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Question:

The pair(s) of reagents that yield paramagnetic species is/are: Na and excess of NH₃, K and excess of O₂, Cu and dilute HNO₃, O₂ and 2-ethylanthraquinol

Na and excess of NH₃

K and excess of O₂

O₂ and 2 - ethylanthraquinol

Cu and dilute HNO₃

Solution:

(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.