Class 11

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 3 Classification of Elements and Periodicity in Properties Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties

PSEB 11th Class Chemistry Guide Classification of Elements and Periodicity in Properties InText Questions and Answers

Question 1.
What is the basic theme of organisation in the periodic table?
Answer:
The basic theme of organisation of elements in the periodic table is to classify the elements in periods and groups according to their properties. This arrangement makes the study of elements and their compounds simple and systematic. In the periodic table, elements with similar properties are placed in the same group.

Question 2.
Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that? ‘
Answer:
Mendeleev arranged the elements in his periodic table ordered by atomic weight or mass. He arranged the elements in periods and groups in order of their increasing atomic weight. He placed the elements with similar properties in the same group.
However, he did not stick to this arrangement for long. He found out that if the elements were arranged strictly in order of their increasing atomic weights, then some elements did not fit within this scheme of classification. Therefore, he ignored the order of atomic weights in some cases. For example, the atomic weight of iodine is lower than that of tellurium. Still Mendeleev placed tellurium (in Group VI) before iodine (in Group VII) simply because iodine’s properties are so similar to fluorine, chlorine and bromine.

Question 3.
What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modem Periodic Law?
Answer:
Mendeleev’s periodic law : It states that the properties of the elements are a periodic function of their atomic weights.
Modern periodic law : It states that the properties of the elements are a periodic function of their atomic numbers.
Thus, change in the base of classification of elements from atomic weight to atomic number is the basic difference between Mendeleev’s periodic law and the modern periodic law.

Question 4.
On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
Answer:
In the periodic table of the elements, a period indicates the value of the principal quantum number (n) for the outermost shells. Each period begins with the filling of principal quantum number (n). The value of n for the sixth period is 6. For n = 6, azimuthal quantum number (0 can have values of 0, 1, 2, 3, 4, 5.
According to Aufbau’s principle, electrons are added to different orbitals in order of their increasing energies. The energy of the 6d subshell is even higher than that of the 7s subshell.

In the 6th period, electrons can be filled in only 6s, 4f, 5d and 6p subshells. Now 6s has one orbital, 4f has seven orbitals, 5d has five orbitals, and 6p has three orbitals. Therefore, there are a total of sixteen (1+ 7 + 5 + 3 = 16) orbitals available. According to Pauli’s exclusion principle, each orbital can accommodate a maximum of 2 electrons. Thus, 16 orbitals can accommodate a maximum of 32 electrons.

Hence, the sixth period of the periodic table should have 32 elements.

Question 5.
In terms of period and group where would you locate the element with Z = 114?
Answer:
₁₁₄Z = ₈₆[Rn] 7s², 5f¹⁴, 6d¹⁰, 7p²
In the periodic table the element with Z = 114 is located in
Block : p-block (as last electron enters in p-subshell).
Period : 7th (as n = 7 for valence shell).
Group : 14th (for p-block elements, group number = 10 + number of electrons in the valence shell).

Question 6.
Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
Answer:
There are two elements in the 1st period and eight elements in the 2nd period. The third period starts with the element with Z = 11. Now, there are eight elements in the third period. Thus, the 3rd period ends with the element with Z = 18 i.e., the element in the 18th group of the third period has Z = 18. Hence, the element in the 17th group of the third period has atomic number Z = 17.

Question 7.
Which element do you think would have been named by
(i) Lawrence Berkeley Laboratory
(ii) Seaborg’s group.
Answer:
(i) Lawrencium (Lr) with Z = 103 and Berkelium (Bk) with Z = 97
(ii) Seaborgium (Sg) withZ = 106 ,

Question 8.
Why do elements in the same group have similar physical and chemical properties?
Answer:
Same group elements have similar electronic configuration therefore, have similar physical and chemical properties.

Question 9.
What does atomic radius and ionic radius really mean to you?
Answer:
Atomic radius is the radius of an atom. It measures the size of an atom. If the element is a metal, then the atomic radius refers to the metallic radius, and if the element is a non-metal, then it refers to the covalent radius.

Metallic radius is calculated as half the internuclear distance separating the metal cores in the metallic crystal. For example, the internuclear distance between two adjacent copper atoms in solid copper is 256 pm. Thus, the metallic radius of copper is taken as \(\frac{256}{2}\) pm = 128 pm.

Covalent radius is measured as the distance between two atoms when they are found together by a single bond in a covalent molecule. For example, the distance between two chlorine atoms in chlorine molecule is 198 pm. Thus, the covalent radius of chlorine is taken as \(\frac{198}{2}\)pm = 99 pm.

Ionic radius means the radius of an ion (cation or anion). The ionic radii can be calculated by measuring the distances between the cations and anions in ionic crystals.
Since a cation is formed by removing an electron from an atom, the cation has fewer electrons than the parent atom resulting in an increase in the effective nuclear charge. Thus, a cation is smaller than the parent atom. For example, the ionic radius of Na+ ion is 95 pm, whereas the atomic radius of Na atom is 186 pm. On the other hand, an anion is larger in size than its parent atom. This is because an anion has the same nuclear charge, but more electrons than the parent atom resulting in an increased repulsion among the electrons and a decrease in the effective nuclear charge. For example, the ionic radius of F^– ion is 136 pm, whereas the atomic radius of F atom is 64 pm.

Question10.
How do atomic radius vary in a period and in a group? How do you explain the variation?
Answer:
Atomic radius generally decreases from left to right across a period. This is because within a period, the outer electrons are present in the same valence shell and the atomic number increases from left to right across a period, resulting in an increased effective nuclear charge. As a result, the attraction of electrons to the nucleus increases.
On the other hand, the atomic radius generally increases down a group. This is because down a group, the principal quantum number (n) increases which results in an increase of the distance between the nucleus and valence electrons.

Question 11.
What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions.
(i) F^– (ii) Ar (iii) Mg²⁺ (iv) Rb⁺
Answer:
Atoms and ions having the same number of electrons but different nuclear charges are called isoelectronic species. In case of isoelectronic species, as the nuclear charge increases their size decreases.
(i) F^– ion has 9+1 = 10 electrons.
(ii) Ar has 18 electrons.
(iii) Mg²⁺ ion has 12 – 2 = 10 electrons.
(iv) Rb⁺ ion has 37 -1 = 36 electrons.

Question12.
Consider the following species :
N^3-, 0^2-, F^–, Na⁺, Mg²⁺ and Al³⁺
(a) What is common in them?
(b) Arrange them in the order of increasing ionic radii.
Ans. (i) All the given species have same number of electrons (10e^–). Therefore, all are isoelectronic.
(ii) The ionic radii of isoelectronic species decreases with increase in atomic number (as magnitude of the nuclear charge increases with increase in atomic number). Therefore, their ionic radii increase in the order.
Isoelectronic ions = Al³⁺ < Mg²⁺ < Na⁺ < F^– < O^2- < N^3-
Atomic number =13 12 11 9 8 7

Question 13.
Explain why cations are smaller and anions larger in radii than their parent atoms?
Answer:
A cation has a fewer number of electrons than its parent atom, while its nuclear charge remains the same. As a result, the attraction of electrons to the nucleus is more in a cation than in its parent atom. Therefore, a cation is smaller in size than its parent atom.

On the other hand, an anion has one or more electrons than its parent atom, resulting in an increased repulsion among the electrons and a decrease in the effective nuclear charge. As a result, the distance between the valence electrons and the nucleus is more in anions than in its parent atom. Hence, an anion is larger in radius than its parent atom.

Question 14.
What is the significance of the terms-‘isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron gain enthalpy?
Answer:
Ionization enthalpy : It is the minimum amount of energy required to remove an electron from an isolated gaseous atom (A) in its ground state.
X(g) → X⁺ (g) + e^–
The force by which an electron is attracted by nucleus is also affected by the presence of other atoms within its molecule or in the neighbourhood. Therefore, ionization enthalpy is determined in gaseous state because in gaseous state interatomic distances are larger and interatomic forces of attractions are minimum. Further more, ionization enthalpy is determined at a low pressure because it is not possible to isolate a single atom but interatomic attractions can be further reduced by reducing pressure. Due to these reasons, the term isolated gaseous atom in ground state has been included in definition of ionization enthalpy.

Electron gain enthalpy : It is the energy released when an isolated gaseous atom (X) in ground state gains an electron to form gaseous anion.
X(g) + e^– → X^– (g)
The most stable state of an atom is ground state. If isolated gaseous atom is in excited state, comparatively lesser energy will be released on addition of an electron. So, electron gain enthalpies of gaseous atoms must be determined in their ground states. Therefore, the terms ground state and isolated gaseous atom (explained above) has been also included in the definition of electron gain enthalpy.

Question 15.
Energy of an electron in the ground state of the hydrogen atom is -2.18 x 10^-18J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol^-1.
Answer:
It is given that the energy of an electron in the ground state of the hydrogen atom is -2.18 x 10^-18 J.
Therefore, the energy required to remove that electron from the ground state of hydrogen atom is 2.18 x 10^-18 J.
∴ Ionization enthalpy of atomic hydrogen = 2.18 x 10^-18 J
Hence, ionization enthalpy per mol of hydrogen atoms
= 2.18 x 10^-18 x 6.022 x 10²³ J mol^-1
= 1.31 x 10⁶ J mol^-1

Question16.
Among the second period elements the actual ionization enthalpies are in the order Li < B < Be < C < Cl < N < F < Ne. Explain why
(i) Be has higher A^H than B
(ii) O has lower AjH than N and F?
Answer:
(i) Δ_iH of Be is higher than that of B .
Electronic configuration of Be is 1s², 2s²
whereas that of B is 1s², 2s², 2p_x¹
In the case of Be, electron has to be removed from an s-orbital whereas in the case of B, it has to be removed from a p-orbital. It is difficult to remove an s-electron because it is closer to the nucleus than a p-electron hence more energy is required to remove an electron from 2s and Be, than 2p-electron in the case of B. Hence, ionization enthalpy of Be is higher than that of B.

(ii) Electronic configuration of O is
1s², 2s², \(2 p_{x}^{2} 2 p_{y}^{1} 2 p_{z}^{1}\) (neither exactly half-filled nor completely filled)
whereas N is 1s², 2s², \(2 p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1}\) (exactly half-filled)
It is difficult to remove an electron from the valence shell of N because its p-subshell is exactly half-filled and so has more stability whereas O has electronic configuration which is neither completely filled nor exactly half-filled. Therefore, it is easier to remove one electron from O atom. F, due to increased nuclear charge, has more ionization enthalpy than either O or N.

Question17.
How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
Answer:
The first electron in both the cases has to be removed from 3s orbital, but nuclear charge of Na is less than that of Mg. Hence ionization enthalpy of Na is lower than that of Mg.

