Question

The correct order of the spin-only magnetic moments of the following complexes is:
$\left(I\right)\left[Cr\right(H_{2}O{)}_6]B{r}_2$
$\left(II\right)N{a}_4\left[Fe\right(CN{)}_6]$
$\left(III\right)N{a}_3\left[Fe\right(C_2{O}_4{)}_3\left]\right({\Delta }_0>P)$
$\left(IV\right)\left(E{t}_{4}N{)}_2\right[CoC{l}_4]$

• A. $\left(III\right)>\left(I\right)>\left(IV\right)>\left(II\right)$
• B. $\left(III\right)>\left(I\right)>\left(II\right)>\left(IV\right)$
• C. $\left(II\right)>\left(I\right)>\left(IV\right)>\left(III\right)$
• D. $\left(I\right)>\left(IV\right)>\left(III\right)>\left(II\right)$

Answer 1

The correct option is D $\left(I\right)>\left(IV\right)>\left(III\right)>\left(II\right)$

$\left[Cr{\left(H_{2}O\right)}_6\right]{Br}_2:{Cr}^{2+}=\left[Ar\right]4s^{0}3d^4$
$t_{2g}^3{eg}^1$
$n=4,\mu =\sqrt{n(n+2)}=\sqrt{24}BM$
${Na}_4\left[Fe\right(CN{)}_6]:{Fe}^{2+}=\left[Ar\right]4s^{0}3d^6$
$t_{2g}^6{eg}^0$
$n=0,\mu =0$
${Na}_3\left[Fe{\left(C_2{O}_4\right)}_3\right]:{Fe}^{3+}=\left[Ar\right]3d^5$
$t_{2g}^5{eg}^0$
$n=1,\mu =\sqrt{3}BM$
${\left({Et}_{4}N\right)}_2\left[{CoCl}_4\right]:{Co}^{2+}=\left[Ar\right]3d^7$
$e_g^4{t}_{2g}^3$
$n=3,\mu =\sqrt{15}BM$