Category: Class 12 Chemistry

NCERT Exemplar Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties are part of NCERT Exemplar Class 11 Chemistry. Here we have given NCERT Exemplar Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties. https://www.cbselabs.com/ncert-exemplar-problems-class-11-chemistry-chapter-3-classification-elements-periodicity-properties/

NCERT Exemplar Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties

Multiple Choice Questions
Single Correct Answer Type

Q1. Consider the isoelectronic species, Na⁺, Mg²⁺, F and O^2-. The correct order of increasing length of their radii is

Sol: (b) Amongst isoelectronic ions, ionic radii decrease with increase in nuclear charge:
Mg²⁺(12) < Na⁺(11) < F (10) < 0^2- (8)

Q2. Which of the following is not an actinoid?
(a) Curium (Z = 96)
(b)Californium (Z = 98)
(c) Uranium (Z = 92)                               
(d)    Terbium (Z = 65)
Sol: (d) Actinoids are elements with Z= 90 – 103. Therefore, Terbium (Z = 65) is not an actinoid. Terbium (Z = 65) is a lanthanoid. Tb: [Xe]4f⁹5d⁰6s²

Q3. The order of screening effect of electrons of s, p, d and/orbitals of a given shell of an atom on its outer shell electrons is
(a) s>p> d>f
(b) f> d> p> s
(c) p < d < s <f
(d) f> p> s> d
Sol: (a) The screening effect of the orbitals follows the order of s >p > d>f as screening effect decreases from s-orbital to f-orbital in an atom on account of shape of the orbital.

Free Enthalpy Calculator is helpful to find the chnage in enthalpy of a reaction.

Q4. The first ionization enthalpies of Na, Mg, A1 and Si are in the order
(a)    Na < Mg > A1 < Si                          
(b)     Na>Mg>Al>Si
(c)    Na < Mg < A1 < Si                          
(d)    Na > Mg > A1 < Si

Sol: (a) The electronic configurations of Na and Mg are:

Na (11): [Ne] 3s¹ and Mg (12): [Ne] 3s² In both the atoms, the electron is to be removed from 3s-orbital but nuclear charge in Na is less than Mg. Thus, ionisation energy of Na is less than Mg (Na < Mg).
The electronic configurations of Mg and Al are:
Mg: [Ne] 3s²; Al: [Ne] 3s² 3p^l
In Mg, the electron is to be removed from 3s-orbital while in Al, it is to be removed from 3p-orbital. Since it is easier to remove an electron from 3p-orbital in comparison to 3s-orbital, the ionization enthalpy of Mg is higher than Al (Mg > Al).
The electronic configurations of Al and Si are:
Al (13): [Ne] 3s² 3p¹ Si (14): [Ne] 3s² 3p²
In both the atoms, the electron is to be removed from 3p-orbital but nuclear charge in Si is more than Al. Thus, ionisation enthalpy of Al is less than Si

Q5. The electronic configuration of gadolinium (Atomic number 64) is


Sol: (c) The electronic configuration of La (Z = 57) is [Xe] 5d^l 6s². Therefore, further addition of electrons occurs in the lower energy 4f-orbital till it is exactly half-filled at Eu (Z = 63) Thus, the electronic configuration of Eu is [Xe] 4f⁷ 6s². Thereafter, addition of next electron does not occur in the more stable exactly half-filled 4f⁷ shell but occurs in the little higher energy 5d-orbital. Thus, the electronic configuration of Gd (Z = 64) is [Xe] 4f⁷ 5d^l 6s².

Q6. The statement that is not correct for periodic classification of elements is
(a) The properties of elements are periodic function of their atomic numbers.
(b) Non-metallic elements are less in number than metallic elements.
(c) For transition elements, the 3d-orbitals are filled with electrons after 3p-orbitals and before 4s-orbitals.
(d) The first ionization enthalpies of elements generally increase with increase in atomic number as we go along a period.
Sol: (c) For transition elements, the 3d-orbitals are filled with electrons after 3p and 4s-orbitals and before 4p-orbitals. The order of filling the orbitals is:
Is 2s 2p 3s 3p 4s 3d4p 5s 4d 5p 6s …

Q7. Among halogens, the correct order of amount of energy released in electron gain (electron gain enthalpy) is
(a) F > Cl > Br > I
(b) F < Cl < Br < I
(c) F < Cl > Br > I
(d) F < Cl < Br > I
Sol: (c) Chlorine has higher electron gain enthalpy than fluorine. This is due to small size of fluorine atom, i.e., the electron density is high which resists the addition of an electron (F < Cl).
In general, electron gain enthalpy decreases as atomic size increases. Thus, electron gain enthalpy follows the order:
Cl > Br > I

Q8. The period number in the long form of the periodic table is equal to
(a) magnetic quantum number of any element of the period.
(b) atomic number of any element of the period.
(c) maximum principal quantum number of any element of the period.
(d) maximum Azimuthal quantum number of any element of the period.
Sol: (c) Since each period starts with the filling of electrons in a new principal quantum number, therefore, the period number in the long form of the periodic table refers to the maximum principal quantum number of any element in the period.
Period number = maximum n of any element
(where, n = principal quantum number).

Q9. The elements in which electrons are progressively filled in 4f-orbital are called
(a) actinoids.
(b) transition elements.
(c) lanthanoids.
(d) halogens.
Sol: (c) In lanthanoids, the electrons are filled in 4f-orbitals.

Q10. Which of the following is the correct order of size of the given species?
(a) I>I>I⁺
(b) I⁺>I^–>I
(c) i>i⁺>i^–
(d) i^– >i>i⁺

Sol: Anion is bigger than the parent atom and cation is smaller than the parent atom.
Thus, I^– > I > I⁺

Q11.  The formation of the oxide ion, 0^2-(g), from oxygen atom requires first an exothermic and then an endothermic step as shown below:
O(g) + e^–→0^– (g), ∆H= -141 kJ mol^-1
0^–(g) + e^–→O² (g), ∆H = +780 kJ mol^-1
Thus process of formation of O^2- ion in gas phase is unfavourable even though O^2- is isoelectronic with neon. It is due to the fact that

(a) Oxygen is more electronegative.
(b) Addition of electron in oxygen results in larger size of the ion.
(c) Electron repulsion outweighs the stability gained by achieving noble gas configuration.
(d) 0^– ion has comparatively smaller size than oxygen atom.
Sol:(c) There is a lot of repulsion when similar charges approach each other as
O^–(g), and electron are both negatively charged. To add an electron under such situation, the force of repulsion is to be overcome by applying external energy.

Q12. Comprehension given below is followed by some multiple choice questions. Each question has one correct option.
In the modem periodic table, elements are arranged in order of increasing atomic numbers which is related to the electronic configuration. Depending upon the type of orbitals receiving the last electron, the elements in the periodic table have been divided into four blocks, viz, s,p, d and f The modem periodic table consists of 7 periods and 18 groups. Each period begins with the filling of a new energy shell. In accordance with the Aufbau principle, the seven periods (1 to 7) have 2, 8, 18, 18, 32 and 32 elements respectively. The seventh period is still incomplete. To avoid the periodic table being too long, the two series of f-block elements, called lanthanoids and actinoids, are placed at the bottom of the main body of the periodic table.
(i) The element with atomic number 57 belongs to
(a) s-block
(b) p-block
(c)  d-block
(d)  f-block

(ii) The last element of the p-block in 6^th period is represented by the outermost electronic configuration.

(a)    7s² 7p⁶
(b)     5f¹⁴ 6d¹⁰ 7s² 7p⁰
(c)     4f¹⁴ 5dⁱ⁰ 6s²6
(d)     4f¹⁴ 5dⁱ⁰     6s² 6p⁴

(iii) Which of the elements whose atomic numbers are given below, cannot be accommodated in the present set up of the long form of the periodic table?

(a) 107
(b) 118
(c) 126               .
(d) 102

(iv) The electronic configuration of the element which is just above the element with atomic number 43 in the same group is________ .

(a) 1s² 2s²2p⁶ 3s² 3p⁶ 3d⁵ 4s²
(b) 1s² 2s² 2p⁶ 3 s² 3p⁶ 3d⁵ 4s³ 4p⁶
(c) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²
(d) ls² 2s² 2p⁶ 3s² 3p⁶ 3d⁷4s²

(v) The element with atomic number 35, 53 and 85 are all_____ .

(a) noble gases
(b) halogens
(c) heavy metals
(d) light metals

Q13. Electronic configurations of four elements A, B, C and D are given below:
(A) 1s² 2s¹2p⁶             
(B)  1 s² 2s² 2p⁴
(C)     1s² 2s² 2p⁶ 3s¹                                  
(D)    Is² 2s² 2p⁵

Which of the following is the correct order of increasing tendency to gain electron?

(a) A < C < B < D
(b)     A < B < C < D
(c)    D < B < C < A                                 
(d)     D < A< B < C

Sol. (a) A – Is² 2s² 2p⁶ – Noble gas configuration

B -1s² 2s² 2p⁴ – 2 electrons short of stable configuration

C – 1s² 2s² 2p⁶ 3.?¹ – Requires one electron to complete 5-orbital

D -1s² 2s² 2p⁵ – Requires one electron to attain noble gas configuration

• Noble gases have no tendency to gain electrons since all their orbitals are completely filled. Thus, element A has the least electron gain enthalpy.
• Since element D has one electron less and element B has two electrons less than the corresponding noble gas configuration, hence, element D has the highest electron gain enthalpy followed by element B.
• Since, element C has one electron in the 5-orbital and hence needs one more electron to complete it, therefore, electron gain enthalpy of C is less than that of element B. Combining all the facts given above, the electron gain enthalpies of the four elements increase in the order A < C < B < D.

