1. Find the oxidation state of Pt in the complex ion

The given compound is $\left[\mathrm{Pt}(\mathrm{NH}_3)_2\mathrm{Cl}(\mathrm{NH}_2\mathrm{CH}_3)\right]\mathrm{Cl}$

The outside $\mathrm{Cl}^-$ is the counter ion, so the complex inside brackets has charge $+1$.

Let oxidation state of Pt be $x$.

Ligand charges:

• $\mathrm{NH}_3$ is neutral
• $\mathrm{NH}_2\mathrm{CH}_3$ (methylamine) is neutral
• coordinated $\mathrm{Cl}^-$ is $-1$

So, $x+0+0+(-1)+0=+1$ $x-1=1$ $x=+2$

Hence, metal ion is $\mathrm{Pt}^{2+}$.

2. Find the electronic configuration of $\mathrm{Pt}^{2+}$

Atomic number of Pt = 78.

Ground-state configuration of Pt: $\mathrm{Pt}: [\mathrm{Xe}]\,4f^{14}5d^96s^1$

For $\mathrm{Pt}^{2+}$, remove two electrons: first from $6s$, then from $5d$.

Thus, $\mathrm{Pt}^{2+}: [\mathrm{Xe}]\,4f^{14}5d^8$

So it is a $d^8$ system.

3. Determine geometry and pairing

The complex is square planar (given in the question).

For a 5d metal like Pt(II), a $d^8$ configuration in square planar field strongly favors complete pairing.

Square planar $d^8$ complexes of Pt(II) are low spin and have no unpaired electrons.

Thus, $n=0$ where $n$ = number of unpaired electrons.

4. Calculate spin-only magnetic moment

Spin-only magnetic moment is $\mu = \sqrt{n(n+2)}\ \text{B.M.}$

Substitute $n=0$: $\mu = \sqrt{0(0+2)}=0\ \text{B.M.}$

Nearest integer: $\boxed{0}$

5. Comparison with stored correct answer

Stored correct answer = 0

Derived answer = 0

So the derived answer agrees with the stored correct answer.