Question

For \([Mn(CO)_6]^{+}\) and \([V(CO)_6]^{-}\) complexes, the correct statement(s) is(are)

• A. Stretching frequency of the CO is higher in the Mn-complex
• B. Metal-carbonyl bond is stronger in the V-complex
• C. Mn-complex does not obey 18e\(^-\) rule
• D. V-complex obeys 18e\(^-\) rule

Hint:

Vanadium follows the 18e\(^-\) rule, resulting in a stronger metal-carbonyl bond compared to manganese complexes, which do not strictly obey the 18e\(^-\) rule.

Answer 1

The Correct Option is B

Step 1: Analysis of bonding in metal-carbonyl complexes.
- For Mn(CO)_6^{+}: Manganese has a lower atomic number than vanadium and typically forms weaker metal-carbonyl bonds. The 18e\(^-\) rule is not strictly followed in Mn-complexes because Mn is in an oxidation state +1 and has a 19-electron configuration.
- For V(CO)_6^{-}: Vanadium is in the +2 oxidation state, forming a more stable complex with a stronger metal-carbonyl bond. It also obeys the 18e\(^-\) rule because it has 18 valence electrons in its coordination sphere.
Step 2: Conclusion.
The metal-carbonyl bond is stronger in the V-complex compared to the Mn-complex. Thus, option (B) is correct.
Final Answer: \[ \boxed{\text{Metal-carbonyl bond is stronger in the V-complex.}} \]