Chemical Bonding & Molecular Structure

$ \text { Correct Lewis representation is } $

Formal charge $=$ Number of valence electrons - number of lone pair electrons

$ -\frac{1}{2} \times \text { number of bond pair electrons } $

So, formal charge on

$ \begin{aligned} & a=4-0-\frac{1}{2} \times 8=4-4=0 \\ & b=6-4-\frac{1}{2} \times 4=6-6=0 \\ & c=6-6-\frac{1}{2} \times 2=6-7=-1 \end{aligned} $

Hence,

$ \begin{aligned} & a: 0 \\ & b: 0 \\ & c:-1 . \end{aligned} $

The first bond between any two atoms is always a sigma bond and the successive bonds are pi-bonds. Therefore, $\mathrm{O}_3$ molecule has 2 sigma and 1 pi-bond alongwith 6 lone pair of electrons. The structure of ozone is shown below.

$ \text { Molecule } $	Shape
$ \mathrm{SnCl}_2 $	Bent or angular
$ \mathrm{PbCl}_2 $	Bent or V shape
$ \mathrm{SO}_2 $	Trigonal planar
$ \mathrm{XeF}_2 $	Linear

Bond order of $\mathrm{CO}_3^{2-}=1.33$

Bond order of $\mathrm{CO}=3$

Bond order of $\mathrm{CO}_2=2$

Bond order $\propto \frac{1}{\text { bond length }}$

Therefore, the correct order of bond length is

$ \mathrm{CO}<\mathrm{CO}_2<\mathrm{CO}_3^{2-} . $

Molecule	Total number of electrons	Electronic configuration	Bond order $(\mathrm{BO})=\frac{\mathrm{a}-\mathrm{b}}{2}$$(\mathrm{BO})=\frac{\mathrm{a}-\mathrm{b}}{2}$(BO)=(a-b)/(2)
${\mathrm{CN}}^{+}$${\mathrm{CN}}^{+}$CN^(+)	12	$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^1=\pi 2p_y^1{\pi }^{\ast }2p_x^0={\pi }^{\ast }2p_y^0$$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^1=\pi 2p_y^1{\pi }^{\ast }2p_x^0={\pi }^{\ast }2p_y^0$sigma1s^(2)sigma^(**)1s^(2)sigma2s^(2)sigma^(**)2s^(2)sigma2p_(z)^(2)pi2p_(x)^(1)=pi2p_(y)^(1)pi^(**)2p_(x)^(0)=pi^(**)2p_(y)^(0)	$\frac{8-4}{2}=2$$\frac{8-4}{2}=2$(8-4)/(2)=2
${\mathrm{NO}}^{-}$${\mathrm{NO}}^{-}$NO^(-)	16	$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2=\pi 2p_y^2{\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^1$$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2=\pi 2p_y^2{\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^1$sigma1s^(2)sigma^(**)1s^(2)sigma2s^(2)sigma^(**)2s^(2)sigma2p_(z)^(2)pi2p_(x)^(2)=pi2p_(y)^(2)pi^(**)2p_(x)^(1)=pi^(**)2p_(y)^(1)	$\frac{10-6}{2}=2$$\frac{10-6}{2}=2$(10-6)/(2)=2
${\mathrm{N}}_2^{2+}$${\mathrm{N}}_2^{2+}$N_(2)^(2+)	12	$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^1=\pi 2p_y^1{\pi }^{\ast }2p_x^0={\pi }^{\ast }2p_y^0$$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^1=\pi 2p_y^1{\pi }^{\ast }2p_x^0={\pi }^{\ast }2p_y^0$sigma1s^(2)sigma^(**)1s^(2)sigma2s^(2)sigma^(**)2s^(2)sigma2p_(z)^(2)pi2p_(x)^(1)=pi2p_(y)^(1)pi^(**)2p_(x)^(0)=pi^(**)2p_(y)^(0)	$\frac{8-4}{2}=2$$\frac{8-4}{2}=2$(8-4)/(2)=2
${\mathrm{O}}_2$${\mathrm{O}}_2$O_(2)	16	$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2=\pi 2p_y^2{\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^1$$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2=\pi 2p_y^2{\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^1$sigma1s^(2)sigma^(**)1s^(2)sigma2s^(2)sigma^(**)2s^(2)sigma2p_(z)^(2)pi2p_(x)^(2)=pi2p_(y)^(2)pi^(**)2p_(x)^(1)=pi^(**)2p_(y)^(1)	$\frac{10-6}{2}=2$$\frac{10-6}{2}=2$(10-6)/(2)=2
${\mathrm{C}}_2$${\mathrm{C}}_2$C_(2)	12	$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^1=\pi 2p_y^1{\pi }^{\ast }2p_x^0={\pi }^{\ast }2p_y^0$$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^1=\pi 2p_y^1{\pi }^{\ast }2p_x^0={\pi }^{\ast }2p_y^0$sigma1s^(2)sigma^(**)1s^(2)sigma2s^(2)sigma^(**)2s^(2)sigma2p_(z)^(2)pi2p_(x)^(1)=pi2p_(y)^(1)pi^(**)2p_(x)^(0)=pi^(**)2p_(y)^(0)	$\frac{8-4}{2}=2$$\frac{8-4}{2}=2$(8-4)/(2)=2
${\mathrm{C}}_2^{+}$${\mathrm{C}}_2^{+}$C_(2)^(+)	11	$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^1=\pi 2p_y^0{\pi }^{\ast }2p_x^0={\pi }^{\ast }2p_y^0$$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^1=\pi 2p_y^0{\pi }^{\ast }2p_x^0={\pi }^{\ast }2p_y^0$sigma1s^(2)sigma^(**)1s^(2)sigma2s^(2)sigma^(**)2s^(2)sigma2p_(z)^(2)pi2p_(x)^(1)=pi2p_(y)^(0)pi^(**)2p_(x)^(0)=pi^(**)2p_(y)^(0)	$\frac{7-4}{2}=1.5$$\frac{7-4}{2}=1.5$(7-4)/(2)=1.5
${\mathrm{B}}_2^{2-}$${\mathrm{B}}_2^{2-}$B_(2)^(2-)	12	$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^1=\pi 2p_y^1{\pi }^{\ast }2p_x^0={\pi }^{\ast }2p_y^0$$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^1=\pi 2p_y^1{\pi }^{\ast }2p_x^0={\pi }^{\ast }2p_y^0$sigma1s^(2)sigma^(**)1s^(2)sigma2s^(2)sigma^(**)2s^(2)sigma2p_(z)^(2)pi2p_(x)^(1)=pi2p_(y)^(1)pi^(**)2p_(x)^(0)=pi^(**)2p_(y)^(0)	$\frac{8-4}{2}=2$$\frac{8-4}{2}=2$(8-4)/(2)=2

