Coordination number of nickel in $${\left[ {Ni{{\left( {{C_2}{O_4}} \right)}_3}} \right]^{4 - }}$$   is 6 because $${C_2}O_4^{2 - }$$  is a bidentate ligand.

Triethoxyaluminium has no $$Al – C$$  linkage

Oxalate is a bidentate ligand hence forms a chelate.

$$\left[ {Co{{\left( {N{H_3}} \right)}_5}Br} \right]S{O_4}$$     gives white precipitate of $$BaS{O_4}$$  with $$BaC{l_2}\left( {aq} \right)$$   whereas $$\left[ {Co{{\left( {N{H_3}} \right)}_5}S{O_4}} \right]Br$$     gives yellow precipitate $$(AgBr)$$   with $$AgN{O_3}\left( {aq} \right).$$

$$SnC{l_4}$$  is obtained by passing chlorine over tin. So $$Sn$$  (IV) chloride is made by dissolving tin solution in concentrated solution statement is incorrect and answer is (B).

$$Co{\left( {N{H_3}} \right)_4}B{r_2}Cl\,{\text{will show both geometrical and ionization isomerism}}{\text{. }}$$.
$$\left[ {Co{{\left( {N{H_3}} \right)}_4}B{r_2}} \right]Cl\,\,{\text{and}}\,\,\left[ {Co{{\left( {N{H_3}} \right)}_4}B{r_2}Cl} \right]Br\,\,{\text{are}}\,{\text{ionization isomers and geometrical isomers are }}$$

The $$CFSE$$   of the ligands is in the order : $${H_2}O < N{H_3} < C{N^ - }$$
Hence, excitation energies are in the order : $${\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }} < {\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$$       $$ < {\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }}$$
From the relation $$E = \frac{{hc}}{\lambda } \Rightarrow E \propto \frac{1}{\lambda }$$
The order of absorption of wavelength of light in the visible region is : $${\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }} > {\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}$$       $$ > {\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }}$$

The formula of dichlorobis (urea) copper (II) is $$\left[ {CuC{l_2}{{\left\{ {O = C{{\left( {N{H_2}} \right)}_2}} \right\}}_2}} \right].$$

In $$\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]S{O_4}$$    primary valency is 2 and secondary valency is 4.

Mole ratio of $$Na, Cu$$   and
$$F = \frac{{1.08}}{{0.539}}:\frac{{0.539}}{{0.539}}:\frac{{2.16}}{{0.539}} = 2:1:4$$
Empirical formula $$ = N{a_2}Cu{F_4}:{\text{Van't}}\,{\text{Hoff}}$$
factor $$\left( i \right) = 3$$   (given)
Hence, formula of the compound :
$$N{a_2}\left[ {Cu{F_4}} \right] \to 2N{a^ + } + {\left[ {Cu{F_4}} \right]^{2 - }}$$