Na and excess of NH₃
K and excess of O₂
O₂ and 2 - ethylanthraquinol
Cu and dilute HNO₃
(A) Na and excess of NH₃ yields solvated electrons which are unpaired electrons. Hence, the species is paramagnetic
(B) K and excess of O₂ forms superoxide KO₂ which contains paramagnetic superoxide ion O₂⁻ with one unpaired electron
(C) Cu reacts with dilute HNO₃ to form cupric nitrate with 1 unpaired electron in Cu²⁺ and also in NO. This results in paramagnetism.
3Cu + 8HNO₃(dil) → 3Cu(NO₃)₂ + 4H₂O + 2NO
(D) Refer image.