Question

Which among the following metal carbonyls are diamagnetic inner orbital complexes?
$\left(I\right)Ni(CO{)}_4$
$\left(II\right)Fe(CO{)}_5$
$\left(III\right)V(CO{)}_6$
$\left(IV\right)Cr(CO{)}_6$
Select the correct answer from the codes given below:

• A. I and II only
• B. II, III and IV only
• C. II and IV only
• D. I, II and IV only

Answer 1

The correct option is C II and IV only

Carbonyl is a strong field ligand. When it approaches the central atom, the splitting caused is high. In other words, one can say it creates a massive energy difference between the $t_{2g}and{e}_g$ orbitals and hence restricts the electrons from making the jump. Simply put, it forces pairing.

$\left(I\right)Ni(CO{)}_4$

Ground state electronic configuration of $Ni$ is $4s^{2}3d^8$. The 4s electrons are also forced to pair in 3d, as a result of which, the final electronic configuration of $Ni$ is $4s^{0}3d^{10}$. Here, all d electrons are paired which means it is a diamagnetic complex.
The four ligands use one 4s and three 4p orbitals for complex formation after this. So, the hybridization is $s{p}^3$. Clearly, $Ni(CO{)}_4$ is an outer orbital complex.
Hence, we can say that $Ni(CO{)}_4$ is a diamagnetic outer orbital complex.

$\left(II\right)Fe(CO{)}_5$

Ground state electronic configuration of $Fe$ is $4s^{2}3d^6$. The 4s electrons are also forced to pair in 3d, as a result of which, the final electronic configuration of $Fe$ is $4s^{0}3d^8$. Here, all d electrons are paired which means it is a diamagnetic complex. The hybridization is $ds{p}^3$, which clearly indicates it is an inner orbital complex.
Hence, we can say that $Fe(CO{)}_5$ is an inner orbital diamagnetic complex.

$\left(III\right)V(CO{)}_6$

Ground state electronic configuration of $V$ is $4s^{2}3d^3$. The 4s electrons are also forced to pair in 3d, as a result of which, the final electronic configuration of $V$ is $4s^{0}3d^5$. Here, four d electrons are paired which means it is a paramagnetic complex. The hybridization is $d^{2}s{p}^3$, which clearly indicates it is an inner orbital complex.
Hence, we can say that $V(CO{)}_6$ is an inner orbital paramagnetic complex.

$\left(IV\right)Cr(CO{)}_6$

Ground state electronic configuration of $Cr$ is $4s^{2}3d^4$. The 4s electrons are also forced to pair in 3d, as a result of which, the final electronic configuration of $Fe$ is $4s^{0}3d^6$. Here, all d electrons are paired which means it is a diamagnetic complex. The hybridization is $d^{2}s{p}^3$, which clearly indicates it is an inner orbital complex.
Hence, we can say that $Cr(CO{)}_6$ is an inner orbital diamagnetic complex.