Coordination Compounds

Key idea Complexes with square planar geometry are optically inactive.

Among the given options, only option (a) has square planar geometry with $d s p^2$-hybridisation. Rest all other option show non-superimposable mirror images, thus their structure are optically active. Hence, [ $\mathrm{RhCl}(\mathrm{CO})(\mathrm{P}. \mathrm{Ph}_3)(\mathrm{NH}_3)]$ is optically inactive and option (a) is the correct.

$\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$ contains $\mathrm{Co}^{3+}$ as central metal ion.

The electronic configuration of $\mathrm{Co}^{3+}$ ion $=[\mathrm{Ar}]_{18} 3 d^6$

The splitting in octahedral field is shown below

$\mathrm{Co}^{3+}=3 d^6\left(\text { or } t_{2 g}^6 e_g\right)$

$\because$ There is no unpaired electron, so it is diamagnetic in nature.

Since, the coordination number of $\mathrm{Mn}^{2+}$ ion in the complex ion is 4 , it will have either tetrahedral ($s p^3$-hybridisation) or square planar ($d s p^2$-hybridisation) geometry. But the fact that the magnetic moment of the complex ion is 5.9 BM suggests that it should have tetrahedral geometry rather than square planar because of the presence of five unpaired electrons in the $d$-orbitals.

The number of unpaired electrons can be determined as follows:

As we know that, magnetic moment $=\sqrt{n(n+2)}$

$5.9=\sqrt{n(n+2)}$

On squaring both sides, the above equation becomes

$\begin{aligned} (5.9)^2 & =n^2+2 n \\ n^2+2 n-35 & =0 \end{aligned}$

On solving the above equation, we get $n=5$

$[\mathrm{Pt}(\mathrm{Py})_4][\mathrm{PtCl}_4]$

Let the oxidation state of $\mathrm{Pt}$ is $x$.

$x+0 \times 4+x+(-1) \times 4=0$

or, $2 x=+4 \text { or, } x=+2$

Thus, it is the complex in which the central metal atom in cationic and anionic parts have same oxidation state.

The IUPAC naming of coordination complexes follows specific rules, where ligands are named in alphabetical order, followed by the central metal and its oxidation state in parentheses.

For the complex $[\mathrm{Pt}\left(\mathrm{NH}_3\right)_3 \mathrm{Br}\left(\mathrm{NO}_2\right) \mathrm{Cl}] \mathrm{Cl}$:

Identify the ligands in alphabetical order: The ligands within the coordination sphere include ammine ($\mathrm{NH}_3$), bromo ($\mathrm{Br}$), chloro ($\mathrm{Cl}$), and nitro ($\mathrm{NO}_2$). Arranging these in alphabetical order: ammine, bromo, chloro, nitro.

Number of each type of ligand:

Ammine: 3

Bromo: 1

Chloro: 1

Nitro: 1

Naming the ligands:

Ammine is represented as "triamine" due to three molecules.

Bromo, chloro, and nitro are prefixed with the number when there are multiple, but here there is only one of each, so no prefix number is necessary, just bromo, chloro, nitro.

Determining the oxidation state of platinum:

The overall charge of the complex is neutral.

Chloride ion outside contributes -1 charge.

Ammine is neutral, bromo is -1, chloro is -1, and nitro is -1.

Let the oxidation state of platinum be $ x $.

Solving for $ x $:

$ x + 0 \times 3 - 1 - 1 - 1 = +1 $

$ x - 3 = +1 $

$ x = +4 $

The name of the complex:

Considering the alphabetical order of ligands and the name of the metal followed by its oxidation state:

$ \text{triamine bromidochloridonitro platinum(IV) chloride} $

Thus, the IUPAC name of the complex is: Option B: triamine bromidochloridonitro platinum (IV) chloride.

$\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_4 \mathrm{Cl}_2\right]^{+}$ shows geometrical isomerism because it is a $M A_4 B_2$ type coordination compound which contains two set of equivalent ligands, four $\mathrm{H}_2 \mathrm{O}$ and $2 \mathrm{Cl}$.

Hence, the possible geometrical isomers are:

Hence, correct option is (a).

Assertion and reason both are correct and reason is the correct explanation of the assertion. Toxic metal ions are removed by chelating ligands. When a solution of chelating ligand is added to solution containing toxic metal, ligands chelate the metal ions by formation of stable complex.