$$\eqalign{ & {\left[ {CuC{l_4}} \right]^{2 - }}:{\text{Tetrahedra}} \cr & {\left[ {Fe{{\left( {N{H_3}} \right)}_6}} \right]^{2 + }}:{\text{Octahedral}} \cr} $$

$$\left[ {Cr{{\left( {{H_2}O} \right)}_6}} \right]C{l_3}$$
$$ \Rightarrow x + 6 \times 0 + \left( { - 1} \right) \times 3 = 0$$
$$ \Rightarrow x = + 3$$
$$\left[ {Cr{{\left( {{C_6}{H_6}} \right)}_2}} \right]$$
$$x + 2 \times 0 = 0;x = 0$$
$${K_2}\left[ {Cr{{\left( {CN} \right)}_2}{{\left( O \right)}_2}\left( {N{H_3}} \right)} \right]$$
$$2 \times 1 + x + 2 \times \left( { - 1} \right) + 2 \times \left( { - 2} \right) + \left( { - 2} \right) + 0 = 0$$
$$x = + 6$$

Key Concept When complex ion is an anion, the name of the metal ends with suffix -ate along with its oxidation number in the complex ion.
$${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }} = $$    Hexacyanoferrate (III) ion

$$\left[ {M{A_5}B} \right]$$   due to absence of symmetry of $$B$$  cannot exist in the form of $$cis - trans - {\text{isomer}}{\text{.}}$$

$$C{r^{3 + }}$$ has $${d^3}$$  configuration and forms an octahedral inner orbitals complex.
The set of degenerate orbitals are $$\left( {{d_{xy}},{d_{yz}}\,{\text{and}}\,{d_{xz}}} \right)$$   and $$\left( {{d_{{x^2} - {y^2}\,}}\,{\text{and}}\,{d_{{z^2}}}} \right).$$

The given compound may have linkage isomerism due to presence of $$N{O_2}$$  group which may be in the form $$ - N{O_2}$$  or $$ - ONO.$$  $$\left[ {Co{{\left( {N{H_3}} \right)}_4}{{\left( {N{O_2}} \right)}_2}} \right]Cl$$     $$\& \left[ {CO{{\left( {N{H_3}} \right)}_4}{{\left( {ONO} \right)}_2}} \right]Cl$$      It may have ionisation isomerism due to presence of two ionisable groups $$ - N{O_2}$$  $$\& - Cl.$$
$$\left[ {Co{{\left( {N{H_3}} \right)}_4}Cl\left( {N{O_2}} \right)} \right]N{O_2}$$      $$\& \left[ {Co\left( {N{H_3}} \right){{\left( {N{O_2}} \right)}_2}} \right]Cl$$      It may have geometrical isomerism in the form of cis trans form as follows :

Dien (Diethylenetriamine) has the following structure.
$${H_2}\ddot N - C{H_2} - C{H_2} - \ddot NH - $$      $$C{H_2} - C{H_2} - \ddot N{H_2}$$

$$N{a_2}\left[ {CdC{l_4}} \right]$$   does not contain any unpaired electron hence colourless

For the reaction of the type $$M + 4L \rightleftharpoons M{L_4},$$    larger the stability constant, the higher the proportion of $$M{L_4}$$  that exists in solution.