More than One Correct Answer Type
Q14. Which of the following elements can show covalency greater than 4?
(a) Be (b) P (c) S (d) B
Sol: (b, c) P and S have d-orbitals in their valence shell and therefore, can accommodate more than 8. electrons in their respective valence shells. Hence they show covalency more than 4.

Q15. Those elements impart colour to the flame on heating in it, the atoms of which require low energy for the ionization (i.e., absorb energy in the visible region of spectrum). The elements of which of the following groups will impart colour to the flame?
(a) 2 (b) 13 (c) 1 (d) 17
Sol: (a, c) The elements of group 1 (alkali metals) and group 2 (alkaline earth metals) have low ionization enthalpies. Therefore, they impart colour to flame.

Q16. Which of the following sequences contain atomic numbers of only representative elements?
(a) 3, 33, 53, 87
(b) 2, 10, 22, 36
(c) 7, 17,25,37,48
(d) 9,35,51,88
Sol: (a, d) Elements of 5 and p-block elements are called representative elements. Elements of f-block (Z=21 – 30; 39 – 48; 57 and 72 – 80; 89 and 104 – 112) are called transition elements while those of f-block (with Z = 58-71 and Z = 90 – 103) are called inner transition elements.
(a) 3 – Group 1, 33 – group 15, 53 – group 17 and 87 – group 1.
(d) 9 – Group 17, 35 – Group 17, 51 – Group 15, 88 – Group 2.

Q17. Which of the following elements will gain one electron more readily in comparison to other elements of their group?
(a) S (g) (b) Na (g) (c) O (g) (d) Cl(g)
Sol:(a, d) Chlorine has highest tendency to gain an electron and its electron gain enthalpy (-ve) is high. O and S belong to group 16 but S has larger tendency to accept electron.

Q18. Which of the following statements are correct?
(a) Helium has the highest first ionization enthalpy in the periodic table.
(b) Chlorine has less negative electron gain enthalpy than fluorine.
(c) Mercury and bromine are liquids at room temperature.
(d) In any period, atomic radius of alkali metal is the highest.
Sol: (a, c, d) Chlorine has more negative electron gain enthalpy than fluorine due to bigger size and lesser electronic repulsion.

Q19. Which of the following sets contain only isoelectronic ions?
(a) Zn²⁺, Ca²⁺, Ga³⁺, Al³⁺                       
(b) K⁺, Ca²⁺, Sc³⁺, Cl^–
(c) P^3-, S^2- Cl^–,K⁺                          
(d) Ti⁴⁺, Ar, Cl³⁺, V⁵⁺
Sol: (b, c)
(a) Zn²⁺ (30 – 2 = 28), Ca²⁺ (20 – 2 = 18), Ga³⁺ (31-3= 28), Al³⁺ (13 – 3 = 10) are not isoelectronic.
(b) K⁺ (19 – 1 = 18), Ca²⁺ (20 – 2 = 18), Sc³⁺ (21 – 3 = 18), Cl^– (17 + 1 = 18) are isoelectronic.
(c) P^3- (15 + 3 = 18), S^2- (16 + 2 = 18), Cl^– (17 + 1 = 18), K⁺ (19 – 1 = 18) are isoelectronic.
(d) Ti⁴⁺ (22 – 4 = 18), Ar (18), Cr³⁺ (24 – 3 = 21), V⁵⁺ (23 – 5 = 18) are not isoelectronic.

Q20. In which of the following options order of arrangement does not agree with the variation of property indicated against it?
(a) Al³⁺ < Mg²⁺ < Na⁺ < F^– (increasing ionic size)
(b) B < C < N < O (increasing first ionization enthalpy)
(c) I < Br < Cl < F (increasing electron gain enthalpy)
(d) Li < Na < K < Rb (increasing metallic radius)
Sol: (b, c) For increasing first ionization enthalpy, the order should be:
B < C < O < N
For increasing electron gain enthalpy, the order should be:
I < Br < F < Cl

Q21. Which of the following have no unit?
(a) Electronegativity (b) Electron gain enthalpy
(c) Ionisation enthalpy (d) Metallic character
Sol: (a, d) Electron gain enthalpy and ionization enthalpy have units of enthalpy.

Q22. Ionic radii vary in
(a) inverse proportion to the effective nuclear charge.
(b) inverse proportion to the square of effective nuclear charge.
(c) direct proportion to the screening effect.
(d) direct proportion to the square of screening effect.
Sol: (a, c) Ionic radii decreases as the effective nuclear charge increases.

Q23. An element belongs to 3^rd period and group-13 of the periodic table. Which of the following properties will be shown by the element?
(a) Good conductor of electricity
(b) Liquid, metallic
(c) Solid, metallic  
(d) Solid, non metallic
Sol:(a, c) The element belonging to 3^rd period and 13^th group is aluminium which is a metal. Hence, it is solid, metallic and good conductor of electricity.

Short Answer Type Questions
Q24. Explain why the electron gain enthalpy of fluorine is less negative than that of chlorine.
Sol: In fluorine, the new electron to be added goes to 2p-subshell while in chlorine, the added electron goes to 3p-subshell. Since the 2p-subshell is relatively small as compared to 3p-subshell, the added electron in small 2p-subshell experiences strong interelectronic repulsions in comparison to that in 3p-subshell in Cl. As a result, the incoming electron does not feel much attraction from the nucleus and therefore, the electron gain enthalpy of F is less negative than that of Cl.

Q25. All transition elements are d-block elements, but all d-block elements are not transition elements. Explain.
Sol: All the elements in between s- and p-block, i.e., between group 2 and 13 are called transition elements. Elements in which the last electron enters the d-orbitals of their respective penultimate shells are called d-block elements. According to this definition, Zn, Cd and Hg cannot be regarded as d-block elements because last electron in these elements enters the s-orbital of their outermost shells rather than d-orbital of their penultimate shells. Therefore, these elements should not be regarded as d-block elements. However, the properties of these elements resemble with cf-block elements. Therefore, to make the study of periodic classification of elements more rational, these are studied along with d-block elements. Thus, on the basis of properties, all the transition elements are d-block elements but on the basis of electronic configuration, all d-block elements are not transition elements.

Q26. Identify the group and valency of the element having atomic number 119. Also predict the outermost electronic configuration and write the general formula of its oxide.
Sol: The present set up of the Long Form of the Periodic Table can accommodate at the maximum 118 elements. After this, according to Aufbau principle, 8.y-orbital should be filled. Therefore, the outer electronic configuration of element having atomic number 119 will be 8s¹. Since it has one electron in the outermost s-orbital, its valency will be 1 and it should belong to group 1 along with alkali metals. The general formula of its oxide will be M₂0, where M represents the element.

Q27. Ionisation enthalpies of elements of second period are given below:
Ionisation enthalpy/kJ mol-1: 520, 899, 801, 1086, 1402, 1314, 1681, 2080. Match the correct enthalpy with the elements and complete the graph given in figure. Also write symbols of elements with their atomic number.

Sol: To match the correct enthalpy with the elements and to complete the graph, the following points are taken into consideration. As we move from left to right across a period, the ionization enthalpy keeps on increasing due to increased nuclear charge and simultaneous decrease in atomic radius. However, there are some exceptions given below-                                                                                                                    –

• In spite of increased nuclear charge, the first ionisation enthalpy of B is lower than that of Be. This is due to the presence of fully filled 2s-orbital of Be [1s²2s²] which is a stable electronic arrangement. Thus, higher energy is required to knock out the electron from fully filled 2.v-orbital. While B [1s² 2s² 2p¹] contains valence electrons in 2s and 2p-orbitals. It can easily lose its one e^– from 2p-orbital in order to achieve noble gas configuration. Thus, first ionisation enthalpy of B is lower than that of Be.
  Since the electrons in 2s-orbital are more tightly held by the nucleus than those present in 2p-orbital, therefore, ionisation enthalpy of B is lower than that of Be.

• The first ionisation enthalpy of N is higher than that of O though the nuclear charge of O is higher than that of N. This is due to the reason that in case of N, the electron is to be removed from a more stable, exactly half-filled electronic configuration (1s² 2s² 2¹_x2p^l_y 2p¹_z) which is not present in O (1s² 2s² 2p²_x 2p¹_y 2p¹_z).
  Therefore, the first ionisation enthalpy of N is higher than that of O. The symbols of elements along with their atomic numbers are given in the following graph

Q28. Among the elements B, Al, C and Si,
(a) which element has the highest first ionization enthalpy
(b) which element has the most metallic character?
Justify your answer in each case.
Sol: Arranging the elements into different groups and periods:

Group	13	14
Period 2	B	C
Period 3	Al	Si

• Ionization enthalpy increases along a period and decreases down a group. Therefore, C has the highest first ionization enthalpy.
• Metallic character decreases along a period and increases down a group. Therefore, Al has the most metallic character.

