NCERT Exemplar: Chemical Bonding and Molecular Structure- 2 | Chemistry for JEE Main & Advanced PDF Download

Short Answer Type Questions

Q.31. Explain the nonlinear shape of H₂S and nonplanar shape of PCl₃ using valence shell electron pair repulsion theory.
Ans. The Lewis structure of H₂S is:
S-atom is surrounded by four electron pairs (two bonded and two lone pairs). These four electron pairs adopt a tetrahedral arrangement. The presence of two lone pairs brings distortion in the molecule on account of repulsion with bonded pairs of electrons. Thus, the shape of the H₂S molecule is V-shaped and not linear.
The Lewis structure of PCI₃ is:
P-atom is surrounded by four electron pairs (3 bonded and one lone pair). These four pairs adopt a tetrahedral geometry. Due to the presence of a lone pair, PCl₃ has a distorted tetrahedral geometry. Thus, it is pyramidal in shape and not non-planar shape.

Q.32. Using molecular orbital theory, compare the bond energy and magnetic character of O⁺₂ and O^–₂ species.
Ans.


The bond energy of O⁺₂ > O^-₂
Both are paramagnetic due to the presence of unpaired electrons.

Q.33. Explain the shape of BrF₅.
Ans. Br-atom has configuration:
1s², 2s²2p⁶ , 3s²3p⁶3d¹⁰, 4s²4p⁵
To get pentavalent, 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.

Q.34. Structures of molecules of two compounds are given below :
(i)
(ii)
(a) Which of the two compounds will have intermolecular 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 the 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 a hydrogen bond with water easily and be more soluble in it.
Ans. (a) Compound I will form an intramolecular hydrogen bond because NO₂ and OH groups are close together whereas it is not so in compound II. Compound II will have intermolecular hydrogen bonding. Bonding is both cases is shown below:
(i)

(ii)

(b) As a large number of molecules can be linked together through intermo­lecular hydrogen bonding, compound II will show a higher melting point.
(c) Due to intramolecular hydrogen bonding, compound I will not be able to form hydrogen bonds with H₂O molecules. Hence, it will be less soluble in water. However, compound II can form hydrogen bonds with H₂O molecules easily, and hence it will be more soluble in water.

Q.35. Why does the type of overlap given in the following figure not result in bond formation?
Ans. In the first figure, the ++ overlap is equal to +- overlap, and therefore, these cancel out and net overlap is zero.
In the second figure, no overlap is possible because the two orbitals are perpendicular to each other.

Q.36. Explain why PCl₅ is trigonal bipyramidal whereas IF₅ is square pyramidal.
Ans. 

• In PCl₅ , the electronic configuration of phosphorus is 1s², 2s², 2p⁶, 3s², 3p³ to form five bonds with chlorine. Electron form s-orbital is excited to the vacant d-orbital due to which phosphorus gains the ability to make 5 bonds.
• Now, it has a total of five unpaired electrons.
• So, the hybridization of phosphorus is sp³d and the geometry will be trigonal bipyramidal.
• Because it has 5 bond pairs and 0 lone pairs.
• PCl₅ – The ground state and the excited state outer electronic configurations of phosphorus (Z = 15) are represented below:
  
• In IF₅, according to the electronic configuration of iodine two electrons from 5p-orbital are excited to move to the 5d-orbital.
• So, now iodine tends to make 5 bonds and along with it have one lone pair.
• The lone pair will cause the repulsion towards the bond pair and increases the bond angles that’s why it is oriented in such a way that it causes less repulsion.
• So, the hybridisation will be sp³d² and the geometry will be square pyramidal.
• The structure is:

Note: The lone pair- lone pair repulsion causes maximum repulsion than lone pair-bond pair and lone pair-bond pair repulsion is more than the bond pair-bond pair repulsion.

Q.37. In both water and dimethyl ether, the oxygen atom is the central atom and has the same hybridization, yet they have different bond angles. Which one has a greater bond angle? Give reason.
Ans. Dimethyl ether has a 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 the water. The carbon of the 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 the carbon atom. Hence, the repulsion between the two CH₃ groups will be more than that between two H atoms.

