Answer: Option A. ->
:
B
Tertiary carbocation is most stable among primary, secondary and tertiary due to more hyper conjugation.

Answer: Option A. ->
:
D
In p-nitrophenol, $N{O}_2$group is an electron withdrawing group but in structure (d) it is shown as electron releasing group.

Answer: Option A. ->
:
B
All free radicals have planar geometry but bicyclo compounds free radical has a pyramidal shape.
Because of steric strain, the carbon atom carrying the unpaired electron cannot assume a planar geometry.

Answer: Option C. -> Acetophenone
:
C

The dipole moment is the highest for Acetophenone because of -M effect.

Answer: Option D. -> None of these
:
D
$\underset{Cl}{\underset{|}{\stackrel{C{H}_2}{\stackrel{||}{CH}}}}+HCl$ $\to$ $\stackrel{C{H}_3}{{\stackrel{|}{CH}}_{\underset{Cl}{╲}}^{\stackrel{Cl}{╱}}}$
(1,1-dichloroethane)
It is an addition reaction which is according to Markownikoff rule.

Answer: Option C. -> Resonance
:
C

All the bonds (C-C) are equal in benzene. The
C-C bond length is 1.39$\stackrel{\circ }{A}$ which is in between
C-C bond (1.54$\stackrel{\circ }{A}$ ) abd C=C (1.34$\stackrel{\circ }{A}$).

Answer: Option C. -> Sp2 in C-3 and C-5 and SP in C-4
:
C
Carbon which has $3\sigma$ bonds and $1\pi$ bond in C-3 and C-5 $s{p}^2$ hybridised.
Carbon which has $3\sigma$ bonds and $1\pi$ bond as in C-4 is sp hybridised.

Answer: Option B. -> sp2
:
B
Look at the structure of benzene.

Every carbon atom in the benzene ring has a double bond with a neighbouring carbon atom and a single bond with another carbon atom. It is also bonded to one hydrogen atom with a single bond.
So, each carbon atom has 3 sigma bonds and 1 pi bond. This is possible only if the carbon atom is$s{p}^2$ hybridized where we get 3 $s{p}^2$ hybrid orbitals to form 3 sigma bonds and 1 unhybridised p orbital to form a pi bond.
Another feature of $s{p}^2$ hybridisation is that it leads to a trigonal planar geometry which is also true in this case.

Answer: Option C. -> Electrophilic substitution reaction
:
C
It so happens that the most common type of reaction in aromatic compounds is an electrophilic substitution reaction.
This is because the benzene molecule is electron-rich.

Answer: Option C. -> (CH3)3COH
:
C