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.

List-$I$ (Complex)	List-$II$ (Spin only Magnetic moment)
$A$. $[CoF_6]^{3-}$	$I$. $0$
$B$. $[Co(C_2O_4)_3]^{3-}$	$II$. $\sqrt{24}$
$C$. $[FeF_6]^{3-}$	$III$. $\sqrt{8}$
$D$. $[Mn(CN)_6]^{3-}$	$IV$. $\sqrt{35}$
	$V$. $\sqrt{15}$

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

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Match the following:

List-$I$ (Complex) List-$II$ (Spin only Magnetic moment)

$A$. $[CoF_6]^{3-}$ $I$. $0$

$B$. $[Co(C_2O_4)_3]^{3-}$ $II$. $\sqrt{24}$

$C$. $[FeF_6]^{3-}$ $III$. $\sqrt{8}$

$D$. $[Mn(CN)_6]^{3-}$ $IV$. $\sqrt{35}$

$V$. $\sqrt{15}$

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