Q.38. Write Lewis structure of the following compounds and show a formal charge on each atom.
HNO₃,  NO₂,   H₂SO₄
Ans. 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]
(i) 
F.C. on O (1) → 6 - 4 -  = 0
F.C. on O (2) → 6 - 4 -  = 0

F.C. on O (3) → 6 - 6 -  = -1
F.C. on N → 5 - 0 - = +1
F.C. on H  → 1 - 0 - =0
(ii) 
F.C. on O (1) → 6 - 6 -  = - 1
F.C. on O (2) → 6 - 4 -  = 0
F.C. on N → 5 - 2 -  = 0
(iii)
F.C. on O (1) and (4) → 6 - 4 - = 0
F.C. on O (2) and (3) → 6 - 4 -  = 0
F.C. on H → 1 - 0 - = 0
F.C. on H → 6 - 0 -  = 0

Q.39. The energy of σ2p_z molecular orbital is greater than π2p_x and π2p_y molecular orbitals in nitrogen molecules. Write the complete sequence of energy levels in the increasing order of energy in the molecule. Compare the relative stability and the magnetic behavior of the following species :
N₂, N⁺₂, N^–₂, N²⁺₂
Ans. The sequence of energy levels

For the N₂ molecule, the M.O. configuration is:

B.O. = 1/2(10 - 4) = 3, diamagnetic

B.O. = 1/2(9 -4) = 2.5, paramagnetic

B.O. = 1/2(10 -5) = 2.5, paramagnetic

B.O. = 1/2(8 -4) = 2.5, diamagnetic
Stability order: N₂ > N^-₂ > N⁺₂ > N²⁺₂
(N^-₂ has more bonding electrons as compared to N⁺₂).

Q.40. What is the effect of the following processes on the bond order in N₂ and O₂?
(i) N₂ → N⁺₂ + e^–
(ii) O₂ → O⁺₂ + e^–
Ans.

Q.41. Give reasons for the following:
(i) Covalent bonds are directional bonds while ionic bonds are nondirectional.
(ii) Water molecule has a bent structure, whereas carbon dioxide molecule is linear.
(iii) Ethyne molecule is linear.
Ans. (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 the presence of two lone pairs of electrons on the oxygen atom in H₂O, the repulsion between Ip-lp is more. CO₂ undergoes sp hybridization resulting in linear shape (O = C = O) while H₂O undergoes sp₃ hybridization resulting in distorted tetrahedral or bent structure.
(iii) In the ethyne molecule carbon undergoes sp hybridization with two unhybridized 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. Unhybridized orbitals form π bonds.

Q.42. What is an ionic bond? With two suitable examples explain the difference between an ionic and a covalent bond?
Ans. An ionic bond is formed as a result of the electrostatic attraction between the positive and negative ions formed by the transfer of electrons from one atom to another.

A covalent bond is formed by the sharing of electrons between atoms. Examples of covalent bonds:
Multiple bonds are formed during covalent bonding while in ionic bond a number of bonds can be formed between different atoms.

Q.43. Arrange the following bonds in order of increasing ionic character giving a reason.
N—H,  F—H,  C—H, and O—H
Ans. Electronegativity difference

Increasing order of electronegativity difference: C-H < N-H < O-H < F-H
Greater is the difference in electronegativity between two bonded atoms, greater is the ionic character.

Q.44. Explain why CO^2–₃ ion cannot be represented by a single Lewis structure. How can it be best represented?
Ans. 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 the same bond length and same bond strength.

Q.45. Predict the hybridization of each carbon in the molecule of the organic compound given below. Also indicate the total number of sigma and pi bonds in this molecule.

Ans.
Number of σ Bonds = 11
Number of π Bonds = 4

Q.46. Group the following as linear and non-linear molecules :
H₂O, HOCl, BeCl₂, Cl₂O
Ans.

Q.47. 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?
Ans. (i)


(ii) Z will be the most electronegative element and hence HZ will have the highest dipole moment.

Q.48. Draw the resonating structure of
(i) Ozone molecule
(ii) Nitrate ion
Ans. Resonating structures:
(i) Ozone molecule
(ii) Nitrate ion

Q.49. Predict the shapes of the following molecules on the basis of hybridization.
BCl₃, CH₄, CO₂, NH₃
Ans. BCl₃ – sp² hybridization – Trigonal planar
CH₄ – sp³ hybridization – Tetrahedral

CO₂ - sp hybridization - Linear
NH₃ – sp³ hybridization – Distorted tetrahedral or Pyramidal

Q.50. All the C—O bonds in carbonate ion (CO^2–₃ ) are equal in length. Explain.
Ans.  Carbonate ion is represented by resonating structures as given below:
The actual structure will be the mixture of all resonating structures, also called a resonance hybrid structure.