After the loss of first electron the electronic configuration of Na⁺ is 1s², 2s², 2p⁶, i.e., that of noble gas which is very stable and hence the removal of second electron from Na+ is very difficult. In the case of Mg after the loss of first electron, electronic configuration of Mg⁺ ion is 1s² , 2s² 2p⁶ , 3s¹ . The second electron to be removed is from 3s orbital which is easier.
Hence, second ionization enthalpy of sodium is much larger than second . ionization enthalpy of Mg.

Question 18.
What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
Answer:
The factors responsible for the ionization enthalpy of the main group elements to decrease down a group are listed below:
(i) Atomic size : On moving down the group atomic size increases due to addition of new higher energy shell. As a result, force of attraction of nucleus for valence electrons decreases and ionization enthalpy also decreases.
(ii) Screening effect : On moving down the group, screening effect or shielding effect increases so ionization enthalpy decreases.

Question19.
The first ionization enthalpy values (in kJ mol^-1) of group 13 elements are :

How would you explain this deviation from the general trend?
Answer:
On moving down a group, ionization enthalpy generally decreases due to an increase in the atomic size and shielding. Thus, on moving down group 13, ionization enthalpy decreases from B to Al. But, Ga has higher ionization enthalpy than Al. Al follows immediately after s-block elements, whereas Ga follows after d-block elements. The shielding provided by d-electrons is not very effective. These electrons do not shield the valence electrons very effectively. As a result, the valence electrons of Ga experience a greater effective nuclear charge than those of Al. Further, moving from Ga to In, the ionization enthalpy decreases due to an increase in the atomic size and shielding. But, on moving from In to Tl, the ionization enthalpy again increases. In the periodic table, Tl follows after 4f and Ad electrons. The shielding provided by the electrons in both these orbitals is not very effective. Therefore, the valence electron is held quite strongly by the nucleus. Hence, the ionization energy of Tl is on the higher side.

Question20.
Which of the following pairs of elements would have more negative electron gain enthalpy?
(i) OorF (ii) F or Cl
Answer:
(i) O and F are present in the same period of the periodic table. An F atom has one proton and one electron more than O and as an electron is being added to the same shell, the atomic size of F is smaller than that of O. As F contains one proton more than O, its nucleus can attract the incoming electron more strongly in comparison to the nucleus of O atom. Also, F needs only one more electron to attain the stable noble gas configuration. Hence, the electron gain enthalpy of F is more negative (- 328 kJ mol^-1) than that of O (-141 kJ mol^-1).

(ii) F and Cl belong to the same group of the periodic table. The electron gain enthalpy usually becomes less negative on moving down a group.
However, in this case, the value of the electron gain enthalpy of Cl is more negative than that of F. This is because the atomic size of F is smaller than that of Cl. In F, the electron will be added to quantum level n = 2, but in Cl, the electron is added to quantum level n = 3. Therefore, there are less electron-electron repulsion in Cl and an additional electron can be accommodated easily. Hence, the electron gain enthalpy of Cl is more negative (- 349kJ mol^-1) than that of F (-328kJ mol^-1).

Question 21.
Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.
Answer:
When an electron is added to O atom to form O^– ion, energy is released. Thus, the first electron gain enthalpy of O is negative.
0(g) + e^– → O^– (g); Δ_egH = -141 kJ mol^-1

On the other hand, when an electron is added to O ion to form O ion, energy has to be given out in order to overcome the strong electronic repulsion. Thus, the second electron gain enthalpy of O is positive.
O^–(g) + e^– → O^2-(g); Δ_egH = +780 kJ mol^-1

Question 22.
What is the basic difference between the terms electron gain enthalpy and electronegativity? •
Answer:
Electron gain enthalpy is the measure of the tendency of an isolated gaseous atom to accept an electron, whereas electronegativity is the measure of the tendency of an atom in a chemical compound to attract a shared pair of electrons.

Question 23.
How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?
Answer:
On Pauling scale, the electronegativity of nitrogen, (3.0) indicates that it is sufficiently electronegative. But it is not correct to say that the electronegativity of nitrogen in all the compounds is 3. It depends upon its state of hybridisation in a particular compound, greater the percentage of s-character, more will be the electronegativity of the element. Thus, the electronegativity of nitrogen increases in moving from sp³ hybridised orbitals to sp hybridised orbitals i.e., as sp³ < sp² < sp.

Question 24.
Describe the theory associated with the radius of an atom as it
(a) gains an electron
(b) loses an electron
Answer:
(a) When an atom gains an electron, its size increases. When an electron is added, the number of electrons goes up by one. This results in an increase in repulsion among the electrons. However, the number of protons remains the same. As a result, the effective nuclear charge of the atom decreases and the radius of the atom increases. For example, ionic radius of Cl^– ion is greater than the radius of its parent atom Cl.

(b) When an atom loses an electron, the number of electrons decreases by one while the nuclear charge remains the same. Therefore, the interelectronic repulsions in the atom decrease. As a result, the effective nuclear charge increases. Hence, the radius of the atom decreases.
For example, ionic radius of Na⁺ is smaller than the radius of its parent atom Na.

Question 25.
Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.
Answer:
The ionization enthalpy of an atom depends on the number of electrons and protons (nuclear charge) of that atom. Now, the isotopes of an element have the same number of protons and electrons. Therefore, the first ionization enthalpy for two isotopes of the same element should be the same.

Question 26.
What are the major differences between metals and non-metals?
Answer:

Metals	Non-metals
1. Metals can lose electrons easily.	Non-metals cannot lose electrons easily.
2. Metals cannot gain electrons easily.	Non-metals can gain electrons easily.
3. Metals generally form ionic compounds.	Non-metals generally form covalent compounds.
4. Metal oxides are basic in nature.	Non-metal oxides are acidic in nature.
5. Metals have low ionization enthalpies.	Non-metals have high ionization enthalpies.
6. Metals have less negative electron gain enthalpies.	Non-metals have high negative electron gain enthalpies.
7. Metals are less electronegative. They are rather electropositive elements.	Non-metals are electronegative.
8. Metals have a high reducing power.	Non-metals have a low reducing power.

Question27.
Use the periodic table to answer the following questions.
(a) Identify an element with five electrons in the outer subshell.
(b) Identify an element that would tend to lose two electrons.
(c) Identify an element that would tend to gain two electrons.
(d) Identify the group having metal, non-metal liquid as well as gas at the room temperature.
Answer:
(a) General electronic configuration of elements having five electrons in the outer sub shell is ns2 np . This configuration belongs to halogen family, i.e., F, Cl, Br, I, At.
(b) Elements of second group are known as alkaline earth metals (Mg, Ca, Sr, Ba, etc). Their general electronic configuration for valence shell is ns2. These elements form dipositive cations by the lose of two electrons easily.
(c) 16th group elements such as O, S, Se, etc., have a tendency to accept two electrons because by the gain of two electrons they attain noble gas configuration. Their general electronic configuration for valence shell is ns² np⁴.
(d) Group 1 or 17 of the periodic table contains metal, non-metal, liquid as well as gas at the room temperature, e.g., H2 is a non-metal and in gaseous state at room temperature. All other elements of this group are metals. Cs is a liquid metal. Similarly, Br₂ is a liquid non-metal while other elements of this group are gaseous non-metals. Iodine can form I⁺ so it consists some metallic properties.

Question 28.
The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas that among group 17 elements is F > Cl > Br > I. Explain.
Answer:
The elements present in group 1 have only 1 valency electron, which they tend to lose. Group 17 elements, on the other hand, need only one electron to attain the noble gas configuration. On moving down group 1, the ionization enthalpies decrease. This means that the energy required to lose the valence electron decreases. Thus reactivity increases on moving down a group. Thus, the increasing order of ( reactivity among group 1 elements is as follows :
Li < Na < K < Rb < Cs In group 17, as we move down the group from Cl to I, the electron gain enthalpy becomes less negative i. e., its tendency to gain electrons decreases on moving down a group. Thus, reactivity decreases down a group. The electron gain enthalpy of F is less negative than Cl. Still, it is the most reactive halogen. This is because of its low bond dissociation energy. Thus, the decreasing order of reactivity among group 17 elements is as follows : F > Cl > Br > I.

Question 29.
Write the general outer electronic configuration of s-, p-, d- and f- block elements.
Answer:

Element	General outer electronic configuration
s-block	ns1-2, where n = 2 – 7
p-block	ns2 np1-6      , where n = 2 – 6
d-block	(n – 1)d1–10ns(0 – 2), where n = 4 – 7
f-block	(n – 2)f1–14 (n – 1)d0- 1ns2, where n = 6 – 7

Question30.
Assign the position of the element having outer electronic configuration
(i) ns² np⁴ for n = 3 (ii) (n – 1 )d²ns² for n = 4, and
(iii) (n – 2)f⁷(n – 1)d¹ns² for n = 6, in the periodic table.
Answer:
(i) ns²np⁴ for n = 3; it is 3s²3p⁴
The complete electronic configuration is 1s², 2s², 2p⁶, 3s², 3p⁴
Atomic number = 2 + 2+ 6+ 2 + 4 = 16
The element is sulphur in the 3rd period and in Group 16 (p-block)

(ii) (n – 1)d²ns² for n = 4; it is 3d²4s²
The complete electronic configuration is
1s², 2s²,2p⁶, 3s², 3p⁶, 3d², 4s² Atomic number is 22, the element is Titanium.
It is a transition element present in the 4th period and in Group 4.

(iii) (n – 2)f⁷(n – l)d¹ns² for n = 6; it is 4f⁷5d¹6s²
Its complete electronic configuration is
1s², 2s², 2p⁶, 3s², 3p⁶, 3d¹⁰, 4s², 4p⁶, 4d¹⁰, 4f⁷, 5s², 5p⁶, 5d¹, 6s²
It is Gadolinium (Gd)
It is an inner transition element, belongs to Lanthanoid series or 4f series. It is an f-block element.

Question 31.
The first (Δ_i,H₁) and the second (Δ_iH₂) ionization enthalpies (in kJ mol^-1) and the (Δ_egH) electron gain enthalpy (in kJ mol^-1) of a few elements are given below:

Which of the above elements is likely to be :
(a) the least reactive element.
(b) the most reactive metal.
(c) the most reactive non-metal
(d) the least reactive non-metal.
(e) the metal which can form a stable binary halide of the formula MX₂ (X = halogen).
(f) the metal which can form a predominantly stable covalent halide of the formula MX (X = halogen)?
Answer:
(a) Element V is likely to be the least reactive element. This is because it has the highest first ionization enthalpy (Δ_i,H₁) and a positive electron gain enthalpy (Δ_eg,H₁).
(b) Element II is likely to be the most reactive metal as it has the lowest first ionization enthalpy (Δ_i,H₁) and the highest negative electron gain enthalpy (AegH).
(c) Element III is likely to be the most reactive non-metal as it has a high first ionization enthalpy Δ_i,H₁) and the highest negative electron gain enthalpy (Δ_eg,H).
(d) Element IV is likely to be the least reactive non-metal since it has a very high first ionization enthalpy (Δ_i,H₂) and a positive electron gain enthalpy (Δ_eg,H).
(e) Element VI has a low negative electron gain enthalpy (Δ_eg,H). Thus it is a metal. Further, it has the lowest second ionization enthalpy (Δ_i,H₂). Hence, it can form a stable binary halide of the formula MX₂ ( X = halogen).
(f) Element I has low Δ_i,H₁ but a very high Δ_i,H₂. It has less negative electron gain enthalpy. So, element (I) is alkali metal. The given values for element I match with Li. Lithium firms predominantly stable covalent halide of the formula Mx.

