, \({ }_{1}^{3} \mathrm{H}\) or T
The mass ratio of protium, deuterium and tritium is 1.008 : 2.014 : 3.016 or 1:2:3.

Question 3.
Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal conditions?
Answer:
The ionization enthalpy of hydrogen atom is very high (1312 kJ mol^-1). Hence, it is very hard to remove its electron. As a result, its tendency to exist in the monoatomic form is rather low. Instead, hydrogen forms a covalent bond with another hydrogen atom and exists as a diatomic (H₂) molecule.

Question 4.
How can the production of dihydrogen, obtained from ‘coal gasification’, be increased?
Answer:
Dihydrogen, produced by coal gasification method as :

The yield of dihydrogen (obtained from coal gasification) can be increased by reacting carbon monoxide (formed during the reaction) with steam in the presence of iron chromate as a catalyst.

This reaction is called the water-gas shift reaction. Carbon dioxide is removed by scrubbing it with a solution of sodium arsenite.

Question 5.
Describe the bulk preparation of dihydrogen by electrolytic method. What is the role of an electrolyte in this process?
Answer:
Electrolysis of acidified water using platinum electrodes gives dihydrogen.

The role of an electrolyte is to make water conducting.

Question 6.
Complete the following reactions :

Answer:

Question 7.
Discuss the consequences of high enthalpy of H-H bond in terms of chemical reactivity of dihydrogen.
Answer:
The ionization enthalpy of H-H bond is very high (1312 kJ mol^-1). This indicates that dihydrogen has a low tendency to form H+ ions. Its ionization enthalpy value is comparable to that of halogens. Hence, it forms diatomic molecules (H₂), hydrides with elements, and a large number of covalent bonds.
Since ionization enthalpy is very high, dihydrogen does not possess metallic characteristics (lustre, ductility, etc.) like metals.

Question 8.
What do you understand by (i) electron-deficient,
(ii) electron-precise, and (iii) electron-rich compounds of hydrogen? Provide justification with suitable examples.
Answer:
(i) An electron-deficient hydride : It has very few electrons, less than that required for representing its conventional Lewis structure e.g., diborane (B₂H₆). In B₂H₆, there are six bonds in all, out of which only four bonds are regular i.e., two electrons are shared by two atoms.
The remaining two bonds are three centered-two electron bonds i.e., two electrons are shared by three atoms. Hence, its conventional Lewis structure cannot be drawn.

(ii) An electron-precise hydride : It has sufficient number of electrons to be represented by its conventional Lewis structure e.g., CH₄. The Lewis structure can be written as :

Four regular bonds are formed where two electrons are shared by two atoms.

(iii) An electron-rich hybride : It contains excess electrons as lone pair e.g., NH₃

There are three regular bonds in all with a lone pair of electrons on the nitrogen atom.

Question 9.
What characteristics do you expect from an electron-deficient hydride with respect to its structure and chemical reactions?
Answer:
These hydrides do not have sufficient number of electrons to form normal covalent bonds, e.g., B in BF₃ has 6 electrons in its valence shell. These hydrides are trigonal planar in shape.

These hydrides act as Lewis acids, i.e., electron pair acceptor e.g.,

To make up the deficiency of electrons, these hydrides exist in polymeric forms e.g., B₂H₆, B₄H₁₀, etc. Electron deficient hydrides are very reactive. These reacts readily with metals and non-metals and their compounds, e.g.,

B₂H₆ + 3O₂(g) → B₂O₃(s) + 3H₂O(g)

Question 10.
Do you expect the carbon hydrides of the type (CraH2jt + 2) to act as ‘Lewis’ acid or base? Justify your answer.
Answer:
Carbon hydrides of the type C_nH_2n+2 are CH₄, C₂H₆ etc. in which number of electrons present are just sufficient to write down their conventional Lewis structures.

C has neither extra electrons nor less electrons. Such compounds of the formula C_nH_2n+2 are called ELECTRON-PRECISE compounds. They will act neither as Lewis acids nor as Lewis bases.

Question 11.
What do you understand by the term “non-stoichiometric hydrides”? Do you expect this type of the hydrides to be formed by alkali metals? Justify your answer.
Answer:
Non-stoichiometric hydrides are hydrogen-deficient compounds formed by the reaction of dihydrogen with d-block and f-block elements. These hydrides do not follow the law of constant composition.

For example :
LaH₂₈₇, YbH_2.55, TiH_{1.5 – 1.8} etc.
Alkali metals form stoichiometric hydrides. These hydrides are ionic in nature. Hydride ions have comparable sizes (208 pm) with alkali metal ions. Hence, strong binding forces exist between the constituting metal and hydride ion. As a result, stoichiometric hydrides are formed. Alkali metals will not form non-stoichiometric hydrides.

Question 12.
How do you expect the metallic hydrides to be useful for hydrogen storage? Explain.
Answer:
Metallic hydrides are hydrogen deficient, i.e., they do not hold the law of constant composition. It has been established that in the hydrides of Ni, Pd, Ce and Ac hydrogen occupies the interstitial position in lattices allowing further absorption of hydrogen on these metals. Metals like Pd, Pt, etc. have the capacity to accommodate a large volume of hydrogen. Therefore, they are used for the storage of hydrogen and serve as a source of energy.

Question 13.
How does the atomic hydrogen or oxy-hydrogen torch function for cutting and welding purposes? Explain.
Answer:
Atomic hydrogen atoms are produced by the dissociation of dihydrogen with the help of an electric arc. This releases a huge amount of energy (435.88 kJ mol^-1). This energy can be used to generate a temperature of 4000 K, which is ideal for welding and cutting metals. Hence, atomic k hydrogen or oxy-hydrogen torches are used for these purposes. For this reason, atomic hydrogen is allowed to recombine on the surface to be welded to generate the desired temperature.

Question 14.
Among NH₃, H₂O and HF, which would you expect to have highest magnitude of hydrogen bonding and why?
Answer:
The extent of hydrogen bonding depends upon electronegativity and the number of hydrogen atoms available for bonding. Among nitrogen, fluorine, and oxygen, the increasing order of their electronegativities are N < O < F. Hence, the order of the extent of hydrogen bonding is HF > H₂O > NH₃.

