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

Paramagnetic character is shown by those atoms or molecules which have unpaired electrons. In the given compounds $$CO$$  is not paramagnetic since, it does not have unpaired electrons. The configuration of $$CO$$  molecule is
$$CO\left( {14} \right) = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},$$      $$\mathop \sigma \limits^* 2{s^2},\sigma 2p_x^2,\pi 2p_y^2 \approx \pi 2p_z^2$$

$$N{H_4}Cl$$  contains both covalent and ionic bonds

∴ Total number of lone pair of electrons is \[9.\]

Paramagnetic species contains unpaired electrons in their molecular orbital electronic configuration.
Molecular orbital configuration of the given species is as
$$CO\left( {6 + 8 = 14} \right)$$
$$ = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},$$      $$\pi 2p_x^2 \approx \pi 2p_y^2,\sigma 2p_z^2$$
( All the electrons are paired so, it is diamagnetic ).
$$O_2^ - \left( {8 + 8 + 1 = 17} \right)$$
$$ = \sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},$$      $$\sigma 2p_z^2,\pi 2p_x^2 \approx \pi 2p_y^2,\mathop \pi \limits^* 2p_x^2 \approx \mathop \pi \limits^* 2p_y^1$$
( It contains one unpaired electron so, it is paramagnetic. )
$$C{N^ - }\left( {6 + 7 + 1 = 14} \right) = {\text{same}}\,{\text{as}}\,CO$$
$$N{O^ + }\left( {7 + 8 - 1 = 14} \right) = {\text{same}}\,{\text{as}}\,CO$$
Thus, among the given species only $$O_2^ - $$  is paramagnetic.

In
there are two types of bonds $$C = C$$  and $$C - H$$  bonds, which are not equal.

$$SO_4^{2 - }$$  and $$BF_4^ - $$  have $$s{p^3}$$  hybridisation and are tetrahedral in shape.

In $${C_2}{H_2},C$$   undergoes, $$sp$$  hybridisation.
Ground state :
Excited state :
In $$S{F_6},S$$  undergoes $$s{p^3}{d^2}$$  hybridisation.
Ground state :
Excited state :
In $$S{O_2},S$$   undergoes $$s{p^2}$$ hybridisation.
In $$I{F_7},I$$  undergoes $$s{p^3}{d^3}$$  hybridisation.
Ground state :
Excited state :

All the molecules have $$O$$ - atom with lone pairs, but in $${H_2}O$$  the $$H $$ - atom has no vacant orbital for $$\pi $$ - bonding. That's why it does not have any $$\pi $$ - bond.
In all other given molecules, the central atom because of the presence of vacant orbitals is capable to form $$\pi $$ - bonds.

$$C{F_4} - s{p^3},$$   tetrahedral but no unpaired electron.
$$S{F_4} - s{p^3}d,$$   trigonal bipyramidal with one unpaired electron.
$$Xe{F_4} - s{p^3}{d^2},$$    square planar, two lone pairs of electrons.

\[\begin{align} & \text{The order of bond angles} \\ & \begin{matrix} B{{F}_{3}} \\ {{120}^{\circ }} \\ \end{matrix}\,\,\begin{matrix} > \\ {} \\ \end{matrix}\begin{matrix} Si{{H}_{4}} \\ {{109}^{\circ }}28' \\ \end{matrix}\begin{matrix} > \\ {} \\ \end{matrix}\begin{matrix} N{{H}_{3}} \\ {{107}^{\circ }} \\ \end{matrix}\,\,\begin{matrix} > \\ {} \\ \end{matrix}\begin{matrix} {{H}_{2}}S \\ {{92.5}^{\circ }} \\ \end{matrix} \\ \end{align}\]