Chemical Bonding & Molecular Structure

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.

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

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.