Question 15.
Saline hydrides are known to react with water violently producing fire. Can CO₂, a well known fire extinguisher, be used in this case? Explain.
Answer:
Saline hydrides (i.e., NaH, LiH, etc.) react with water to form a base and hydrogen gas. The chemical equation used to represent the reaction can be written as:
NaH(s) + H₂O(7) → NaOH(aq) + H₂(g)
This reaction is violent and produces fire.
This type of fire cannot be extinguished by CO₂ because it gets reduced by the hot metal hydride to form sodium format.
NaH + CO₂ → HCOONa

Question 16.
Arrange the following
(i) CaH₂, BeH₂ and TiH₂ in order of increasing electrical conductance.
(ii) LiH, NaH and CsH in order of increasing ionic character.
(iii) H-H, D-D and F-F in order of increasing bond dissociation enthalpy.
(iv) NaH, MgH₂ and H₂O in order of increasing reducing property.
Answer:
(i) BeH₂ < CaH₂ < TiH₂
(ii) LiH < NaH < CsH
(iii) F—F < H—H < D—D
(iv) H₂O < MgH₂ < NaH

Question 17.
Compare the structures of H₂O and H₂O₂.
Answer:
In gaseous phase, water molecule has a bent form with a bond angle of 104.5°. The O—H bond length is 95.7 pm. The structure can be shown as:

Hydrogen peroxide has a non-planar structure both in gas and solid phase. The dihedral angle in gas and solid phase is 111.5° and 90.2° respectively.

Question 18.
What do you understand by the term ‘auto-protolysis’ of water? What is its significance?
Answer:
Auto-protolysis (self-ionization) of water is a chemical reaction in which two water molecules react to produce a hydroxide ion (OH^–) and a hydronium ion (H₃O⁺).

The reaction involved can be represented as :

Auto-protolysis of water indicates its amphoteric nature i.e., its ability to act as an acid as well as a base. The acid-base reaction can be written as :

Question 19.
Consider the reaction of water with F₂ and suggest, in terms of oxidation and reduction, which species are oxidized/reduced.
Answer:

In these reactions, water acts as a reducing agent and hence itself gets oxidised to either oxygen or ozone. Fluorine acts as an oxidising agent and hence itself reduced to F^– ion.

Question 20.
Complete the following chemical reactions.
(i) PbS(s) + H₂O₂(ciq) →
(ii) MnO^–₄(aq) + H₂O₂(aq) →
(iii) CaO(s) + H₂O(g) →
(iv) AlCl₃(g) + H₂O(Z) →
(v) Ca₃N₂(s) + H₂O(l) →
Classify the above into (a) hydrolysis, (b) redox and (c) hydration reactions.
Answer:
(i) PbS(s) + 4H₂O₂(aq) → PbSO₄(s) + 4H₂O(l)
H₂O₂ is acting as an oxidizing agent in the reaction. Hence, it is a redox reaction.

(ii) 2MnO₄^– (aq) + 5H₂O₂(Z) + 6H⁺(aq) → 2Mn²⁺(aq) + 8H₂O(Z) + 5O₂(g)
H₂O₂ is acting as a reducing agent in the acidic medium, thereby oxidizing MnO^–₄(aq). Hence, the given reaction is a redox reaction.

(iii) CaO(s) + H₂O(g) → Ca(OH)₂(aq)
The reactions in which a compound reacts with water to produce other compounds are called hydrolysis reactions. The given reaction is hydrolysis reaction.

(iv) AlCl₃(g) + 6H₂O(Z) → [Al(OH₂)₆]³⁺ (aq) + 3Cl^–(aq)
It is a hydration reaction, because A1C13 is hydrated to [Al(OH₂)₆]³⁺.

(v) Ca₃N₂(s) + 6H₂O(Z) → 3Ca(OH)₂(aq) + 2NH₃(aq)
The reactions in which a compounds reacts with water to produce other compounds are called hydrolysis reactions. The given reaction represents hydrolysis of Ca₃N₂.

Question 21.
Describe the structure of the common form of ice.
Answer:
Ice is the crystalline form of water. It takes a hexagonal form if crystallized at atmospheric pressure, but condenses to cubic form if the temperature is very low.
The three-dimensional structure of ice is represented as :

The structure is highly ordered and has hydrogen bonding. Each oxygen atom is surrounded tetrahedrally by four other oxygen atoms at a distance of 276 pm. The structure also contains wide holes that can hold molecules of appropriate sizes interstitially.

Question 22.
What causes the temporary and permanent hardness of water?
Answer:
Temporary hardness of water is due to the presence of soluble salts of magnesium and calcium in the form of hydrogen carbonates (MHCO₃, where M = Mg, Ca) in water.
Permanent hardness of water is due to the presence of soluble salts of calcium and magnesium in the form of chlorides in water.

Question 23.
Discuss the principle and method of softening of hard water by synthetic ion-exchange resins.
Answer:
The process of treating permanent hardness of water using synthetic resins is based on the exchange of cations (e.g., Na⁺, Ca²⁺, Mg²⁺ etc.) and anions (e.g.,Cl^– , \(\mathrm{SO}_{4}^{2-}\), \(\mathrm{HCO}_{3}^{-}\) etc.) present in water by H⁺ and OH^– ions respectively.

Synthetic resins are of two types :
1. Cation exchange resins
2. Anion exchange resins

Cation exchange resins are large organic molecules that contain the -SO₃H group. The resin is firstly changed to RNa (from ROS₃H) by treating it with NaCl. This resin then exchanges Na⁺ ions with Ca²⁺ and Mg²⁺ ions, thereby making the water soft.

2RNa + M²⁺ (aq) → R₂M(s) + 2Na⁺ (aq)

There are cation exchange resins in H+ form. The resins exchange H⁺ ions for Na⁺, Ca²⁺ and Mg²⁺ ions.

2RH + M²⁺ (aq) ⇌ MR₂(s) + 2H⁺(aq)

Anion exchange resins exchange OH- ions for anions like Cl^–, \(\) and \(\) present in water.