As it performs resonance so because of its resonating property, all the bonds have partial double bond character and results into equal bond lengths. Carbonate ion is a symmetric, trigonal planar molecule. All three carbon-oxygen bond distances are about 1.28 Angstroms long.

Q.51. What is meant by the term average bond enthalpy? Why is there a difference in bond enthalpy of O—H bond in ethanol (C₂H₅OH) and water?
Ans. All the similar bonds in a molecule do not have the same bond enthalpies, e.g., in CH₄ the four C - H bonds do not have the same bond enthalpies because after breaking of the bonds one by one, the electronic environment around carbon changes. Hence, the actual bond enthalpy of the C - H bond is taken as the average value.
O - H bond in ethanol and that in water do not have a similar electronic environment around oxygen atom. Hence, their O - H bond enthalpies are different.

Q.52. Match the species in Column I with the type of hybrid orbitals in Column II.

Ans. (i) → c; (ii) → a; (iii) → e; (iv) → d
(i) SF₄ = number of bp (4) + number of lp (1) = sp³d hybridisation
(ii) IF₅ = number of bp (5) + number of lp (1) = sp³d² hybridisation
(iii) NO⁺₂ = number of bp (2) + number of lp (0) = sp hybridisation
(iv) NH⁺₄ = number of bp (4) + number of Ip (0) = sp³ hybridisation

Q.53. Match the species in Column I with the geometry/shape in Column II.

Ans. (i) → e; (ii) → a; (iii) → b; (iv) → c
(i) H₃O⁺ → (e) Pyramidal
(ii) HC ≡ CH → (a) Linear
(iii) ClO^-₂ → (b) Angular
(iv) NH⁺₄ → (c) Tetrahedral

Q.54. Match the species in Column I with the bond order in Column II.

Ans. (i) → c; (ii) → d; (iii) → a; (iv) → b
NO → B.O. = 1/2(10 - 5) = 2.5
CO → B.O. = 1/2(10 4) = 3
O^-₂ → B.O. = 1/2(10-7)= 1.5
O₂ → B.O. = 1/2(10-6) = 2

Q.55. Match the items given in Column I with examples given in Column II.

Ans. (i) → d; (ii) → e; (iii) → b; (iv) → a
(i) H - F : Hydrogen bond
(ii) O₃ : Resonance
(iii) LiF: Ionic solid
(iv) C: Covalent solid

Q.56. Match the shape of molecules in Column I with the type of hybridization in Column II.

Ans. (i) →c; (ii) → a; (iii) → b
sp³ hybridization – Tetrahedral shape
sp² hybridization – Trigonal shape
sp hybridization – Linear shape

Assertion and Reason Type

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.

Q.57. 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.
Ans. (a)
Solution: Sodium chloride (Na⁺Cl^–) is a stable ionic compound because both Na⁺ and Cl^– ions have a complete octet in the outermost shell.

Q.58. Assertion (A): Though the central atom of both NH₃ and H₂O molecules are sp³ hybridized, yet H–N–H bond angle is greater than that of H–O–H.
Reason (R): This is because the nitrogen atom has one lone pair and the 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.
Ans. (a)
Solution: 

H₂O has two lone pairs while NH₃ has single lone pair, hence, H₂O involves greater lone pair-bond pair repulsion.

Q.59. Assertion (A): Among the two O–H bonds in the H₂O molecule, the energy required to break the first O–H bond and the other O–H bond is the same.
Reason (R): This is because the electronic environment around oxygen is the same even after the 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.
Ans. (d)
Solution: Bond energy of two (O – H) bonds in H₂O will be different.

