Question

The Crystal field stabilization energy (CFSE) and magnetic moment (spin-only) of an octahedral aqua complex of a metal ion $\left(M^{Z+}\right)$ are $-0.8{\Delta }_o$ and $3.87BM$ , respectively. Identify $\left(M^{Z+}\right)$ :

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• A. $C{o}^{2+}$
• B. $V^{3+}$
• C. $C{r}^{3+}$
• D. $M{n}^{4+}$

Answer 1

The correct option is A $C{o}^{2+}$

$C{o}^{2+}=\left[Ar\right]3d^{7}4s^o$

$H_{2}O$ will behave as weak field ligand

$C{o}^{2+}=t_{2g}^5,e_g^2$

$CFSE=(-0.4\times 5+2\times 0.6){\Delta }_o$

$=-0.8{\Delta }_o$

$C{o}^{2+}$ has 3 unpaired $e^{-},\mu =3.87BM$

$C{r}^{3+}\to t_{2g}^3{e}_g^0=-1.2{\Delta }_o,\mu =3.87BM$
$V^{3+}\to t_{2g}^2{e}_g^0=-0.8{\Delta }_o,\mu =2.82BM$
$M{n}^{4+}\to t_{2g}^3{e}_g^0=-1.2{\Delta }_o,\mu =3.87BM$

Hence, option (d) is correct.