Paramagnetism arises due to the presence of unpaired electrons. Let us check electronic configurations:
(A) Zn$^{2+$:}
Zn (Z = 30): [Ar] 3d$^{10}$4s$^2$
Zn$^{2+}$ → [Ar] 3d$^{10}$ (all paired) → diamagnetic.
(B) Ni$^{2+$:}
Ni (Z = 28): [Ar] 3d$^8$4s$^2$
Ni$^{2+}$ → [Ar] 3d$^8$ → has 2 unpaired electrons → paramagnetic.
(C) Cu$^{2+$:}
Cu (Z = 29): [Ar] 3d$^{10}$4s$^1$
Cu$^{2+}$ → [Ar] 3d$^9$ → has 1 unpaired electron → also paramagnetic.
(D) Ag$^{+$:}
Ag (Z = 47): [Kr] 4d$^{10}$5s$^1$
Ag$^{+}$ → [Kr] 4d$^{10}$ → all paired → diamagnetic.
Step 2: Comparing.
- Ni$^{2+}$ and Cu$^{2+}$ are both paramagnetic.
- But among the given options, Ni$^{2+}$ (3d$^8$) is the most typical example considered in such questions.
Step 3: Final Answer.
Hence, the correct paramagnetic ion is:
\[
\boxed{\text{Ni}^{2+}}
\]