Long Answer Type Questions

Q.60. (i) Discuss the significance/ applications of dipole moment.
(ii) Represent diagrammatically the bond moments and the resultant dipole moment in CO₂, NF_3, and CHCl₃.
Ans. (i) Dipole moment plays a very important role in understanding the nature of the chemical bonds.
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, especially in a diatomic molecule. The greater the dipole moment, the greater is the polarity in such a molecule.
(c) Shape of molecules: In the 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, the 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 the 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 a 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 usually has a 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.
(ii)


Q.61. 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.
Ans. Nitrogen molecule (N₂): Electron configuration of N-atom (Z = 7) is Is² 2s² 2p¹_x 2p¹_y 2p¹_z. 

The total number of electrons present in the N₂ molecule is 14, 7 from each N-atom. The electron configuration of the N₂ molecule will be:
Bond = 1/2 (10 - 4) = 3 (triple bond)
Fluorine molecule (F₂): Electronic configuration of F-atom (Z = 9) is 1s² 2s² 2p²_x 2p²_y 2p¹_z. The total number of electrons in the F₂ molecule is 18, 9 from each F-atom. Thus, it has two electrons more than O₂ molecules. These two additional electrons will enter into π*(2p_x) and π*(2p_y) orbitals, one in each of them so that the pairing of electrons takes place in these orbitals. Thus, the electronic configuration of F₂ will be:

Bond Order = 1/2 (10 - 8) = 1 (single bond)
Ne₂:
Total number of electrons in Ne₂ = 20.
The electron configuration of the N₂ molecule will be

No bond is formed between two Ne atoms or in other words, Ne₂ does not exist.
Bond Order = 1/2(10 – 10) = 0 (No bond).

Q.62. 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?
Ans. 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 the 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 bond formation.
(iii) During the formation of a bond, only the valence electrons from each bonded atom lose their identity. The other electrons remain unaffected.
(iv) The stability of the bond is accounted for by the fact that the formation of the bond is accompanied by the release of energy. The molecule has minimum energy at a certain distance between the atoms known as internuclear distance. The larger the decrease in energy, the stronger will be the bond formed.

Valence bond Treatment of Hydrogen Molecule:
Consider two hydrogen atoms A and B approaching each other having nuclei 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 the nucleus of an atom and the 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 atoms closer to each other atomic orbitals begin to interact.
(c) Attractive and repulsive forces in hydrogen atoms when interaction begins.
In the 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 atoms 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 the formation of hydrogen molecules:
When two hydrogen atoms are at a farther distance, there is no force operating between them, when they start coming closer to each other, the 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 the force of repulsion comes into play and molecules start becoming unstable.

Q.63. Describe hybridization in the case of PCl₅ and SF₆. The axial bonds are longer as compared to equatorial bonds in PCl₅ whereas in SF₆ both axial bonds and equatorial bonds have the same bond length. Explain.
Ans. Formation of PCI₅ (sp³d hybridization): The ground state and the excited state outer electronic configurations of phosphorus (Z = 15) are represented below:

Trigonal bipyramidal geometry of PCl₅ molecule

Three P—Cl bonds lie in one plane and make an angle of 120° with each other; these bonds are termed equatorial bonds. The remaining two P—Cl bonds—one lying above and the other lying below the equatorial plane, make an angle of 90⁰ with the plane. These bonds are called axial bonds. 

As the axial bond pairs suffer more repulsive interaction than the equatorial bond pairs, therefore axial bonds have been found to be slightly longer and hence slightly weaker than the equatorial bonds; which makes the PCl₅ molecule more reactive.
Formation of SF₆ (sp³d² hybridization): Molecule has a regular octahedral geometry.

Octahedral geometry of SF₆ molecule
Q.64. (i) Discuss the concept of hybridization. What are its different types of a carbon atom?
(ii) What is the type of hybridization of carbon atoms marked with a star.
(a)
(b)
(c)
(d)
(e)
Ans. (i) Hybridization: In order to explain the characteristic geometrical shapes of polyatomic molecules like CH₄, NH_3, and H₂O, etc., Pauling introduced the concept of hybridization. According to him, the atomic orbitals combine to form a new set of equivalent orbitals known as hybrid orbitals. 

Unlike pure orbitals, hybrid orbitals are used in bond formation. The phenomenon is known as hybridization, which can be defined as the process of intermixing the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of a new set of orbitals of equivalent energies and shape.