Q29. Write four characteristic properties of p-block elements.
Sol: The four important characteristic properties of p-block elements are the following:
(a) p-Block elements include both metals and non-metals but the number of non-metals is much higher than that of metals. Further, the metallic character increases from top to bottom within a group and non-metallic character increases from left to right along a period in this block.
(b) Their ionization enthalpies are relatively higher as compared to s-block elements.
(c) They mostly form covalent compounds.
(d) Some of them show more than one (variable) oxidation states in their compounds. Their oxidizing character increases from left to right in a period and reducing character increases from top to bottom in a group.

Q30. Choose the correct order of atomic radii of fluorine and neon (in pm) out of the options given below and justify your answer.
(i) 72,160 (b) 160,160 (c) 72,72 (d) 160,72
Sol:(i) Atomic radius decreases as we move from left to right in a period in the periodic table. Fluorine has the smallest atomic radius. As we move to neon
in the same period, the atomic radius increases as it has van der Waals radius and van der Waals radii are bigger than covalent radii.

Q31. Illustrate by taking examples of transition elements and non-transition elements that oxidation states of elements are largely based on electronic configuration.
Sol: The oxidation state of an element is based on its electronic configuration. The various oxidation states of a transition metal are due to the involvement of (n-1)d and outer ns electrons in bonding.
For example, Ti (22, electronic configuration [Ar]3d²4s²) can show three oxidation states (+2, +3 and +4) in various compounds like Ti0₂ (+4), Ti₂0₃ (+3) and TiO (+2).
The non-transition elements, mainly thep-block elements can show a number of oxidation states from +n to (n – 8) where, n is the number of electrons present in the outermost shell. For example, phosphorus can show -3, +3 and +5 oxidation states.
Lower oxidation states are ionic as the atom accepts the electron or electrons to achieve stable configuration while higher oxidation states are achieved by unpairing the paired electrons and shifting the electrons to vacant d-orbital.

Q32. Nitrogen has positive electron gain enthalpy whereas oxygen has negative. However, oxygen has lower ionization enthalpy than nitrogen explain.
Sol: The outermost electronic configuration of nitrogen is 2s²2p¹_x2p¹_y2p¹_z. It is stable because it has exactly half filled 2p-subshell. Therefore, it has no tendency to accept extra electron and energy has to be supplied to add additional electron. Thus, electron gain enthalpy of nitrogen is slightly positive. On the other hand, the outermost electronic configuration of O is 2s²2p²_x2p¹_y2p¹_z. It has higher positive charge (+8) than nitrogen (+7) and lower atomic size than N. Therefore, it has a tendency to accept an extra electron. Thus, electron gain enthalpy of O is negative. However, oxygen has four electrons in the 2p subshell and can lose one electron to acquire stable half filled configuration and therefore, it has low ionization enthalpy. Because of stable configuration of N, it cannot readily lose electron and therefore, its ionization enthalpy is higher than that of O.

Q33. First member of each group of representative elements (i.e., s and p-block elements) shows anomalous behaviour. Illustrate with two examples.
Sol:First member of each group of s- and p-block elements shows anomalous behaviour due to the following reasons:
(i) Small size
(ii) High ionization enthalpy
(iii) High electronegativity
(iv) Absence of d-orbitals
Examples: Li in the first group shows different properties from the rest of elements like covalent nature of its compounds, formation of nitrides.

Similarly, beryllium, the first element of second group differs from its own group in the following ways:

• Beryllium carbide reacts With water to produce methane gas while carbides of other elements give acetylene.
• Beryllium shows a coordination number of four while other elements show a coordination number of six.

Q34. p-Block elements form acidic, basic and amphoteric oxides. Explain each property by giving two examples and also write the reactions of these oxides with wa

 

Q35. How would you explain the fact that first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
Sol: Electronic configuration of Na is 1s²2s²2p⁶3s¹. After losing one electron from its outermost shell, sodium easily attains stable electronic configuration (1s²2s²2p⁶ ), while magnesium does not lose its electron easily due to presence of two electrons in s-orbital (ls²2s²2p⁶3s²). Hence first ionisation energy of sodium is less than magnesium.
When one electron is removed from Na and Mg, their configurations become ls²2s²2p⁶ and 1s²2s²2p⁶3s¹ respectively. Now it is easier to remove one electron from 3s of Mg⁺ than 2p⁶ of Na⁺. Hence, second ionisation energy of Mg is less than Na.

Q36. What do you understand by exothermic reaction and endothermic reaction? Give one example of each type.
Sol:Exothermic reaction: The reaction which is accompanied with evolution of heat is known as exothermic reaction.
CaO + C0₂ → CaC0₃, ΔH= -178.3 kJ mol^-1
Endothermic reaction: The reaction which is accompanied with absorption of energy is known as endothermic reaction, e.g.,
2NH₃ → N₂ + 3H₂; ΔH = +91.8 kJ mol^-1

Q37. Arrange the elements N, P, O and S in the order of
(i) increasing first ionisation enthalpy.
(ii) increasing non-metallic character.
Give reason for the arrangement assigned.
Sol: (i) S < P < O < N
Ionisation enthalpy increases from left to right in a period and decreases down the group. N has higher ionisation enthalpy than O due to extra stability of half-filled orbitals. Similarly, P has higher ionisation enthalpy than S due to half-filled orbitals.
(ii) P < S < N < O
Non-metallic character decreases down the group and increases from left to right in a period.

Q38. Explain the deviation in ionization enthalpy of some elements from the general trend by using the given figure.

Sol: Ionisation enthalpy increases in a period with increase in atomic number. The graph shows few exceptions and not the linear relationship. Ionisation enthalpy of Be is greater than that of B due to filled s-orbital in Be (Be – l.s² 1s², B – 1s² 2s² 2p^l).
Ionisation enthalpy of N is greater than that of O due to half-filled/?-orbitals in nitrogen.
(N – Is² 2s ²2p³ O – ls ² 2s² 2p⁴).

Q39. Explain the following:
(a) Electronegativity of elements increases on moving from left to right in the periodic table.
(b) Ionisation enthalpy decreases in a group from top to bottom.
Sol: (a) The electronegativity generally increases on moving across a period from left to right (e.g., from Li to F in the second period). This is due to decrease in atomic size and increase in effective nuclear charge. As a result of increase in effective nuclear charge, the attraction for the outer electron and the nucleus increases in a period and therefore, electronegativity also increases.
(b) On moving down a group there is a gradual decrease in ionisation enthalpy. The decrease in ionisation enthalpy down a group can be explained in terms of net effect of the following factors:
(i) In going from top to bottom in a group, the nuclear charge increases.
(ii) There is a gradual increase in atomic size due to an additional main energy shell (n).
(iii) There is increase in shielding effect on the outermost electron due to increase in the number of inner electrons.
The effect of increase in atomic size and the shielding effect is much more than the effect of increase in nuclear charge. As a result, the electron becomes less and less firmly held to the nucleus as we move down the group. Hence, there is a gradual decrease in the ionization enthalpies in a group.

Q40. How does the metallic and non metallic character vary on moving from left to right in a period?
Sol: As we move from left to right in a period, the number of valence electrons increases by one at each succeeding element but number of shells remains same. Due to this, effective nuclear charge increases. More is the effective nuclear charge, more is the attraction between the nucleus and electron.
Hence, the tendency of the element to lose electrons decreases. This results in decrease in metallic character. Furthermore, the tendency of an element to gain electrons increases with increase in effective nuclear charge, so non- metallic character increases on moving from left to right in a period.

Q41. The radius of Na⁺ cation is less than that of Na atom. Give reason.
Sol: The radius of Na⁺ is less than Na atom because Na⁺ is formed by losing one energy shell.
Na – 1s² 2s² 2p⁶ 3s¹
Na⁺– 1s²2s² 2p⁶

Q42. Among alkali metals which element do you expect to be least electronegative and why?  
Sol: Cs is the most electropositive or least electronegative element among the alkali metals since electronegative character decreases in a group.

Matching Column Type Questions
Q43. Match the correct atomic radius with the element.

Column I (Element)	Column II (Atomic radius (pm)
Be	74
C	88
0	111
B	77
N	66

Sol: All the given elements are of same period and along period, atomic radii decrease because effective nuclear charge increases. Thus, the order of atomic radii is
O < N < C < B < Be or, Be = 111 pm, O = 66 pm, C = 77 pm, B = 88 pm, N = 74 pm.

Q44. Match the correct ionization enthalpies and electron gain enthalpies of the following elements.

Solu:

Q45. Electronic configurations of some elements are given in column I and their electron gain enthalpies are given in column II. Match the electronic configuration with electron gain enthalpy.

Column I (Electronic configuration)	Column II ’ (Electron gain enthalpy/kJ moL-1
(i) 1s2 2s2 2p6	A. -53
(ii) ls22s22p63sl	B. -328
(iii) ls22s22p5	C. -141
(iv) 1s2 2S2 2p4	D. +48

Assertion and Reason Type Questions
In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.
Q46. Assertion (A): Generally, ionization enthalpy increases from left to right in a period.
Reason (R): When successive electrons are added to the orbitals in the same principal quantum level, the shielding effect of inner core of electrons does not increase very much to compensate for the increased attraction of the electron to the nucleus.
(a) Assertion is correct statement and reason is wrong statement.
(b) Assertion and reason both are correct statements and reason is correct explanation of assertion.
(c) Assertion and reason both are wrong statements.
(d) Assertion is wrong statement and reason is correct statement.
Sol: (b) Ionisation enthalpy increases from left to right across the period due to decrease in atomic size.
The electrons present within the subshell has almost same effective nuclear charge.