Question32.
Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.
(a) Lithium and oxygen
(b) Magnesium and nitrogen
(c) Aluminium and iodine
(d) Silicon and oxygen
(e) Phosphorus and fluorine
(f) Element 71 and fluorine
Answer:
(a) Li₂0
(b) Mg₃N₂
(c) AlI₃
(d) SiO₂
(e) PF₃ or PF₅
(f) The element with the atomic number 71 is Lutetium (Lu). It has valency 3. Hence, the formula of the compound is LuF₃.

Question 33.
In the modem periodic table, the period indicates the value of:
(a) atomic number
(b) atomic mass
(c) principal quantum number
(d) azimuthal quantum number.
Answer:
(c) The value of the principal quantum number (n) for the outermost shell or the valence shell indicates a period in the Modern periodic table.

Question 34.
Which of the following statements related to the modem periodic table is incorrect?
(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell.
(b) The df-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.
(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
(d) The block indicates value of azimuthal quantum number (l) for the last subshell that received electrons in building up the electronic configuration.
Answer:
The statement (b) is incorrect. The correct statement (b) is that the d-block has 10 columns because a maximum of 10 electrons can occupy all the orbitals in a d-subshell.
All other given statements are correct.

Question 35.
Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?
(a) Valence principal quantum number (n)
(b) Nuclear charge (Z)
(c) Nuclear mass
(d) Number of core electrons.
Answer:
(c) Nuclear mass (protons + neutrons) does not affect the valence shell, only protons i. e., nuclear charge affects the valence shell.

Question 36.
The size of isoelectronic species F^–, Ne and Na⁺ is affected by
(a) Nuclear charge (Z)
(b) Valence principal quantum number (n)
(c) Electron-electron interaction in the outer orbitals.
(d) None of the factors because their size is same.
Answer:
(a) The size of an isoelectronic species increases with a decrease in the nuclear charge (Z).

Question 37.
Which one of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.
Answer:
(d) Electrons in orbitals bearing a lower n value are more attracted to the nucleus than electrons in orbitals bearing a higher n value. Hence, the removal of electrons from orbitals bearing a higher n value is easier than the removal of electrons from orbitals having a lower n value.

Question 38.
Considering the elements B, Al, Mg and K the correct order of their metallic character is:
(a) B > Al > Mg > K
(b) Al > Mg > B > K
(c) Mg > Al > K > B
(d) K > Mg > Al > B
Answer:
(d) In a group, metallic character increases from top to bottom as ; ionisation energy decreases and in a period metallic character decreases from left to right as tendency to lose electron decreases. Therefore, the correct order is K > Mg > Al > B.

Question 39.
Considering the elements B, C, N, F and Si the correct order of their non-metallic character is
(a) B > C > Si > N > F
(b) Si > C > B > N > F
(c) F > N > C > B > Si
(d) F > N > C > Si > B
Answer:
(c) The non-metallic character of elements increases from left to right across a period. Thus, the decreasing order of non-metallic character is F > N > c > B.
Again, the non metallic character of elements decreases down a group. ;
Thus, the decreasing order of non-metallic characters of C and Si are C > Si. However, Si is less non-metallic than B i.e., B > Si.
Hence, the correct order of non-metallic characters is F > N > C > B > Si.

Question 40.
Considering the elements F, Cl, O and N the correct order of their chemical reactivity in terms of oxidizing property is :
(a) F > Cl > O > N
(b) F > O > Cl > N
(c) Cl > F > O > N
(d) O > F > N > Cl
Answer:
(b) In a group, oxidising power decreases from top to bottom as the size increases but when we move left to right in a period it increases because size decreases.
Therefore, among F, Cl, O and N, the oxidising power decreases in the order F > O > Cl > N.

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 6 Thermodynamics Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 6 Thermodynamics

Very Short Answer Type Questions

Question 1.
Identify the state functions and path functions out of the following.
Enthalpy, entropy, heat, temperature, work, free energy.
Answer:
State function Enthalpy, entropy, temperature, free energy.
Path function Heat, work

Question 2.
At 1 atm will the Δ_fH⁰ be zero for Cl₂(g) and Br₂(g)? Explain.
Answer:
Δ_fH⁰ for Cl₂(g) will be zero but Δ_fH⁰ for Br₂(g) will not be zero because liquid bromine state is elementary state not gaseous.

Question 3.
Why for predicting the spontaneity of a reaction, free energy criteria is better than the entropy criteria?
Answer:
Criteria of free energy change is better because it requires free energy change of the system only whereas the entropy change requires the total entropy change of the system and the surroundings.

Question 4.
Water can be lifted into the water tank at the top of the house with the help of a pump. Then why is it not considered to be spontaneous?
Answer:
A spontaneous process should occur continuously by itself after initiation. But this is not so in the given case because water will go up so long as the pump is working.

Question 5.
Given that ΔH = 0 for mixing of two gases. Explain whether the diffusion of these gases into each other in a closed container is a spontaneous process or not?
Answer:
It is a spontaneous process because although ΔH = 0, i.e., energy factor has no role to play but randomness increases, i.e., randomness factor favours the process.

Question 6.
Under what condition, the heat evolved or absorbed in a reaction is equal to its free energy change?
Answer:
As ΔG = ΔH – TΔS. Thus, ΔG = ΔH only when either the reaction is carried out at 0 K or the reaction is not accompanied by any entropy change, i.e., ΔS = 0.

Question 7.
In the equation, N₂(g) + 3H₂(g) ⇌ 2NH₃(g) what would be the sign of work done?
Answer:
The sign of work done will be positive, i.e., work will be done on the system due to decrease in volume.

Question 8.
The molar enthalpy of vaporisation of acetone is less than that of water. Why?
Answer:
Enthalpy of vaporisation of water is more than that of acetone because there is strong hydrogen bonding in H₂O molecules.

Question 9.
One mole of acetone requires less heat to vaporise than 1 mole of water. Which of the two liquids has higher enthalpy of vaporisation?
Answer:
Less the heat required to vaporise 1 mole of a liquid, less is its enthalpy of vaporisation. Hence, water has higher enthalpy of vaporisation.

Question 10.
Which quantity out of Δ_rG and Δ_rG° will be zero at equilibrium?
Answer:
Δ_rG = Δ_rG° + RT In K
At equilibrium, 0 (zero) = Δ_rG° + RT In K
(v Δ_rG = 0)
or Δ_rG° = -RT In It;
Δ_rG° = 0 when it = 1
For all other values of K, Δ_rG° will be non-zero.

Short Answer Type Questions

Question 1.
Define the following :
(i) First law of thermodynamics.
(ii) Standard enthalpy of formation.
Answer:
(i) First law of thermodynamics : It states that energy can neither be created nor be destroyed. The energy of an isolated system is constant. ΔU = q + w
(ii) Standard Enthalpy of Formation : It is defined as the amount of heat evolved or absorbed when one mole of the compound is formed from its constituent elements in their standard states.

Question 2.
Give reason for the following:
(i) Neither q nor w is a state function but q + w is a state function.
(ii) A real crystal has more entropy than an ideal crystal.
Answer:
(a) q + w = ΔU
As ΔU is a state function hence, q + w is a state function.
(b) A real crystal has some disorder due to the presence of defects in its structural arrangement whereas ideal crystal does not have any disorder. Hence, a real crystal has more entropy than an ideal crystal.

Question 3.
Represent the potential energy/enthalpy change in the following processes graphically
(i) Throwing a stone from the ground to roof.
(ii) \(\frac{1}{2}\)H₂(g) + \(\frac{1}{2}\)Cl₂(g) ⇌ HCl(g); Δ_rH^s = – 92.32kJ mol^-1
In which of the processes potential energy/enthalpy change is contributing factor to the spontaneity?
Answer:

Energy increases in (a) and it decreases in (b). Hence, in process (b), enthalpy change is the contributing factor to the spontaneity.

Question 4.
A man takes a diet equivalent to 10000 kJ per day and does work, by expending his energy in all forms equivalent to 12500 kJ per day. What is change in internal energy per day? If the energy lost was stored as sucrose (1632 kJ per 100 g), how many days should it take to lose 2 kg of his weight? (Ignore water loss)
Answer:
Energy taken by a man = 10000 kJ
Change in internal energy per day = 12500 -10000 = 2500 kJ
The energy is lost by the man as he expends more energy than he takes.
Now 100 g of sugar corresponds to energy = 1632 kJ loss in energy.
2000 g of sugar corresponds to energy = \(\frac{1632 \times 2000}{100}\) = 32640 kJ
∴ Number of days required to lose 2000 g of weight or 32640 kJ of energy = \(\frac{32640}{2500}\) = 13 days

Question 5.
Give the appropriate reason :
(i) It is preferable to determine the change in enthalpy rather than the change in internal energy.
(ii) It is necessary to define the ‘standard state’.
(iii) It is necessary to specify the phases of the reactants and products in a thermochemical equation.
Answer:
(i) Because it is easier to make measurement under constant pressure than under constant volume conditions.
(ii) Enthalpy change depends upon the conditions in which a reaction is carried out. For making the comparison of results obtained by different people meaningful, the reaction conditions must be well-defined.
(iii) Because enthalpy depends upon the phase of reactants and products.

Long Answer Type Questions

Question 1.
(i) A cylinder of gas supplied by a company is assumed to contain 14 kg of butane. If a normal family requires 20000 kJ of energy per day for cooking, how long will the cylinder last?
(ii) If the air supplied to the burner is insufficient, a portion of gas escapes without combustion. Assuming that 25% of the gas is wasted due to this inefficiency, how long will the cylinder last (Heat of combustion of butane = 2658kJ!mol.)?
Answer:
(i) Molecular formula of butane = C₄H₁₀
Molecular mass of butane = 4 x 12 +10 x 1 = 58
Heat of combustion of butane 2658 kJ mol^-1
1 mole.or 58 g of butane on complete combustion gives heat = 2658 kJ
∴ 14 x 10³ g of butane on complete combustion will give heat
= \(\frac{2658 \times 14 \times 10^{3}}{58}\) = 641586 kJ
The family needs 20000 kJ of heat per day.
∴ 20000 kJ of heat is used for cooking by a family in 1 day.
∴ 641586 kJ of heat will be used for cooking by a family in
= \(\frac{641586}{20000}\) = 32days
The cylinder will last for 32 days

(ii) 25 per cent of the gas is wasted due to inefficiency. This means that only 75% of butane gets combusted. Therefore, the energy produced by
75% combustion of butane = \(\frac{641586 \times 75}{100}\) = 481189.5 kJ
∴ The number of days the cylinder will last = \(\frac{481189.5}{20000}\) = 24 days.

