Atomic numbers of Cr and Fe are 24 and 26 res | vedclass

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(A) $1$. $[Cr(NH_3)_6]^{3+}$: $Cr$ is in $+3$ oxidation state. Electronic configuration of $Cr^{3+}$ is $[Ar] 3d^3$. It has $3$ unpaired electrons,so

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

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(A) $1$. $[Cr(NH_3)_6]^{3+}$: $Cr$ is in $+3$ oxidation state. Electronic configuration of $Cr^{3+}$ is $[Ar] 3d^3$. It has $3$ unpaired electrons,so it is paramagnetic.$2$. $[Fe(CO)_5]$: $Fe$ is in $0$ oxidation state. $CO$ is a strong field ligand,causing pairing of electrons. $Fe(0)$ is $3d^6 4s^2$. All electrons are paired,so it is diamagnetic.$3$. $[Fe(CN)_6]^{4-}$: $Fe$ is in $+2$ oxidation state. $Fe^{2+}$ is $3d^6$. $CN^-$ is a strong field ligand,causing pairing of electrons. All electrons are paired,so it is diamagnetic.$4$. $[Cr(CO)_6]$: $Cr$ is in $0$ oxidation state. $Cr(0)$ is $3d^5 4s^1$. $CO$ is a strong field ligand,causing pairing of all electrons. It is diamagnetic.Therefore,$[Cr(NH_3)_6]^{3+}$ is paramagnetic.

Complexes	No. of unpaired electrons
$A. [CrF_6]^{4-}$	$P. 5$
$B. [MnF_6]^{4-}$	$Q. 2$
$C. [Cr(CN)_6]^{4-}$	$R. 1$
$D. [Mn(CN)_6]^{4-}$	$S. 4$

Atomic numbers of $Cr$ and $Fe$ are $24$ and $26$ respectively. Which of the following is paramagnetic?

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The correct match is:

Complexes No. of unpaired electrons

$A. [CrF_6]^{4-}$ $P. 5$

$B. [MnF_6]^{4-}$ $Q. 2$

$C. [Cr(CN)_6]^{4-}$ $R. 1$

$D. [Mn(CN)_6]^{4-}$ $S. 4$

The magnetic moment is measured in Bohr Magneton $(BM)$. Spin only magnetic moment of $Fe$ in $[Fe(H_2O)_6]^{3+}$ and $[Fe(CN)_6]^{3-}$ complexes respectively is:

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