Chemical Bonding & Molecular Structure

Li₂O $\Rightarrow$ 2Li⁺ O^2$-$

CaO $\Rightarrow$ Ca²⁺ O^2$-$

Na₂O₂ $\Rightarrow$ 2Na⁺ O$_2^{2 - }$

KO₂ $\Rightarrow$ O$_2^{- }$

MgO $\Rightarrow$ Mg²⁺ O^2$-$

K₂O $\Rightarrow$ 2K⁺ O^2$-$

O^2$-$ $\Rightarrow$ Complete octet, diamagnetic

$O_2^{2−}$ has 18 electrons.

Moleculer orbital configuration of $O_2^{2−}$ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $

Here is no unpaired electron so it is diamagnetic.

$O_2^{−}$ has 17 electrons.

Moleculer orbital configuration of $O_2^{−}$ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ * $

Here is 1 unpaired electron so it is paramagnetic.

Species	(Number of lone pairs of electrons on the central atom)
$Xe{F_2}$
$Xe{O_2}{F_2}$
$Xe{O_3}{F_2}$
$Xe{F_4}$

A. NaCl is diamagnetic because all electrons are paired in Na⁺ and in Cl⁻ . So, it shows diamagnetism.

B. Fe₃O₄ is ferrimagnetic because of the presence of the unequal alignment of magnetic moment in opposite direction.

C. O₂ molecule has two unpaired electrons. So, it is paramagnetic.

D. In MnO, the orientation of the electrons of Mn and O is such that they cancel their effects, hence it is antiferromagnetic.

$O_2^ - = {({\sigma _{1s}})^2}{(\sigma _{1s}^*)^2}{({\sigma _{2s}})^2}{(\sigma _{2s}^*)^2}{({\sigma _{2{p_z}}})^2}$$\left( {\pi _{2{p_x}}^2 = \pi _{2{p_y}}^2} \right)\left( {\pi _{2{p_x}}^{*2} = \pi _{2{p_y}}^{*1}} \right)$

Bond order = ${{10 - 7} \over 2} = 1.5$

and paramagnetic.

Note :

(1) $\,\,\,\,$ Bond strength $ \propto $ Bond order

(2) $\,\,\,\,$ Bond length $ \propto $ ${1 \over {Bond\,\,order}}$

(3) $\,$ Bond order $ = {1 \over 2}$ [N_b $-$ N_a]

N_b = No of electrons in bonding molecular orbital

N_a $=$ No of electrons in anti bonding molecular orbital

(4) $\,\,\,\,$ upto 14 electrons, molecular orbital configuration is



Here N_a = Anti bonding electron $=$ 4 and N_b = 10

(5) $\,\,\,\,$ After 14 electrons to 20 electrons molecular orbital configuration is - - -



Here N_a = 10

and N_b = 10

In O atom 8 electrons present, so in O₂, 8 $ \times $ 2 = 16 electrons present.

Then in $O_2^ + $ no of electrons = 15

in $O_2^ - $ no of electrons = 17

in $O_2^{2 - }$ no of electrons = 18

$\therefore\,\,\,\,$ Molecular orbital configuration of O₂ (16 electrons) is

${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$ ${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $

$\therefore\,\,\,\,$N_a = 6

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 6} \right] = 2$

Molecular orbital configuration of O$_2^ + $ (15 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 5

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 5} \right]$ = 2.5

Molecular orbital configuration of $O_2^ - $ (17 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 7

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 7} \right]$ = 1.5

Molecular orbital configuration of O $_2^{2 - }$ (18 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 8

$\therefore\,\,\,\,$ BO = ${1 \over 2}$ [ 10 $-$ 8] = 1

As Bond strength $ \propto $ Bond order so, correct order is

$O_2^{2 - } < O_2^ - < {O_2} < O_2^ + $

Among SO₃, O₂, SO₂ and CO$_3^{2 - }$, only O₂ and CO$_3^{2 - }$ has only one $\pi$-bond.

(a) SF₄ - sp³d hybridization

(b) IF₅ - sp³d² hybridization

(c) NO$_2^ + $ - sp hybridization

(d) NH$_4^ + $ - sp³ hybridization

Hybridisation = Number of sigma bonds + number of lone pair of electrons + number of coordinate bond.

For $NO_2^ - $ hybridisation = 2 sigma NO bonds + 1 lone pair on N = 3; so $NO_2^ - $ is sp²-hybridised.


For $NO_2^ + $ hybridisation = 2 sigma NO bonds = 2; so $NO_2^ + $ sp-hybridised.