Q47. Assertion (A): Boron has a smaller first ionization enthalpy than beryllium. Reason (R): The penetration of a 2s electron to the nucleus is more than the 2p electron, hence, 2p electron is more shielded by the inner core of electrons that the 2s electrons.
(a) Assertion and reason both are correct statements but reason is not correct explanation of assertion.
(b) Assertion is correct statement but reason is wrong statement.
(c) Assertion and reason both are correct statements and reason is correct explanation of assertion.
(d) Assertion and reason both are wrong statement.

Sol: (c) Boron (1s²2s²2p¹) has less first ionisation enthalpy than beryllium (1s²2s²) because beryllium has fully filled .s-subshell.
2s-electrons are nearer to the nucleus as compared to 2s-electrons.

Q48. Assertion (A): Electron gain enthalpy becomes less negative as we go down a group.
Reason (R): Size of the atom increases on going down the group and the added electron would be farther from the nucleus.
(a) Assertion and reason both are correct statements but reason is not correct explanation of assertion.
(b) Assertion and reason both are correct statements and reason is correct explanation of assertion.
(c) Assertion and reason both are wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
Sol:(b) On moving down the group, the electron gain enthalpy becomes less negative because on moving down the group the atomic size increases and the added electron lies away from the nucleus.

Long Answer Type Questions
Q49. Discuss the factors affecting electron gain enthalpy and the trend in its variation in the periodic table.
Sol: Factors affecting electron gain enthalpy:
(i) Nuclear charge: The electron gain enthalpy become more negative as the nuclear charge increases. This is due to greater attraction for the incoming electron if nuclear charge is high.
(ii) Size of the atom: With the increase in size of the atom, the distance between the nucleus and the incoming electron increases and this results in lesser attraction. Consequently, the electron gain enthalpy become less negative with increase in size of the atom of the element.
(iii) Electronic configuration: The elements having stable electronic configurations of half filled and completely filled valence subshells show very small tendency to accept additional electron and thus electron gain enthalpies are less negative.

Variation of electron gain enthalpies in periodic table:
Electron gain enthalpy, in general, becomes more negative from left to right in a period and becomes less negative as we go from top to bottom in a group.
(a) Variation down a group: On moving down a group, the size and nuclear charge increases. But the effect of increase in atomic size is much more pronounced than that of nuclear charge and thus the additional electron feels less attraction by the large atom. Consequently, electron gain enthalpy becomes less negative. This is clear from decrease of electron gain enthalpy in going from chlorine to bromine and to iodine.
(b) Variation along a period: On moving across a period, the size of the atom decreases and nuclear charge increases. Both these factors result in greater attraction for the incoming electron, therefore, electron gain enthalpy, in general, becomes more negative in a period from left to right. However, certain irregularities are observed in the general trend. These are mainly due to the stable electronic configurations of certain atoms.
Important Trends in Electron Gain Enthalpies:
There are some important features of electron gain enthalpies of elements. These are:
(i) Halogens have the highest negative electron gain enthalpies.
(ii) Electron gain enthalpy values of noble gases are positive while those of Be, Mg, N and P are almost zero.
(iii) Electron gain enthalpy of fluorine is unexpectedly less negative than that of chlorine.

Q50. Define ionization enthalpy. Discuss the factors affecting ionisation enthalpy of the elements and its trends in the periodic table.
Sol: Ionisation enthalpy: It is defined as the minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom.

M(g) →M⁺(g) + e^–I₁  = First ionisation enthalpy

Similarly, second and third electrons are also removed by providing successive ionisation enthalpies.

Factors on which Ionisation Enthalpy Depends:
(i) Size of the atom: The larger the atomic size, smaller is the value of ionisation enthalpy. In a larger atom, the outer electrons are far away from the nucleus and thus force of attraction with which they are attracted by the nucleus is less and hence can be easily removed.

Ionisation enthalpy ∝ 1/ Atomic size

(ii) Screening effect: Higher the screening effect, the lesser is the value of ionisation enthalpy as the screening effect reduces the force of attraction towards nucleus and hence the outer electrons can be easily removed.

Ionisation enthalpy ∝  1/ Screening effect

(iii) Nuclear charge: As the nuclear charge increases among atoms having same number of energy shells, the ionisation enthalpy increases because the force of attraction towards nucleus increases.

Ionisation enthalpy ∝ Nuclear charge

(iv) Half filled and fully filled orbitals: The atoms having half filled and fully filled orbitals are comparatively more stable, hence more energy is required to remove the electron from such atoms. The ionization enthalpy is rather higher than the expected value in case of such an atom.

Ionisation enthalpy ∝ Stable electronic configuration

(v) Shape of orbital: The s-orbital is more close to nucleus than the p-orbital of the same orbit. Thus, it is easier to remove electron from a p-orbital in comparison to s-orbital. In general, the ionisation enthalpy follows the following order

(s>p> d>f) orbitals of the same orbit.

Variation of ionisation enthalpy in the periodic table
ln general, the ionisation energy decreases down the group due to increase in atomic size. On the other hand, the ionisation energy increases across the period from left to right, again due to decrease in atomic size from left to right.

Q51. Justify the given statement with suitable examples “the properties of the elements are a periodic function of their atomic numbers”.
Sol. The similarities in the properties arise due to the same distribution of electrons in the outermost orbitals or electronic configuration which depends upon the atomic number. The elements present in a group or period exhibit similar chemical properties which depend upon the atomic number.

Q52. Write down the outermost electronic configuration of alkali metals. How will you justify their placement in group 1 of the periodic table?

Q53. Write the drawbacks in Mendeleev’s periodic table that led to its modification.
Sol: Drawbacks of Mendeleev’s periodic table:
1. Position of hydrogen: Hydrogen is placed in group I. However, it resembles the elements of group I (alkali metals) as well as the elements of group VILA, (halogens). Therefore, the position of hydrogen in the periodic table is not correctly defined.
2. Anomalous pairs: In certain pairs of elements, the increasing order of atomic masses was not obeyed. In these cases, Mendeleev placed elements according to similarities in their properties and not in increasing order of their atomic masses. For example, argon (Ar, atomic mass 39.9) is placed before potassium (K, atomic mass 39.1). Similarly, cobalt (Co, atomic mass 58.9) is placed before nickel (Ni, atomic mass 58.6) and tellurium (Te, atomic mass 127.6) is placed before iodine (I, atomic mass 126.9). These positions were not justified.
3. Position of isotopes: Isotopes are the atoms of the same element having different atomic masses but same atomic number. Therefore, according to Mendeleev’s classification, these should be placed at different places depending upon their atomic masses. For example, isotopes of hydrogen with atomic masses 1,2 and 3 should be placed at three places. However, isotopes have not been given separate places in the periodic table.
4. Some similar elements are separated and dissimilar elements are grouped together: In the Mendeleev’s periodic table, some similar elements were placed in different groups while some dissimilar elements
had been grouped together. For example, copper and mercury resembled
in their properties but they had been placed in different groups. At the same time, elements of group IA such as Li, Na and K were grouped with copper (Cu), silver (Ag) and gold (Au), though their properties are quite different.
5. Cause of periodicity: Mendeleev did not explain the cause of periodicity among the elements.
6. Position of lanthanoids (or lanthanides) and actinoids (or actinides):
The fourteen elements following lanthanum (known as lanthanoids, from atomic number 58-71) and the fourteen elements following actinium (known as actinoids, from atomic number 90 – 103) have not been given separate places in Mendeleev’s table.
In order to cover more elements, Mendeleev modified his periodic table.

Q54. In what manner is the long form of periodic table better than Mendeleev’s
periodic table? Explain with examples.
Sol: Superiority of the Long form of the Table over Mendeleev’s Table:
(i) This table is based on a more fundamental property, i.e., atomic number.
(ii) It correlates the position of the elements with their electronic configurations more clearly.
(iii) The completion of each period is more logical. In a period as the atomic number increases, the energy shells are gradually filled up until an inert gas configuration is reached. It eliminates the even and odd series of IV, V, VI and VII periods of Mendeleev’s periodic table.
(iv) The position of VIIIth group  is appropriate in this table. All the transition elements have been brought in the middle as the properties of transition elements are intermediate between s- and p-block elements.
(v) Due to separation of two subgroups, dissimilar elements do not fall together. One vertical column accommodates elements with same electronic configuration thereby showing same properties.
(vi) In this table, a complete separation between metals and non-metals has been achieved. The non-metals are present in the upper right comer of the periodic table.
(vii) There is a gradual change in properties of the elements with increase in their atomic numbers, i.e., periodicity of properties can be easily visualized. The same properties o recurrence in properties occur after the intervals of 2, 8, 8, 18, 18 and 32 elements which indicates the capacity of various periods of the table.
(viii) This arrangement of elements is easy to remember and reproduce.

Q55. Discuss and compare the trend in ionization enthalpy of the elements of group 1 with those of group 17 elements.
Sol: The ionization enthalpies decrease regularly as we move down a group from one element to the other. This is evident from the values of the first ionisation enthalpies of the elements of group 1 (alkali metals) and group 17 elements as given in table and figure.

