The atomic numbers of Cr and Fe are 24 and 26 | vedclass

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(D) $1$. $[Cr(CO)_6]$: $Cr$ is in $0$ oxidation state. Configuration is $3d^5 4s^1$. $CO$ is a strong field ligand,causing pairing. All electrons are

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

Vedclass · Answer

(D) $1$. $[Cr(CO)_6]$: $Cr$ is in $0$ oxidation state. Configuration is $3d^5 4s^1$. $CO$ is a strong field ligand,causing pairing. All electrons are paired. Diamagnetic.$2$. $[Fe(CO)_5]$: $Fe$ is in $0$ oxidation state. Configuration is $3d^6 4s^2$. $CO$ is a strong field ligand,causing pairing. All electrons are paired. Diamagnetic.$3$. $[Fe(CN)_6]^{4-}$: $Fe$ is in $+2$ oxidation state $(3d^6)$. $CN^-$ is a strong field ligand,causing pairing. All electrons are paired. Diamagnetic.$4$. $[Cr(NH_3)_6]^{3+}$: $Cr$ is in $+3$ oxidation state. Configuration is $3d^3$. There are $3$ unpaired electrons in the $t_{2g}$ orbitals. Paramagnetic.

The atomic numbers of $Cr$ and $Fe$ are $24$ and $26$ respectively. Which of the following is paramagnetic in terms of electron spin?

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