For $NO_4^ + $ Hybridisation = 4 sigma NH bonds = 4; so $NO_4^ + $ is sp³-hybridised.


Therefore, hybridisation of N in $NO_2^ - $, $NO_2^ + $, $NO_4^ + $ are sp², sp, sp³ respectively.

sp³d hybridised
T-shaped

due to lp - lp repulsion bond angle decrease.

Bond order $ = {1 \over 2}$ [N_b $-$ N_a]

N_b = No of electrons in bonding molecular orbital

N_a $=$ No of electrons in anti bonding molecular orbital

(4) $\,\,\,\,$ upto 14 electrons, molecular orbital configuration is



Here N_a = Anti bonding electron $=$ 4 and N_b = 10

(5) $\,\,\,\,$ After 14 electrons to 20 electrons molecular orbital configuration is - - -



Here N_a = 10

and N_b = 10

(a) Molecular orbital configuration of Ne₂ (20 electrons) is

${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$ ${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^2}^ * \,\, = \pi _{2p_y^2}^ * $$\sigma _{2p_z^2}^*$

$\therefore\,\,\,\,$N_a = 10

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 10} \right] = 0$

(Note : All inert gases has BO = 0 and it does not exist as molecule. Here Ne is also an inert gas.)

(b) Moleculer orbital configuration of $N_2$ (14 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$

$\therefore\,\,\,\,$N_a = 4

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 4} \right] = 3$

(d) Molecular orbital configuration of F₂ (18 electrons) is

${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$ ${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^2}^ * \,\, = \pi _{2p_y^2}^ * $

$\therefore\,\,\,\,$N_a = 8

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 8} \right] = 1$

(d) Molecular orbital configuration of O₂ (16 electrons) is

${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$ ${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $

$\therefore\,\,\,\,$N_a = 6

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 6} \right] = 2$

Dipole - Dipole are not only the interaction responsible for hydrogen bond formation. Ion-dipole can also be responsible for hydrogen bond formation.

F is most electronegative element and anhydrous HF in solid phase has symmetrical hydrogen bonding.

(a)   XeF₄  is sp³d² hybridised with 4 bond pairs and 1 lone pair and structure is square planar. Here all the bond lengths are equal.



(b) $\,\,\,\,$ BF$_4^ - $, 4 bond pair present so angle is 109^o 28^' and sp³ hybridised. So structure is regular tetrahedral. Here all the bond lengths are equal.



(c) SiF₄ is sp³ hybridisation and regular tetrahedral geometry. Here all the bond lengths are equal.



(d)   SF₄ is sp³d hybridised with 4 bond pairs and 1 lone pair and its expected trigonal bipyramidal geometry gets distorted due to presence of a lone pair of electrons and it becomes distorted tetrahedral or see-saw with the bond angles equal to < 120^o and 179^o instead of the expected angles of 120^o and 180^o respectively. Here axial and equitorial both bonds are presents. And we know axial bonds are longer and weaker.

Note :

According to molecules orbital theory, when a molecule have bond order = 0 then that molecule does not exist.

We know, Bond order $ = {1 \over 2}$ [N_b $-$ N_a]

N_b = No of electrons in bonding molecular orbital

N_a $=$ No of electrons in anti bonding molecular orbital

(4) $\,\,\,\,$ upto 14 electrons, molecular orbital configuration is



Here N_a = Anti bonding electron $=$ 4 and N_b = 10

(5) $\,\,\,\,$ After 14 electrons to 20 electrons molecular orbital configuration is - - -



Here N_a = 10

and N_b = 10

Molecular orbital configuration of O $_2^{2 - }$ (18 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 8

$\therefore\,\,\,\,$ BO = ${1 \over 2}$ [ 10 $-$ 8] = 1

$\,\,\,$ Configuration of $He_2^ + $ (3 electrons) is = ${\sigma _{1{s^2}}}\,\,\sigma _{1{s^1}}^ * $

$\therefore\,\,\,$ Bond order = ${1 \over 2}$ (2 $-$1) = 0.5

$\,\,\,$ Configuration of $He_2^ - $ (5 electrons) is = ${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * {\sigma _{2{s^1}}}\,$

$\therefore\,\,\,$ Bond order = ${1 \over 2}$ (3 $-$2) = 0.5

$\,\,\,$ Configuration of $Be_2 $ (4 electrons) is = ${\sigma _{1{s^2}}}$ $\sigma _{1{s^2}}^ * $

$\therefore$ Bond order = ${1 \over 2}$ (2 $-$ 2) = 0

Correct option is i.e. LiF > LiCl; MgO > NaCl. Melting point is directly proportional to lattice energy. Lattice energy is the energy required to separate a mole of an ionic solid into gaseous ions. It depends upon charge of ions and size of ions.