Group ‘ 1	First ionisation enthalpies (kJ moL-1)	Group 17	First ionisation enthalpies (kJ mol-1)
H	1312	F	1681
Li	520	Cl	1255
Na	496	Br	1142
K	419	I	1009
Rb	403	At	917
Cs	374

Given trend can be easily explaihed on the basis of increasing atomic size and screening effect as follows
(i) On moving down the group, the atomic size increases gradually due to the addition of one new principal energy shell at each succeeding element. Hence, the distance of the valence electrons from the nucleus increases.
Consequently, the force of attraction by the nucleus for the valence electrons decreases and hence the ionisation enthalpy should decrease.
(ii) With the addition of new shells, the shielding or the screening effect increases. As a result, the force of attraction of the nucleus for the valence electrons further decreases and hence the ionisation enthalpy should decrease.
(iii) Nuclear charge increases with increase in atomic number. As a result, the force of attraction by the nucleus for the valence electrons should increase and accordingly the ionisation enthalpy should increase.
The combined effect of the increase in the atomic size and the screening effect more than compensates the effect of the increased nuclear charges. Consequently, the valence electrons become less and less firmly held by the nuclear and hence the ionisation enthalpies gradually decrease as move down the group.

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NCERT Exemplar Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

Multiple Choice Questions
Single Correct Answer Type

Q1. Isostructural species are those which have the same shape and hybridization. Among the given species identify the isostructural pairs.
(a) [NF₃ and BF₃]
(b) [BF^–₄andNH⁺₄]
(c) [BC1₃ and BrCl₃]                             
(d) [NH₃ and N0^–₃ ]
Sol: (b) NF₃ is pyramidal whereas BF₃ is planar triangular.
BF^–₄ and NH⁺₄ ions both are tetrahedral.
BC1₃ is triangular planar whereas BrCl₃ is pyramidal.
NH₃ is pyramidal whereas N0₃ is triangular planar.

Q2. Polarity in a molecule and hence the dipole moment depends primarily on electronegativity of the constituent atoms and shape of a molecule. Which of the following has the highest dipole moment?
(a) C0₂
(b) HI                      
(c) H₂0                     
(d) S0₂

Q3. The types of hybrid orbitals of nitrogen in N0₂, N0₃ and NH₄ respectively are expected to be
(a) sp, sp³ and sp²                                  
(b) sp, sp² and sp³
(c) sp², sp and sp³                                   
(d) sp², sp³ and sp
Sol: (b) The number of orbitals involved in hybridization can be determined by the application of formula:

where H = number of orbitals involved in hybridization
V= valence electrons of central atom
M= number of monovalent atoms linked with central atom
C = charge on the cation
A = charge on the anion

Q4. Hydrogen bonds are formed in many compounds e.g., H₂0, HF, NH₃. The boiling point of such compounds depends to a large extent on the strength of hydrogen bond and the number of hydrogen bonds: The correct decreasing order of the boiling points of above compounds is
(a) HF>H₂0>NH₃                        
(b) H₂0>HF>NH₃
(c) NH₃>HF>H₂0                       
(d) NH₃>H₂0>HF
Sol: (b) Strength of H-bonding depends on the electronegativity of the atom which follows the order: F > O > N .
Strength of H-bond is in the order:
H……. F > H…….. O > H…….. N
But each H₂0 molecule is linked to 4 other H₂0 molecules through H-bonds whereas each HF molecule is linked only to two other HF molecules.
Hence, correct decreasing order of the boiling points is H_zO > HF > NH₃.

Q5. In PO₄^3- ion the formal charge on the oxygen atom of P – O bond is
(a) +1                      (b) -1                        (c) -0.75                    (d) +0.75
Sol: (b) Formal charge of the atom in the molecule or ion = (Number of valence electrons in free atom) – (Number of lone pair electrons + 1/2 Number of bonding electrons)

Q6. In N0^–₃ ion, the number of bond pairs and lone pairs of electrons on nitrogen atom are
(a) 2, 2                    (b) 3, 1                      (c) 1,3                      (d) 4, 0
Sol: (d) In N-atom, number of valence electrons = 5
Due to the presence of one negative charge, number of valence electrons = 5 + 1 = 6. One O-atom forms two bonds (= bond) and two O-atom are shared with two electrons of N-atom.
Thus, 3 O-atoms are shared with 8 electrons of N-atom.
Number of bond pairs (or shared pairs) = 4
Number of lone pairs = 0

Q7. Which of the following species has tetrahedral geometry?

Q9. Which molecule/ion out of the following does not contain unpaired electrons?
(a) N⁺₂
(b) 0₂                        
(c) O₂^2-                    
(d) B₂

Q10. In which of the following molecule/ion all the bonds are not equal?
(a) XeF₄                 
(b) BF^–₄                    
(c) C₂H₄                  
(d) SiF₄
Sol: (c) C₂H₄ has one double bond and four single bonds. Bond length of double bond (C = C) is smaller than single bond (C – H).

Q11. In which of the following substances will hydrogen bond be strongest?
(a) HCl
(b) H₂0                     
(c) HI                      
(d) H₂S
Sol: (b) HC1, HI and H₂S do not from H-bonds. Only H₂0 forms hydrogen bonds. One H₂0 molecule forms four H-bonding.

Q12. If the electronic configuration of an element is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s² , the four electrons involved in chemical bond formation will be .
(a) 3p⁶
(b 3p⁶, 4s²          
(c)  3p⁶, 3d²         
(d) 3d¹,4s²

Sol:(d) In transition elements (n -1 )d and ns orbitals take part in bond formation.

Q13. Which of the following angle corresponds to sp² hybridisation?
(a) 90° (b)     120°                (c)     180°               (d) 109°
Sol: (b) sp² hybridisation gives three sp² hybrid orbitals which are planar triangular forming an angle of 120° with each other.

The electronic configurations of three elements A, B and C are given below.
Answer the questions from 14 to 17 on the basis of these configurations.
A ls²2s²2p⁶
B ls²2s²2p⁶3s²3p³
C ls²2s²2p⁶3s²3p^s

Q14. Stable form of A may be represented by the formula
(a) A
(b) A₂                    
(c) A₃                   
(d)  A₄
Sol: (a) The given electronic configuration shows that A represents noble gas because the octet is complete. A is neon which has 10 atomic number.

Q15. Stable form of C may be represented by the formula
(a) C
(b) C₂                     
(c) C₃                   
(d)  C₄
Sol: (b) The electronic configuration of C represent chlorine. Its stable form is dichlorine (Cl₂), i.e., C₂.

Q16. The molecular formula of the compound formed from B and C will be
(a) BC
(b) B₂C                 
(c)  BC₂                
(d)  BC₃
Sol: (d) B represent P and C represents Cl. The stable compound is PC1₃ i.e., BC₃.

Q17. The bond between B and C will be
(a) ionic
(b) covalent         
(c) hydrogen
(d) coordinate
Sol: (b) Both B and C are non-metals and therefore, bond formed between them will be covalent.

Q18. Which of the following order of energies of molecular orbitals of N₂ is correct?

Q19. Which of the following statements is not correct from the view point of molecular orbital theory?
(a) Be₂ is not a stable molecule.
(b) He₂ is not stable but He⁺₂ is expected to exist.
(c) Bond strength of N₂ is maximum amongst the homonuclear diatomic molecules belonging to the second period.

Thus, dinitrogen (N₂) molecule contains triple bond and no other molecule of second period have more than double bond. Hence, bond strength of N₂ is maximum amongst the homonuclear diatomic molecules belonging to the second period.
(d) It is incorrect. The correct order of energies of molecular orbitals in N₂ molecule is

Q20. Which of the following options represents the correct bond order?
(a) o₂>o₂>o⁺₂
(b) o^–₂<o₂<o₂
(c) o₂>o₂<o₂
(d) o^–₂<o₂>o⁺₂

Q21. The electronic configuration of the outermost shell of the most electronegative element is
(a) 2s²2p⁵
(b) 3s²3p⁵                               
(c) 4s²4p⁵                
(d) 5s²5p^s
Sol: (a) The electronic configuration represents
2s²2p⁵= fluorine = most electronegative element
3s²3p⁵ = chlorine
4s²4p⁵ = bromine
5s²5p⁵ = iodine

Q22. Amongst the following elements, whose electronic configurations are given below, the one having the highest ionization enthalpy is
(a) [Ne]3s²3p¹
(b) [Ne]3s²3p³
(c) [Ne]3s²3p²
(d) [Ar]3d¹⁰4s²4p³
Sol: (b) (b) and (d) have exactly half-filled p-orbitals but (b) is smaller in size than Hence, (b) has highest ionization enthalpy.

More than One Correct Answer Type

Q23. Which of the following have identical bond order?
(a) CN^– (b) NO⁺                      (c) 0^–₂                        (d) 0₂^2-
Sol: (a, b) CN^– (number of electrons = 6 + 7 + 1 = 14)
NO⁺ (number of electrons = 7 + 8 – 1 = 14)
0^–₂ (number of electrons = 8 + 8 + 1 = 17)
0₂^2- (number of electrons = 8 + 8 + 2 = 18)
Thus, CN and NO⁺ because of the presence of same number of electrons, have same bond order.

Q24. Which of the following attain the linear structure?
(a) BeCl₂
(b) NCO⁺                  
(c) N0₂                     
(d) CS₂
Sol: (a, d) BeCl₂ (Cl – Be – Cl) and CS₂ (S = C = S) both are linear, NCO⁺ is non-linear. However, remember that NCO(N = C = O) is linear because it is isoelectronic with C0₂.
N0₂ is angular with bond angled 132° and each O – N bond length of 1.20A^o (intermediate between single and double bond).

