The spin magnetic moment of cobalt in Hg[Co(S | vedclass

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Question

(C) The spin-only magnetic moment is calculated using the formula $\mu_{eff} = \sqrt{n(n+2)} \ B.M.$In the complex $Hg[Co(SCN)_4]$,$Hg$ is in the $+2$

Vedclass · Accepted Answer

$\sqrt{15}$

Vedclass · Answer

(C) The spin-only magnetic moment is calculated using the formula $\mu_{eff} = \sqrt{n(n+2)} \ B.M.$In the complex $Hg[Co(SCN)_4]$,$Hg$ is in the $+2$ oxidation state,so the complex ion is $[Co(SCN)_4]^{2-}$.Let the oxidation state of $Co$ be $x$. Then $x + 4(-1) = -2$,which gives $x = +2$.The electronic configuration of $Co^{2+}$ is $[Ar] 3d^7$.In a tetrahedral field,$3d^7$ configuration results in $3$ unpaired electrons $(n=3)$.Substituting $n=3$ into the formula: $\mu_{eff} = \sqrt{3(3+2)} = \sqrt{15} \ B.M.$Thus,option $C$ is correct.

List-$I$	List-$II$
$I. [FeF_6]^{3-}$	$A. 1.73 \ BM$
$II. [Ti(H_2O)_6]^{3+}$	$B. 5.93 \ BM$
$III. [Cr(NH_3)_6]^{3+}$	$C. 0.00 \ BM$
$IV. [Ni(H_2O)_6]^{2+}$	$D. 2.83 \ BM$
$V. [Fe(CN)_6]^{4-}$	$E. 3.88 \ BM$

The spin magnetic moment of cobalt in $Hg[Co(SCN)_4]$ is

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Match List-$I$ with List-$II$ and select the correct answer using the codes given below:
List-$I$ List-$II$
$I. [FeF_6]^{3-}$ $A. 1.73 \ BM$
$II. [Ti(H_2O)_6]^{3+}$ $B. 5.93 \ BM$
$III. [Cr(NH_3)_6]^{3+}$ $C. 0.00 \ BM$
$IV. [Ni(H_2O)_6]^{2+}$ $D. 2.83 \ BM$
$V. [Fe(CN)_6]^{4-}$ $E. 3.88 \ BM$

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