Match the List-I with List-II:List-I (Complex | vedclass

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

Vedclass · Accepted Answer

$A-III, B-I, C-IV, D-II$

Vedclass · Answer

(D) The spin-only magnetic moment $(mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} 	ext{ BM}$,where $n$ is the number of unpaired electrons.$A$. $[Cr(H_2O)_6]^{2+}$: $Cr^{2+}$ has a $d^4$ configuration. Number of unpaired electrons $(n)$ = $4$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 	ext{ BM}$ $(III)$.$B$. $[Co(H_2O)_6]^{2+}$: $Co^{2+}$ has a $d^7$ configuration. Number of unpaired electrons $(n)$ = $3$. $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 	ext{ BM}$ $(I)$.$C$. $[Cu(H_2O)_6]^{2+}$: $Cu^{2+}$ has a $d^9$ configuration. Number of unpaired electrons $(n)$ = $1$. $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 	ext{ BM}$ $(IV)$.$D$. $[Mn(H_2O)_6]^{2+}$: $Mn^{2+}$ has a $d^5$ configuration. Number of unpaired electrons $(n)$ = $5$. $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 	ext{ BM}$ $(II)$.Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

List-$I$ (Complex ion)	List-$II$ (Calculated spin only magnetic moment $(BM)$)
$A$. $[Cr(H_2O)_6]^{2+}$	$I$. $3.87$
$B$. $[Co(H_2O)_6]^{2+}$	$II$. $5.92$
$C$. $[Cu(H_2O)_6]^{2+}$	$III$. $4.90$
$D$. $[Mn(H_2O)_6]^{2+}$	$IV$. $1.73$

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