During the complete process, water first passes through the cation exchange process. The water obtained after this process is free from mineral cations and is acidic in nature.
This acidic water is then passed through the anion exchange process where OH^– ions neutralize the H⁺ ions and de-ionize the water obtained.

Question 24.
Write chemical reactions to show the amphoteric nature of water.
Answer:
Water is amphoteric in character. It behaves both as an acid as well as a base. With acids stronger than itself, it behaves as a base and with bases stronger than itself, it acts as an acid.

Question 25.
Write chemical reactions to justify that hydrogen peroxide can function as an oxidizing as well as reducing agent.
Answer:
Hydrogen peroxide, H₂O₂ acts as an oxidizing as well as a reducing agent in both acidic and alkaline media. Reactions involving oxidizing actions are :

Question 26.
What is meant by ‘demineralised’ water and how can it be obtained?
Answer:
Water which does not contain cations and anions is called ‘demineralised’ water. It is soft water. Demineralised water is obtained the same way as soft water is obtained from hard water. Demineralised or deionised water is obtained by passing hard water first through a cation exchange resin (RCOOH or RSO₃H) which removes Ca²⁺ and Mg²⁺ ions from hard water by exchanging them with H⁺ ions and then passing through an anion exchange resin which removes Cl^– and \(\mathrm{SO}_{4}^{2-}\) ions present in hard water by exchanging them with OH^– ions.

Question 27.
Is demineralised or distilled water useful for drinking purposes? If not, how can it be made useful?
Answer:
Water is an important part of life. It contains several dissolved nutrients that are required by human beings, plants and animals for survival.
Demineralised water is free from all soluble minerals. Hence, it is not fit for drinking.
It can be made useful only after the addition of desired minerals in specific amounts, which are important for growth.

Question 28.
Describe the usefulness of water in biosphere and biological systems.
Answer:
Water is essential for all forms of life. It constitutes around 65% of the human body and 95% of plants. Water plays an important role in the biosphere owing to its high specific heat, thermal conductivity, surface tension, dipole moment and dielectric constant.
The high heat of vaporization and heat of capacity of water helps in moderating the climate and body temperature of all living beings.
It acts as a carrier of various nutrients required by plants and animals for various metabolic reactions.
Water is also required for photosynthesis in plants which releases O₂ into the atmosphere.

Question 29.
What properties of water make it useful as a solvent? What types of compound can it (i) dissolve, and (ii) hydrolyse?
Answer:
Water because of its high dielectric constant (78.39) has the ability to dissolve most of the inorganic (ionic) compounds and is, therefore, regarded as a universal solvent. Whereas the solubility of ionic compounds takes place due to ion-dipole interactions (i.e., solvation of ions) the solubility of covalent compounds such as alcohols, amines, urea, glucose, sugar etc. takes place due to tendency of these molecules to form hydrogen bonds with water.

(i) It can dissolve both ionic compounds as well as covalent compounds which can form hydrogen bonds with water.
Ionic compounds whose lattice energy is lower than hydration energy get dissolved in water.
(ii) Water can hydrolyse many oxides (metallic and non-metallic), hydrides, carbides, nitrides, phosphides and many other salts e.g.,

CaO(s) + H₂O(l) → Ca(OH)₂
SO₂(g) + H₂O(l) → H₂SO₃(aq)
CaH₂(s) + 2H₂O(l) → Ca(OH)₂(uq) + 2H₂(g)
SiCl₄(l) + 4H₂O(l) → SiO₂ -2H₂O(s) + 4HCl(aq)
Al₄C₃ + 12H₂O(l) → 4Al(OH)₃ + 3CH₄
Ca₃P₂(s) + 6H₂O → 3Ca(OH)₂ + 2PH₃

Question 30.
Knowing the properties of H20 and DaO, do you think that D20 can be used for drinking purposes?
Answer:
Heavy water (D₂O) acts as a moderator, i.e., it slows the rate of a reaction. Due to this property of D₂O, it cannot be used for drinking purposes because it will slow down anabolic and catabolic reactions takes place in the body and lead to a casualty.

Question 31.
What is the difference between the terms ‘hydrolysis’ and ‘hydration’?
Answer:
Hydrolysis is defined as a chemical reaction in which hydrogen and hydroxide ions (H+ and OH- ions) of water molecule react with a compound to form products. For example :
NaH + H₂O → NaOH + H₂
Hydration is defined as the addition of one or more water molecules to ions or molecules to form hydrated compounds. For example :
CUSO₄ + 5H₂O → CUSO₄ . 5H₂O

Question 32.
How can saline hydrides remove traces of water from organic compounds? ’
Answer:
Saline hydrides are ionic in nature. They react with water to form a metal hydroxide along with the liberation of hydrogen gas. The reaction of saline hydrides with water can be represented as :
AH(s) + H20(Z) → AOH(aq) + H2(g)
(where, A= Na, Ca, )

When added to an organic solvent, they react with water present in it. Hydrogen escapes into the atmosphere leaving behind the metallic hydroxide.Then, the dry organic solvent distills over.

Question 33.
What do you expect the nature of hydrides is, if formed by elements of atomic number 15, 19, 23 and 44 with dihydrogen? Compare their behaviour towards water,
Answer:
The elements of atomic number 15, 19, 23 and 44 are nitrogen, potassium, vanadium and ruthenium respectively.
1. Hydride of nitrogen
Hydride of nitrogen (NH₃) is a covalent molecule. It is an electron-rich hydride owing to the presence of excess electrons as a lone pair on nitrogen.

2. Hydride of potassium
Dihydrogen forms an ionic hydride with potassium owing to the high electropositive nature of potassium. It is crystalline and non-volatile in nature.

3. Hydrides of Vanadium and Ruthenium
Both vanadium and ruthenium belong to the d-block of the periodic table. The metals of d-block form metallic or non-stoichiometric hydrides. Hydrides of vanadium and ruthenium are therefore, metallic in nature having a deficiency of hydrogen.

4. Behaviour of hydrides towards water
Potassium hydride reacts violently with water as :
KH(s) + H₂O(aq) → KOH(aq) + H₂(g)
Ammonia (NH₃) behaves as a Lewis base and reacts with water as :
H₂O(Z) + NH₃(aq) ⇌ OH^–(aq) + \(\mathrm{NH}_{4}^{+}\)(aq)
Hydrides of vanadium and ruthenium do not react with water. Hence, the increasing order of reactivity of the hydrides is as: H < NH₃ < KH.

