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

$$M.O.{\text{ configuration of }}\,O_2^ + :$$
$$\left( {\sigma 1{s^2}} \right)\left( {{\sigma ^ * }1{s^2}} \right)\left( {\sigma 2{s^2}} \right)\left( {{\sigma ^ * }2{s^2}} \right)$$     $$\left( {\sigma 2p_z^2} \right)\left( {\pi 2p_x^2 = \pi 2p_y^2} \right)\left( {{\pi ^ * }2p_x^1} \right)$$
$$B.O. = \frac{{10 - 5}}{2} = 2.5$$
$$B.O.\,\,{\text{of}}\,\,{O_2} = 2\left[ {\mathop {O_2^ + }\limits_{2.5} > \mathop {{O_2}}\limits_2 > \mathop {O_2^ - }\limits_{1.5} > \mathop {O_2^{2 - }}\limits_1 } \right]$$

$$OS{F_2}:\frac{N}{2} = \frac{{6 + 2}}{2} = 4$$     It has $$1$$  lone pair.

The shapes of $$S{O_3},Br{F_3}$$   and $$SiO_3^{2 - }$$  are triangular planar respectively.

The structure of $${H_2}O$$  is angular or $$V - {\text{shaped}}$$   and has $$s{p^3} - $$  hybridisation and $${104.5^ \circ }$$  bond angle. Thus, its dipole moment is positive or more than zero.

But in $$Be{F_2},$$  structure is linear due to $$sp$$  hybridisation $$\left( {\mu = 0} \right)$$

Thus, due to $$\mu > 0,{H_2}O$$   is dipolar and due to $$\mu = 0.$$
$$Be{F_2}$$  is non-polar.

In $$N_2^ - $$  total electrons = 14 + 1 = 15
Electronic configuration of $$N_2^ - $$  is
$$\sigma 1{s^2},\mathop \sigma \limits^* 1{s^2},\sigma 2{s^2},\mathop \sigma \limits^* 2{s^2},\sigma 2p_x^2,$$      $$\pi 2p_y^2 \approx \pi 2p_z^2,\mathop \pi \limits^* 2p_y^1$$
Due to presence of one unpaired electron, it shows paramagnetic character.

No explanation is given for this question. Let's discuss the answer together.

Water is a polar solvent and have dielectric constant 80. As $$NaCl$$  is a polar compound and like dissolves like so, forces of attraction between $$N{a^ + }$$  and $$C{l^ - }$$  $$ion$$  will reduced to $$\frac{1}{{80}}$$  in water.

In $$s{p^3}$$  hybridisation bond angle is $${109^ \circ }$$ $${28^,}.$$ 
In $$s{p^2}$$  hybridisation bond angle is $${120^ \circ }.$$
In $$sp$$  hybridisation bond angle is $${180^ \circ }.$$

In octahedral structure $$M{X_6},$$  the six hybrid orbitals $$\left( {s{p^3}{d^2}} \right)$$  are directed towards the cornes of a regular octahedral with an angle of $${90^ \circ }.$$  According to following structure of $$M{X_6},$$  the number of $$X - M - X$$    bonds at $${180^ \circ }$$  must be three.

$${H_2}O$$  have bent structure in which the two $$O - H$$  bonds are oriented at an angle of $${104.5^ \circ },$$  so water have a net dipole moment whereas $$Be{F_2}$$  have linear geometry, so the dipole moment of one bond is cancelled by another bond, so it have zero dipole moment.

Ionic compounds are more soluble in water or in aqueous medium.
According to Fajans’ rule,
Size of the cation increases the ionic character also increases.
$${\text{Ionic}}\,{\text{character}} \propto {\text{size}}\,{\text{of}}\,{\text{cation}}\left( {{\text{if}}\,{\text{anion}}\,{\text{is}}\,{\text{same}}} \right)$$
The order of size of cation is
$$N{a^ + } > Z{n^{2 + }} > C{u^{2 + }}$$
∴ The order of ionic character and hence, of solubility in water is as
$$N{a_2}S > ZnS > CuS$$