Among Co^{3+},Ti^{2+},V^{2+} and Cr^{2+} ions | vedclass

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(B) The standard electrode potential $E^{\circ}$ for the reduction of $H^{+}$ to $H_{2}$ is $0.00 \ V$.$A$ metal ion can liberate $H_{2}$ from a dilut

Vedclass · Accepted Answer

$5$

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(B) The standard electrode potential $E^{\circ}$ for the reduction of $H^{+}$ to $H_{2}$ is $0.00 \ V$.$A$ metal ion can liberate $H_{2}$ from a dilute mineral acid if its standard reduction potential $E^{\circ}(M^{n+}/M)$ is negative.Among the given ions,$Co^{3+}$ has a high positive reduction potential $(E^{\circ}(Co^{3+}/Co^{2+}) = +1.82 \ V)$,meaning it is a strong oxidizing agent and will not liberate $H_{2}$ from acid; instead,it is reduced to $Co^{2+}$.The electronic configuration of $Co^{3+}$ $(Z=27)$ is $[Ar] 3d^{6}$.In the $3d^{6}$ configuration,the number of unpaired electrons $(n)$ is $4$.The spin-only magnetic moment is calculated as $\mu = \sqrt{n(n+2)} \ B.M.$$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.899 \ B.M.$The nearest integer value is $5$.

Among $Co^{3+}$,$Ti^{2+}$,$V^{2+}$ and $Cr^{2+}$ ions,one if used as a reagent cannot liberate $H_{2}$ from dilute mineral acid solution,its spin-only magnetic moment in gaseous state is $.....$ $B.M.$ (Nearest integer)

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