Question 34.
Do you expect different products in solution when aluminium (III) chloride and potassium chloride treated separately with
(i) normal water (ii) acidified water, and (iii) alkaline water? Write equations wherever necessary.
Answer:
Potassium chloride (KCl) is the salt of a strong acid (HCl) and strong base (KOH). Hence, it is neutral in nature and does not undergo hydrolysis in normal water. It dissociates into ions as follows :

In acidified and alkaline water, the ions do not react and remain as such. Aluminium (III) chloride is the salt of a strong acid (HCl) and weak base [Al(OH)₃]. Hence, it undergoes hydrolysis in normal water.

In acidified water, H⁺ ions react with Al(OH)₃ forming water and giving Al³⁺ ions. Hence, in acidified water, AlCl₃ will exist as Al³⁺(aq) and Cl^–(aq) ions.

In alkaline water the following reaction takes place :

Question 35.
How does H₂O₂ behave as a bleaching agent?
Answer:
H₂O₂ acts as a bleaching agent due to the nascent oxygen.
H₂O₂ → H₂O + O
Coloured matter + [O] → Colourless matter
It bleaches materials like silk, hair, ivory, cotton, wool, etc.

Question 36.
What do you understand by the terms :
(i) hydrogen economy (ii) hydrogenation (iii) ‘syngas’ (iv) water-gas shift reaction (v) fuel-cell?
Answer:
(i) Hydrogen economy : Hydrogen economy is a technique of using dihydrogen in an efficient way. It involves transportation and storage of dihydrogen in the form of liquid or gas.
Dihydrogen releases more energy than petrol and is more eco-friendly. Hence, it can be used in fuel cells to generate electric power. Hydrogen economy is about the transmission of this energy in the form of dihydrogen.

(ii) Hydrogenation : It refers to the addition of dihydrogen to another reactant. This process is used to reduce a compound in the presence of a suitable catalyst. For example, hydrogenation of vegetable oil using nickel as a catalyst gives edible fats such as vanaspati ghee etc.

(iii) Syngas : Syngas is a mixture of carbon monoxide and dihydrogen. Since the mixture of the two gases is used for the synthesis of methanol, it is called syngas, synthesis gas, or water gas.

Syngas is produced on the action of steam with hydrocarbons or coke at a high temperature in the presence of a catalyst.

(iv) Water gas shift reaction : The production of hydrogen by reacting carbon monoxide (CO) of syngas mixtures with steam in the presence of iron chromate as catalyst is called water-gas shift reaction.

CO₂ is removed by scrubbing with sodium arsenite solution.
(v) Fuel cell: It is a device which converts the energy produced during the combustion of a fuel directly into electrical energy. One such fuel cell is hydrogen-oxygen fuel cell. It does not cause any pollution. Fuel cells generated electricity with conversion efficiency of 70-85%.

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 10 The s-Block Elements Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 10 The s-Block Elements

PSEB 11th Class Chemistry Guide The s-Block Elements InText Questions and Answers

Question 1.
What are the common physical and chemical features of alkali metals?
Answer:
Physical properties of alkali metals are as follows :

1. They are quite soft and can be cut easily. Sodium metal can be easily cut using a knife.
2. They are light coloured and are mostly silvery white in appearance.
3. They have low density because of the large atomic sizes. The density increases down the group from Li to Cs. The only exception to this is K, which has lower density than Na.
4. The metallic bonding present in alkali metals is quite weak. Therefore they have low melting and boiling points.
5. Alkali metals and their salts impart a characteristic colour to flames. This is because the heat from the flame excites the electron present in the outermost orbital to a high energy level. When this excited electron reverts back to the ground state, it emits excess energy as radiation that falls in the visible region.
6. They also display photoelectric effect. When metals such as Cs and K are irradiated with light, they lose electrons.

Chemical properties of alkali metals are as follows :

Alkali metals are highly reactive due to their low ionization enthalpy. As we move down the group, the reactivity increases.
(1) They react with water to form respective oxides or hydroxides. As we move down the group, the reaction becomes more and more spontaneous.

(2) They react with water to form their respective hydroxides and dihydrogens. The general reaction for the same is given as :
2M + 2H²O → 2M⁺ + 2OH^– + H₂

(3) They react with dihydrogen to form metal hydrides. These hydrides are ionic solids and have high melting points.
2M + H₂ → 2M⁺ H^–

(4) Almost all alkali metals, except Li, react directly with halogens to form ionic halides.
2M + Cl₂ → 2MCl (M = Li, K, Rb, Cs)

Since Li⁺ ion is very small in size, it can easily distort the electron cloud around the negative halide ion. Therefore lithium halides are covalent in nature.

(5) They are strong reducing agents. The reducing power of alkali metals increases on moving down the group.
However, lithium is an exception. It is the strongest reducing agent among the alkali metals. It is because of its high hydration energy.

(6) They dissolve in liquid ammonia to form deep blue coloured solutions. Therefore, these solutions are conducting in nature.
M + (x + y) NH₃ → [M(NH₃)_x]⁺[M(NH₃)_y]^–

The ammoniated electrons cause the blue colour of the solution. These solutions are paramagnetic and if allowed to stand for some time, then they liberate hydrogen. This results in the formation of amides.
M⁺ + e^– + NH₃(l) → MNH + \(\frac{1}{2}\)H₂(g)

In a highly concentrated solution, the blue colour changes to bronze and the solution becomes diamagnetic.