According to Fajan's rule, smaller is the size of cation and more positive charge it has, stronger is the polarising power of that cation. Hence, that ion has a larger tendency to distort the electron cloud of anion. Hence, it has more tendency to form covalent compound.

The metal ions of $\mathrm{FeF}_3, \mathrm{VF}_5, \mathrm{VF}_2$ and $\mathrm{TiF}_2$ all belong to fourth period of periodic table. The metal ions exist as $\mathrm{Fe}^{3+}, \mathrm{V}^{5+}, \mathrm{V}^{2+}$ and $\mathrm{Ti}^{+2}$ respectively. The anions are same in all the compounds. The $\mathrm{V}^{5+}$ has the smallest size and largest charge. So the molecule $\mathrm{VF}_5$ has largest covalent character.

$\begin{aligned} & \text { FC }(\text { oxygen } 1)=6-6-\frac{1}{2} \times 2=-1 \\ & \text { FC }(\text { oxygen } 2)=6-4-\frac{1}{2} \times 4=6-6=0 \\ & \text { FC }(\text { oxygen } 3)=6-6-\frac{1}{2} \times 2=-1 \end{aligned}$

Bond order $(\mathrm{BO})=\frac{1}{2}$ (number of electron in bonding MO's $-$ number of electron in anti-bonding MO's)

$\mathrm{C}_2^{2-}$ : Total number of electrons $=6+6+2=14$

Electronic configuration

$\begin{aligned} & \text { Electronic configuration } \\ & \sigma\left(1 s^2\right), \sigma\left(1 s^2\right), \sigma\left(2 s^2\right), \sigma^2\left(2 s^2\right), \pi 2 p_x^2=\pi 2 p_y^2, \sigma 2 p_z^2 \\ & \mathrm{BO}=\frac{1}{2} \times(10-4)=\frac{1}{2} \times 6=3 \end{aligned}$

$\mathbf{N}_{\mathbf{2}}$ : Total number of electrons $=7+7=14$

$\begin{aligned} & \sigma\left(1 s^2\right), \sigma^{\circ}\left(\left(s^2\right), \sigma\left(2 s^2\right), \sigma^*\left(2 s^2\right), \pi 2 p_x^2=\pi 2 p_y^2, \sigma 2 p_z^2\right. \\ & \mathrm{BO}=\frac{1}{2}(10-4)=\frac{6}{2}=3 \end{aligned}$

$\begin{aligned} & \begin{array}{l} \mathrm{O}_2^{2-}: \text { Total number of electron }=8+8+2=18 \\ \sigma\left(l s^2\right), \sigma^{\prime}\left(1 s^2\right), \sigma\left(2 s^2\right), \sigma^{\prime}\left(2 s^2\right), \sigma 2 p_z^2, \pi 2 p_x^2=\pi 2 p_y^2, \\ \pi^2 2 p_x^2=\pi^{\prime} 2 p_y^2 \end{array} \\ & \begin{aligned} \mathrm{BO} & =\frac{1}{2}(10-8) \\ & =\frac{1}{2} \times 2=1 \end{aligned} \end{aligned}$

Option (a) is incorrect.

