Which of the following pairs will have the sa | vedclass

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(B) The effective magnetic moment $(\mu_{eff})$ is calculated using the formula $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired e

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$Cr^{2+}$ and $Fe^{2+}$

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(B) The effective magnetic moment $(\mu_{eff})$ is calculated using the formula $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons.For $Cr^{2+}$: Electronic configuration is $[Ar] 3d^4$. Number of unpaired electrons $(n)$ = $4$.For $Fe^{2+}$: Electronic configuration is $[Ar] 3d^6$. Number of unpaired electrons $(n)$ = $4$.Since both $Cr^{2+}$ and $Fe^{2+}$ have $4$ unpaired electrons,they will have the same effective magnetic moment.Therefore,the correct option is $B$.

List-$I$ Species	List-$II$ Electronic distribution
$A$. $Cr^{2+}$	$I$. $3d^8$
$B$. $Mn^{+}$	$II$. $3d^3 4s^1$
$C$. $Ni^{2+}$	$III$. $3d^4$
$D$. $V^{+}$	$IV$. $3d^5 4s^1$

Which of the following pairs will have the same effective magnetic moment?

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Match List-$I$ with List-$II$

List-$I$ Species List-$II$ Electronic distribution

$A$. $Cr^{2+}$ $I$. $3d^8$

$B$. $Mn^{+}$ $II$. $3d^3 4s^1$

$C$. $Ni^{2+}$ $III$. $3d^4$

$D$. $V^{+}$ $IV$. $3d^5 4s^1$

Choose the correct answer from the options given below:

Which element from the following exhibits the highest number of various different possible oxidation states?

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