Question 2.
10 moles of an ideal gas expand isothermally and reversibly from a pressure of 5 atm to 1 atm at 300 K. What is the largest mass that can be lifted through a height of 1 m by this expansion?
Answer:
W_exp = -2.303 nRT log \(\frac{p_{1}}{p_{2}}\)
= -2.303(10) x (8.314)(300) log \(\frac{5}{1}\) = – 40.15 x 10³ J
If M is the mass that can be lifted by this work through a height of 1 m, then work done = Mgh
40.15 x 10³ J = M x 9.81 ms^-1 x 1 m
or M = \(\frac{40.15 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}}{9.81 \mathrm{~m} \mathrm{~s}^{-2} \times 1 \mathrm{~m}}\) [∵ J = kg m²s^-2]
= 4092.76 kg

Punjab State Board Syllabus PSEB 11th Class Agriculture Book Solutions Guide Pdf in English Medium and Punjabi Medium are part of PSEB Solutions for Class 11.

PSEB 11th Class Agriculture Guide | Agriculture Guide for Class 11 PSEB in English Medium

Agriculture Guide for Class 11 PSEB | PSEB 11th Class Agriculture Book Solutions

PSEB 11th Class Agriculture Book Solutions in Hindi Medium

• Chapter 1 ख़रीफ़ की फसलें
• Chapter 2 ख़रीफ़ की सब्ज़ियाँ
• Chapter 3 फूलों की खेती
• Chapter 4 कृषि उत्पादों का मंडीकरण
• Chapter 5 बीज, खाद और कीड़ेमार दवाइयों का क्वालिटी कन्ट्रोल
• Chapter 6 पशु पालन
• Chapter 7 दूध और दूध से बनने वाले पदार्थ
• Chapter 8 मुर्गी पालन
• Chapter 9 सूअर, भेड़ें/बकरियां और खरगोश पालना
• Chapter 10 मछली पालन
• Chapter 11 कुछ नए कृषि विषय
• Class 11 Agriculture Objective Questions

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 5 States of Matter Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 5 States of Matter

Very Short Answer Type Questions

Question 1.
Name two intermolecular forces that exist between HF molecules in liquid state.
Answer:
HF are polar covalent molecules. In liquid state, there are dipole-dipole interactions and H-bonding.

Question 2.
Explain why Boyle’s law cannot be used to calculate the volume of a real gas when it is converted from its initial state to final state by an adiabatic expansion.
Answer:
During adiabatic expansion, temperature is lowered and therefore, Boyle’s law cannot be applied.

Question 3.
Boyle’s law states that at constant temperature, if pressure is increased on a gas, volume decreases and vice-versa. But when we fill air in a balloon, volume as well as pressure increase. Why?
Answer:
The law is applicable only for a definite mass of the gas. As we fill air into the balloon, we are introducing more and more air into the balloon.
Thus, we are increasing the mass of air inside. Hence, the law is not applicable.

Question 4.
What will be the molar volume of nitrogen and argon at 273.15 K and 1 atm?
Answer:
Every gas has 22.4 L molar volume at 273.15 K and 1 atm pressure (STP).

Question 5.
A gas that follows Boyle’s law, Charles’ law and Avogadro’s law is called an ideal gas. Under what conditions a real gas would behave ideally?
Answer:
At low pressure and high temperature, a real gas behaves as an ideal gas.

Question 6.
Explain why temperature of a boiling liquid remains constant?
Answer:
This is because at the boiling point, the heat supplied is used up in breaking off the intermolecular forces of attraction of the liquid to change it into vapour and not for raising the temperature of the liquid.

Question 7.
Assuming C02 to be van der Waals’ gas, calculate its Boyle temperature.
Given a = 3.59 L² atm mol^-2 and b = 0.0427 L mol^-1.
\(T_{b}=\frac{a}{R b}=\frac{3.59 \mathrm{~L}^{2} \mathrm{~atm} \mathrm{} \mathrm{mol}^{-2}}{\left(0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)\left(0.0427 \mathrm{~L} \mathrm{~mol}^{-1}\right)}\) = 1025.3 K

Question 8.
Name two phenomena that can be explained on the basis of surface tension.
Answer:
Surface tension can explain
(i) capillary action, i.e., rise or fall of a liquid in capillary,
(ii) spherical shape of small liquid drops.

Question 9.
Why are the gases helium and hydrogen not liquefied at room temperature by applying very high pressure?
Answer:
Because their, critical temperature is lower than room temperature. (Gases cannot be liquefied above the critical temperature by applying even very high pressure).

Question 10.
What would have happened to the gas if the molecular collisions were not elastic?
Answer:
On every collision, there would have been loss of energy. As a result, the molecules would have slowed down and ultimately settle down in the vessel. Moreover, the pressure would have gradually reduced to zero.

Short Answer Type Questions

Question 1.
(i) What do you mean by ‘Surface Tension’ of a liquid?
(ii) Explain the factors which can affect the surface tension of a liquid.
Answer:
(i) Surface tension : It is defined as the force acting per unit length perpendicular to the line drawn on the surface. It’s unit is Nm-1.
(ii) Surface tension of a liquid depends upon the following factors :
(a) Temperature : Surface tension decreases with rise in temperature. As the temperature of the liquid increases, the average kinetic energy of the molecules increases. Thus, there is a decrease in intermolecular force of attraction which decrease the surface tension.
(b) Nature of the liquid : Greater the magnitude of
intermolecular forces of attraction in the liquid, greater will be the value of surface tension.

Question 2.
A neon-dioxygen mixture contains 70.6 g dioxygen and 167.5g neon. If pressure of the mixture of gases in the cylinder is 25 bar.
What is the partial pressure of dioxygen and neon in the mixture?
Answer:

Alternatively, mole fraction of neon = 1 – 0.21 = 0.79
Partial pressure of a gas = mole fraction x total pressure
⇒ Partial pressure of oxygen = 0.21 x (25bar) = 5.25bar
Partial pressure of neon = 0.79 x (25bar) = 19.75bar

Question 3.
Give reasons for the following:
(i) The size of weather balloon becomes larger and larger as it ascends into higher altitudes.
(ii) Tyres of automobiles are inflated to lesser pressure in summer than in winter.
Answer:
(i) As we go to higher altitudes, the atmospheric pressure decreases.
Thus, the pressure outside the balloon decreases. To regain equilibrium with the external pressure, the gas inside expands to decrease its pressure, Hence, the size of the balloon increases.
(ii) In summer, due to higher temperature, the average kinetic energy of the air molecules inside the tyre increases, i.e., molecules start moving faster. Hence, the pressure on the walls of the tube increases. If pressure inside is not kept low at the time of inflation, at higher temperature, the pressure may become so high that the tyre may burst.

Question 4.
On the basis of intermolecular forces and thermal energy, explain why .
(i) a solid has rigidity but liquids do not have rigidity?
(ii) gases have high compressibility but liquids and solids have poor compressibility?
Answer:
(i) It is because in solids, the intermolecular forces are very strong and predominate over thermal energy but in liquid, these forces are no longer strong enough.
(ii) Because of very weak intermolecular forces and high thermal energy, molecules of gases are far apart. That is why gases are highly compressible.

Question 5.
A gas is enclosed in room. The temperature, pressure, density and number of moles respectively are t°C,p atm, g cm^-3 and n moles.
(i) What will be the pressure, temperature, density and number of moles in each compartment, if room is partitioned into four equal compartments?
(ii) What will be the value of pressure, temperature, density and number of moles in each compartment if the walls between the two compartments (say 1 and 2) are removed?
(iii) What will be the values of pressure, temperature, density and number of moles, if an equal volume of gas at pressure
(p) and temperature (t) is let inside the same room? .
Answer:
(i) (a) Pressure in each compartment is same, (p atm)
(b) Temperature will remain same (t°C).
(c) Density will remain same (d g cm^-3).
(d) Because of partition, volume of each compartment becomes 1/4 and the number of molecules also become 1/4. The number of moles in each compartment will be n/4.

(ii) (a) Pressure will remain same (p atm).
(b) Temperature will remain same (t°C).
(c) Density will remain same (d g cm^-3).
(d) The number of moles in each compartment will be n/2.

(iii) (a) Pressure will be doubled (2p atm).
(b) Temperature will remain same.
(c) Density will remain same (d g cm^-3) ,
(d) Number of moles will be doubled i.e., 2n.

Long Answer Type Questions

Question 1.
Explain the following:
(i) The boiling point of a liquid rises on increasing pressure. ;
(ii) Drops of liquid assume spherical space.
(iii) The boiling point of water (373 K) is abnormally high when compared to that of H₂S (211.2 K).
(iv) The level of mercury in capillary tube is lower than the s level outside when a capillary tube is inserted in the mercury.
(v) Tea or coffee is sipped from a saucer when it is quite hot.
Answer:
(i) A liquid boils when its vapour pressure becomes equal to the atmospheric pressure. An increase in pressure on liquid, therefore, causes a rise in the boiling temperature of the liquids.
(ii) Liquids have a property, called surface tension, due to which liquids tend to contract (to decrease the surface area). For a given volume of a liquid, since a sphere has the least surface area, hence the liquids tend to form spherical droplet.
(iii) The extensive hydrogen bonding in water gives a polymeric structure. This makes the escape of molecules from the liquid more difficult. Therefore, water requires higher temperature to bring its vapour pressure equal to the atmospheric pressure.
On the other hand, sulphur being less electronegative, does not form hydrogen bonds with H of H2S. As a result, H2S has low boiling point.
(iv) The cohesive forces in mercury are much stronger than the force of adhesion between glass and mercury. Therefore, mercury-glass contract angle is greater than 90°C.
As a result, the vertical component of the surface tension forces acts vertically downward, thereby lowering the level of mercury column in the capillary tube.
(v) Evaporation causes cooling and the rate of evaporation increases with an increase in the surface area. Since, saucer has a large surface area, hence tea/coffee taken in a saucer cools quickly.