Q25. CO is isoelectronic with
(a) NO⁺
(b) N₂                       
(c) SnCl₂                  
(d) N0₂
Sol: (a, b) Number of electrons in CO =14
Number of electrons in NO⁺ =14
Number of electrons in N₂ = 14
Number of electrons in SnCl₂ = 84
Number of electrons in N0^–₂ = 24

Q26. Which of the following species have the same shape?
(a) C0₂
(b) CC1₄                 
(c) 0₃                        
(d) N0^–₂
Sol: (c, d) C0₂ →Linear, CC1₄ → Tetrahedral, 0₃ →Angular (V-shaped), N0^–₂ →Angular (V-shaped)

Q27. Which of the following statements are correct about CO₃^2- ?
(a) The hybridization of central atom is sp³.
(b) Its resonance structure has one C – O single bond and two C = O double bonds.
(c) The average formal charge on each oxygen atom is 0.67 units.
(d) All C – O bond lengths are equal.

Q28. Diamagnetic species are those which contain no unpaired electrons. Which among the following are diamagnetic?
(a) N₂                     
(b) N₂^2- 
(c) 0₂                      
(d) o₂^2-

Q29. Species having same bond order are
(a) N₂                      
(b) N^–₂                       
(_C) F⁺₂                      
(d) o^–₂

Q30. Which of the following statements are not correct?
(a) NaCl being an ionic compound is a good conductor of electricity in the solid state.
(b) In canonical structures there is a difference in the arrangement of atoms.
(c) Hybrid orbitals form stronger bonds than pure orbitals.
(d) VSEPR theory can explain the square planar geometry of XeF₄.
Sol: (a, b)
(a) Ionic compounds are good conductors only in molten state or aqueous solution since ions are not furnished in solid state.
(b) In canonical structures there is a difference in arrangement of electrons.

Short Answer Type Questions
Q31. Explain the non linear shape of H₂S and non planar shape of PCl₃ using valence shell electron pair repulsion theory.
Sol: The Lewis structure of H₂S is:

Q32. Using molecular orbital theory, compare the bond energy and magnetic character of 0⁺₂ and O^–₂

Q33. Explain the shape of BrF₅.
Sol: Br-atom has configuration:
1s², 2s²2p⁶ , 3s²3p⁶3d¹⁰, 4s²4p⁵
To get pentavalency, two of the p-orbitals are unpaired and electrons are shifted to 4d-orbitals.

In this excited state, sp³d²-hybridisation occurs giving octahedral structure. Five positions are occupied by F atoms forming sigma bonds with hybrid bonds and one position occupied by lone pair, i.e., the molecule as a square pyramidal shape.

Q34. Structures of molecules of two compounds are given below:

(a) Which of the two compounds will have intermolccular hydrogen bonding and which compound is expected to show intramolecular hydrogen bonding?
(b) The melting point of a compound depends on. among other things, the extent of hydrogen bonding. On this basis explain which of the above two compounds will show higher melting point.
(c) Solubility of compounds in water depends on power to form hydrogen bonds with water. Which of the above compounds will form hydrogen bond with water easily and be more soluble in it?
Sol:(a) Compound I will form intramolecular hydrogen bond because N02 and OH groups are close together whereas it is not so in compound II. Compound II will have intermolecular hydrogen bonding. Bonding in both the cases is shown below:

(b) As a large number of molecules can be linked together through intermo­lecular hydrogen bonding, compound II will show higher melting point.

(c) Due to intramolecular hydrogen bonding, compound I will not be able to form hydrogen bonds with H₂0 molecules. Hence, it will be less soluble in water. However, compound II can form hydrogen bonds with H₂0 molecules easily and hence it will be more soluble in water.

Q35. Why does type of overlap given in the following figure not result in the bond formation?

Sol: In first figure, the ++ overlap is equal to +- overlap and therefore, these cancel out and net overlap is zero.
In second figure, no overlap is possible because the two orbitals are perpendicular to each other.

Q36. Explain why PC1₅ is trigonal bipyramidal whereas IF₅ is square pyramidal.
Sol: PC1₅ – The ground state and the excited state outer electronic configurations of phosphorus (Z = 15) are represented below

Sol: Dimethyl ether has larger bond angle than water. This is because there is more repulsion between bond pairs of CH₃ groups attached in ether than between bond pairs of hydrogen atoms attached to oxygen in water. The carbon of CH₃ group in ether is attached to three hydrogen atoms through c bonds and electron pairs of these bonds add to the electron charge density on carbon atom. Hence, repulsion between two -CH₃ groups will be more than that between two H atoms.

Q38. Write Lewis structure of the following compounds and show formal charge on each atom. HN0₃, No₂, H₂so₄
Sol: Formal charge on an atom in a Lewis structure
= [total number of valence electrons in free atom] – [total number of non-bonding (lone pairs) electrons]
—1/2 [total number of bonding or shared electrons]

Q39. The energy of σ2p_z: molecular orbital is greater than 2p_x and 2p_v molecular orbitals in nitrogen molecule. Write the complete sequence of energy levels in the increasing order of energy in the molecule. Compare the relative stability and the magnetic behaviour of the following species:
N₂, N⁺₂, N^–₂, N²₂⁺

Q40. What is the effect of the following processes on the bond order in N-, and 0₂?
(i) N₂ → N⁺₂ + e^– (ii) 0₂ → O⁺₂ + e^–

Q41. Give reasons for the following: ‘
(a) Covalent bonds are directional bonds while ionic bonds are non- directional.
(b) Water molecule has bent structure whereas carbon dioxide molecule is linear.
(c) Ethyne molecule is linear.
Sol: (i) Since the covalent bond depends on the overlapping of orbitals between different orbitals, the geometry of the molecule is different. The orientation of overlap is different. The orientation of overlap is the factor responsible for their directional nature.
(ii) Due to presence of two lone pairs of electrons on oxygen atom in HiO the repulsion between Ip-lp is more. C0₂ undergoes sp hybridization resulting in linear shape (O = C = O) while H₂0 undergoes .sp³ hybridisation resulting in distorted tetrahedral or bent structure.

(iii) In ethyne molecule carbon undergoes sp hybridization with two unhybridised orbitals. One sp hybrid orbital of one carbon atom overlaps axially with sp hybrid orbital of the other carbon atom to form C – C sigma bond while the other hybridized orbital of each carbon atom overlaps axially with S orbitals of hydrogen atoms forming σ bonds. Unhybridised orbitals form π bonds

Q42. What is an ionic bond? With two suitable examples explain the difference between an ionic and covalent bond?
Sol: An ionic bond is formed as a result of the electrostatic attraction between the positive and negative ions formed by transfer of electrons from one atom to another.

Q43. Arrange the following bonds ‘in order of increasing ionic character giving reason.
N-H, F-H, C-H and O-H

Q44. Explain why CO^2-₃ ion cannot be represented by a single Lewis structure. How can it be best represented?
Sol: A single Lewis structure of CO^2-₃  ion cannot explain all the properties of this ion. It can be represented as a resonance hybrid of the following structures:

If it were represented only by one structure, there should be two types of bonds, i.e., C = O double bond and C – O single bonds but actually all bonds are found to be identical with same bond length and same bond strength.

Q45. Predict the hybridization of each carbon in the molecule of organic compound given below. Also indicate the total number of sigma and pi bonds in this molecule.

Q46. Group the following in linear and non-linear molecules: H₂0, HOC1, BeCl₂ C1₂0

Q47. Elements X, Y and Z have 4, 5 and 7 valence electrons respectively, (i) Write the molecular formula of the compounds formed by these elements individually with hydrogen, (ii) Which of these compounds will have the highest dipole moment?

Q48. Draw the resonating structure of (i) Ozone molecule (ii) Nitrate ion

Q49. Predict the shapes of the following molecules on the basis of hybridization. BC1₃, ch₄, co₂, nh₃
Sol: BCl₃ – sp² hybridisation – Trigonal planar
CH₄ – sp³ hybridisation – Tetrahedral .
NH₃ – sp³ hybridisation – Distorted tetrahedral or Pyramidal

Q50. All the C – O bonds in carbonate ion (CO^2-₃) are equal in length. Explain.
Sol: Carbonate ion is represented by resonating structures as given below:

Q51. What is meant by the term average bond enthalpy? Why there is difference in bond enthalpy of O – H bond in ethanol (C₂H₅OH) and water?

Matching Column Type Questions
Q52. Match the species in Column I with the type of hybrid orbitals in Column II.

Column I	Column II
(i) SF4	(a) sp3cf
(ii) if5	(b) d2sp3
(iii) NO2+	(c) sp3 d
(iv) NH4	(d) sp3
	(e) sp

 

Q53. Match the species in Column I with the geometry/shape in Column II.

Column I	Column II
(i) H30+	(a) Linear
(ii) HC = CH	(b) Angular
(iii) Cl0–2	(c) Tetrahedral
(iv) NH+4	(d) Trigonal bipyramidal
–	(e) Pyramidal

Q54.  Match the species in Column I with the bond order in Column II.

Column I	, . Column II
(i) NO	(a) 1.5
(ii) CO	(b) 2.0
(iii) o–2	(c) 2.5
(iv) 02	(d) 3.0

Q55. Match the items given in Column I with examples given in Column II.