Question 2.
Discuss the general characteristics and gradation in properties of alkaline earth metals.
Answer:
Physical and atomic properties of alkaline earth metals are as follows:

• The general electronic configuration of alkaline earth metals is (noble gas) ns².
• These metals lose two electrons to acquire the nearest noble gas configuration. Therefore, their oxidation state is +2.
• These metals have atomic and ionic radii smaller than that of alkali metals. Also when moved down the group, the effective nuclear charge decreases and this causes an increase in their atomic radii and ionic radii.
• Since the alkaline earth metals have large size, their ionization enthalpies are found to be fairly low. However, their first ionization enthalpies are higher than the corresponding group 1 metals.
• These metals are lustrous and silvery white in appearance. They are relatively less soft as compared to alkali metals.
• Atoms of alkaline earth metals are smaller than that of alkali metals. Also they have two valence electrons forming stronger metallic bonds. These two factors cause alkaline earth metals to have high melting and boiling points as compared to alkali metals.
• They are highly electropositive in nature. This is due to their low ionization enthalpies. Also the electropositive character increases on moving down the group from Be to Ba.
• Ca, Sr, and Ba impart characteristic colours to flames.
  Ca – Brick red Sr – Crimson red Ba – Apple green
  In Be and Mg, the electrons are too strongly bound to be excited. Hence, these do not impart any colour to the flame.

Chemical properties of alkaline earth metals are as follows :
The alkaline earth metals are less reactive than alkali metals and their reactivity increases on moving down the group.

(i) Reaction with air and water : Be and Mg are almost inert to air and water because of the formation of oxide layer on their surface.
(a) Powdered Be burns in air to form BeO and Be₃N₂.
(b) Mg, being more electropositive, burns in air with a dazzling sparkle to form MgO and Mg₃N₂.
(c) Ca, Sr, and Ba react readily with air to form respective oxides and nitrides.
(d) Ca, Ba, and Sr react vigorously even with cold water.

(ii) Alkaline earth metals react with halogens at high temperatures to form halides.
M + X₂ → MX₂ (X = F, Cl, Br, I)
(iii) All the alkaline earth metals, except Be, react with hydrogen to form hydrides.
(iv) They react readily with acids to form salts and liberate hydrogen gas.
M + 2HCl → MCl₂ + H₂(g) ↑
(v) They are strong reducing agents. However, their reducing power is less than that of alkali metals. As we move down the group, the reducing power increases.
(vi) Similar to alkali metals, the alkaline earth metals also dissolve in liquid ammonia to give deep blue coloured solutions.
M + (x – y) NH₃ → [M (NH₃)_x]²⁺ + 2 [e (NH₃)_y]^–

Question 3.
Why are alkali metals not found in nature?
Answer:
Alkali metals include lithium, sodium, potassium, rubidium, cesium, and francium. These metals have only one electron in their valence shell, which they lose easily, owing to their low ionization energies. Therefore, alkali metals are highly reactive and are not found in nature in their elemental state.

Question 4.
Find out the oxidation state of sodium in Na₂O₂.
Answer:
Let the oxidation state of Na be x. The oxidation state of oxygen, in case of peroxides, is -1.
Therefore, 2(x) + 2(-1) = 0 ⇒ 2x – 2 = 0 ⇒ 2x = 2 ⇒ x = ±1
Therefore, the oxidation state of sodium in Na₂O₂ is +1.

Question 5.
Explain why is sodium less reactive than potassium?
Answer:
In alkali metals, on moving down the group, the atomic size increases and the effective nuclear charge decreases.
Because of these factors, the outermost electron in potassium can be lost easily as compared to sodium. Hence, potassium is more reactive than sodium.

Question 6.
Compare the alkali metals and alkaline earth metals with respect to (i) ionization enthalpy (ii) basicity of oxides and (iii) solubility of hydroxides.
Answer:

• Ionization enthalpies : The first ionization enthalpies of the alkaline earth metals are higher than those of the corresponding alkali metals. This is due to their small size as compared to the corresponding alkali metals. But second ionization enthalpies of the alkaline earth metals are smaller than those of the corresponding alkali metals.
• Basicity of oxides : The oxides of the alkali and alkaline earth metals dissolves in water to form basic hydroxides. The alkaline earth metal hydroxides are however less basic and less stable than alkali metal hydroxides.
• Solubility of hydroxides : The solubility of hydroxides of alkaline earth metals is relatively less than their corresponding alkali metal hydroxides.

Question 7.
In what ways lithium shows similarities to magnesium in its chemical behaviour?
Answer:
Similarities between lithium and magnesium are as follows :
(i) Both Li and Mg react slowly with cold water.
(ii) The oxides of both Li and Mg are much less soluble in water and their hydroxides decompose at high temperature.
im 1

(iii) Both Li and Mg react withN2 to form nitrides.
im 2

(iv) Neither Li nor Mg form peroxides or superoxides.
(v) The carbonates of both are covalent in nature. Also, these decompose on heating.
im 3

(vi) Li and Mg do not form solid bicarbonates.
(vii) Both LiCl and MgCl₂ are soluble in ethanol owing to their covalent nature.
(viii) Both LiCl and MgCl₂ are deliquescent in nature. They crystallize from aqueous solutions as hydrates, for example, LiCl.₂H₂O and MgCl₂ . 8H₂O.

Question 8.
Explain why can alkali and alkaline earth metals not be obtained by chemical reduction methods?
Answer:
In the process of chemical reduction, oxides of metals are reduced using a stronger reducing agent. Alkali metals and alkaline earth metals are among the strongest reducing agents and the reducing agents that are stronger than them are not available. Therefore, they cannot be obtained by chemical reduction of their oxides.

Question 9.
Why are potassium and caesium, rather than lithium used in photoelectric cells?
Answer:
Potassium and caesium have much lower ionization enthalpy than that of lithium. Therefore, these metals on exposure to light emit electrons easily but lithium does not. That’s why K and Cs rather than Li are used in photoelectric cells.

Question 10.
When an alkali metal dissolves in liquid ammonia the solution can acquire different colours. Explain the reasons for this type of colour change.
Answer:
When an alkali metal is dissolved in liquid ammonia, it results in the formation of a deep blue coloured solution.
M + (x + y) NH₃ → [M (NH₃)_x]⁺ + [e^-1 (NH₃)_y] [Ammoniated electron]
The ammoniated electrons absorb energy corresponding to red region of visible light. Therefore, the transmitted light is blue in colour.
At a higher concentration (3 M), clusters of metal ions are formed. This causes the solution to attain a copper-bronze colour and a characteristic metallic lustre.

