Question

Atomic number of $Cr$ and $Fe$ are respectively $24$ and $26$, which of the following is paramagnetic with the spin of electron :-

• A. $[Cr(CO{)}_6]$
• B. $[Fe(CO{)}_5]$
• C. $[Fe(CN)6{]}^{4-}$
• D. $[Cr(N{H}_3{)}_6{]}^{3+}$

Answer 1

Answer: (D)

Atoms, ions or molecules having unpaired electrons are paramagnetic. $In\left[Cr{\left(N{H}_3\right)}_6^{3+}\right],Cr$ is present as $Cr$ (III) or $C{r}^{3+}$ So electronic configuration is
${}_{24}Cr=1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^5,4s^1$ ground state
$C{r}^{3+}=1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^3$

Number of unpaired electrons $=3$
$In\left[Cr(CO{)}_6\right],(O\cdot N\cdot \text{of}Cr=0)$
${}_{24}Cr=1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^5,4s^1$

Number of unpaired electron $=0$
$In\left[Fe(CO{)}_5\right],(ON$ of $Fe=0)$
$26Fe=1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^6,4s^2$

Number of unpaired electron $=0$
In ${\left[Fe(CN{)}_6\right]}^4,(ONo$ of $Fe=+2)$
$F{e}^{2+}=1s^2,2s^{2}2p^6,3s^{2}3p^{6}3d^6$

Number of unpaired electron $=0$
Hence, in above complex ion paramagnetic character is in ${\left[Cr{\left(N{H}_3\right)}_6\right]}^{3+}$ as it contains three unpaired electrons.