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

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

Punjab PMETPunjab PMET 2007Coordination Compounds

Solution:

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
$Cr ^{3+}=1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{3}$
image
Number of unpaired electrons $=3$
$In\left[Cr( C O )_{6}\right],( O \cdot N \cdot \text{of} Cr =0)$
${ }_{24} Cr =1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{5}, 4 s^{1}$
image
Number of unpaired electron $=0$
$In\left[ Fe ( CO )_{5}\right],( ON$ of $Fe =0)$
$26 Fe =1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6} 3 d^{6}, 4 s^{2}$
image
Number of unpaired electron $=0$
In $\left[ Fe ( CN )_{6}\right]^{4},( O No$ of $Fe =+2)$
$Fe ^{2+}=1 s^{2}, 2 s^{2} 2 p^{6}, 3 s ^{2} 3 p^{6} 3 d^{6}$
image
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.