Chemical Bonding & Molecular Structure

Average bond order = 6/4 = 1.5

Average bond order = 6/3 = 2

Average bond order = 4/3 = 1.33

Correct sequence of bond order is

SO$_3$ > SO$_4^{2-}$ > S$_2$O$_3^{2-}$

The sulphur dioxide molecule have dipole moment value of 161 debye. It is a bent molecule with a sulphur atom placed in the centre and two atoms of oxygen on the side, one oxygen atom is connected by a double bond and the oxygen ion by a single bond.

Hence, among all SO$_2$, have highest dipole moment.

Bond order (BO) = $\mathrm{{{(Number\,of\,electron\,in\,bonding\,MO) - (Number\,of\,electron\,in\,anti - bonding\,MO)} \over 2}}$

S. No.	Molecule	BO
1.	H$_2^+$	$1-0/2=0.5$
2.	He$_2^ + $	$2-1/2=0.5$
3.	He$_2^ - $	$2-1/2=0.5$
4.	B$_2^ + $	$3-2/2=0.5$
5.	F$_2^ - $	$4-2/2=1$
6.	Be$_2^{2 - }$	$3-1/2=1$

Hence, among all of these only $\mathrm{H_2^ + ,He_2^ + ,He_2^ - ,B_2^ +}$ give 0.5 bond order.

S. No.	Molecule	Steric no.	Hybridisation	Structure	Shape
(A)	$\mathrm{SnCl_2}$	3	$sp^2$		Angular or bent (1)
(B)	$\mathrm{XeF_4}$	6	$sp^3d^2$		Square planar (5)
(C)	$\mathrm{CIF_3}$	5	$sp^3d$		T-shape (4)
(D)	$\mathrm{IF_5}$	6	$sp^3d^2$		Square pyramidal (3)

Diborane $\left(\mathrm{B}_2 \mathrm{H}_6\right)$ is one of the simplest boron hydride.

The structure of diborane molecule consist four $\mathrm{H}$-atom and two boron atoms coming on the same plane. The boron atom present in $s p^3$ hybridised state and from these four hybrid orbitals, three of orbitals have one electron each, and one is an empty orbital. It will form $\mathrm{B}-\mathrm{H}-\mathrm{B}$ bonds called banana bond.

(a) $\mathrm{NCl}_5$ does not exist while $\mathrm{PCl}_5$ does. This is correct due to $\mathrm{N}$ does not have vacant $d$-orbitals and it cannot expand its oxidation state but $\mathrm{P}$ has vacant $d$-orbitals.

(b) $\mathrm{Pb}$ prefer to form tetravalent compound. This is wrong as to lead prefers to form bivalent compounds due to inert pair effect.

(c) The three $\mathrm{C}-\mathrm{O}$ bond are equal in carbonate ion. This is because the bonds are in resonance.

(d) Both O$_2^+$ and NO are paramagnetic as they contain unpaired electrons.

O$_2^+$ $\Rightarrow$ 15e$^-$ (odd) paramagnetic

NO $\Rightarrow$ 15e$^-$ (odd) paramagnetic

Hence, only (b) option is incorrect.

$p \pi-p \pi$ bonding is weak bonding. Due to larger atomic size of phosphorus. It is unstable to form $\pi-\pi$ bond and so, it is tetra-atomic in which each $p$-atom is linked with 3 others P-atom by $3 \sigma$-bonds. In nitrogen, due to smaller atomic size $\mathrm{N}$ forms $1 \sigma$ and $2 \pi$-bonds i.e. triple bonds with other $\mathrm{N}$ atom and exists as diatomic molecules. Hence, the bonding is responsible for that. Phosphorus from $\mathrm{P}_4$ because in $\mathrm{P}_2, p \pi$-$p \pi$ bond is weaker.

In $\mathrm{SF}_6$, the sulphur atom is shielded by 6 fluorine atoms which hinders kinetically any reaction with water, alkali hydroxide, ammonia or strong acid, as a result, it remains inert to these reagents.

$\mathrm{SF}_6$ contain 6 steric number and $s p^3 d^2$ hybridisation. $\mathrm{S}_2 \mathrm{O}_3^{2-}$ is similar look like ammonia shape i.e. tetrahedral.

There are two $\pi$-bond present in $\mathrm{SO}_4^{2-} \cdot \mathrm{SO}_4^{2-}$ is a polyatomic ion as well as a non-polar covalent compound.

According to Mulliken equation,

$\mathrm{EN}=\frac{\mathrm{IE}+\mathrm{EA}}{5.6},$ ..... (i)

when both IE and EA are taken in $\mathrm{eV}$.

(Here, IE = Ionisation energy

EA = Electron gain enthalpy

EN = Electros negativity)

IE for chlorine $=13 \mathrm{eV}$

EA for chlorine $=4 \mathrm{eV}$

Put value in Eq. (i)

$\begin{aligned} \mathrm{EN} & =\frac{13+4}{5.6}=\frac{17}{5.6} \\ & =3.03 \approx 3.0 \mathrm{eV} \end{aligned}$

Dipole moment is a measure of polarity of a bond. It is the product of charges and the distance between partial charge. Dipole moment of polar molecule containing lone pairs is vector sum of dipole of lone pair and net dipole moments of bonds.