Question 2.
Nitrogen molecule (N₂) has radius of about 0.2 nm. Assuming that nitrogen molecule is spherical in shape, calculate
(i) volume of a single molecule of N₂.
(ii) the percentage of empty space in one mole of N₂ gas at STP.
Answer:
(i) The volume of a sphere = \(\frac{4}{3}\)πr³ nr where Volume of a molecule of N₂
= \(\frac{4}{3} \times \frac{22}{7}\) x (2 x 10^-8)³ cm³ – 3.35 x 10^-23 cm³

(ii) To calculate the empty space, let us first find the total volume of 1 mole (6.022 x 10²³ molecules) of N₂.
Volume of 6.022 x 10²³ molecules of N₂
= 3.35 x 10^-23 x 6.022 x 10²³ = 20.17 cm³
Now, volume occupied by 1 mole of gas at STP
= 22.4 litre = 22400 cm³
Empty volume = Total volume of gas – Volume occupied by molecules
= (22400 – 20.17) cm³ – 22379.83 cm³
∴ Percentage of empty space = \(\frac{Empty space}{Total volume}\) x 100
= \(\frac{22379.83}{22400}\) x 100 = 99.9%
Thus, 99.9% of space of 1 mole of N₂ at STP is empty.

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 11 Transport in Plants Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 11 Transport in Plants

PSEB 11th Class Biology Guide Transport in Plants Textbook Questions and Answers

Question 1.
What are the factors affecting the rate of diffusion?
Answer:
Factors affecting the rate of diffusion are as follows:

• Gradient of concentration
• Permeability of membrane
• Temperature
• Pressure

Question 2.
What are porins? What role do they play in diffusion?
Answer:
The porins are proteins that form huge pores in the outer membranes of the plastids, mitochondria and some bacteria allowing molecules up to the size of small proteins to pass through. Porins facilitate diffusion.

Question 3.
Describe the role played by protein pumps during active transport in plants.
Answer:
Protein pumps use energy to carry substances across the cell membrane. These pumps can transport substances from a low concentration to a high concentration (‘uphill’ transport). Transport rate reaches a maximum when all the protein transporters are being used or are saturated. Like enzymes, the carrier protein is very specific in what it carries across the membrane. These proteins are sensitive to inhibitors that react with protein side chains.

Question 4.
Explain why pure water has the maximum water potential.
Answer:
Water molecules possess kinetic energy. In liquid and gaseous form, they are in random motion that is both rapid and constant. The greater the concentration of water in a system, the greater is its kinetic energy or ‘water potential’. Hence, pure water will have the greatest water potential. Water potential is denoted by the Greek symbol psi or \p and is, expressed in pressure units such as pascals (Pa).

Question 5.
Differentiate between the following:
(a) Diffusion and Osmosis
(b) Transpiration and Evaporation
(c) Osmotic pressure and Osmotic Potential
(d) Imbibition and Diffusion
(e) Apoplast and Symplast Pathways of Movement of Water in Plants
(f) Guttation and Transpiration

(a) Differences between Diffusion and Osmosis

Diffusion	Osmosis
1. It is a movement of molecules from high concentration to low concentration.	It is a movement of molecules from high concentration to low concentration
2. It does not require any driving force.	It occurs in response to a driving force.

(b) Differences between Transpiration and Evaporation

Transpiration	Evaporation
1. It is the loss of water through the aerial parts of plants.	It is the loss of water from free surface of water.
2. It occurs in living tissues.	It occurs in non-living surfaces.
3. It is both physical and physiological process.	It is only a physical process, controlled by environmental factors.

(c) Differences between Osmotic Pressure and Osmotic Potential

Osmotic Pressure	Osmotic Potential
1. It is the pressure required to stop the movement of water molecules through a semipermeable membrane.	It is the amount by which water potential is reduced by the presence of solute.
2. Osmotic pressure is the positive pressure.	Osmotic potential is negative.

(d) Differences between Imbibition and Diffusion

Imbibition	Diffusion
It is a special type of diffusion, where water is absorbed by solids-colloids causing them to increase in volume. For example, absorption of water by dry seeds and dry wood.	In diffusion, molecules move in a random fashion. It is not dependent on a living system.

(e) Differences between Apoplast and Symplast Pathways of Movement of Water in Plants

Apoplast	Symplast
1. It is the system of adjacent cell walls that is continuous throughout the plant except casparian strips of the endodermis of the roots.	It is the system of interconnected protoplast.
2. Water moves through the intercellular spaces and the walls of cells.	Water travels through the cytoplasm
3. Movement does not involve crossing the cell membrane.	Water has to move in cells through the cell membrane.

(f) Differences between Guttation and Transpiration

Guttation	Transpiration
1. It occurs through hydathodes, present at the vein ends.	It occurs through general surface stomata and lenticles.
2. It occurs in leaves only.	It can occur through all aerial parts.
3. It does not occur in deficient water conditions and never leads to wilting.	It can occur in water deficient conditions leading to wilting.
4. It is regulated by humidity, temperature and presence of water in soil.	It is regulated by a number of external and internal factors such as relative humidity, temperature, opening and closing of stomata, etc.

Question 6.
Briefly describe water potential. What are the factors affecting it?
Answer:
Water potential is the potential energy of water relative to pure free water (e.g., deionised water). It quantifies the tendency of water to move from one area to another due to osmosis, gravity, mecanical pressure or matrix effects including surface tension. Water potential is measured in units of pressure and is commonly represented by the Greek letter (psi). This concept has proved especially useful in understanding water movement within plants, animals and soil.

Water potential of a cell is affected by both solute and pressure potential. The relationship between them is as follows:
Ψ_w = Ψ_s + Ψ_p

Question 7.
What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?
Answer:
If a pressure greater than atmospheric pressure is applied to pure water or a solution its water potential increases. It is equivalent to pumping water from one place to another. Pressure can be build up in a plant system when water enters a plant cell due to diffusion causing a pressure build up against the cell wall. It makes the cell turgid, this increases the pressure potential. Pressure potential is usually positive. It is denoted by Ψ_s.

Question 8.
(a) With the help of well-labelled diagrams, describe the process of plasmolysis in plants, giving appropriate examples.
(b) Explain what will happen to a plant cell if it is kept in a solution having higher water potential.
Answer:
(a) Plasmolysis occurs when water moves out of the cell and the cell membrane of a plant cell shrinks away from its cell wall. This occurs when the cell is kept in a solution that is hypertonic (has more solutes) to the protoplasm. Water moves out from the cell through diffusion and causes the protoplasm to shrink away from the walls. In such situation, cell becomes plasmolysed.

When the cell is placed in an isotonic solution. There is not flow of water towards the inside or outside. If the external solution balances the osmotic pressure of the cytoplasm, it is said to be isotonic. When the water flow into the cell and out of the cells are in equilibrium the cell is called flaccid.

(b) When the plant cell is kept in a solution having high water potential (hypotonic solution or dilute solution as compared to cytoplasm), water diffuses into the cell causing the cytoplasm to build up a pressure against the wall, called turgor pressure. The pressure exerted by the protoplasts due to entry of water against the rigid walls is called pressure potential (Ψ_p). Because of the rigidity of the cell wall, the cell does not rupture. This turgor pressure is ultimately responsible for enlargement of cells.

Question 9.
How is the mycorrhizal association helpful in absorption of water and minerals in plants?
Answer:
A mycorrhiza is a symbiotic association of a fungus with a root system. The fungal filaments form a network around the young root or they penetrate the root cells. The hyphae have a very large surface area that absorb mineral ions and water from the soil from a much larger volume of soil that perhaps a root cannot do. The fungus provides minerals and water to the roots, in turn the roots provide sugars and N-containing compounds to the mycorrhizae.

Question 10.
What role does root pressure play in water movement in plants?
Answer:
Root pressure can provide only modest push during water transport in plants. The main role of root pressure is in re-establishing the continuous chain of water molecules in the xylem. The continuous chain often breaks due to enormous tension created by transpiration pull.

Question 11.
Describe transpiration pull model of water transport in plants. What are the factors influencing transpiration? How is it useful to plants?
Answer:
Transpiration occurs mainly through the stomata in the leaves. As water evaporates through the stomata, since the thin film of water over the cells is continuous, it results in pulling of water, molecule by molecule, into the leaf from the xylem. Also, because of lower concentration of water vapour in the atmosphere as compared to the substomatal cavity and intercellular spaces, water diffuses into the surrounding air. This creates a transpiration pull.

Factors Affecting Transpiration: Temperature, light, humidity and wind speed.
Importance of Transpiration: Transport of liquids and minerals is facilitated because of transpiration.

Question 12.
Discuss the factors responsible for ascent of xylem sap in plants.
Answer:
The transpiration driven ascent of xylem sap depends mainly on the following physical properties of water:

• Cohesion: Mutual attraction between water molecules.
• Adhesion: Attraction of water molecules to polar surfaces (such as the surface of tracheary elements).
• Surface Tension: Water molecules are attracted to each other in the liquid phase more than to water in the gas phase.

These properties give water high tensile strength, i.e., an ability to resist a pulling force and high capillarity, i.e., the ability to rise in thin tubes. In plants, capillarity is aided by the small diameter of the tracheary elements, the tracheids and vessel elements.

Question 13.
What essential role does the root endodermis play during mineral absorption in plants?
Answer:
The endodermis of roots have many transport proteins embedded in their plasma membrane. They let some solutes cross the membrane but not all. Transport proteins in endodermis cells enable plant cells lo adjust the quantity and types of solutes to be absorbed from the soil. It regulates the quantity and type of minerals and ions, that reach the xylem tissue of plants.

Question 14.
Explain why xylem transport is unidirectional and phloem transport bi-directional?
Answer:
The source sink (food making tissue-tissue which stores food) relationship is variable in plants so, the direction of movement in the phloem can be upwards downwards, i.e., bi-directional. It is opposite to xylem, where the movement is always unidirectional. Hence, unlike one-way flow of water in transpiration, food in phloem sap can be transported in any required direction so long there is a source of sugar and a sink is able to use, store or remove the sugar. Here, in case of unidirectional flow in xylem tissue, it is important to note that root endodermis because of the layer of suberin has the ability to actively transport ions in one direction only.

Question 15.
Explain pressure flow hypothesis of translocation of sugars in plants.
Answer:
The Pressure Flow or Mass Flow Hypothesis: The accepted mechanism used for the translocation of sugars from source to sink is called the pressure flow hypothesis. As glucose is prepared at the source (by photosynthesis) it is converted to sucrose (a disaccharide). The sugar is then moved in the form of sucrose into the companion cells and then into the living phloem sieve tube cells by active transport.

As osmotic pressure builds up the phloem sap will move to areas of lower pressure. At the sink, osmotic pressure must be reduced. Again active transport is necessary to move the sucrose out of the phloem sap and into the cells which will use the sugar converting it into energy, starch or cellulose. As sugars are removed, the osmotic pressure decreases and water moves out of the phloem.

Hydrostatic pressure in the phloem sieve tube increases, pressure flow begins and the sap moves through the phloem. Meanwhile, at the sink, incoming sugars are actively transported out of the phloem and removed as complex carbohydrates. The loss of solute produces a high water potential in the phloem and water passes out, returning eventually to xylem.