Column I	Column II
(i) Hydrogen bond	(a) C
(ii) Resonance	(b) LiF
(iii) Ionic solid	(c) H2
(iv) Covalent solid	(d) HF
	(e) 03

Q56. Match the shape of molecules in Column I with the type of hybridization in Column II.

Column I	Column II
(i) Tetrahedral	(a) sp2
(ii) Trigonal	(b) sp
(iii) Linear	(c) sp3

Sol: (i) →c; (ii) → a; (iii) —> b

sp³ hybridisation – Tetrahedralshape
sp² hybridisation – Trigonal shape
sp hybridization – Linear shape

Assertion and Reason Type Questions

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Q57. Assertion (A): Sodium chloride formed by the action of chlorine gas on sodium metal is a stable compound.
Reason (R): This is because sodium and chloride ions acquire octet in sodium chloride formation.
(a) A and R both are correct, and R is the correct explanation of A.
(b) A and R both are correct, but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A and R both are false.
Sol: (a) Sodium chloride (Na⁺CL^–) is stable ionic compound because both Na⁺ and CL^– ions have complete octet in outermost shell.

Q58. Assertion (A): Though the central atom of both NH₃ and H₂0 molecules are sp³ hybridised, yet H – N – H bond angle is greater than that of H – O – H.
Reason (R): This is because nitrogen atom has one lone pair and oxygen atom has two lone pairs.
(a) A and R both are correct, and R is the correct explanation of A.
(b) A and R both are correct, but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A and R both are false.

Sol:(a)

H₂0 has two lone pairs while NH₃ has single lone pair, hence, H₂0 involves greater lone pair-bond pair repulsion.

Q59. Assertion (A): Among the two O – H bonds in H₂0 molecule, the energy required to break the first O – H bond and other O – H bond is the same.
Reason (R): This is because the electronic environment around oxygen is the same even after breakage of one O – H bond.
(a) A and R both are correct, and R is the correct explanation of A.
(b) A and R both are correct, but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A and R both are false.
Sol: (d) Bond energy of two (-O – H) bonds in H₂0 will be different.

Long Answer Type Questions
Q60. (i) Discuss the significance/applications of dipole moment.
(ii) Represent diagrammatically the bond moments and the resultant dipole moment in C0₂, NF₃ and CHCl₃.

Sol: (i) Dipole moment plays very important role in understanding the nature of chemical bond. A few applications are given below:
(a) Distinction between, polar and non-polar molecules. The measurement of dipole moment can help us to distinguish between polar and non-polar molecules. Non-polar molecules have zero dipole
moment while polar molecules have some value of dipole moment.
(b) Degree of polarity in a molecule. Dipole moment measurement also gives an idea about the degree of polarity specially in a diatomic molecule. The greater the dipole moment, the greater is the polarity in such a molecule.
(c) Shape of molecules. In case of molecules containing more than two atoms, the dipole moment not only depends upon the individual dipole moments of the bonds but also on the arrangement of bonds.
.Thus, dipole moment is used to find the shapes of molecules.
(d) Ionic character in a molecule. Knowing the electronegativities of atoms involved in a molecule, it is possible to predict the nature of chemical bond formed. If the difference in electronegativities of two atoms is large, the bond will be highly polar. As an extreme case, when the electron is completely transferred from one atom to another, an ionic bond is formed. Therefore, the ionic bond is regarded as an extreme case of covalent bond. The greater the difference in electronegativities of the bonded atoms, the higher is the ionic character
(e) Distinguish between cis- and trans- isomers. Dipole moment measurements help to distinguish between cis- and trans- isomers because ds-isomer has usually higher dipole moment than trans isomer.
(f) Distinguish between ortho, meta and para isomers. Dipole moment measurements help to distinguish between o-, m- and p-isomers because the dipole moment of p-isomer is zero and that of o-isomers is more than that of m-isomer.

Q61. Use the molecular orbital energy level diagram to show that N₂ would be expected to have a triple bond, F₂ a single bond and Ne₂ no bond.

No bond is formed between two Ne atoms or in other words, Ne₂ does not exist.
Bond Order = ½ (10 – 10) = 0 (No bond)

Q62. Briefly describe the valence bond theory of covalent bond formation by taking an example of hydrogen. How can you interpret energy changes taking place in the formation of dihydrogen?
Sol: The valence bond theory was put forward by Heitler and London in 1927. It was later improved and developed by L. Pauling and J.C. Slater in 1931. The valence bond theory is based on the knowledge of atomic orbitals and electronic configurations of elements, overlap criteria of atomic orbitals and stability of molecule.
The main points of valence bond theory are
(i) Atoms do not lose their identity even after the formation of the molecule.
(ii) The bond is formed due to the interaction of only the valence electrons as the two atoms come close to each other. The inner electrons do not participate in the bond formation.
(iii) During the formation of bond, only the valence electrons from each bonded atom lose their identity. The other electrons remain unaffected.
(iv) The stability of bond is accounted by the fact that the formation of bond is accompanied by release of energy. The molecule has minimum energy at a certain distance between the atoms known as intemuclear distance. Larger the decrease in energy, stronger will be the bond formed.

Valence bond Treatment of Hydrogen Molecule:
Consider two hydrogen atoms A and B approaching each other havingnuclei H_a and H_B and the corresponding electrons e_A and e_B respectively.
When atoms come closer to form molecules new forces begin to operate.
(a) The force of attraction between nucleus of atom and electron of another atom.
(b) The force of repulsion between two nuclei of the atom and electron of two atoms.

Fig. (a) Two hydrogen atoms at a large distance and hence, no interaction, (b) Two hydrogen atom closer to each other atomic orbitals begin to interact, (c) Attractive and repulsive forces in hydrogen atoms when interaction begins. In case of hydrogen: Figure ‘a’ shows that two hydrogen atoms are at farthest distances and their electron distribution is absolutely symmetrical.
(a) When two hydrogen atom start coming closer to each other, the electron cloud becomes distorted and new attractive and repulsive forces begin to operate as shown in figure ‘c’
(b) In figure ‘c’ dotted lines show attractive forces present in atom already and bold lines show the new attractive and repulsive forces.
(c) It has been found experimentally that the magnitude of net attractive forces is more than net repulsive forces. Thus stable hydrogen molecule is formed.

Potential energy diagram for formation of hydrogen molecules:
When two hydrogen atoms are at farther distance, there is no force operating between them, when they start coming closer to each other, force of attraction comes into play and their potential energy starts decreasing. As they come closer to each other potential goes on decreasing, but a point is reached, when potential energy acquires minimum value.
Note:
(a) This distance corresponding to this minimum energy value is called the distance of maximum possible approach, i.e. the point which corresponds to minimum energy and maximum stability.
(b) If atoms come further closer than this distance of maximum possible approach, then potential energy starts increasing and force of repulsion comes into play and molecules starts becoming unstable.

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NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers

Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 15 Polymers:

Section Name	Topic Name
15	Polymers
15.1	Classification of Polymers
15.2	Types of Polymerisation Reactions
15.3	Molecular Mass of Polymers
15.4	Biodegradable Polymers
15.5	Polymers of Commercial Importance

NCERT INTEXT QUESTIONS

15.1. What are polymers?
Ans: Polymers are high molecular mass substances (10³ — 10⁷u) consisting of a very large number of simple repeating structural units joined together through covalent bonds in a linear fashion. They are also called macromolecules. Ex: polythene, nylon 6,6, bakelite, rubber, etc.

15.2. How are polymers classified on the basis of structure?
Ans: On the basis of structure, polymers are classified into three types. These are linear chain polymers, branched chain polymers and crossedlinkedpolymers.

1. Linear chain polymers: In this case, the monomer units are linked to one another to form long linear chains. These linear chains are placed one above the other and are closely packed in space. The close packing results in high densities, tensile strength and also high melting and boiling points. High density polyethene is a very common example of this type. Nylon, polyesters and PVC are also linear chain polymers.

2. Branched chain polymers: In this type of polymers, the monomer units are linked to form long chains which have also side chains or branched chains of different Lengths attached to them. As a result of branching, these polymers are not closely packed in space. They have low densities, low tensile strength as well as low melting and boiling points. Some common examples of such polymers are ; low density polyethene, amylopectin, starch, glycogen etc.

3. Cross: linked polymers. In these polymers, also called net—work polymers, the monomer units are linked together to form three dimensionaL net—work as shown in the figure. These are expected to be quite hard, rigid and brittle. Examples of cross linked polymers are bakelite, glyptal. melamine formaldehyde polymer etc.

15.3. Write the names of the monomers of the following polymers:

Ans: (i) Hexamethylene diamine NH₂-(CH₂)6NH₂ and adipic acid HOOC – (CH₂)₄ – COOH
(ii) Caprolactum
(iii) Tetrafluoroethene F₂C = CF₂

15.4. Classify the following as addition and condensation polymers:
Terylene, Bakelite, Polyvinyl chloride,Polythene
Ans: Addition polymers: Polyvinyl chloride, Polythene
Condensation polymers : Terylene, bakelite.

15.5. Explain the difference between Buna-N and Buna-S.
Ans: Both Buna-N and Buna-S are synthetic rubber and are co-polymers in nature. They differ in their constituents.
Buna-N: Constituents are : buta-1, 3-diene and acrylonitrile.
Buna-S: Constituents are : buta-1, 3-diene, and styrene. They condense in the presence of Na.

