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

The boiling point of $$p$$ - nitrophenol is higher than that of $$o$$ - nitrophenol because $$p$$ - nitrophenol have intermolecular hydrogen bonding whereas $$o$$ - nitrophenol have intramolecular $$H$$ - bonding as given below

Due to electrostatic force of attraction, the ions do not move when electric current is applied in solid state. The ions become mobile when melted or dissolved in a polar solvent and conduct electricity.

The molecular orbital configuration of the molecules given is
Total no. of electrons in $$NO = 7\left( N \right) + 8\left( O \right) = 15$$
Hence $$E.C.$$  of $$NO$$
$$\eqalign{ & = KK\sigma {\left( {2s} \right)^2}{\sigma ^ * }{\left( {2s} \right)^2}\sigma {\left( {2{p_z}} \right)^2} \cr & \pi {\left( {2{p_x}} \right)^2} = \pi {\left( {2{p_y}} \right)^2}{\pi ^ * }{\left( {2{p_x}} \right)^1} \cr & = \pi * {\left( {2{p_y}} \right)^o} \cr} $$
Due to presence of one unpaired electron $$NO$$  is paramagnetic.
Except $$NO$$  all are diamagnetic due to absence of unpaired electrons.

$$Se{F_4}$$  has distorted tetrahedral geometry while $$C{H_4}$$  has tetrahedral geometry.

(A)
(B)
(C) Itisa pale blue gase. At $$ - {249.7^ \circ },$$   it forms violet black crystals.
(D) It is diamagnic in nature due to presence of paired electrons.

Fluorine is the most electronegative element and its configuration is $$1{s^2}2{s^2}2{p^5}.$$

$$N{i^{ + 2}} + 4C{l^ - } \to \mathop {{{\left[ {NiC{l_4}} \right]}^{2 - }}}\limits_{s{p^3}} $$
$${\left[ {NiC{l_4}} \right]^{2 - }} = 3{d^8}$$    configuration with nickel in $$+ 2$$  oxidation state, $$C{l^ - }$$  being weak field ligand does not compel for pairing of electrons.
So,
$${\left[ {NiC{l_4}} \right]^{2 - }}$$

Hence, complex has tetrahedral geometry

$$N{i^{ + 2}} + 4C{N^ - } \to {\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$
$${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }} = 3{d^8}$$    configuration with nickel in $$ + \,\,\,2$$  oxidation state, $$C{N^ - }$$  being strong field ligand compels for pairing of electrons.
So,
$${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{ - 2}}$$

Hence, complex has square planar geometry.

$$N{i^{ + 2}} + 6{H_2}O \to {\left[ {Ni{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }}$$
$$\left[ {Ni{{\left( {{H_2}O} \right)}_6}} \right] = 3{d^8}$$    configuration with nickel in $$ + 2$$  oxidation state. As with $$3{d^8}$$   configuration two $$d - $$  orbitals are not available for $${d^2}s{p^3}$$  hybridisation. So, hybridisation of $$Ni\left( {{\text{II}}} \right)$$  is $$s{p^3}{d^2}$$  and $$Ni\left( {{\text{II}}} \right)$$  with six co-ordination will have octahedral geometry.

Note : With water as ligand, $$Ni\left( {{\text{II}}} \right)$$  forms octahedral complexes.

The formula to find the hybridisation of central atom is
$$Z = \frac{1}{2}$$    [Number of valence electrons on central atom $$ + $$  No. of monovalent atom altached to it $$ + $$  negative charge if any $$ - $$  positive charge if any]
For $$N{O_3} - Z = \frac{1}{2}\left[ {5 + 0 + 1 - 0} \right] = 3$$

For $$NO_2^ + ,Z = \frac{1}{2}\left[ {5 + 0 + 0 - 1} \right] = 2$$

For $$NH_4^ + ,\,Z = \frac{1}{2}\left[ {5 + 4 + 0 - 1} \right] = 4$$

Bond order in $${N_2}$$  and $$O_2^{2 + }$$ is 3 ( calculate by energy level diagram )

Due to electrostatic force of attraction, the melting points and boiling points of ionic compounds are high.