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

Species with $$s{p^2}$$  hybridisation are planar triangular in shape. Among the given species $$NO_3^ - $$  is $$s{p^2}$$  hybridised with no lone pair of electrons on central atom, $$N$$ . Whereas, $${N_3},NO_2^ - $$  and $$C{O_2}$$  are $$sp$$  hybridised with a linear shape.

Bond order $$ = \frac{{{\text{No}}{\text{. of bonding electrons}} - {\text{No}}{\text{. of antibonding electrons}}}}{2}$$
Bond order in $$O_2^ + = \frac{{10 - 5}}{2} = 2.5$$
Bond order in $$O_2^ - = \frac{{10 - 7}}{2} = 1.5$$
Bond order in $$O_2^{2 - } = \frac{{10 - 8}}{2} = 1\,$$
Bond order in $$\,O_2^{2 + } = \frac{{10 - 4}}{2} = 3$$
$${\text{Since Bond order }} \propto \frac{1}{{{\text{Bond}}\,\,{\text{length}}}}$$
∴ Bond length is shortest in $$O_2^{2 + }.$$

Bonding molecular orbital has lower energy than the antibonding molecular orbital.

$$\eqalign{ & P - O\,\,{\text{bond order}} \cr & = \frac{{{\text{Total Number of bonds in all possible direction}}\,{\text{between two atoms }}}}{{{\text{Total number of resonating structures }}}} \cr & = \frac{{2 + 1 + 1 + 1}}{4} \cr & = \frac{5}{4} \cr & = 1.25 \cr & \therefore {\text{Bond}}\,{\text{order}} = 1.25 \cr & {\text{Resonating structures are}} \cr} $$

$$\eqalign{ & {\text{Total charge on}}\,\,PO_4^{3 - }\,\,{\text{ion}}\,\,{\text{is}}\,\, - 3 \cr & = \frac{{{\text{Total}}\,\,{\text{charge}}}}{{{\text{Total}}\,\,{\text{entity}}\,\,{\text{of}}\,\,O - {\text{atom}}}} \cr} $$
So, the average formal charge on each $${O - {\text{atom}}}$$   is
$$\eqalign{ & = - \frac{3}{4} \cr & = - 0.75 \cr} $$

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$${H_3}\mathop C\limits^4 - \mathop C\limits^3 \equiv \mathop C\limits^2 - \mathop C\limits^1 {H_3}$$     is linear because $${C_2}$$  and $${C_3}$$  are $$sp$$  hybridised carbon atom.

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TIPS/Formulae :
Molecule having $$s{p^3}$$  hybridisation and one lone pair of electron will have pyramidal structure.
$$\left( {\text{i}} \right)CO_3^{2 - }$$  and $$NO_3^ - $$  have tetrahedron structure.
$$\left( {{\text{ii}}} \right)$$ In $$PC{l_3},P$$   is $$s{p^3}$$  hybridised and has one lone pair of electrons, hence it is pyramidal in shape.

\[KCl\]  is an ionic compound while other \[(P{H_3},{O_2},{B_2}{H_6},{H_2}S{O_4})\]      are covalent compounds.