$\begin{aligned} & \mathrm{O}_2^{+} \text {Total number of electrons }=8+8-1=15 \\ & \sigma\left(s^2\right), \sigma^*\left(1 s^2\right), \sigma\left(2 s^2\right), \sigma^*\left(2 s^2\right), \sigma\left(2 p_z^2\right), \pi\left(2 p_x^2\right) \\ & =\pi\left(2 p_y^2\right), \pi^{\cdot} 2 p_x^1 \\ \end{aligned}$

$\mathrm{BO}=\frac{1}{2}(10-5)=\frac{1}{2} \times 5=2.5$

$\mathrm{O}_2^{-}$ Total number of electrons $=8+8+1=17$

$\sigma\left(1 s^2\right), \sigma^*\left(1 s^2\right), \sigma\left(2 s^2\right), \sigma^2\left(2 s^2\right), \sigma\left(2 p_z^2\right), \pi\left(2 p_x^2\right)$

$\begin{aligned} & \sigma\left(1 s^2\right), \sigma\left(1 s^2\right), \sigma\left(2 s^2\right), \sigma\left(2 s^2\right), \sigma\left(2 p_z^2, \pi\left(22_x^2\right)\right. \\ & =\pi\left(2 p_y^2\right), \pi^*\left(2 p_x^2\right)=\pi^*\left(2 p_y^1\right) \\ & \mathrm{BO}=\frac{1}{2}(10-7)=\frac{1}{2} \times 3=1.5 \end{aligned}$

$\mathbf{N}_2^{+}$ Total number of electron $=7+7-1=13$

$\begin{aligned} & \mathbf{N}_2 \text { Total number } \\ & \sigma\left(1 s^2\right), \sigma\left(1 s^2\right), \sigma\left(2 s^2\right), \sigma\left(2 s^2\right), \pi\left(2 p_x^2\right)=\pi\left(2 p_y^2\right), \sigma\left(2 p_z^1\right) \\ & \mathrm{BO}=\frac{1}{2}(9-4)=\frac{1}{2} \times 5=2.5 \end{aligned}$

Option (b) is correct.

Similarly, bond order of $\mathrm{O}_2=\frac{1}{2}(10-6)=\frac{1}{2} \times 4=2$

Bond order of $\mathrm{Li}_2=\frac{1}{2}(4-2)=1$

Bond order of $\mathrm{H}_2^{+}=\frac{1}{2}(1-0)=0.5$

Bond order of $C_2=\frac{1}{2}(8-4)=\frac{1}{2} \times 4=2$

Rest other options are incorrect, as they don't have bond order in fraction.

Lewis dot structure of $\mathrm{SF}_6$ is Central atom, S has total 12 electrons. So, it does not obey octet rule.

Lewis dot structure of $\mathrm{PCl}_5$ is central atom, P has total 10 electrons. So, it also does not obey octet rule.

Lewis dot structure of $\mathrm{XeF}_2$ is central atom, Xe has total 10 electrons. So, it does not obey octet rule. Hence, the set of molecules that does not obey octet rule-is $\mathrm{SF}_6, \mathrm{PCl}_5$ and $\mathrm{XeF}_2$.

Formal charge (FC) can be calculated by using formula : $\mathrm{FC}=$ (number of electrons in neutral atom) $-\frac{1}{2}$ (number of electrons in covalent bonds) - (number of electrons in lone pairs)

Formal charge on 1st atom i.e., oxygen is

$\mathrm{FC}=6-\frac{1}{2} \times 4-4=6-2-4=0$

Formal charge on 2 nd atom i.e., nitrogen is

$\mathrm{FC}=5-\frac{1}{2} \times 8-0=5-4=+1$

Formal charge on 3rd atom i.e., oxygen is

$\mathrm{FC}=6-\frac{1}{2} \times 4-4=6-2-4=0$

Hence, formal charge of atoms 1, 2 and 3 in the ion $\mathrm{{[\mathop O\limits_1 = \mathop N\limits_2 = \mathop O\limits_3 ]^ + }}$ is 0, +1 and 0 respectively.

In graphite, each carbon atom combines with three other carbon atoms with three sigma bonds. So, hybridisation is $s p^2$. In diamond, each carbon atom combines with four other carbon atoms with four sigma bonds. So, hybridisation is $s p^3$. In fullerene, hybridisation of carbon atoms is $s p^2$. Hence, the hybridisation of carbon in graphite, diamond and fullerene, $\mathrm{C}_{60}$ are $s p^2, s p^3$ and $s p^2$ respectively.

