Question

The correct order of bond length between carbon and oxygen atoms in the metal carbonyls given in the options is

• A. Na$_2$[Fe(CO)$_4$] $>$ Na[Co(CO)$_4$] $>$ [Ni(CO)$_4$]
• B. [Ni(CO)$_4$] $>$ Na[Co(CO)$_4$] $>$ Na$_2$[Fe(CO)$_4$]
• C. Na$_2$[Fe(CO)$_4$] $>$ [Ni(CO)$_4$] $>$ Na[Co(CO)$_4$]
• D. [Ni(CO)$_4$] $>$ Na$_2$[Fe(CO)$_4$] $>$ Na[Co(CO)$_4$]

Hint:

More negative charge on metal increases $\pi$-back bonding, which weakens and lengthens the C–O bond

Answer 1

The Correct Option is A

Step 1: Link back bonding to bond length.
More electron density on the metal means stronger back bonding into CO, which weakens and lengthens the C to O bond.

Step 2: Compare the metal charge.
Iron in Na$_2$[Fe(CO)$_4$] is most negative, then cobalt in Na[Co(CO)$_4$], then neutral nickel in Ni(CO)$_4$.

Step 3: Translate to bond length.
The most electron rich metal gives the longest C to O bond.

Step 4: Answer.
\[ \boxed{\text{Na}_2[\text{Fe(CO)}_4] > \text{Na}[\text{Co(CO)}_4] > \text{Ni(CO)}_4} \]