Solution

(A) $$Na$$ and excess of $$NH_3$$ yields solvated electrons which are unpaired electrons. Hence, the species is paramagnetic.

(B) $$K$$ and excess of $$O_2$$ forms superoxide $$KO_2$$ which contains paramagnetic superoxide ion $$O_2^-$$ with one unpaired electron.

(C) $$Cu$$ reacts with dilute $$HNO_3$$ to form cupric nitrate with 1 unpaired electron in $$Cu^{2+}$$ and also in $$NO$$. This results in paramagnetism.

$$3Cu+8HNO_3 (dil) \rightarrow 3Cu(NO_3)_2 +4H_2O + 2NO$$

(D) Refer image.