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(D) The 'spin only' magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired ele

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$B$ and $D$ only

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(D) The 'spin only' magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.

	
		
			Ion and Configuration
			No. of unpaired electrons $(n)$
		
		
			$Ti^{3+}: 3d^{1}$
			$1$
		
		
			$Cr^{2+}: 3d^{4}$
			$4$
		
		
			$Mn^{2+}: 3d^{5}$
			$5$
		
		
			$Fe^{2+}: 3d^{6}$
			$4$
		
		
			$Sc^{3+}: 3d^{0}$
			$0$
		
	


Since $Cr^{2+}$ and $Fe^{2+}$ both have $n = 4$ unpaired electrons,they have the same 'spin only' magnetic moment.

'Spin only' magnetic moment is same for which of the following ions?
$A$. $Ti^{3+}$ $B$. $Cr^{2+}$ $C$. $Mn^{2+}$ $D$. $Fe^{2+}$ $E$. $Sc^{3+}$
Choose the most appropriate answer from the options given below.

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'Spin only' magnetic moment is same for which of the following ions? $A$. $Ti^{3+}$ $B$. $Cr^{2+}$ $C$. $Mn^{2+}$ $D$. $Fe^{2+}$ $E$. $Sc^{3+}$ Choose the most appropriate answer from the options given below.