Chemical Bonding and Molecular Structure MCQ Questions & Answers in Inorganic Chemistry | Chemistry

Valence electrons in $$P$$  is 2; Valence electrons in $$Q$$  is 6.
$$\eqalign{ & P - 2{e^ - } \to {P^{2 + }} \cr & Q + 2{e^ - } \to {Q^{2 - }} \cr} $$
Since both show a valency of 2, the formula for the compound will be $$PQ.$$

The star marked carbon has a valency of 5 and hence this formula is not correct.

According to $$VSEPR$$   theory,
The order of repulsion of electrons is $$lp{\text{ - }}lp$$  repulsion > $$lp{\text{ - }}bp$$  repulsion > $$bp{\text{ - }}bp$$  repulsion
Higher the number of lone pair of electrons ( greater is repulsion ) lower is the bond angle and vice-versa. In $$NO_2^ - $$  one lone pair of electron is present while in $$NO_2^ + $$  no lone pair of electron is present. Hence, the correct order of bond angles is $$NO_2^ - < N{O_2} < NO_2^ + $$

$$B{e_2} = \sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}$$
$${B_2} = \,\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}{\sigma ^2}p_z^2$$
$${C_2} = \sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}{\sigma ^2}p_z^2\,\,\pi 2p_x^1\,\pi 2p_y^1$$
$${N_2} = \sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\,\pi 2p_x^2\,\pi 2p_y^2$$
Thus only $${C_2}$$ will be paramagnetic

$$Br{F_5}$$  and $$XeO{F_4}$$  have $$s{p^3}{d^2}$$  hybridisation hence their shape is square pyramidal.

Key Idea Molecules with zero bond order, do not exist.
According to molecular orbital theory,
$$\left( {\text{A}} \right)Be_2^ + \left( {4 + 4 - 1 = 7} \right)$$     $$ = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^1}$$
$$\eqalign{ & {\text{Bond}}\,\,{\text{order}}\left( {BO} \right) = \frac{{4 - 3}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.5 \cr} $$
$$\left( {\text{B}} \right)B{e_2}\left( {4 + 4 = 8} \right) = $$     $$\sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2}$$
$$BO = \frac{{4 - 4}}{2} = 0$$
$$\left( {\text{C}} \right){B_2}\left( {5 + 5 = 10} \right)$$     $$ = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},$$     $$\pi 2p_x^1 \approx \pi 2p_y^1$$
$$\eqalign{ & {\text{Bond}}\,\,{\text{order}}\left( {BO} \right) = \frac{{6 - 4}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 \cr} $$
$$\left( {\text{D}} \right)L{i_2}\left( {3 + 3 = 6} \right)$$     $$ = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2}$$
$${\text{BO}} = \frac{{4 - 2}}{2} = 1$$
Thus, $$B{e_2}$$  does not exist under normal conditions.

The proportion of covalent character in an ionic bond is decided by polarisability of the metal cation as well as the electrongativity of both elements involved in bonding. Polarisability is further decided by the density of positive charge on the metal cation. $$AlC{l_3}$$ is considered to show maximum covalent character among the given compounds. This is because $$A{l^{3 + }}$$ bears $$3$$ unit of positive charge and shows strong tendency to distort the electron cloud, thus the covalent character in $$Al{\text{ - }}Cl$$  bond dramatically increases.

Ionic bond or electrovalent between $$C{u^{2 + }}$$  and $$SO_4^{2 - }\,,$$
covalent and coordinate in $$SO_4^{2 - };$$ 
ion.

In $${N_2},$$  similar atoms are linked to each other and thus there is no polarity.

In $$NH_4^ + $$  bond angle is maximum ( nearer $${109^ \circ }$$ ) due to its tetrahedral geometry.