Question 16.
What causes the opening and clog” T of guard cells of stomata during transpiration?
Answer:
The immediate cause of the opening or closing-of the stomata is a change in the turgidity of the guard cells. The inner wall of each guard cell, towards the pore or stomatal, aperture, is thick and elastic. When, turgidity increases within the two guard cells flanking each stomatal aperture or pore, the thin outer walls bulge out and force the inner walls into a crescent shape. The opening of the stoma is also aided due to the orientation of the microfibrils in the cell walls of the guard cells. Cellulose microfibrils are oriented radially rather than longitudinally making it easier for the stoma to open. When the guard cells lose turgor, due to water loss (or water stress) the elastic inner walls regain their original shape, the guard cells become flaccid and the stoma closes.

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 10 Cell Cycle and Cell Division Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division

PSEB 11th Class Biology Guide Cell Cycle and Cell Division Textbook Questions and Answers

Question 1.
What is the average cell cycle span for a mammalian cell ?
Answer:
The average cell cycle span for a mammalian cell is 24 hours.

Question 2.
Distinguish cytokinesis from karyokinesis?
Answer:
Cytokinesis is the division of cytoplasm, whereas karyokinesis is the division of nucleus of the cell.

Question 3.
Describe the events taking place during interphase.
Answer:
The interphase, though called the resting phase, is the time during which the cell is preparing for division by undergoing cell growth and DNA replication. The interphase is divided into three further phases
(i) G₁ – phase (Gap 1)
(ii) S – phase (Synthesis)
(iii) G₂ – phase (Gap 2)
G₁ – phase corresponds to the interval between mitosis and initiation of DNA replication. During Ga-phase the cell is metabolically active and continuously grows but does not replicate its DNA. S or synthesis phase marks the period during, which, DNA synthesis or replication takes place. During this time, the amount of DNA per cell doubles.

During the G₂ – phase, proteins are synthesised in preparation for mitosis, while cell growth continues.
Cells in this stage remain metabolically active but no longer proliferate unless called on to do so depending on the requirement of the organism.

Question 4.
What is G₀ (quiescent phase) of cell cycle?
Answer:
G₀ is the quiescent stage of the cell cycle. It is also known as inactive stage of the cell cycle. Cells in this stage remain metabolically remain active, but no longer proliferate unless called on to do so depending on the requirement of the organisms.

Question 5.
Why is mitosis called equational division?
Answer:
Mitosis is the kind of cell division in which daughter cells have the same number and kind of chromosomes as that of parent cell. Mitosis is therefore, also known as equational division.

Question 6.
Name the stage of cell cycle at which one of the following events occur:
(i) Chromosomes are moved to spindle equator.
(ii) Centromere splits and chromatids separate.
(iii) Pairing between homologous chromosomes takes place.
(iv) Crossing over between homologous chromosomes takes place.
Answer:
(i) Prophase,
(ii) Anaphase,
(iii) Zygotene stage of meiosis-I,
(iv) Pachytene stage.

Question 7.
Describe the following:
(i) Synapsis
(ii) Bivalent
(iii) Chiasmata
Draw a diagram to illustrate your answer.
Answer:
(i) Synapsis: During meiosis-I, the process of pairing of two homologous chromosomes is known as synapsis. It is so exact that pairing is not merely* between corresponding chromosomes but between corresponding individual units.

(ii) Bivalent: The complex formed by a pair of synapsed homologous chromosomes is called a bivalent or a tetrad. It consists of four chromatids.

(iii) Chiasmata: The chiasmata formation is the indication of completion of crossing over and beginning of separation of chromosomes. The chiasma is formed when the chromosomal parts begin to repel each other except in the region where these are in contact. Thus, chiasmata formation is necessary for the separation of homologous chromosomes.

Question 8.
How does cytokinesis in plant cells differ from that in animal cells?
Answer:
Cytokinesis in an Animal Cell: It occurs by a process called cleavage. A cleavage furrow appears and at the site of the cleavage furrow the cytoplasm has a ring of microfilaments made of actin associated with molecules of the protein myosin. This ring of proteins will contract causing the furrow to deepen, which will then in turn pinch the cell into two separate cells.

Cytokinesis in a Plant Cell: In a plant cell, vesicles containing cell wall material collect at the middle of the parent cell. They will fuse and form a membranous cell plate. This plate will grow outward and it accumulates more cell wall material. Eventually, the plate will fuse with the plasma membrane and the cell plate contents will join the parental cell wall. This results in two different daughter cells.

Question 9.
Find examples where the four daughter cells from meiosis are equal in size and where they are found unequal in size.
Answer:
The four daughter haploid cells may or may not be equal in size at the end of meiosis-II with telophase-II

Question 10.
Distinguish anaphase of mitosis from anaphase-1 of meiosis. Ans. Differences between anaphase of mitosis and anaphase-I of meiosis.
Answer:

Anaphase of Mitosis	Anaphase-I of Meiosis
1. Each chromosome arranged at the metaphase plate is split simultaneously and the two daughter chromatids, begin to migrate towards the two opposite poles.	The spindle fibres contract and pull the centromeres of homologous chromosomes towards the opposite poles.
2. The centromere of each chromosome is towards the pole with arms of chromosome trailing behind.	The centromere is not divided, so half of the chromosomes of parent nucleus go to one pole and the remaining half in the opposite pole.
3. In this stage (a) centromeres split and chromatids separate. (b) chromatids move to opposite poles.	In this stage (a) homologous chromosomes separate. (b) sister chromatids remain associated at their centromere.

Question 11.
List the main differences between mitosis and meiosis.
Answer:

Mitosis	Meiosis
1. The division occurs in somatic cells.	It occurs in reproductive cells.
2. It is a single division.	It is a double division.
3. The daughter cells resemble each other as well as their mother cell.	The daughter cells neither resemble one another nor their mother cell.
4. Replication of chromosomes occurs before every mitotic division.	Replication of chromosomes occurs only once though meiosis is a double division.
5. Mitosis does not introduce variations.	Meiosis introduces variations.
6. Mitosis is required for growth, repair and healing.	Meiosis is involved in sexual reproduction.

Question 12.
What is the significance of meiosis?
Answer:
Meiosis is the mechanism by which conservation of specific chromosome number of each species is achieved across generations in sexually reproducing organisms.
It also increases the genetic variability in the population of organisms from one generation to the next. Variations are very important for the process of evolution.

Question 13.
Discuss with your teacher about:
(i) haploid insects and lower plants where cell-division occurs, and
(ii) some haploid cells in higher plants where cell-division does
not occur.
Answer:
(i) In some insects and lower plants fertilization is immediately followed by zygotic meiosis, which leads to the production of haploid organisms. This type of life cycle is known as haplontic life cycle.

(ii) The phenomenon of polyploidy can be observed in some haploid cells in higher plants in which cell division does not occur. Polyploidy is a state in which cells contain multiple pairs of chromosomes than the basic set. Polyploidy can be artificially induced in plants by applying colichine to cell culture.

Question 14.
Can there be mitosis without DNA replication in S-phase?
Answer:
No, DNA replication is required for doubling the chromosome number.

Question 15.
Can there be DNA replication without cell division?
Answer:
Yes, DNA replication takes place during the synthesis stage of interphase in the cell cycle. It is totally independent of cell division. After G₂ -phase a cell may or may not enter into the M-phase.

Question 16.
Analyse the events during every stage of cell cycle and notice how the following two parameters change?
(i) Number of chromosomes (n) per cell
(ii) Amount of DNA content (C) per cell.
Answer:
(i) The number of chromosomes (n) is double in the prophase, metaphase and anaphase stage of an organism. In the telophase stage, during daughter cell formation, the number of chromosomes get reduced to half.

(ii) Amount of DNA content (C) per cell also changes during different phases of cell division. It is double in the prophase, metaphase and anaphase stage of an organism. In the telophase stage, during daughter cell formation, the number of chromosomes get reduced to half than that of their parents. ’

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 14 Respiration in Plants Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 14 Respiration in Plants

PSEB 11th Class Biology Guide Respiration in Plants Textbook Questions and Answers

Question 1.
Differentiate between:
(a) Respiration and Combustion
(b) Glycolysis and Krebs’ cycle
(c) Aerobic respiration and Fermentation
Answer:
(a) Differences between Respiration and Combustion

Respiration	Combustion
1. It is the breakdown of complex compounds through oxidation within the cells, leading to release of considerable amount of energy.	Combustion is the complete burning of glucose, which produces CO2 and H2O and yields energy which is given out as heat.
2. It is a controlled biochemical process.	It is an uncontrolled physicochemical process.
3. Many chemical bonds break simultaneously by releasing large amounts of energy.	Chemical bonds break one after another to release energy.
4. Enzymes are involved.	Enzymes are not involved.

(b) Differences between Glycolysis and Krebs’ Cycle

Glycolysis	Krebs’ Cycle
1. It is also called as Embden Meyerhof-Parnas (EMP) pathway.	It is also known as citric acid cycle or tricarboxylic acid cycle.
2. It occurs in the cytosol of prokaryotes as well as eukaryotes.	In eukaryotes, it occurs in matrix of mitochondria, and in prokaryotes, it occurs in cytoplasm.
3. It starts with the oxidation of glucose.	It starts with the oxidation of pyruvic acid.
4. It is an enzyme-controlled 10 steps process by which glucose, fructose or sucrose is reduced to form pyruvic acid with the production of ATP and NADH2.	It involves 8 steps to oxidize two molecules of acetyl CO-A

(c) Differences between Aerobic Respiration and Fermentation

Aerobic Respiration	Fermentation
1. The presence of O2 is required for complete oxidation of organic substances.	Incomplete oxidation of glucose occurs in the absence of oxygen.
2. It releases CO2, water, and a large amount of energy present in the substrate.	Pyruvic acid is converted to CO2 and ethanol and less amount of energy is released.
3. Only two molecules of ATP are produced.	Many molecules of ATP are produced.
4. NADH is oxidized to NAD+ slowly.	This reaction is very vigorous under aerobic respiration.

Question 2.
What are respiratory substrates? Name the most common respiratory substrate.
Answer:
The compounds oxidized during the process of respiration are called respiratory substrates. Carbohydrates, especially glucose, act as respiratory substrates. Fats, proteins, and organic acids also act as respiratory substrates.

Question 3.
Give the schematic representation of glycolysis.
Answer:
Glycolysis

• The scheme of reactions was given by Embden, Meycrhof and Pumas, hence it is also called EMP Pathway; it occurs in the cytoplasm of cells.
• In this process, glucose is partially oxidized/ converted into two molecules of pyruvic acid.
• Glucose is phosphorylated with the help of ATP into Glucose-6-Phosphate, catalyzed by the enzyme hexokinase.
• Glucose-6-Phosphate is converted into its isomer, Fructose-6-Phosphate, catalyzed by isomerase.
• Fructose-6-Phosphate is then phosphorylated with the use of ATP into Fructose 1,6-bisphosphate, catalyzed by
  phosphofructokinase.
• Fructose 1,6-bisphosphate is split into one molecule of 3-pho sphoglyceraldchyd and one molecule of
  dihydroxy-acetone phosphate; these two products are interconvertible.
• 3-phosphoglyceraldehyde is oxidized to 1, 3 bi phosphoglycerate, where NAD is reduced to NADH.
• 1,3-bisphosphoglycerate is split into 3-phosphoglycerate along with the formation of ATP, catalyzed by phosphoglycerate kinase.

