Identify transition metal complexes which are not octahedral in shape.

$\begin{array}{l}\left[\mathrm{Co}\right({\mathrm{NH}}_3\left)6\right]{}^{3+}\\ \left[\mathrm{Ni}\right(\mathrm{CO}\left)4\right]\\ \left[\mathrm{CoCl}\right({\mathrm{NH}}_3\left)5\right]{}^{2+}\\ \left[\mathrm{CoCl}{}_2\right({\mathrm{NH}}_3\left)4\right]+\\ \left[\mathrm{PtCl}{}_4\right]{}^{2-}\end{array}$

$\begin{array}{l}{Choose\; the\; correct\; answer\; from\; the\; options\; given\; below:}^{}\end{array}$

$\begin{array}{l}{<br>}^{}\end{array}$

The correct answer is (B) and (E) only.

To determine which transition metal complexes are not octahedral in shape, let us examine the coordination number (the number of donor atoms directly bonded to the central metal atom or ion) for each of the given complexes:

1. ${\left[\mathrm{Co}\right({\mathrm{NH}}_3\left)6_{}\right]}^{3+}$ (Complex A):
Here, the cobalt ion is bonded to six monodentate ammine (${\mathrm{NH}}_3$) ligands. The coordination number is 6, which corresponds to an octahedral geometry.

2. $\left[\mathrm{Ni}\right(\mathrm{CO}\left)4_{}\right]$ (Complex B):
In this complex, the nickel atom is bonded to four carbonyl ($\mathrm{CO}$) ligands. The coordination number is 4. Nickel is in the zero oxidation state ($d^{10}$ configuration), and with the strong field carbonyl ligands, it undergoes ${\mathrm{sp}}^3$ hybridization. Therefore, the shape is tetrahedral (not octahedral).

3. ${\left[\mathrm{CoCl}\right({\mathrm{NH}}_3\left)5_{}\right]}^{2+}$ (Complex C):
Here, the cobalt ion is bonded to one chloride ligand and five ammine ligands, giving a total of six ligands. The coordination number is 6, which results in an octahedral geometry.

4. ${\left[{\mathrm{CoCl}}_2\right({\mathrm{NH}}_3\left)4_{}\right]}^{+}$ (Complex D):
Here, the cobalt ion is bonded to two chloride ligands and four ammine ligands. The coordination number is 6, which also corresponds to an octahedral geometry.

5. ${\left[{\mathrm{PtCl}}_4\right]}^{2-}$ (Complex E):
In this complex, the platinum ion (${\mathrm{Pt}}^{2+}$, a $d^8$ transition metal ion of the 5d series) is coordinated to four chloride ligands. Due to the high crystal field splitting energy of 5d elements, it forms a low-spin complex with ${\mathrm{dsp}}^2$ hybridization. The shape is square planar (not octahedral).

Thus, the complexes which are not octahedral in shape are (B) and (E) only.