Chemical Bonding & Molecular Structure

On finding number of bond pairs in each

$ \begin{aligned} (A)\,& \mathrm{H}_2 \mathrm{CO}_3 \frac{2 \times 1+4+3 \times 6}{8}=\frac{24}{8}=3 \\ & \Rightarrow s p^2 \text { hybridised } \end{aligned} $

(B) $\mathrm{SiF}_4 \frac{4+4 \times 7}{8}=\frac{32}{8}=4 \Rightarrow s p^3$

hybridised

(C) $\mathrm{BF}_3 \frac{3+3 \times 7}{8}=\frac{24}{8}=3 \Rightarrow s p^2$

hybridised

(D) $\mathrm{HClO}_2 \frac{1+7+2 \times 6}{8}=\frac{20}{8}$ ( 2 lone

pair +4 bond pair $s p^3$ hybridised

A and C only are having $s p^2$ hybridised central atom.

Hybridisation $=\frac{1}{2}$ (Number of valence electron + number of monovalent atoms other than central atom + negative charge - positive charge)

$ =\frac{1}{2}(7+2+1) $

$ \begin{aligned} &=\frac{10}{2}\\ &=5 \cong s p^3 d . \end{aligned} $

$ =5 \cong s p^3 d . $

Also it can be determined by sum of number of lone pairs and number of $\sigma$-bonds central metal atom will form. As I is further linked to two iodine atoms and number of lone pair on each will be 3 .

$ \therefore 3+2=5 \cong s p^3 d \text { hybridised. } $

So, shape is linear and hybridisation is $s p^3 d$.

Dipole moment of the given compounds along with their structures are given below.

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

(a) $\mathrm{PF}_6^{-}\left(6\right.$ bond pairs) and $\mathrm{SF}_6$ ( 6 bond pairs) are isostructural.

(b) $\mathrm{IO}_3^{-}(4$ pairs, i.e. 3 bond pair +1 lone pair) and $\mathrm{XeO}_3$ (4 pairs, i.e. 3 bond pairs +1 lone pair) are isostructural.

(c) $\mathrm{BH}_4^{-}$(4 bond pairs) and $\mathrm{NH}_4^{+}$( 4 bond pairs) are isostructural.

(d) $\mathrm{BrF}_5$ ( 5 bond pairs) and $\mathrm{XeF}_4$ ( 4 bond pairs and 1 lone pair) are not isostructural.

Hydrogen fluoride (HF) has higher boiling point as it possess strongest hydrogen bonding due to high electronegativity of fluorine. Hence, both $(A)$ and $(R)$ are true and $(\mathrm{R})$ is the correct explanation of (A).

Intramolecular hydrogen bonding is present in 2-hydroxybenzoic acid.

Intramolecular hydrogen bonding is not present.

The correct order of bond angle is

$ \mathrm{BeCl}_2>\mathrm{BF}_3>\mathrm{SiCl}_4>\mathrm{SF}_6 $

In $\mathrm{SF}_6$, the molecule geometry is octahedral where bond angle is $90^{\circ}$.

In $\mathrm{SiCl}_4$, the geometry is tetrahedral and the bond angle is $109.5^{\circ}$.

In $\mathrm{BF}_3$, the geometry is trigonal planar where bond angle is $120^{\circ}$.

In $\mathrm{BeCl}_2$, the geometry is linear and bond angle is $180^{\circ}$.

If the bond order of species is zero, then species will not exist.

$ =\frac{\text { Bond order bonding molecular orbital electrons - antibonding molecular orbital electrons }}{2} $

$ \begin{aligned} & \mathrm{H}_2^{+}-\sigma \mathrm{l} s^1 \quad \therefore \text { Bond order }=\frac{1-0}{2}=0.5 \\ & \mathrm{He}_2^{2+}-\sigma l s^2 \quad \therefore \text { Bond order }=\frac{2-0}{2}=1 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \mathrm{Li}_2^{2-}-\sigma l s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2 \\ & \begin{aligned} \therefore & \text { Bond order }=\frac{4-4}{2} \\ & =0(\text { do not exist }) \end{aligned} \\ & \begin{aligned} \mathrm{Ne}_2: \sigma l s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2 \\ \begin{aligned} & 2 p_y^2, \pi^* 2 p_x^2 \approx \pi^* 2 p_y^2, \sigma^* 2 p_z^2 \\ & \therefore \text { Bond order }=\frac{10-10}{2} \\ &=0 \text { (do not exist) } \\ & \mathrm{Be}_2^{-}-\sigma l s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^1 \\ & \approx \pi 2 p_y^0 \end{aligned} \\ \therefore \mathrm{Bond} \mathrm{order}^2=\frac{5-4}{2}=0.5 \\ \mathrm{He}_2-\sigma \mathrm{l} s^2, \sigma^* 1 s^2 \\ \therefore \text { Bond order }=\frac{2-2}{2}=0 \text { (do not exist) } \end{aligned} \end{aligned}\\ &\text { Hence, } \mathrm{Li}_2^{2-}, \mathrm{Ne}_2 \text { and } \mathrm{He}_2 \text { will not exist. } \end{aligned} $

Molecular orbital electronic configurations are

$ \begin{aligned} & \mathrm{O}_2 \rightarrow \sigma l s^2 \sigma^* 1 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 \pi^* 2 p_x^1 \pi^* 2 p_y^1 \end{aligned} $

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

paramagnetic

$ \begin{aligned} & \mathrm{O}_2^{+} \rightarrow \sigma 1 s^2 \sigma^* 1 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 \pi^* 2 p_x^1 \end{aligned} $

Bond order $\rightarrow \frac{10-5}{2}=\frac{5}{2}=2.5$,

paramagnetic

$ \begin{array}{r} \mathrm{O}_2^{2-} \rightarrow \sigma l s^2 \sigma^* 1 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 \quad \pi^* 2 p_x^2 \pi^* 2 p_y^2 \end{array} $

Bond order $=\frac{10-8}{2}=\frac{2}{2}=1$,

diamagnetic

$ \begin{aligned} \mathrm{O}_2^{-} \rightarrow \sigma l s^2 \sigma^* 1 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 \pi^* 2 p_x^2 \pi^* 2 p_y^1 \end{aligned} $

$ \text { Bond order }=\frac{10-7}{2}=1.5=\frac{3}{2}=1.5, $

paramagnetic

The species with $(A)$, the highest bond order is $\mathrm{O}_2^{+}$and ( $B$ ), diamagnetic character is $\mathrm{O}_2^{2-}$. Hence, option (b) is correct.

$ \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.

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-} . $

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

$X_3>X_1>X_2$

This is one unpaired electron in (O$-$F), hence, paramagnetic, thus (a) is correct.

As bond order of O$-$F > F$-$F thus, bond energy of O$-$F > F$-$F.

Hence, (a) and (b) both are correct.

Hydrogen peroxide (H₂O₂) is a non-planar molecule with twisted symmetrical structure. It has non-linear structure with an open book structure.

Larger the bond order, stronger the bond formed.

CH₄ ; bp = 4; sp³-hybridisation

NF₃ ; bp = 3; $l$p = 1; sp³-hybridisation

NH$_4^ + $ ; bp = 4; $lp$ = 0; sp³-hybridisation

H₂O; bp = 2; lp = 2; sp³-hybridisation

Hence, all have the same slope.

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.

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 )