Identify the metal ions among Co^{2+}, Ni^{2+ | vedclass

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(A) The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.For $\mu > 3.0 \ BM$,we need $\

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$7$

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(A) The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.For $\mu > 3.0 \ BM$,we need $\sqrt{n(n+2)} > 3.0$,which implies $n(n+2) > 9$. This holds for $n \geq 3$.Electronic configurations:$V^{3+} (3d^2): n = 2, \mu = \sqrt{2(4)} = 2.83 \ BM$$Ti^{2+} (3d^2): n = 2, \mu = \sqrt{2(4)} = 2.83 \ BM$$Ni^{2+} (3d^8): n = 2, \mu = \sqrt{2(4)} = 2.83 \ BM$$Fe^{2+} (3d^6): n = 4, \mu = \sqrt{4(6)} = 4.90 \ BM$$Co^{2+} (3d^7): n = 3, \mu = \sqrt{3(5)} = 3.87 \ BM$Thus,$Fe^{2+}$ and $Co^{2+}$ have $\mu > 3.0 \ BM$.In high spin octahedral complexes:$Fe^{2+} (3d^6)$ has $4$ unpaired electrons.$Co^{2+} (3d^7)$ has $3$ unpaired electrons.Sum of unpaired electrons = $4 + 3 = 7$.

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

Identify the metal ions among $Co^{2+}, Ni^{2+}, Fe^{2+}, V^{3+}$ and $Ti^{2+}$ having a spin-only magnetic moment value more than $3.0 \ BM$. The sum of unpaired electrons present in the high spin octahedral complexes formed by those metal ions is . . . . . . .

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

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

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

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