Question

The CORRECT order of p$K_a$ for the compounds I to IV in water at 298 K is 

• A. I>II>III>IV
• B. IV>III>II>I
• C. IV>II>III>I
• D. I>III>II>IV

Hint:

Replacing $\pi$-acceptor ligands (like CO) with $\sigma$-donors (like PPh$_3$) increases electron density on the metal, decreasing acidity (increasing p$K_a$).

Answer 1

The Correct Option is C

Step 1: Relation between acidity and p$K_a$.
Lower p$K_a$ corresponds to higher acidity. In metal hydride complexes, acidity decreases as electron density on the metal center increases.
Step 2: Ligand effects.
CO is a strong $\pi$-acceptor, withdrawing electron density from the metal and increasing acidity (lowering p$K_a$).
PPh$_3$ is a $\sigma$-donor and weak $\pi$-acceptor, increasing metal electron density and reducing acidity (raising p$K_a$).
Step 3: Comparing the complexes.
\[ \text{I: HCo(CO)}_4 \rightarrow \text{most acidic (lowest p}K_a) \] \[ \text{IV: HCo(CO)}_2(\text{PPh}_3)_2 \rightarrow \text{least acidic (highest p}K_a) \] Thus, the p$K_a$ order is IV>III>II>I.
Step 4: Conclusion.
The correct order of p$K_a$ values is IV>III>II>I.