Question 11.
Beryllium and magnesium do not give colour to flame whereas other alkaline earth metals do so. Why?
Answer:
When an alkaline earth metal is heated, the valence electrons get excited to a higher energy level. When this excited electron comes back to its lower energy level, it radiates energy, which belongs to the visible region. Hence, the colour is observed. In Be and Mg, the electrons are strongly bound. The energy required to excite these electrons is very high. Therefore, when the electron reverts back to its original position the energy released does not fall in the visible region. Hence, no colour in the flame is seen.

Question 12.
Discuss the various reactions that occur in the Solvay process.
Answer:
In Solvay ammonia process, CO₂ is passed through brine, (a concentrated solution of NaCl) saturated with ammonia. The process involves the formation of a sparingly soluble sodium bicarbonate.
NaCl + NH₃ + CO₂ + H₂O → NaHCO₃↓ + NH₄Cl .
Sodium bicarbonate thus formed is filtered, dried and heated to obtain sodium carbonate.
im 4
CO₂ used in carbonating tower is prepared by heating calcium carbonate and the quicklime, CaO thus formed is dissolved in water to form slaked lime, Ca(OH)₂.
im 5
In ammonia recovery tower, NH3 is prepared by heating NH₄Cl with Ca(OH)₂.
im 6

Question 13.
Potassium carbonate cannot be prepared by Solvay process. Why?
Answer:
Solvay process cannot be used to prepare potassium carbonate. This is because unlike sodium bicarbonate, potassium bicarbonate is fairly soluble in water and does not precipitate out.

Question 14.
Why is Li₂CO₃ decomposed at a lower temperature whereas Na₂CO₃ at higher temperature?
Answer:
As we move down the alkali metal group, the electropositive character increases. This causes an increase in the stability of alkali carbonates.
However, lithium carbonate is not so stable to heat. This is because lithium carbonate is covalent. Lithium ion, being very small in size, polarizes a large carbonate ion, leading to the formation of more stable lithium oxide.
im 7
Therefore, lithium carbonate decomposes at a low temperature while a stable sodium carbonate decomposes at a high temperature.

Question 15.
Compare the solubility and thermal stability of the following compounds of the alkali metals with those of the alkaline earth metals (a) Nitrates (b) Carbonates (c) Sulphates.
Answer:
(a) Nitrates of alkali metals and alkaline earth metals :
(i) Nitrates of alkali metals are thermally not stable and decompose on heating to give MNO₂ and O₂ (except LiN03) whereas nitrates of alkaline earth metals decompose on heating give their oxides, nitrogen dioxide and oxygen gas.

im 8

(ii) Nitrates of alkali metals are highly soluble in water whereas alkaline earth metal nitrates are sparingly soluble and crystallise with six molecules of water.
(b) Carbonates of alkali metals and alkaline earth metals:
(i) Carbonates of alkali metals except Li are quite stable upto 1273 K and do not decompose, whereas carbonates of alkaline earth metals decompose at different temperatures, to give their oxides and carbon dioxide.

im 9

The thermal stability of carbonates of alkaline earth metals increase down the group.
BeCO₃ is least stable and BaCO₃ is most stable.

(ii) All the carbonates of alkali metals are generally soluble in water and their solubility increases rapidly on descending the group. This is due to the reason that their lattice energies decrease more rapidly than their hydration energies down the group. In the case of carbonates of alkaline earth metals they are sparingly soluble in water and their solubility decreases down the group from Be to Ba. For example, MgCO₃ is slightly soluble in water, but BaC03 is almost insoluble.

(c) Sulphates of alkali metals and alkaline earth metals :
(i) The sulphates of alkali metals are thermally quite stable whereas the sulphates of alkaline earth metals decompose on heating to give oxides and SO₃. The temperature of decomposition increases
down the group.
im 10

(ii) The sulphates of alkali metals Na and K are soluble in water. As far as the solubility of sulphates of alkaline earth metals in water is concerned, BeSO₄ and MgSO₄ are highly soluble, CaSO₄ is sparingly soluble, but the sulphates of Sr, Ba and Ra are virtually insoluble. Thus, the solubility of their sulphates in water decreases down the group.
BeSO₄ > MgSO₄ > CaSO₄ > SiSO₄ > BaSO₄

Question 16.
Starting with sodium chloride how would you proceed to prepare (i) sodium metal (ii) sodium hydroxide (iii) sodium peroxide (iv) sodium carbonate?
Answer:
(i) Sodium metal from sodium chloride: Sodium is prepared from fused (molten) sodium chloride. Sodium chloride is mixed with CaCl₂ and KF [to lower the M.Pt. of NaCl to 850-875 K] and subjected to electrolysis (in DOWN’S CELL) when the following reactions occur :

im 11

Sodium, liberated at the cathode, is collected in kerosene oil and chlorine gas is liberated at the anode.
(ii) Sodium hydroxide from sodium chloride: Sodium hydroxide (caustic soda) is generally prepared by the electrolysis of brine solution (NaCl solution in water) in Castner Kellner cell. A mercury cathode and carbon anode are used. Sodium metal discharged at the cathode combines with mercury to form sodium amalgam. Cl₂ gas is evolved at the anode.

im 12

The amalgam is treated with water to give sodium hydroxide and hydrogen gas.
2Na – amalgam + 2H₂O → 2NaOH + 2Hg + H₂↑

(iii) Sodium peroxide from sodium chloride: Sodium metal obtained, by the electrolysis of molten sodium chloride is heated with O₂ at about 575 K when sodium forms mainly sodium peroxide.

im 13

(iv) Sodium carbonate from sodium chloride: Sodium carbonate is prepared from an aqueous solution of NaCl by SOLVAY PROCESS. In this process, CO₂ is passed through NaCl solution saturated with ammonia, when following reactions occur :

2NH₃ + H₂O + CO₂ → (NH₄)₂ CO₃
(NH₄)₂ CO₃ + H₂O + CO₂ → 2NH₄HCO₃
NH₄HCO₃ + NaCl → NH₄Cl + NaHCO₃
im 14

Question 17.
What happens when (i) magnesium is burnt in air (ii) quick lime is heated with silica (iii) chlorine reacts with slaked lime (iv) calcium nitrate is heated?
Answer:
(i) Magnesium bums in air with a dazzling light to form MgO and Mg3N₂.
im 15

(ii) Quick lime (CaO) combines with silica (SiO₂) to form slag.
im 16

(iii) When chlorine is added to slaked lime, it gives bleaching powder.
im 17

(iv) Calcium nitrate, on heating decomposes to give calcium oxide.
im 18

Question 18.
Describe two important uses of each of the following (i) caustic soda (ii) sodium carbonate (iii) quicklime.
Answer:
(i) Uses of caustic soda
(a) It is used in soap industry.
(b) It is used as a reagent in laboratory.
(ii) Uses of sodium carbonate
(a) It is generally used in glass and soap industry.
(b) It is used as a water softener.
(iii) Uses of quick lime
(a) It is used as a starting material for obtaining slaked lime.
(b) It is used in the manufacture of glass and cement.

