Chemical Bonding & Molecular Structure

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

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

$X_3>X_1>X_2$

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.

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.

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

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

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.