$M.P. \propto L.E. \propto {{Charge} \over {Size}}$

$\matrix{ {Li \to + 1} & {Li \to + 1} \cr {F = - 1} & {Cl \to - 1} \cr } $

Both LiF and LiCl having same charge, so melting point will depend on size.

Larger the size of anion, lesser the lattice energy and hence, melting point order is LiF > LiCl.

Similarly, $\matrix{ {MgO} & {NaCl} \cr {Mg \to 2 + } & {Na \to 1 + } \cr {O \to 2 - } & {Cl \to 1 - } \cr } $

MgO having + 2 charge which is greater than NaCl (+ 1) charge. So, greater the charge on the ions greater will be lattice energy and hence, melting point order is MgO > NaCl.

Hybridisation of central $I$ in $I_3^ - $ is sp³d with 3 lone pair and 2 bond pair.

Shape Linear

Lone pair 3 lone pair

Bond angle 180$^\circ$ (for linear molecule)

Isostructural compounds are those compounds which have same structure as well as same hybridisation.

Formula for find hybridisation : H = (Lone pair + Sigma bond + coordinate bond )

If number of sigma bond ($\sigma $), co-ordinate bond and lone pair are same for given pairs, they are isostructural.

CH₄ - Bond angle = 109^o28'

H₂S - Bond angle = 92^o

NH₃ - Bond angle = 107^o

H₂O - Bond angle = 104^o5'

Potential energy curve for H₂ molecules is

In solid state PCl₅ exist in Ionpair i.e. [PCl₄] ⁺ and [PCl₆] ^–

[PCl₄]⁺ is tetrahedral.
[PCl₆]^– is octahedral.

To determine the option in which the hybridisation of the underlined atom is affected, we need to look at each reaction and predict the changes in hybridisation that might be occuring with the reactants and products.

Option A:

In the reaction $ \underline {Xe}F_4 + SbF_5 \to $, xenon tetrafluoride (XeF₄) reacts with antimony pentafluoride (SbF₅). XeF₄ is a square planar molecule, and the central Xe atom is $sp^3d^2$ hybridised. The reaction with SbF₅ can lead to the formation of a complex where the hybridisation of Xe changes if it coordinates with SbF₅. Thus, the hybridisation of Xe may indeed be affected in this case.

$\underline {Xe} $F4	+	SbF5	$ \to $	[XeF3]+[SbF6]-
sp3d2				sp3d   sp3d2

Option B:

In the reaction $ H_2\underline{S}O_4 + NaCl \buildrel {420K} \over \longrightarrow $, sulfuric acid (H₂SO₄) reacts with sodium chloride (NaCl). The underlined atom here is sulfur (S). In sulfuric acid, sulfur is hybridised as $sp^3$. Upon heating, there is typically no change in the hybridisation of the sulfur atom within sulfuric acid as part of this reaction. Therefore, the hybridisation of S is likely not affected.

Option C:

In the reaction $ H_3\underline{P}O_2 \buildrel {Disproportionation} \over \longrightarrow $, phosphorous acid (H₃PO₂) is undergoing a disproportionation reaction. The underlined atom is phosphorus (P). In H₃PO₂, the hybridisation of the phosphorus atom is $sp^3$. A typical disproportionation reaction of phosphorous acid would result in phosphine (PH₃) and orthophosphoric acid (H₃PO₄). The phosphorus atom in PH₃ will have $sp^3$ hybridisation while in H₃PO₄ the phosphorus atom is $sp^3$ hybridised. Given this information, the hybridisation of phosphorus may or may not change, depending on the resulting state of the phosphorus atom in the products. It's not a clear-cut case without knowing the specific products.

Option D:

In the reaction $ \underline{N}H_3 \buildrel {{H^+}} \over \longrightarrow $, ammonia (NH₃) is reacting with a proton (H⁺). The underlined atom is nitrogen (N). In NH₃, the nitrogen is $sp^3$ hybridised. When ammonia accepts a proton, it forms ammonium ion (NH₄⁺), which is also $sp^3$ hybridised. Therefore, there is no change in the hybridisation of nitrogen in this case.

Given this information, the option in which the hybridisation of the underlined atom is most clearly affected is Option A, where the reaction can lead to coordination changes to the Xe atom, potentially affecting its hybridisation.

From the given graph, potential energy of A-B molecule is minimum.

Thus A-B bond is most stable as it requires more energy to dissociate and have strongest bond amongst these.

