JEE Main Chemistry question

Which combination is odd with respect to oxidation numbers of S, Cr, N and H respectively:

(A)
(B)K_2Cr_2O_7, K_2CrO_4, CrO_5, CrO_2Cl_2
(C)Correct
(D)CaH_2, NaH, LiH, MgH_2

Answer: (C)

Solution

In (A) all four compounds have S at +6; in (B) all four have Cr at +6; in (D) all four have H at −1 — each of these sets is internally consistent. In (C), however, nitrogen is −3

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Question ID VK-136802