• 3-Phosphoglycerate is subsequently converted into 2-phosphoglycerate and then into Phosphoenol Pyruvate (PEP).
• PEP is converted into pyruvate along with the formation of ATP, catalyzed by the enzyme pyruvate kinase.

In this pathway, ATP molecules are formed in two ways :

1. Direct/substrate phosphorylation of ADP to ATP.
2. Oxidation of NADH to NAD ’ and ATP formation through electron transport (oxidative phosphorylation).

Four molecules of ATP are formed directly in the following steps:

• When 1, 3 bi phosphoglyceric acids is converted into 3-phosphoglyceric acid and
• When phosphoenolpyruvate is converted into pyruvic acid.

Two molecules of ATP are consumed, one in each of the following steps:

1. Phosphorylation of glucose to glucose-6- phosphate and
2. Conversion of fructose-6-phosphate into fructose 1, 6 bisphosphates.

When 3-phosphoglyceraldehyde (PGAL) is converted into 1, 3 bi phosphoglycerates, two redox equivalents are removed in the form of two hydrogen atoms from PGAL and transferred to NAD⁺, which is reduced to NADH ⁺ H⁺.
Oxidation of two molecules of NADH to NAD yields six molecules of ATP during electron transport. The end products of glycolysis are two molecules each of pyruvic acid, NADH, and ATP.

Question 4.
What are the main steps in aerobic respiration? Where does it take place?
Answer:
The main steps in aerobic respiration are as follows :

• Glycolytic breakdown of glucose into pyruvic acid.
• Oxidative decarboxylation of pyruvic acid to acetyl Co-A (acetyl coenzyme-A).
• Krebs’ cycle.
• Terminal oxidation and phosphorylation in respiratory chain.
  It occurs inside the mitochondrial matrix.

Question 5.
Give the schematic representation of an overall view of Kreb’s cycle.
Answer:
Tricarboxylic Acid Cycle or Kreb’s Cycle:

• It is known as tricarboxylic acid cycle, since the first product, citric acid (hence citric acid cycle) is a tricarboxylic acid (has three COOH groups).
• The sequence of reactions in citric acid cycle was first followed by Sir Hans Kreb; hence it is also known as Kreb’s cycle.
• During this cycle of reactions, 3 molecules of NAD and one molecule of FAD are reduced to NADH and FADH respectively.
• Respiration in Plants 115:
• One ATP molecule is produced directly from GTP formed by substrate-level phosphorylation.

The reactions of aerobic oxidation of Pyruvic acid can be summed up as :
Pyruvic acid + 4NAD+ + FAD+ + 2H₂O + ADP + Pi → 3CO₂ + 4NADH + 4H⁺ + FADH₂ + ATP

Question 6.
Explain ETS.
Answer:
Electron Transport System (ETS):

• ETS occurs in the electron transport particles (ETP) on the inner surface of the inner membrane of mitochondria.
• NADH formed in glycolysis and citric acid cycle are oxidized by NADH-dehydrogenase (complex I) and the electrons are transferred to ubiquinone, via FMN.
• Ubiquinone also receives reducing equivalents via FADH generated during the oxidation of succinate by succinate dehydrogenase (complex II).
• The reduced ubiquinone called ubiquinol is then oxidized by transfer of electrons to cytochrome c, via cytochrome be bc₁ -complex (complex III).
• Cytochrome c acts as a mobile carrier between complex III and complex IV.
• Complex IV refers to cytochrome c oxidase complex containing cytochromes a and a3 and two copper centres.
• When the electrons are shunted over the carriers via complex I to IV in the electron transport chain, they are coupled to ATP synthetase (complex V) for the formation of ATP from ADP and IP.
• Oxygen functions as the terminal acceptor of electrons and is reduced to water along with the hydrogen atoms.

Question 7.
Distinguish between the following:
(a) Aerobic respiration and anaerobic respiration.
(b) Glycolysis and fermentation.
(c) Glycolysis and citric acid cycle.
Answer:
(a) Differences between Aerobic and Anaerobic Respiration

Aerobic Respiration	Anaerobic Respiration
1. Presence of oxygen is essential.	Occurs without oxygen.
2. Complete oxidation of substrates occurs into CO2 and H2O.	Incomplete degradations of substrates.
3. End products are non-toxic.	End products are toxic when accumulated in large amount.
4. Involves glycolysis, Krebs’ cycle, and terminal oxidation.	Involves only glycolysis followed by incomplete breakdown of pyruvic acid.
5. 36 ATP molecules are produced from each glucose molecule.	Only two molecules of ATP are produced from each glucose molecule.

(b) Differences between Glycolysis and Fermentation:

Glycolysis	Fermentation
1. it occurs in cytoplasm of the cell and in all living things.	it occurs in yeast and muscle cells in animals when oxygen is not sufficient for cellular respiration.
2. Glucose undergoes partial oxidation to form two molecules of pyruvic acid.	The enzyme-like pyruvic acid decarboxylase and alcohol dehydrogenase catalyze these reactions.
3. The end products are CO2, H2O and energy.	The end products are CO2 and ethanol and in animal cells lactic acid is released.

(c) Differences between Glycolysis and Krebs’ Cycle

Glycolysis	Krebs’ Cycle
1. It is also called as Embden Meyerhof-Parnas (EMP) pathway.	It is also known as citric acid cycle or tricarboxylic acid cycle.
2. It occurs in the cytosol of prokaryotes as well as eukaryotes.	In eukaryotes, it occurs in matrix of mitochondria and in prokaryotes, it occurs in cytoplasm.
3. It starts with the oxidation of glucose.	It starts with the oxidation of pyruvic acid.
4. It is an enzyme-controlled 10 steps process by which glucose, fructose or sucrose is reduced to form pyruvic acid with the production of ATP and NADH2.	It involves 8 steps to oxidize two molecules of acetyl CO-A

Question 8.
What are the assumptions made during the calculation of net gain of ATP?
Answer:
The calculations of net gain of ATP can be made only on certain assumptions :

• There is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another.
• The NADH synthesized in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
• None of the intermediates in the pathway are utilised to synthesize any other compound.
• Only glucose is being respired-no other alternative substrates are entering in the pathway at any of the intermediary stages.

But this kind of assumptions are not really valid in a living system. All pathways work simultaneously and do not take place one after another. Substrates enter the pathways and are withdrawn from it as end when necessary; ATP is utilized as and when needed; enzymatic rates are controlled by multiple means. In overall steps, there is a net gain of 36 ATP molecules during aerobic respiration of one molecule of glucose.

Question 9.
Discuss ‘the respiratory pathway is an amphibolic pathway.’
Answer:
Aniphibolic Pathway: Glucose is the favored substrate for respiration. All carbohydrates are usually first converted into glucose before they are used for respiration. Other substrates can also be respired but then they do not enter the respiratory pathway at the first step.

Since respiration involves breakdown as well as synthesis of substrates, the respiratory process involves both catabolism and anabolism. That is why respiratory pathway is considered to be an amphibolic pathway rather than as a catabolic one.

Question 10.
Define RQ. What is its value for fats?
Answer:
Respiratory quotient (RQ) or respiratory ratio can be defined as the ratio of the volume of CO₂ evolved to the volume of O₂ consumed during respiration. The value of respiratory quotient depeñds on the type of respiratory substrate. Its value is one for carbohydrates. However, it is always less than one for fats as fats consume more oxygen for respiration than carbohydrates.

It can be illustrated through the example of tripalmitin fatty acid, which consumes 145 molecules of O₂ for respiration while 102 molecules of CO₂ are evolved.The RQ value for tripalmitin is 0.7.

Question 11.
What is oxidative phosphorylation?
Answer:
It refers to the formation of ATP in the mitochondria, utilizing the energy obtained by the oxidation of organic molecules.

Question 12.
What is the significance of step-wise release of energy in respiration?
Answer:
During oxidation within a cell, all the energy contained in respiratory substrates is not released free in a single step. It is released in a series of slow step-wise reactions controlled by enzymes and it is trapped as chemical energy in the form of ATP. The significance of this step-wise release of energy is that some energy is used to synthesize ATP. This ATP is stored for the later utilization wherever required. Hence, ATP acts as the energy currency of cell. The energy

stored in ATP can be utilized for following:

• In various energy-requiring processes of organisms.
• The carbon skeleton produced during respiration is used as precursors for the synthesis of other molecules.

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

PSEB Solutions for Class 11 Biology Chapter 9 Biomolecules

PSEB 11th Class Biology Guide Biomolecules Textbook Questions and Answers

Question 1.
What are macromolecules? Give examples.
Answer:
Chemical compounds, which are found in the acid insoluble fraction are called macromolecules or biomacromolecules. For example, proteins, lipids and carbohydrate, etc.

Question 2.
Illustrate a glycosidic, peptide and a phosphodiester bond.
Answer:
Glycosidic Bond: A glycosidic bond is a type of functional group that joins a carbohydrate (sugar) molecule to another group, which may or may not be another carbohydrate.

Peptide Bond: A peptide bond (amide bond) is a chemical bond formed between two molecules when the carboxyl group of one molecule reacts with the amine group of the other molecule, thereby releasing a molecule of water (H20).
H₂O.

Phosphodiester Bond: A phosphodiester bond is a group of strong covalent bonds between a phosphate group and two other molecules over two ester bonds. In DNA and RNA, the phosphodiester bond is the linkage between the 3′ carbon atom of one sugar molecule and the 5’carbon of another, deoxyribose in DNA and ribose in RNA.

Question 3.
What is meant by tertiary structure of proteins?
Answer:
Tertiary Structure of Protein; The overall shape of a single protein molecule; the spatial relationship of the secondary structures to one another. Tertiary structure is generally stabilized by non-local interactions, most commonly the formation of a hydrophobic core, but also through salt bridges, hydrogen bonds, disulfide bonds,1 and even post-translational modifications. The term “tertiary structure” is often used as synonymous with the term fold. The tertiary structure is what controls the basic function of the protein.

