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

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$$F$$  is more electronegative than $$N,$$  therefore direction of bond is from $$N$$  to $$F$$  whereas $$N$$  is more electronegative than $$H,$$  the direction of the bond is from $$H$$  to $$N.$$  Thus whereas resultant moment of $$N - H$$  bonds adds up to the bond moment of lone pair, that of $$3N - F$$  bonds partly cancel the resultant moment of lone pair. Hence, the net dipole moment of $$N{F_3}$$  is less than that of $$N{H_3}.$$

(A) It has ionic and non-polar
(B) $$H - C = N - $$   it has ionic and polar covalent bonds.
(C)
It has polar and non polar both type of covalent bonds.
(D)
It has non polar covalent bonds only.

The molecular orbital configuration of the given molecules is
$${H_2} = \sigma {\left( {1s} \right)^2}$$    ( no electron anti-bonding )
$$L{i_2} = \sigma {\left( {1s} \right)^2}{\sigma ^ * }{\left( {1s} \right)^2}\sigma {\left( {2s} \right)^2}$$
( two anti-bonding electrons )
$$\eqalign{ & {B_2} = \sigma {\left( {1s} \right)^2}{\sigma ^ * }{\left( {1s} \right)^2}\sigma {\left( {2s} \right)^2}{\sigma ^ * }{\left( {2s} \right)^2} \cr & \pi {\left( {2{p_x}} \right)^1} = \pi {\left( {2{p_y}} \right)^1} \cr} $$
( 4 anti-bonding electrons )
Though the bond order of all the species are same $$(B.O = 1)$$   but stability is different. This is due to difference in the presence of no. of anti-bonding electron.
Higher the no. of anti-bonding electron lower is the stability hence the correct order is $${H_2} > L{i_2} > {B_2}$$

Smaller the size and higher the charge more will be polarising power of cation. Since the order of the size of cation is $${K^ + }\, > C{a^{ + + \,}} > M{g^{ + + }} > B{e^{ + + }}.$$       So the correct order of polarising powder is $${K^ + } < C{a^{2 + }} < M{g^{2 + }} < B{e^{2 + }}$$

Dipole moment = ( Distance between opposite charges ) × ( charge, $$q$$  )
$$\mu = q \times d$$
So, greater the distance between the opposite charges higher the dipole. Due to the resonance the greater charge separation occurs between charges due to linearity.

According to Fajan's rule, higher charge on the ions, more covalent is the compound.

TIPS/Formulae:
$$\eqalign{ & H = \frac{1}{2}\left[ {V + M - C + A} \right] \cr & {\text{Hybridisation of }}N{\text{ in }}N{H_3} \cr & = \frac{1}{2}\left[ {5 + 3 - 0 + 0} \right] = 4\,\,\therefore s{p^3} \cr & {\text{Hybridisation of }}Pt{\text{ in }}{\left[ {PtC{l_4}} \right]^{2 - }} \cr & = \frac{1}{2}\left[ {2 + 4 - 0 + 2} \right] = 4\,\,\therefore ds{p^2} \cr & {\text{Hybridisation of }}P\,{\text{in}}\,\,PC{l_5} \cr & = \frac{1}{2}\left[ {5 + 5 - 0 + 0} \right] = 5\,\,\therefore s{p^3}d \cr & {\text{Hybridisation of }}B{\text{ in }}BC{l_3} \cr & = \frac{1}{2}\left[ {3 + 3 - 0 + 0 = 3} \right]\,\,\therefore s{p^2} \cr} $$

It has $$5\pi $$  bonds and $$19\sigma $$  bonds.