Chemistry - Complete d-block Question with Solution | TestHub

To determine the magnetic nature and magnetic moment, we count the number of unpaired electrons ($n$).

Magnetic moment $\mu =\sqrt{n(n+2)}$ BM.

Species with $n=0$ are diamagnetic ($\mu =0$). Species with $n>0$ are paramagnetic ($\mu >0$).

Let's analyze each option:

A. ${\mathrm{Fe}}^{+3}$ ($3d^5$, $n=5$), ${\mathrm{Cu}}^{+}$ ($3d^{10}$, $n=0$), ${\mathrm{Ag}}^{+}$ ($4d^{10}$, $n=0$). Different $n$.

B. ${\mathrm{Co}}^{+3}$ ($3d^6$, $n=4$), ${\mathrm{Cr}}^{+3}$ ($3d^3$, $n=3$), ${\mathrm{Ni}}^{+2}$ ($3d^8$, $n=2$). Different $n$.

C. ${\mathrm{Cu}}^{+}$ ($3d^{10}$, $n=0$), ${\mathrm{Zn}}^{+2}$ ($3d^{10}$, $n=0$), ${\mathrm{Cd}}^{+2}$ ($4d^{10}$, $n=0$). All have $n=0$, so they are diamagnetic and have a magnetic moment of 0 BM.

D. ${\mathrm{Mn}}^{+2}$ ($3d^5$, $n=5$), ${\mathrm{Fe}}^{+2}$ ($3d^6$, $n=4$), ${\mathrm{As}}^{+3}$ ($3d^{10}4s^{2}4p^0$, $n=0$). Different $n$.

Therefore, option C is the correct set.