$$C{N^ - }$$  is a strong field ligand because it is an example of pseudohalide. Pseudohalide ions are stronger coordinating ligands and they have the ability to form $$\sigma $$  and $$\pi $$ - binds.

$$F{e^{3 + }}:3{d^5},$$   since $$C{N^ - }$$  is a strong field ligand, pairing will take place.
  $$\left[ {\because n = 1} \right]$$
$$\eqalign{ & \mu = \sqrt {n\left( {n + 2} \right)} \cr & \,\,\,\, = \sqrt {1\left( {1 + 2} \right)} \cr & \,\,\,\, = \sqrt 3 \cr & \,\,\,\, = 1.732\,\,B.M. \cr} $$
In $${\left[ {Co{F_6}} \right]^{3 - }},$$   oxidation state of $$Co = + 3$$
$$C{o^{3 + }}:3{d^6},$$   since $${F^ - }$$ is a weak field ligand, pairing will not occur.
  $$\left[ {\because n = 4} \right]$$
$$\eqalign{ & \mu = \sqrt {4\left( {4 + 2} \right)} \cr & \,\,\,\,\, = \sqrt {24} \cr & \,\,\,\,\, = 4.89\,B.M. \cr} $$
In $${\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }},$$    oxidation state of $$Fe = + 3$$
$$F{e^{3 + }}:3{d^5},$$   since $${H_2}O$$  is a weak field ligand, pairing will not take place.
  $$\left[ {\because n = 5} \right]$$
$$\eqalign{ & \mu = \sqrt {5\left( {5 + 2} \right)} \cr & \,\,\,\,\, = \sqrt {35} \cr & \,\,\,\,\, = 5.92\,B.M. \cr} $$
In $${\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }},$$    oxidation state of $$Co=+3.$$
$$C{o^{3 + }}:3{d^6},$$   since $$N{H_3}$$  is a strong field ligand, pairing will take place.
  $$\left[ {\because n = 0} \right]$$
$$\mu = \sqrt {0\left( {0 + 2} \right)} = 0$$

$$\left[ {Co{{\left( {N{H_3}} \right)}_3}C{l_3}} \right]$$    will not exhibit optical isomerism due to presence of plane of symmetry.

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

$$P:C{r^{3 + }}$$  has $$3{d^3}$$  configuration, with 3 unpaired electrons. Hence, it shows paramagnetic behaviour. Complex of the type $$M{a_4}{b_2}$$  shows $$cis-trans$$   isomerism.
$$Q:T{i^{3 + }}$$  has $$3{d^1}$$  configuration, hence shows paramagnetic behaviour. Complex gives $$C{l^ - }$$ and $$NO_3^ - $$  ions in solution hence, shows ionisation isomerism.
$$R:P{t^{2 + }}$$  has $$3{d^8}$$  configuration but ligands are strong field ligands hence, it forms square planar complex. Thus, all electrons are paired and it also exhibits ionisation isomerism.
$$S:C{o^{3 + }}$$  has $$3{d^6}$$  configuration. But, ligands present are strong enough to cause electron pairing, hence, it shows diamagnetic behaviour and exhibits $$cis-trans$$   isomerism as it is $$M{a_4}{b_2}$$  type complex.

$${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$$   obeys the $$EAN$$  rule.

Key Idea For the absorption of visible light, presence of unpaired $$d$$ - electrons is the necessity.
$$\left( {\text{A}} \right){\text{In}}{\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }},Ni$$     is present as $$N{i^{2 + }}.$$
$$N{i^{2 + }} = \left[ {Ar} \right]3{d^8}\,4{s^0}$$
$$\therefore {\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }} = $$

( Pairing occurs because $$C{N^ - }$$  is a strong field ligand ).
Since, in $${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }},$$    no unpaired electron is present in $$d$$ - orbitals so it does not absorb visible light.
$$\left( {\text{B}} \right){\text{In}}{\left[ {Cr{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }},Cr$$      is present as $$C{r^{3 + }}.$$
$$C{r^{3 + }} = \left[ {Ar} \right]3{d^3}4{s^0}$$    ( Three unpaired electrons )
$$\left( {\text{C}} \right){\text{In}}{\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }},Fe$$     is present as $$F{e^{2 + }}.$$
$$\left[ {Ar} \right]3{d^6}\,4{s^0}$$    ( Four unpaired electrons )
$$\left( {\text{D}} \right){\text{I}}n{\left[ {Ni{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }},Ni$$      is present as $$N{i^{2 + }}.$$
$$N{i^{2 + }} = \left[ {Ar} \right]3{d^8}4{s^0}$$    ( Two paired electrons )
The complex given in option (B), (C), (D) have unpaired electrons, thus absorb visible light.
NOTE
In presence of $$N{H_3}$$  and $${H_2}O$$  pairing does not occur as they are strong field ligand.

Thus $$L, M, O$$   and $$P$$ are diamagnetic.

Higher the number of ions in the solution, higher is the conductivity.
No. of ions : $$\left[ {Co{{\left( {N{H_3}} \right)}_3}C{l_3}} \right] = 0;$$     $$\left[ {Co{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]Cl = 2$$
$$\left[ {Co{{\left( {N{H_3}} \right)}_5}Cl} \right]C{l_2} = 3;$$     $$\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]C{l_3} = 4$$

The IUPAC name of $$\left[ {Pt{{\left( {N{H_3}} \right)}_3}\left( {Br} \right)\left( {N{O_2}} \right)Cl} \right]Cl$$       is triamminebromochloronitroplatinum (IV) chloride.
( Oxidation number of $$Pt = +4,$$   and ligands are arranged in alphabetical order )