Buna – S: It is a co—polymer of 1. 3 – butadiene and styrene and is prepared by the polymerisation of these components in the
ratio of 3 : 1 in the presence of sodium.

Buna-N (Nitrile rubber):  h is a co-polymer of buta-1. 3-diene and acrylonitrile. It is formed as follows:

15.6. Arrange the following polymers in increasing order of their intermolecuiar forces.
(i) Nylon 6,6, Buna-S, Polythene
(ii) Nylon 6, Neoprene, Polyvinyl chloride
Ans: On the basis of intermolecuiar forces, polymers
are classified as elastomers, fibres and plastics. The increasing order of intermolecuiar forces is: Elastomer < Plastic < fibre.
Thus, we have
(i)Buns-S < Polythene < Nylon 6,6
(ii)Neoprene < Polyvinyl chloride < Nylon 6.

NCERT EXERCISES

15.1. Explain the terms polymer and monomer.
Ans: Polymers are high molecular mass substances consisting of a very large number of simple repeating structural units joined together through covalent bonds in a regular fashion. Polymers are also called macromolecules. Some examples are polythene, nylon-66, bakelite, rubber, etc. Monomers are the. simple and reactive molecules from which the polymers are prepared either by addition or condensation polymerisation. Some examples are ethene, vinyl chloride, acrylonitrile, phenol and formaldehyde etc.

15.2. What are natural and synthetic polymers ? Give two examples of each.
Ans:
1. Natural polymers: The polymers which occur in nature mostly in plants and animals are called natural polymers. A few common examples are starch, cellulose, proteins, rubber nucleic acids, etc. Among them, starch and cellulose are the polymers of glucose molecules. Proteins are formed from amino acids which may be linked in different ways. These have been discussed in detail in unit 15 on biomolecules. Natural rubber is yet another useful polymer which is obtained from the latex of the rubber tree. The monomer units are of the unsaturated hydrocarbon 2-methyl-i, 3-butadiene, also called isoprene.
Example of natural polymers: Natural rubber, cellulose, nucleic acids, proteins etc.

2. Synthetic polymers: The polymers which are prepared in the laboraroiy are called synthetic polymers. These are also called man made polymers and have been developed in the present century to meet the ever increasing demand of the modem civilisation.
Example of synthetic polymers: Dacron (or terylene), Bakelite, PVC, Nylon-66, Nylon-6 etc.

15.3. Distinguish between the terms homopolymer and copolymer and give an example of each.
Ans: Polymers whose repeating structural units are derived from only one type of monomer units are called homopolymers, e.g., PVC polyethene, PAN, teflon, polystyrene, nylon- 6 etc.
Polymers whose repeating structural units are derived from two or more types of monomer molecules are copolymers, e.g., Buna-S, Buna-N, nylon-66, polyester, bakelite.

15.4. How do you explain the functionality of a monomer?
Ans: Functionality of a monomer implies the number of bonding sites present in it. For example, monomers like propene, styrene, acrylonitrile have functionality of one which means that have one bonding site.
Monomers such as ethylene glycol, hexamethylenediamine, adipic acid have functionality of two which means that they have two bonding sites.

15.5. Define the term polymerisation?
Ans: It is a process of formation of a high molecular Sol. mass polymer from one or more monomers by linking together a large number of repeating structural units through covalent bonds.

15.6. Is (-NH — CHR—CO-)_n a homopolymer or copolymer?
Ans: It is a homopolymer because the repeating structural unit has only one type of monomer, i.e., NH₂—CHR—COOH.

15.7. In which classes, are the polymers classified on the basis of molecular forces?
Ans: Polymers are classified into four classes on the basis of molecular forces. These are:
elastomers, fibres, thermoplastic polymers and thermosetting polymers.

1. Elastomers: In these polymers, the intermolecular forces are the weakest. As a result, they can be readily stretched by applying small stress and regain their original shape when the stress is removed. The elasticity can be further increased by introducing some cross – links in the polymer chains. Natural rubber is the most popular example of elastomers. A few more examples are of: buna-S, buna-N and neoprene.

2. Fibres: Fibres represent a class of polymers which are thread-like and can be woven into fabrics in a number of ways. These are widely used for making clothes, nets, ropes, gauzes etc. Fibres possess high tensile strength because the chains possess strong intermolecular forces such as hydrogen bonding. These forces are also responsible for close packing of the chains. As a result, the fibres are crystalline in nature and have aJso sharp melting points. A few common polymers belonging to this class are nylon – 66, terylene and polyacrylonitrile etc.

3. Thermoplastics: These are linear polymers and have weak van der Waals forces acting in the various chains and are intermediate of the forces present in the elastomers and in the fibres. When heated, they melt and form a fluid which sets into a hard mass on cooling, Thus, they can be cast into different shapes by using suitable moulds. A few common examples are polyethene and polystyrene polyvinyls etc. These can be used for making toys, buckets, telephone apparatus, television cabinets etc.

4. Thermosetting plastics: These are normally semifluid substances with low molecular masses. When heated, they become hard and infusible due to the cross-linking between the polymer chains. As a result, they also become three dimensional in nature. They do not melt when heated. A few common thermosetting polymers are bakelite, melamine-formaldehyde, urea-formaldehyde and polyurethane etc.

15.8. How can you differentiate between addition and condensation polymerisatiop?
Ans: In addition polymerization, the molecules of the same or different monomers simply add on to one another leading to the formation of a macromolecules without elimination of small molecules like H₂O, NH₃ etc. Addition polymerization generally occurs among molecules containing double and triple bonds. For example, formation of polythene from ethene and neoprene from chloroprene, etc. In condensation polymerisation, two or more bifunctional trifimctional molecules undergo a series of independent condensation reactions usually with the elimination of simple molecules like water, alcohol, ammonia, carbon dioxide and hydrogen chloride to form a macromolecule. For example, nylon-6,6 is a condensation polymer of hexamethylenediamine and adipic acid formed by elimination of water molecules.

15.9. Explain the term copolymerisation and give two examples.
Ans: When two or more different monomers are allowed to polymerise together the product formed is called a copolymer, and the process in called copolymerisation. Example, Buna-S and Buna-N. Buna- S is a copolymer of 1, 3- butadiene and styrene while Buna-N is a copolymer of 1,3-butadiene and acrylonitrile.

15.10. Write the free radical mechanism for the polymerisation of ethene.
Ans:

15.11. Define thermoplastics and thermo setting polymers with two examples of each
Ans: Thermoplastics polymers are linear polymer which can be repeatedly melted and moulded again and again on heating without any change in chemical composition and mechanical strength. Examples are polythene and polypropylene.
Thermosetting polymers, on the other hand, are permanently setting polymers. Once on heating in a mould, they get hardened and set, and then cannot be softened again. This hardening on heating is due to cross- linking between different polymeric chains to give a three dimensional network solid. Examples are bakelite, melamine-foimaldehyde polymer etc.

15.12. Write the monomers used for gettingThe following polymers:
(i) Polyvinylchloride
(ii) Teflon (iii) Bakelite
Ans:

15.13. Write the name and structure of one of the common initiators used in free radical addition polymerisation.
Ans:

15.14. How does the presence of double bonds in rubber molecules influence their structure and reactivity?
Ans: Natural rubber is cis-polyisoprene and is obtained by 1, 4-polymerization of isoprene units. In this polymer, double bonds are located between C₂ and C₃ of each isoprene unit. These cis-double bonds do not allow the polymer chains to come closer for effective interactions and hence intermolecular forces are quite weak. As a result, natural rubber, i.e., cis-polyisoprene has a randomly coiled structure not the linear one and hence show elasticity.

15.15. Discuss the main purpose of vulcanisation of rubber.
Ans: Natural rubber has the following disadvantages:
(a) It is soft and sticky and becomes even more so at high temperatures and brittle at low temperatures. Therefore, rubber is generally used in a narrow temperature range (283-335 K) where its elasticity is maintained.
(b)It has large water absorption capacity, has low tensile strength and low resistance to abrasion.
(c)It is not resistant to the action of organic solvents.
(d)It is easily attacked by oxygen and other oxidising agents. .
To improve all these properties, natural rubber is vulcanised by heating it with about 5% sulphur at 373-415 K. The vulcanized rubber thus obtained has excellent elasticity over a larger range of temperature, has low water absorption tendency and is resistant to the action of organic solvents and oxidising agents.

15.16. What are the monomeric repeating units of Nylon-6 and Nylon 6,6?
Ans:

15.17. Write the names and structures of the monomers of the following polymers:
(i) Buna-S (ii) Buna-N (iii) Dacron (iv) Neoprene
Ans:

15.18. Identify the monomer in the following polymeric structures:

Ans:

15.19. How is dacron obtained from ethylene glycol and terephthalic acid?
Ans: Dacron is obtained by condensation polymerization of ethylene glycol and terephthalic acid with the elimination of water molecules. The reaction is carried out at 420 – 460 K in presence of a catalyst consisting of a mixture of zinc acetate and antimony trioxide.

15.20. What is a biodegradable polymer ? Give an example of a biodegradable aliphatic polyester.
Ans: Polymers which disintegrate by themselves over a period of time due to environment degradation by bacteria, etc., are called biodegradable polymers. Example is PHBV, i. e., Poly-β-Hydroxybutyrate-co-β- Hydroxyvalerate.

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