Question

For the conversion of [Pt(L)Cl\(_3\)]\(^-\) to \(\textit{trans}\)-[Pt(L)Cl\(_2\)(H\(_2\)O)], the\( \textit{trans}\)-effect is LEAST when the ligand L is

• A. H\(_2\)O
• B. NH\(_3\)
• C. DMSO
• D. C\(_2\)H\(_4\)

Hint:

A handy fragment of the \textit{trans}-effect series: \(\text{CO}, \text{C}_2\text{H}_4 \gg \text{PR}_3 \gg \text{I}^- \gg \text{Br}^- \gg \text{Cl}^- \gg \text{NH}_3 \gtrsim \text{H}_2\text{O}\).
When unsure, remember: hard, weak \(\sigma\)-donors (e.g., H\(_2\)O) \(\Rightarrow\) least \textit{trans}-effect; soft/\(\pi\)-acceptor ligands \(\Rightarrow\) strong \textit{trans}-effect.

Answer 1

The Correct Option is A

Step 1: Recall the trans-effect. In square-planar Pt(II) complexes, ligands differ in their ability to accelerate substitution of the ligand positioned \(\textit{trans}\) to them (kinetic\(\textit{trans}\)-effect). Strong \(\pi\)-acceptors and soft donors show a large \(\textit{trans}\)-effect; hard, weak \(\sigma\)-donors show a small one.  
Step 2: Rank the given ligands qualitatively. \[ \text{(weak)}\quad \text{H}_2\text{O} \;<\; \text{NH}_3 \;<\; \text{DMSO (via S)} \;<\; \text{C}_2\text{H}_4 \quad \text{(strong)} \] 

\(\mathbf{H_2O}\): very weak \(\sigma\)-donor (hard base) \(\Rightarrow\) least \( \textit{trans}\)-effect. 
\(\mathbf{NH_3}\): slightly stronger \(\sigma\)-donor than water \(\Rightarrow\) larger effect than H\(_2\)O. 
\(\mathbf{DMSO}\) (S-bound): softer donor, moderate \(\textit{trans}\)-effect. 
\(\mathbf{C_2H_4}\): \(\pi\)-acceptor olefin, strong \(\textit{trans}\)-effect. 
Step 3: Conclusion. Since H\(_2\)O has the smallest \(\textit{trans}\)-effect in this set, the conversion will be slowest when \(L = \text{H}_2\text{O}\). Hence, option (A) is correct.