$\text{The Spin only magnetic moment value of square planar complex } [\text{Pt(NH}_3\text{)}_2\text{Cl(NH}_2\text{CH}_3\text{)}]\text{Cl is } \_\_\_\_\_\_ \text{ B.M. (Nearest integer)} \\ \text{(Given atomic number for Pt = 78)}$
Correct Answer: 0
Approach Solution - 1
The square planar complex $[\text{Pt(NH}_3\text{)}_2\text{Cl(NH}_2\text{CH}_3\text{)}]\text{Cl}$ involves a Pt(II) center. For Pt(II), we remove two electrons from the neutral Pt atom, which has the electronic configuration: $$[\text{Xe}]4f^{14}5d^96s^1.$$ Removing two electrons gives: $$5d^8.$$ In a square planar complex like this one, particularly with a d8 configuration, all electrons are paired due to strong field ligands causing large splitting, which leaves the complex diamagnetic.
The spin-only magnetic moment $\mu_s$ is given by $\mu_s=\sqrt{n(n+2)}$ where $n$ is the number of unpaired electrons.
Since the complex is diamagnetic $(n=0)$, the magnetic moment $\mu=0$ B.M.
Checking the range (0,0).
Approach Solution -2
The complex \([ \text{Pt(NH}_3)_2 \text{Cl(NH}_2\text{CH}_3) ] \text{Cl}\) contains \(\text{Pt}^{2+}\) in a square planar geometry.
\(\text{Pt}^{2+}\) has a \(d^8\) electronic configuration. In square planar complexes, the \(d\)-electrons pair up in such a way that no unpaired electrons remain. As a result, the magnetic moment is \(0 \, \text{B.M.}\) (Bohr Magnetons).
The Correct answer is: 0
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