An ion with a definite magnetic moment (spin | vedclass

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(D) The magnetic moment (spin only) is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$Sc^{3+}$ $([Ar]

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Both $(b)$ and $(c)$

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(D) The magnetic moment (spin only) is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.$Sc^{3+}$ $([Ar] 3d^0)$: $n = 0$,so $\mu = 0 \ BM$.$Ti^{3+}$ $([Ar] 3d^1)$: $n = 1$,so $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.$Cu^{2+}$ $([Ar] 3d^9)$: $n = 1$,so $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \ BM$.Since both $Ti^{3+}$ and $Cu^{2+}$ have $n=1$,they possess a definite non-zero magnetic moment.

List-$I$ ($\text{Transition metal ion}$)	List-$II$ ($\text{Spin only magnetic moment (B.M.)}$)
$A$. $Ti^{3+}$	$I$. $3.87$
$B$. $V^{2+}$	$II$. $0.00$
$C$. $Ni^{2+}$	$III$. $1.73$
$D$. $Sc^{3+}$	$IV$. $2.84$

An ion with a definite magnetic moment (spin only) is:

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Match List-$I$ with List-$II$:
List-$I$ ($\text{Transition metal ion}$) List-$II$ ($\text{Spin only magnetic moment (B.M.)}$)
$A$. $Ti^{3+}$ $I$. $3.87$
$B$. $V^{2+}$ $II$. $0.00$
$C$. $Ni^{2+}$ $III$. $1.73$
$D$. $Sc^{3+}$ $IV$. $2.84$

Choose the correct answer from the options given below:

Which of the following is a diamagnetic ion?

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