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: Understand bonding in metal carbonyls.
Metal carbonyl bonding involves $\sigma$-donation from CO to metal and $\pi$-back bonding from metal to COThe extent of $\pi$-back bonding affects the C–O bond strength

Step 2: Effect of $\pi$-back bonding on C–O bond length.
Stronger $\pi$-back bonding weakens the C–O bond and increases its bond lengthThus, more electron-rich metal centers give longer C–O bonds

Step 3: Analyze oxidation states.
\[ \text{Na}_2[\text{Fe(CO)}_4] \Rightarrow \text{Fe}^{-2} \] \[ \text{Na}[\text{Co(CO)}_4] \Rightarrow \text{Co}^{-1} \] \[ \text{Ni(CO)}_4 \Rightarrow \text{Ni}^0 \]

Step 4: Compare electron density on metal.
Fe$^{2-}$ is most electron-rich, followed by Co$^{-1}$, and then Ni$^{0}$

Step 5: Compare $\pi$-back bonding.
\[ \text{Fe}^{2-} > \text{Co}^{-1} > \text{Ni}^0 \]
Thus, $\pi$-back bonding decreases in the same order

Step 6: Relate to C–O bond length.
Greater $\pi$-back bonding $\Rightarrow$ weaker C–O bond $\Rightarrow$ longer bond length
\[ \text{Na}_2[\text{Fe(CO)}_4] > \text{Na}[\text{Co(CO)}_4] > \text{Ni(CO)}_4 \]

Step 7: Conclusion.
\[ \boxed{\text{Na}_2[\text{Fe(CO)}_4] > \text{Na}[\text{Co(CO)}_4] > \text{Ni(CO)}_4} \]