For example, when one 2s and three 2p-orbitals of carbon hybridize, there is the formation of four new sp³ hybrid orbitals.
The main features of hybridization are as under:
1. The number of hybrid orbitals is equal to the number of atomic orbitals that get hybridized.
2. The hybridized orbitals are always equivalent in energy and shape.
3. The hybrid orbitals are more effective in forming stable bonds than the pure atomic orbitals.
4. These hybrid orbitals are directed in space in some preferred direction to have minimum repulsion between electron pairs and thus a stable arrangement.
Therefore, the type of hybridization indicates the geometry of the molecules.
Different types of hybridization in carbon atom are:

• In carbon compounds, if carbon is linked to carbon through the triple bond, e.g. Alkynes, triple bonded carbon is sp hybridized.
• In carbon compounds, if carbon is linked to carbon through a double bond (C=C), e.g., Alkenes, double-bonded carbon is sp² hybridized.
• In carbon compounds, if carbon is linked to carbon through a single bond (C—C), e.g., Alkanes, single-bonded carbon is sp³ hybridized.

(a)
Both the starred C atoms are sp² hybridized.
(b)
The starred C atom is sp³ hybridized.
(c)
The starred C atom is sp² hybridized.
(d)
The starred C atom is sp³ hybridized.
(e)
The starred C atom is sp² hybridized.

Instructions: The comprehension given below is followed by some multiple-choice questions. Each question has one correct option. Choose the correct option.
Molecular orbitals are formed by the overlap of atomic orbitals. Two atomic orbitals combine to form two molecular orbitals called bonding molecular orbital (BMO) and antibonding molecular orbital (ABMO). The energy of the antibonding orbital is raised above the parent atomic orbitals that have combined and the energy of the bonding orbital is lowered than the parent atomic orbitals. Energies of various molecular orbitals for elements hydrogen to nitrogen increase in the order:
and for oxygen and fluorine order of energy of molecular orbitals is given below :

Different atomic orbitals of one atom combine with those atomic orbitals of the second atom which have comparable energies and proper orientation. Further, if the overlapping is head-on, the molecular orbital is called ‘Sigma’, (σ)  and if the overlap is lateral, the molecular orbital is called ‘pi’, (π). The molecular orbitals are filled with electrons according to the same rules as followed for filling of atomic orbitals. However, the order for filling is not the same for all molecules or their ions. Bond order is one of the most important parameters to compare the strength of bonds.

Q.65. Which of the following statements is correct?
(a) In the formation of dioxygen from oxygen atoms, 10 molecular orbitals will be formed.
(b) All the molecular orbitals in the dioxygen will be completely filled.
(c) Total number of bonding molecular orbitals will not be the same as the total number of antibonding orbitals in dioxygen.
(d) Number of filled bonding orbitals will be the same as the number of filled antibonding orbitals.
Ans. (a)
Solution: O₂ = (8 + 8) = 16.


Q.66. Which of the following molecular orbitals has a maximum number of nodal planes?
(a) σ*1 s
(b) σ*2p_z
(c) π2p_x
(d) π*2p_y
Ans. (b)
Solution: a*2p_z has maximum number of nodal planes.

Q.67. Which of the following pair is expected to have the same bond order?
(a) O₂, N₂
(b) O⁺₂, N^–₂
(c) O^–₂, N⁺₂
(d) O^–₂, N^–₂
Ans. (b) Bond order = (N_b — N_a)/2
In O₂, the number of electrons in bonding orbitals is 10, and in antibonding are 5.

Bond order = (10-5)/2 =2.5
In N₂, the number of electrons in bonding orbitals is 10, and in anti-bonding are 5.

Bond order = (10- 5) / 2 = 2.5.

Q.68. In which of the following molecules, σ2p_z molecular orbital is filled after π2p_x and π2p_y  molecular orbitals?
(a) O₂
(b) Ne₂
(c) N₂
(d) F₂
Ans. (c)
Solution: 

We have seen that 1s atomic orbitals on two atoms form two molecular orbitals designated as, In the same manner, the 2s and 2p atomic orbitals (eight atomic orbitals on two atoms) give rise to the following eight molecular orbitals.
Antibonding:
Bonding:
It has been observed experimentally that for molecules such as B₂, C₂, N_2, etc., the increasing order of energies of various molecular orbitals is: 
The important characteristic feature of this order is that the energy of molecular orbital is higher than that of and molecular orbitals.