$$\left[ {Co{{\left( {N{H_3}} \right)}_5}C{O_3}} \right]Cl$$     Pentaamminecarbonatocobalt(III) chloride

$${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$
Let oxidation state of $$Ni$$  in $${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$   is $$x.$$
$$\eqalign{ & \therefore \,\,x - 4 = - 2 \cr & {\text{or}}\,\,\,\,x = 2 \cr} $$
Now, $$N{i^{2 + }} = \left[ {Ar} \right]3{d^8}4{s^0}$$

$$\because \,\,C{N^ - }$$   is a strong field ligand. Hence, all unpaired electrons are paired up.

∴ Hybridisation of $${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$$   is $$ds{p^2}$$

Hence $$C{r^{2 + }}$$  has magnetic moment of the order of $$5\,B.M.$$

$$\left[ {Co{{\left( {N{H_3}} \right)}_5}Br} \right]S{O_4} \rightleftharpoons $$       $${\left[ {Co{{\left( {N{H_3}} \right)}_5}Br} \right]^{2 + }} + SO_4^{2 - }$$
$$SO_4^{2 - } + BaC{l_2} \to \mathop {BaS{O_4}}\limits_{{\text{white ppt}}{\text{.}}} + 2C{l^ - }$$

Isomers
$$\mathop {{{\left[ {Ru{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]}^ + }}\limits_{{\text{cis}}\,\,{\text{and}}\,{\text{trans}}} ,\mathop {{{\left[ {Co{{\left( {N{H_3}} \right)}_5}Cl} \right]}^{2 + }}}\limits_{{\text{none}}} ,$$
$$\mathop {{{\left[ {Ir{{\left( {P{R_3}} \right)}_2}H\left( {CO} \right)} \right]}^{2 + }}}\limits_{{\text{cis}}\,\,{\text{and}}\,{\text{trans}}} ,$$  $$\mathop {{{\left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]}^ + }}\limits_{{\text{cis}}\,\,{\text{and}}\,{\text{trans and}}\,{\text{optical}}\,{\text{isomers}}} $$

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

In the presence of oxygen, $$Ag$$  metal forms a water soluble complex $$Na\left[ {Ag{{\left( {CN} \right)}_2}} \right]$$    with dilute solution of $$NaCN$$
$$4Ag + 8NaCN + 2{H_2}O + {O_2} \to \mathop {4Na\left[ {Ag{{\left( {CN} \right)}_2}} \right]}\limits_{\left( {{\text{Soluble}}} \right)} + 4NaOH$$

$$\left[ {Pt{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]B{r_2} \to $$     $${\left[ {Pt{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]^{2 + }} + 2B{r^ - }$$
$$B{r^ - } + AgN{O_3} \to \mathop {AgBr + NO_3^ - }\limits_{{\text{pale yellow}}} $$
$$\left[ {Pt{{\left( {N{H_3}} \right)}_4}B{r_2}} \right]C{l_2} \to $$     $${\left[ {Pt{{\left( {N{H_3}} \right)}_4}B{r_2}} \right]^{2 + }} + 2C{l^ - }$$
$$C{l^ - } + AgN{O_3} \to \mathop {AgCl}\limits_{{\text{white}}} + NO_3^ - $$

$${C_2}O_4^{2 - }$$  (oxalate) is didentate ligand and forms chelate complexes which are more stable than similar complexes containing unidentate ligands.

For a given metal ion, weak field ligands create a complex with smaller $$\vartriangle ,$$  which will absorbs light of longer $$\lambda $$  and thus lower frequency. Conservely, stronger field ligands create a larger $$\vartriangle ,$$  absorb light of shorter $$\lambda $$  and thus higher vi.e. higher energy.
$$\mathop {{\text{Red}}}\limits_{\lambda = 650nm} < \mathop {{\text{Yellow}}}\limits_{570\,nm} < \mathop {{\text{Green}}}\limits_{490\,nm} < \mathop {{\text{Blue}}}\limits_{450\,nm} $$
So order of ligand strength is
$${L_1} < {L_3} < {L_2} < {L_4}$$