Electronegativity of F is more than that of N, thus the direction of dipole moment of N-F bond will be form F to N. So, net dipole moment of NF$_3$ found be 0.24 D.

Due to less electronegativity of H in ammonia, it has maximum dipole moment. Dipole moment order :

$\matrix{ {N{H_3}} & < & {NB{r_3}} & < & {NC{l_3}} & < & {N{F_3}} \cr {(1.4D)} & {} & {(0.8D)} & {} & {(0.6D)} & {} & {(0.24D)} \cr } $

Here, D = Debye or coulomb metre.

S. No.	Molecule	Steric number	Hybridisation	Shape	Structure
1	$\mathrm{BF_3}$	$\frac{3+3}{2}=3$	$sp^2$	Trigonal planar
2.	$\mathrm{NO_2^-}$	$\frac{5+1}{2}=3$	$sp^2$	Bent shape
3.	$\mathrm{NH_2^-}$	$\frac{5+2+1}{2}=4$	$sp^3$	Angular or V-shape
4.	$\mathrm{H_2O}$	$\frac{6+2}{2}=4$	$sp^3$	V-shape or bent

Hence, BR$_3$ and NO$_2^-$ are sp$^2$ hybridised.

For p-dichlorobenzene and p-cyanobenzene, the dipole moment of individual bonds cancel each other as they are equal in magnitude and opposite in direction. Hence, resultant dipole moment of molecules is zero.

For p-hydroquinone and p-benzenedithiol, the O$-$H and S$-$H bond dipoles in these molecules do not cancel each other as they are not in opposite directions due to existence in different conformation.

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

Hence, option (c) is the correct answer, i.e.

(i) $=\pi^*$

(ii) $=\sigma$

(iii) $=\sigma^*$

(iv) $=\pi$, molecular orbitals

∵ Repulsive intraction of electron pairs is as follows :

The electron pairs place themselve as for apart as possible in the space in order to have minimum force of repulsion.

The magnitude of different types of electronic repulsions follows the order given below:

(I) Lone pair-lone pair > (II) Lone pair-bond pair

>(III) Bond pair-bond pair

Hence, (a) is the correct answser, i.e. (I) $>$ (II) $>$ (III).

(i) $\mathrm{XeOF}_4$ has one oxygen atom, which will form double bond with Xe .

(ii) $\mathrm{XeOF}_4$ has four fluorine atoms, which form four single bond with Xe .

So, it has square pyramidal geometry.

Note The shape is to minimise the repulsion. Hence, option (d) is the correct answer.

We know, bond length $\propto \frac{1}{\text { Bond order (BO) }}$

where, $\mathrm{BO}=\frac{N_b-N_a}{2}$

I. $\mathrm{O}_2\left(16 e^{-}\right) \Rightarrow \mathrm{BO}=\frac{10-6}{2}=2$;

II. $\mathrm{O}_2^{+}\left(15 e^{-}\right) \Rightarrow \frac{10-5}{2}=2.5$;

III. $\mathrm{O}_2^{-}\left(17 e^{-}\right) \Rightarrow \frac{10-7}{2}=1.5$;

IV. $\mathrm{O}_2^{2-}\left(18 e^{-}\right) \Rightarrow \frac{10-8}{2}=1$

In, the order of bond length

$ \text { (IV) } \mathrm{O}_2^{2-}>\text { (III) } \mathrm{O}_2^{-}>\text {(I) } \mathrm{O}_2>\text { (II) } \mathrm{O}_2^{+} $

In acetylene, we have $2 p_z$ of $\mathrm{C}_1$ and $2 p_z$ of $\mathrm{C}_2$ along $z$-axis, because both orbitals are $p$-orbitals and are in same direction, i.e. $z$-direction.

The formula of acetylene is $\mathrm{CH} \equiv \mathrm{CH}$. Hence, option (d) is the correct answer.

If all the electrons are paired, the molecule is diamagnetic, while if one or more electron in a molecule is unpaired, the species is paramagnetic. $\mathrm{O}_2^{2-}$ is not paramagnetic because in $\mathrm{O}_2^{2-}$ all the electrons are paired, i.e. Configuration of $\mathrm{O}_2^{2-}$ (Total electrons $=18)$

$(\sigma l s)^2,(\sigma l s)^{2^*},(\sigma 2 s)^2,(\sigma 2 s)^{2^*},\left(\sigma 2 p_z\right)^2$,

$\left.\left(\pi 2 p_y\right)^2,\left(\pi 2 p_x\right)^2\right),\left(\pi 2 p_x\right)^{2^*},\left(\pi 2 p_y\right)^{2^*}$.

Hence, $\mathrm{O}_2^{2-}$ is diamagnetic.

