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. $[Cr(NH_3)_6]^{3+}$

Answer 1

The Correct Option is D

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

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

Number of unpaired electron $=0$ $In\left[ Fe ( CO )_{5}\right],( ON$ of $Fe =0)$ $26 Fe =1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{6}, 4 s^{2}$

Number of unpaired electron $=0$ In $\left[ Fe ( CN )_{6}\right]^{4},( O No$ of $Fe =+2)$ $Fe ^{2+}=1 s^{2}, 2 s^{2} 2 p^{6}, 3 s ^{2} 3 p^{6} 3 d^{6}$

Number of unpaired electron $=0$ Hence, in above complex ion paramagnetic character is in $\left[ Cr \left( NH _{3}\right)_{6}\right]^{3+}$ as it contains three unpaired electrons.