Question 19.
Draw the structure of (i) BeCl₂ (vapour) (ii) BeCl₂ (solid).
Answer:
(i) In the vapour state, BeCl₂ exists as a monomer with a linear structure.
im 19

(ii) In the solid state, BeCl₂ exists as a polymer in condensed phase.
im 20

Question 20.
The hydroxides and carbonates of sodium and potassium are easily soluble in water while the corresponding salts of magnesium and calcium are sparingly soluble in water. Explain.
Answer:
The atomic size of sodium and potassium is larger than that of magnesium and calcium. Thus, the lattice energies of carbonates and hydroxides formed by calcium and magnesium are much more than those of sodium and potassium.
Hence, carbonates and hydroxides of sodium and potassium dissolve readily in water whereas those of calcium and magnesium are only sparingly soluble.

Question 21.
Describe the importance of the following (i) limestone (ii) cement (iii) plaster of paris.
Answer:
(i) Importance of limestone
(a) It is used in the preparation of lime and cement.
(b) It is used as a flux during the smelting of iron ores.

(ii) Importance of cement
(a) It is used in plastering and in construction of bridges.
(b) It is used in concrete.

(iii) Importance of plaster of Paris
(a) It is used in surgical bandages.
(b) It is also used for making casts and moulds.

Question 22.
Why are lithium salts commonly hydrated and those of the other alkali ions usually anhydrous?
Answer:
Because of its smallest size among alkali metals, Li+ has the maximum degree of hydration. That’s why lithium salts are commonly hydrated and those of other alkali metal ions usually anhydrous.
im 21

Question 23.
Why is LiF almost insoluble in water whereas LiCl soluble not only in water but also in acetone?
Answer:
LiF is insoluble in water. On the contrary, LiCl is soluble not only in water, but also in acetone. This is mainly because of the greater ionic character of LiF as compared to LiCl. The solubility of a compound in water depends on the balance between lattice energy and hydration energy. Since fluoride ion is much smaller in size than chloride ion, the lattice energy of LiF is greater than that of LiCl. Also there is not much difference between the hydration energies of fluoride ion and chloride ion. Thus, the net energy change during the dissolution of LiCl in water is more exothermic than that during the dissolution of LiF in water. Hence, low lattice energy and greater covalent character are the factors making LiCl soluble not only in water, but also in acetone.

Question 24.
Explain the significance of sodium, potassium, magnesium and calcium in biological fluids.
Answer:
(i) Sodium (Na):
Sodium ions are found primarily in the blood plasma. They are also found in the interstitial fluids surrounding the cells.
(a) Sodium ions help in the transmission of nerve signals.
(b) They help in regulating the flow of water across the cell membranes.
(c) They also help in transporting sugars and amino acids into the cells.

(ii) Potassium (K):
Potassium ions are found in the highest quantity within the cell fluids.
(a) K⁺ ions help in activating many enzymes.
(b) They also participate in oxidising glucose to produce ATP.
(c) They also participate in transmitting nerve signals.

(iii) Magnesium (Mg) and calcium (Ca) :
Magnesium and calcium are referred to as macro-minerals. This term indicates their higher abundance in the human body system.
(a) Mg helps in relaxing nerves and muscles.
(b) Mg helps in building and strengthening bones.
(c) Mg maintains normal blood circulation in the human body system.
(d) Ca helps in the coagulation of blood.
(e) Ca also helps in maintaining homeostasis.

Question 25.
What happens when
(i) sodium metal is dropped in water?
(ii) sodium metal is heated in free supply of air?
(iii) sodium peroxide dissolves in water?
Answer:
(i) When sodium metal is dropped in water, it reacts violently to form sodium hydroxide and hydrogen gas. The chemical equation involved in the reaction is:
2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)
(ii) On being heated in air, sodium reacts vigorously with oxygen to form sodium peroxide. The chemical equation involved in the reaction is:
2Na(s) + O₂(g) → Na₂O₂(s)
(iii) When sodium peroxide is dissolved in water, it is readily hydrolysed to form sodium hydroxide and water. The chemical equation involved in the reaction is:
Na₂O₂(s) + 2H₂O(7) → 2NaOH(aq) + H₂O₂(aq)

Question 26.
Comment on each of the following observations:
(a) The mobilities of the alkali metal ions in aqueous solution are Li⁺ < Na⁺ < K⁺ < Rb⁺ < Cs⁺
(b) Lithium is the only alkali metal to form a nitride directly.
(c) \(\mathbf{H}^{\ominus}\) for M²⁺ (aq) + 2e^– → M(s) (where M = Ca, Sr or Ba) is nearly constant.
Answer:
(a) On moving down the alkali group, the ionic and atomic sizes of the metals increase. The given alkali metal ions can be arranged in the increasing order of their ionic sizes as:
Li⁺ < Na⁺ < K⁺ < Rb⁺ < Cs⁺

Smaller the size of an ion, the more highly is it hydrated. Since, Li⁺ is the smallest, it gets heavily hydrated in an aqueous solution. On the other hand, Cs+ is the largest and so it is the least hydrated. The given alkali metal ions can be arranged in the decreasing order of their hydrations as:
Li⁺ > Na⁺ > K⁺ > Rb⁺ > Cs⁺
Greater the mass of a hydrated ion, the lower is its ionic mobility. Therefore, hydrated Li⁺ is the least mobile and hydrated Cs⁺ is the most mobile. Thus, the given alkali metal ions can be arranged in the increasing order of their mobilities as:
Li⁺ < Na⁺ < K⁺ < Rb⁺ < Cs⁺

(b) Unlike the other elements of group 1, Li reacts directly with nitrogen to form lithium nitride. This is because Li⁺ is very small in size and so its size is the most compatible with the N^3- ion. Hence, the lattice energy released is very high. This energy also overcomes the high amount of energy required for the formation of the N^3- ion.