According to Fajan's rule; covalent character in ionic compound depends on

(i) size and charge of cation

(ii) size and charge of anion

(iii) pseudo-inert gas configuration

(a) Size of iodide $(\mathrm{I}^{-})$ is more than $\mathrm{F}^{-}$, so $\mathrm{I}^{-}$ has more polarisibility. Thus, KI is more covalent than KF . So, option (a) is incorrect.

(b) Charge on Sn in $\mathrm{SnCl}_4$ is +4 and in $\mathrm{SnCl}_2$ is +2 $\mathrm{So}, \mathrm{Sn}$ has more polarising power in $\mathrm{SnCl}_4$. Thus, $\mathrm{SnCl}_4$ is more covalent than $\mathrm{SnCl}_2$. So, option (b) is incorrect.

(c) Size of $\mathrm{Li}^{+}$ is less than $\mathrm{K}^{+}$. So, it has more polarising power. Thus, LiF is more covalent than KF. Hence, option (c) is correct.

(d) Charge on Zn is +2 and on Na is +1 . So, Zn has more polarising power. Thus, $\mathrm{ZnCl}_2$ is more covalent than NaCl . Hence, option (d) is incorrect.

$\text { Bond length } \propto \frac{1}{\text { Bond order (BO) }}$

Electronic configuration of $\mathrm{C}_2$ i.e. $X_1$ is $\sigma 1 s^2 \sigma 1 s^{* 2} \sigma 2 s^2 \sigma 2 s^{* 2} \pi 2 p_x^2 \pi 2 p_y^2$

So, bond order

$=\frac{\mathrm{Number of bonding orbital electrons -Number of antibonding orbital electrons}}{2}$

$=\frac{8-4}{2}=\frac{4}{2}=2$

Electronic configuration of i.e. $\mathrm{N}_2$ i.e. $X_2$ is

$\sigma 1 s^2 \sigma l s^{* 2} \sigma 2 s^2 \sigma 2 s^{* 2} \sigma 2 p_z^2 \pi 2 p_x^2 \pi 2 p_y^2$

Bond order $=\frac{10-4}{2}=\frac{6}{2}=3$

Electronic configuration of $\mathrm{B}_2$ i.e. $X_3$ is

$\sigma l s^2 \sigma 1 s^{* 2} \sigma 2 s^2 \sigma 2 s^{* 2} \pi 2 p_x^1 2 \pi 2 p_y^1$

$\text { Bond order }=\frac{6-4}{2}=\frac{2}{2}=1$

So, order of bond order $(\mathrm{BO})$ is

$\mathrm{BO}_2>\mathrm{BO}_1>\mathrm{BO}_3$

Now, since bond order and bond lengths are inversely proportional. Hence, order of bond length is

$X_3>X_1>X_2$

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.

Dipole moment $=1.5 \times 10^{-29} \mathrm{C} . \mathrm{m}$ (given)

Bond length $=150 \mathrm{~pm}=150 \times 10^{-12} \mathrm{~m}$

Charge $=1.602 \times 10^{-19} \mathrm{C}$

$\begin{aligned} \text { Ionic character } & =\frac{\text { Dipole moment }}{\text { Bond length } \times \text { Charge }} \\ & =\frac{1.5 \times 10^{-29}}{150 \times 10^{-12} \times 1.602 \times 10^{-19}} \\ & =0.625 \end{aligned}$

Percentage ionic character

$\begin{aligned} & =0.625 \times 100 \\ & =62.5 \% \end{aligned}$

$\mathrm{SeF}_4$

$\mathrm{Se}=[\mathrm{Ar}] 3 d^{10} 4 s^2 4 p^4 $

Selenium (Se) undergoes $s p^3 d$ hybridisation and form five $s p^3 d$ hybridised orbitals.

Out of five $s p^3 d$ orbitals one is completely filled and four are half filled.

Therefore, geometry of H$_2$O and NH$_3$ are different.

Element with maximum bond energy is carbon because $\mathrm{C}-\mathrm{C}$ bond is strongest as carbon atoms are smaller in size.

Correct order of electronegativity of hybrid orbitals of carbon is

$sp > sp^2 > sp^3$

Electronegativity of carbon increases as the s-character of hybrid orbital increases.

Hybridisation	% of s
$sp$	50%
$sp^2$	33%
$sp^3$	25%

Higher percentage of s-character indicates that hybrid orbital is more closer to the nucleus and that’s why it has higher electronegativity.

Bond length is inversely proportional to the bond order. Higher bond order results in stronger bond, which are accompanied by stronger forces of attraction holding the atoms together.

Since, electronegativity difference is highest in case of N $-$ N bonds, therefore NH$_3$ has highest dipole moment,

As dipole moment is vector quantity. Due to electron density in opposite direction, net dipole moment cancel out each other.