Question 4.
Find and write down structures of 10 interesting small molecular weight biomolecules. Find if there is any industry which manufactures the compounds by isolation. Find out who are the buyers?
Answer:

Question 5.
Proteins have primary structure. If you are given a method to know which amino acid is at either of the two termini (ends) of a protein, can you connect this information to purity or homogeneity of a protein?
Answer:
The sequence of amino acids, i.e., the positional information in a protein which is the first amino acid, which is second and so on is called the primary structure of a protein. The first amino acid is also called as N-terminal amino acid. The last amino acid is called the C-terminal amino acid. Yes, we can connect this information to purity or homogeneity of a protein. Based on number of amino and carboxyl groups, there are acidic (e.g., glutamic acid), basic (lysine) and neutral (valine) amino acids, proteins may be acidic, basic and neutral.

Question 6.
Find out and make a list of proteins used as therapeutic agents. Find other applications of proteins (e. g, cosmetics etc.)
Answer:
Some proteins and their functions are as follows:

Proteins	Functions
1. Collagen	Intercellular ground substance
2. Trypsin	Enzyme
3. Insulin	Hormone
4. Antibody	Fights against infections
5. Receptors	Sensory reception (example-taste)
6. Glut-4	Enables glucose transport in cells
7. Keratolytic protein	Used to soften hard skin
8. Egg protein	Used for skin tightening

Question 7.
Explain the composition of triglyceride.
Answer:
Triglycerides are composed of two types of molecules, i.e., glycerol i (3 carbon molecules) and fatty acids which attach to the glycerol at the alcohol unit. The following is a structural representation of a triglyceride at the molecular level.

Fatty acids are chains of hydrocarbons 4-22 (or more) carbons a long with a carboxyl group at one end. If each carbon has two hydrogen atoms, the fatty acid is saturated. If two carbon atoms are double-bonded, so that there is less hydrogen in the fatty acid, it is unsaturated (monounsaturated). If more than two carbon atoms are unsaturated, the fatty acid is polyunsaturated.

Question 8.
Can you describe what happens when milk is converted into curd or yoghurt, from your understanding of proteins.
Answer:
Milk contains a protein called casein. This protein gives milk its characteristic white colour. It is of high nutritional value because it contains all the essential amino acids required by man’s body. The curd forms because of the chemical reaction between lactic acid bacteria and casein. When curd is added to milk, the lactic acid bacteria present in it cause coagulation of casein and thus, convert it into curd.

Question 9.
Can you attempt building models of biomolecules using commercially available atomic models (Ball and Stick models).
Answer:
Yes, we can make models of biomolecules using commercially available atomic models.

Question 10.
Attempt titrating an amino acid against a weak base and discover the number of dissociating (ionisable) functional groups in the amino acid.
Answer:
When an amino acid is titrated against a weak base, it dissociates and gives two functional groups:
(i) -COOH group (carboxylic group)
(ii) Amino group (NH_2/sub>)

Question 11.
Draw the structure of the amino acid, alanine.
Answer:

Question 12.
What are gums made of? Is fevicol different?
Answer:
Gums are made of carbohydrates, i. e., L-rhamnose, D-galactose and D-galacturonic acid, etc. Fevicol is different from natural gums. It is a synthetic product.

Question 13.
Find out a qualitative test for proteins, fats and oils, amino acids and test any Fruit juice, saliva, sweat and urine for them.
Answer:

• Test for Proteins: Biuret test if Biuret’s reagent added to protein, then the colour of the reagent changes light blue to purple.
• Test for Fats and Oils: Grease or test.
• Test for Amino Acids: Ninhydrin test. If Ninhydrin reagent is added to the solution, then the colourless solution changes to pink, blue or purple depending upon the amino acid.

Question 14.
Find out how much cellulose is made by all the plants in the biosphere and compare it with how much of paper is manufactured by man and hence what is the consumption of plant material by man annually. What a loss of vegetation!
Answer:
Approximately, 100 billion tonnes of cellulose are made per year by all the plants in the biosphere and it takes 17 full grown trees to make one ton of paper. Trees are also used to fulfil the other requirements of man such as for timber, food, medicines, etc. Hence, it is difficult to calculate the annual consumption of plant material by man.

Question 15.
Describe the important properties of enzymes.
Answer:
Important properties of enzymes are given below :

• Enzymes are proteins which catalyse biochemical reactions in the cells.
• They are denatured at high temperatures.
• Enzymes generally function in a narrow range of temperature and pH. Each enzyme shows its highest activity at a particular temperature and pH called the optimum temperature and optimum pH.
• With the increase in substrate concentration, the velocity of the eyzymatic reaction rises at first. The reaction ultimately reaches a maximum velocity (V_max) which is not exceeded by any further rise in concentration of the
  substrate.
• The activity of an enzyme is also sensitive to the presence of specific chemicals that bind to the enzyme.
• Enzymes are substrate specific in their action.

Punjab State Board PSEB 11th Class Biology Book Solutions Chapter 12 Mineral Nutrition Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Biology Chapter 12 Mineral Nutrition

PSEB 11th Class Biology Guide Mineral Nutrition Textbook Questions and Answers

Question 1.
‘All elements that are present in a plant need not be essential to its survival’. Comment.
Answer:
All elements that are present in a plant need not be essential to its survival because they do not directly involved in the composition of their body. However, if the concentration of micronutrients such as Fe, Mn, Cu, Zn, Cl, etc., rise above their critical values, they appear to be toxic for the plant.

Question 2.
Why is purification of water and nutrient salts so important in studies involving mineral nutrition using hydroponics?
Answer:
It is to know the essentiality of a mineral element in the life cycle of a plant. Further, it helps in improving the deficiency symptoms of the plants. The nutrient solution must be adequetly aerated to obtain the optimal growth.

Question 3.
Explain with examples : macronutrients, micronutrients, beneficial nutrients, toxic elements and essential elements.
Answer:
(i) Macronutrients: These are generally present in plant tissues in large amount (in excess 10 m mole kg’1 of dry matter). The macronutrients include carbon, hydrogen, oxygen, nitrogen, phosphorus, sulphur, potassium, calcium and magnesium.

(ii) Micronutrients: Micronutrients or trace elements, are needed in very small amount (less than 10m mole kg~: of dry matter). These include iron, manganese, copper, molybdenum, zinc, boron, chlorine and nickel.

(iii) Beneficial Nutrients: The elements which are not essential for plants, but their presence are beneficial for the growth and development. Such, elements are called beneficial elements.

(iv) Toxic Elements: Any mineral ion concentration in tissues, that reduces the dry weight of tissues by about 10 % is considered toxic. For example, Mn inhibit the absorption of other elements.

(v) Essential Elements: The macronutrients including carbon, hydrogen, oxygen, nitrogen, phosphorus, sulphur, potassium, calcium and magnesium, which are require directly for the growth and metabolism of the plants and whose deficiency produces certain symptoms in the plants are known as essential elements.

Question 4.
Name at least five different deficiency symptoms in plants. Describe them and correlate them with the concerned mineral deficiency.
Answer:
The kind of deficiency symptoms shown in plants include chlorosis, necrosis, stunted plant growth, premature fall of leaves and buds, and inhibition of cell division.

• Chlorosis is the loss of chlorophyll leading to yellowing in leaves. This symptom is caused by the deficiency of elements N, K, Mg, S, Fe, Mn, Zn and Mo.
• Necrosis or death of tissue, particularly leaf tissue, is due to the deficiency of Ca, Mg, Cu, K.
• Lack or low level of N, K, S, Mo causes an inhibition of cell division.
• Some elements like N, S, Mo delay flowering if their concentration in plants is low.

Question 5.
If a plant shows a symptom which could develop due to deficiency of more than one nutrient, how would you find out experimentally, the real deficient mineral element?
Answer:
Every element shows certain characteristic deficiency symptoms in the plants. The deficiency of any one element cannot be met by supplying some other element. So, by absorbing the type of deficiency symptom, we can determine the real deficient mineral element.

Question 6.
Why is that in certain plants deficiency symptoms appear first in younger parts of the plant, while in others they do so in mature organs?
Answer:
For elements that are actively mobilised within the plants and exported to young developing tissues, the deficiency symptoms tend to appear first in the older tissues. For example, the deficiency symptoms of nitrogen, potassium and magnesium are visible first in the senescent leaves. In the older leaves, biomolecules containing these elements are broken down, making these elements available for mobilising to younger leaves.

The deficiency symptoms tend to appear first in the young tissues, whenever the elements are relatively immobile and are not transported out of the mature organs. For example, elements like sulphur and calcium are a part of the structural component of the cell and hence are not easily released.

Question 7.
How are the minerals absorbed by the plants?
Answer:
Mechanism of Absorption of Minerals: The process of absorption can occur into following two main phases :
(i) In the first phase, an initial rapid uptake of ions into the ‘free space’ or ‘outer space’ of cells the apoplast is passive.

(ii) In the second phase of uptake, the ions are taken in slowly into the ‘inner space’ the symplast of the cells. The passive movement of ions into the apoplast usually occurs through ion-channels, the trans-membrane proteins that function as selective pores. On the other hand, the entry or exit of ions to and from the symplast requires the expenditure of metabolic energies. The movement of ions is usually called the inward movement into the cells is influx and the outward movement, efflux.

Question 8.
What are the conditions necessary for fixation of atmospheric nitrogen by Rhizobium ? What is their role in Nitrogen-fixation?
Answer:
The first essential condition for nitrogen fixation is legume-bacteria relationship. Rhizobium bacteria cause nodule formation for this association. The enzyme nitrogenase is highly sensitive to the molecular oxygen. The nodules protect these enzymes by an oxygen scavenger called leghaerrloglobin.
Rhizobium bacteria are free living in soil. They are symbionts, which can fix atmospheric nitrogen for plants.

Question 9.
What are the steps involved in formation of a root nodule?
Answer:
Steps in Nodule Formation: Nodule formation involves a sequence of multiple interactions between Rhizobium and roots of the host plant. Principal stages in the nodule formation are given below:

• Rhizobia multiply and colonise the surroundings of roots and get attached to epidermal and root hair cells.
• The root-hairs curl and the bacteria invade the root-hair.
• An infection thread is produced carrying the bacteria into the cortex of the root, where they initiate the nodule formation in the cortex of the root. Then the bacteria are released from the thread into the cells which leads to the differentiation of specialised nitrogen fixing cells.
• The nodule thus formed, establishes a direct vascular connection with the host for exchange of nutrients.

Question 10.
Which of the following statements are true? If false, correct them:
(a) Boron deficiency leads to stout axis.
(b) Every mineral element that is present in a cell is needed by the cell.
(c) Nitrogen as a nutrient element, is highly immobile in the plants.
(d) It is very easy to establish the essentiality of micronutrients because they are required only in trace quantities.
Answer:
(a) True
(b) All the mineral elements present in a cell are not needed by the cell. For example, plants growing near radioactive mining sites tend to accumulate large amounts of radioactive compounds. These compounds are not essential for the plants.

(c) Nitrogen as a nutrient element is highly mobile in plants. It can be mobilised from the old and mature parts of a plant to its younger parts.
(d) True