(c) Electrode potential (\(\mathbf{E}^{\ominus}\)) of any M²⁺/M electrode depends upon three factors :
(i) Ionisation enthalpy
(ii) Enthalpy of hydration
(iii) Enthalpy of vaporisation
The combined effect of these factors is approximately the same for Ca, Sr, and Ba. Hence, their electrode potentials are nearly constant.

Question 27.
State as to why
(a) a solution of Na₂CO₃ is alkaline?
(b) alkali metals are prepared by electrolysis of their fused chlorides?
(c) sodium is found to be more useful than potassium?
Answer:
(a) When sodium carbonate is added to water, it hydrolyses to give sodium bicarbonate and sodium hydroxide (a strong base). As a result, the solution becomes alkaline.
Na₂CO₃ + H₂O → NaHCO₃ + NaOH

(b) It is not possible to prepare alkali metals by the chemical reduction of their oxides as they themselves are very strong reducing agents. They cannot be prepared by displacement reactions either (wherein one element is displaced by another). This is because these elements are highly electropositive. Electrolysis of aqueous solutions can neither be used to extract these elements. This is because the liberated metals react with water. Hence, to overcome these difficulties, alkali metals are usually prepared by the electrolysis of their fused chlorides.

(c) Blood plasma and the interstitial fluids surrounding the cells are the regions where sodium ions are primarily found. Potassium ions are located withimthe cell fluids. Sodium ions are involved in the transmission of nerve signals, in regulating the flow of water across the cell membranes, and in transporting sugars and amino acids into the cells. Hence, sodium is found to be more useful than potassium.

Question 28.
Write balanced equations for reactions between
(a) Na₂O₂ and water
(b) KO₂ and water
(c) Na₂O and CO₂
Answer:
(a) The balanced chemical equation for the reaction between Na202 and water is:
Na₂O₂(s) + 2H₂O(l) → 2NaOH(aq) + H₂O₂(aq)

(b) The balanced chemical equation for the reaction between K02 and water is:
2KO₂(s) + 2H₂O(l) → 2KOH(aq) + H₂O₂(aq) + O₂(g)
or 4KO₂(s) + 2H₂O(l) → 4KOH(aq) + 3O₂(g)

(c) The balanced chemical equation for the reaction between Na₂O and CO₂ is:
Na₂O(s) + CO₂(g) + Na₂CO₃

Question 29.
How would you explain the following observations?
(i) BeO is almost insoluble but BeSO₄ is soluble in water, ‘
(ii) BaO is soluble but BaSO₄ is insoluble in water,
(iii) Lil is more soluble than KI in ethanol.
Answer:
(i) BeO is almost insoluble in water and BeSO₄ is soluble in water. Be²⁺ is a small cation with a high polarising power and O^2- is a small anion. The size compatibility of Be²⁺ and O^2- is high. Therefore, the lattice energy released during their formation is also very high. When BeO is dissolved in water, the hydration energy of its ions is not sufficient to overcome the high lattice energy. Therefore, BeO is insoluble in water. On the other hand, \(\mathrm{SO}_{4}^{2-}\) ion is a large anion. Hence, Be²⁺ can easily polarise \(\mathrm{SO}_{4}^{2-}\) ions, making BeSO₄ unstable. Thus, the lattice energy of BeSO₄ is not very high and so it is soluble in water.

(ii) BaOis soluble in water, but BaSO₄ is not. Ba²⁺ is a large cation and O^2- is a small anion. The size compatibility of Ba²⁺ and O^2- is not high. As a result, BaO is unstable. The lattice energy released during its formation is also not very large. It can easily be overcome by the hydration energy of the ions. Therefore, BaO is, soluble in water. In BaSO₄, Ba+ and \(\mathrm{SO}_{4}^{2-}\) are both large-sized. The lattice energy released is high. Hence, it is not soluble in water.

(iii) Lil is more soluble than KI in ethanol. As a result, due to its small size, the lithium ion has a higher polarising power than the potassium ion. It polarises the electron cloud of the iodide ion to a much greater extent than the potassium ion. This causes a greater covalent character in Li than in KI. Hence, Lil is more soluble in ethanol.

Question 30.
Which of the alkali metal is having least melting point?
(a) Na (b) K (c) Rb (d) Cs
Answer:
(d) Atomic size increases as we move down the alkali group. As a result, the binding energies of their atoms in the crystal lattice decrease. Also, the strength of metallic bonds decreases on moving down a group in the periodic table.
This causes a decrease in the melting point. Among the given metals, Cs is the largest and has the least melting point.

Question 31.
Which one of the following alkali metals gives hydrated salts?
(a) Li (b) Na (c) K (d) Cs
Answer:
(a) Smaller the size of an ion, the more highly is it hydrated. Among the given alkali metals, Li is the smallest in size. Also, it has the highest charge density and highest polarising power. Hence, it attracts water molecules more strongly than the other alkali metals. As a result, it forms hydrated salts such as LiCl • 2 H20. The other alkali metals are larger than Li and have weaker charge densities. Hence, they usually do not form hydrated salts.

Question 32.
Which one of the alkaline earth metal carbonates is thermally the most stable?
(a) MgCO₃
(b) CaCO₃
(c) SrCO₃
(d) BaCO₃
Answer:
(d) Thermal stability increases with the increase in the size of the cation present in the carbonate. The increasing order of the cationic size of the given alkaline earth metals is
Mg < Ca < Sr < Ba
Hence, the increasing order of the thermal stability of the given alkaline earth metal carbonates is
MgCO₃ < CaCO₃ < SrCO₃ < BaCO₃