Question

For the conversion of [Pt(L)Cl\(_3\)]\(^-\) to trans-[Pt(L)Cl\(_2\)(H\(_2\)O)], the 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 {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) \(⇒\) least {trans}-effect; soft/\(\pi\)-acceptor ligands \(⇒\) strong {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 trans to them (kinetic trans-effect). Strong \(\pi\)-acceptors and soft donors show a large trans-effect; hard, weak \(\sigma\)-donors show a small one.
Step 2: Rank the given ligands qualitatively. \[ \text{(weak)} \text{H}_2\text{O} < \text{NH}_3 < \text{DMSO (via S)} < \text{C}_2\text{H}_4 \text{(strong)} \] \(\mathbf{H_2O}\): very weak \(\sigma\)-donor (hard base) \(⇒\) least trans-effect.
\(\mathbf{NH_3}\): slightly stronger \(\sigma\)-donor than water \(⇒\) larger effect than H\(_2\)O.
\(\mathbf{DMSO}\) (S-bound): softer donor, moderate trans-effect.
\(\mathbf{C_2H_4}\): \(\pi\)-acceptor olefin, strong trans-effect.
Step 3: Conclusion. Since H\(_2\)O has the smallest trans-effect in this set, the conversion will be slowest when \(L = \text{H}_2\text{O}\). Hence, option (A) is correct.