B = Most electronegative

D = Least electronegative

A - B = Shortest bond length

A - B = Largest bond enthalpy

Note :

(1) $\,\,\,\,$ Bond strength $ \propto $ Bond order

(2) $\,\,\,\,$ Bond length $ \propto $ ${1 \over {Bond\,\,order}}$

(3) $\,$ Bond order $ = {1 \over 2}$ [N_b $-$ N_a]

N_b = No of electrons in bonding molecular orbital

N_a $=$ No of electrons in anti bonding molecular orbital

(4) $\,\,\,\,$ upto 14 electrons, molecular orbital configuration is



Here N_a = Anti bonding electron $=$ 4 and N_b = 10

(5) $\,\,\,\,$ After 14 electrons to 20 electrons molecular orbital configuration is - - -



Molecular orbital configuration of NO (15 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 5

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 5} \right]$ = 2.5

Similarly Molecular orbital configuration of NO⁺ (14 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,$

$\therefore\,\,\,\,$ N_b = 10

N_a = 4

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 4} \right]$ = 3

Similarly Molecular orbital configuration of NO²⁺ (13 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^1}}\,$

$\therefore\,\,\,\,$ N_b = 9

N_a = 4

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {9 - 4} \right]$ = 2.5

Molecular orbital configuration of NO^- (16 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^1}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 6

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 6} \right]$ = 2

As Bond strength $ \propto $ Bond order

$ \therefore $ NO^– will have minimum bond strength.

Ion-ion interaction energy $ \propto $ ${1 \over r}$

Dipole-dipole interaction energy $ \propto $ ${1 \over {{r^3}}}$

London dispersion $ \propto $ ${1 \over {{r^6}}}$

[XeF₅]^– $ \to $ 5BP + 2LP $ \to $ sp³d³ hybridisation $ \to $ Pentagonal planar

XeO₃F₂ $ \to $ 5BP + 0LP $ \to $ sp³d hybridisation $ \to $ Trigonal bipyramidal

Here 4$\sigma $ bonds +1 lone pair

Electronic geometry is Trigonal bipyramidal. Moleculer Geometry or Moleculer Shape is see-saw.

Note : The difference between Electronic Geometry and Moleculer Geometry(Moleculer Shape) is that in Electronic Geometry we include lone pair but in Moleculer Geometry(Moleculer Shape) we do not include lone pair.

Compound	Electronic Geometry	Moleculer Geometry or Moleculer Shape	Lone Pair	Polarity
AB4	Tetrahedral	Tetrahedral	0	Nonpolar
AB4	Trigonal Bipyramidal	see-saw	1	Polar
AB4	Octahedral (Square Bipyramidal)	Square Planar	2	Nonpolar

Note : The difference between Electronic Geometry and Moleculer Geometry(Moleculer Shape) is that in Electronic Geometry we include lone pair but in Moleculer Geometry(Moleculer Shape) we do not include lone pair.

Here in this question it is not clear which geometry they are talking about, correct answer should be either Trigonal Bipyramidal or see-saw.

Each carbon atom is sp² hybrid

$ \therefore $ 3 sp² hybrid orbitals are formed by each carbon atom

Total sp² orbitals = 6 × 3 = 18

Magnetic moment = 1.73 BM

$ \therefore $ Unpaired electron = 1

(1) $O_2$ has 16 electrons.

Moleculer orbital configuration of $O_2$ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^1}^ * $

Here 2 unpaired electron present.

(2)   $O_2^{+}$ has 15 electrons.

Moleculer orbital configuration of $O_2^{+}$ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $

Here 1 unpaired electron present.

(3)   $O_2^{-}$ has 17 electrons.

Moleculer orbital configuration of $O_2^{-}$ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * = \,\,\pi _{2p_y^1}^ * $

Here 1 unpaired electron present.

Hence $O_2^{-}$ & $O_2^{+}$ have one unpaired electron.

Bond length order in carbon halogen bonds are

in the order of C – F $<$ C – Cl $<$ C – Br $<$ C – I

Hence, Bond energy order

C – F $>$ C – Cl $>$ C – Br $>$ C – I

    CN^- has 14 electrons.

Moleculer orbital configuration of CN^- is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$

Here is no unpaired electron so it is diamagnetic.
N_b = 10

N_a = 4

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 4} \right]$ = 3

Ionic interactions are stronger as compared to van der waal interactions.

So, correct order is

ion-ion $>$ ion-dipole $>$ dipole-dipole

CHCl₃ is polar while CH₄ and CCl₄ are nonpolar.

So, dipole moment order is :

CH₄ = CCl₄ < CHCl₃