Question

The atomic number of Cr and Fe are respectively 24 and 26, which of the following is paramagnetic with the spin of the electron:

• A. $\left[Cr\right(CO{)}_6]$
• B. $\left[Fe\right(CO{)}_5]$
• C. $\left[Fe\right(CN{)}_6{]}^{-4}$
• D. $\left[Cr\right(N{H}_3{)}_6{]}^{+3}$

Answer 1

The correct option is D $\left[Cr\right(N{H}_3{)}_6{]}^{+3}$

Out of $CO$, $C{N}^{-}$ and $N{H}_3$, increasing field strength is $N{H}_3<C{N}^{-}<CO$

$Cr$ has $24$ electrons, $Fe\to 26$ electrons.

$F{e}^{2+}$ has $24$ electrons and $C{r}^{+3}$ has $21$ electrons.

The $4$ unpaired electrons in $Cr$ and $Fe$ can pair up because of strong field ligand ${}^{\prime }C{O}^{\prime }$. $C{N}^{-}$ also strong field ligand. So, $4$ electrons in $F{e}^{2+}$ also pair up. Though $N{H}_3$ is a strong field ligand, only one electron remains in $C{r}^{+3}$. So no case of pairing here and $\left[Cr\right(N{H}_3{)}_6{]}^{3+}$ is paramagnetic.