$\mathrm{BF}_3$ has trigonal planar geometry. B has 3 bond-pairs and 0 lone pair of electrons.

$\mathrm{ClF}_3$ has T-shape geometry. Cl has 3 bond pairs and 2 lone-pairs of electrons. Electronic geometry will be trigonal bipyramidal.

$\mathrm{NH}_3$ has trigonal pyramidal geometry. N has 3 bond pairs and 1 lone-pair of electrons.

$\mathrm{NH}_4^{+}$has tetrahedral geometry. N has 4 bond pairs.

Hence, correct match is

$ \mathrm{A} \rightarrow \mathrm{II}, \mathrm{~B} \rightarrow \mathrm{III}, \mathrm{C} \rightarrow \mathrm{IV}, \mathrm{D} \rightarrow \mathrm{I} $

$\mathrm{Be}_2$ molecules does not exist according to molecular orbital theory. Because bond order of $\mathrm{Be}_2$ is zero. Hence, it does not exist.

Many simple compounds of elements such as $\mathrm{AlCl}_3, \mathrm{GaCl}_3$ and $\mathrm{InCl}_3$ are covalent while anhydrous but in aqueous solution, these are ionic in nature.

In anhydrous condition, the (charge/radius) ratio, i.e. polarisability of $\mathrm{Al}^{3+}$ is high and hence, according to Fajans' rule, $\mathrm{Al}^{3+}$ polarises $\mathrm{Cl}^{-}$ions to large extent, there introducing covalent character in the compound, i.e. $\mathrm{AlCl}_3$ behaves as a covalent compound in anhydrous conditions.

In aqueous medium the ions get hydrated, because the amount of hydration enthalpy released exceeds, the sum total of ionisation enthalpy required.

Since, the (charge/radius ratio of hydrated aluminium ion is much smaller as compared to that of $\mathrm{Al}^{3+}$, the tendency of $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ to polarise hydrated $\mathrm{Cl}^{\ominus}$ ion decreases and the resulting hydrated compound is ionic in nature.

In (CH₃)₃N, one lone pair and 3 sigma bond present. So it's shape is pyramidal.

But in (SiH₃)₃N, due to backbonding lone pair of N shift to vacant 3d orbital of Si. So now N have only 3 sigma bond and hybridization is sp², that is why it has triangular planar geometry.

As in (SiH₃)₃N. nitrogen(N) don't have any lone pair to donate but in (CH₃)₃N, lone pair of nitrogen(N) can easily be donated. That is why (CH₃)₃N is more basic.

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}^ * $

Molecular orbital configuration of O₂^- (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}^ * $

So, the incoming electron goes to $\pi _{2p_x}^ *$ or $\pi _{2p_y}^ *$

Here you can see, H₂S₂O₇ does not have S – S linkage.

Those species which have unpaired electrons are called paramagnetic species.

And those species which have no unpaired electrons are called diamagnetic species.

(a)    CO has 14 electrons.

Moleculer orbital configuration of CO 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.

(b)   B₂ has 10 electrons.

Molecular orbital configuration of B₂ is

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

Here two unpaired electrons present. So it is paramagnetic.

(c)   $O_2$ 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, so it is paramagnetic.

(d)   NO has 15 electrons.

Moleculer orbital configuration of NO 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 is 1 unpaired electron, So it is Paramagnetic.

Due to strong H-bonding between HF molecules, HF has highest boiling point among the hydrogen halides.

C₂ has 12 electrons.

Moleculer orbital configuration of C₂ is

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

Bond order of C₂ = ${{8 - 4} \over 2}$ = 2

C₂^- has 13 electrons.

Moleculer orbital configuration of C₂^- is

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

Bond order of C₂^- = ${{9 - 4} \over 2}$ = 2.5

As we know, higher is the bond order more will be the stability of the molecule.

So we can say, stability of C₂ increases when it forms C₂^- ion.

According to Fajan’s rule, when the size of cation is less then the covalent character would be more for the cation.

Since Be²⁺ has minimum size among given cation, therefore, BeX₂ would be most covalent among given alkaline with metal halides

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

Then 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 = ${1 \over 2}\left[ {10 - 6} \right] = 2$

Here 2 unpaired electrons are present so it is paramagnetic.

(D) 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

Here no unpaired electrons are present so it is diamagnetic.

(B)   $C_2^{2 - }$ has 14 electrons.

Moleculer orbital configuration of $C_2^{2 - }$ 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 = ${1 \over 2}\left[ {10 - 4} \right] = 3$

Here no unpaired electron present, so it is diamagnetic.

(C)   $N_2^{2 - }$ has 16 electrons.

Moleculer orbital configuration of $N_2^{2 - }$ is

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

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

N_b = 10

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

Here 2 unpaired electron present, so it is paramagnetic.

As Bond length $ \propto $ ${1 \over {Bond\,\,order}}$

so among two diamagnetic ions $C_2^{2 - }$ and $O_2^{2 - }$, bond order of $C_2^{2 - }$ is more so it will have shorter bond length.