Consider following compounds A to E.
(A) XeFn
(B) XeF(n+1)+
(C) XeF(n+1)-
(D) XeF(n+2)
(E) XeF(n+4)2-
If value of n is 4, then calculate value of "p q" here, 'p' is total number of b

Question

Consider following compounds A to E.
(A) XeFn
(B) XeF(n+1)+
(C) XeF(n+1)-
(D) XeF(n+2)
(E) XeF(n+4)2-
If value of n is 4, then calculate value of "p $\div$ q" here, 'p' is total number of bond pair and 'q' is total number of lone pair on central atoms of compounds A to E.

Prashant Kotian · Accepted Answer

Correct answer: 4 / COMPOUNDS ARE XEF4, XEF5+, XEF5-, XEF6, XEF82-. TOTAL BOND PAIRS (P) = 4+5+5+6+8 = 28. TOTAL LONE PAIRS ON CENTRAL ATOM (Q) = 2+1+2+1+1 = 7. P $\DIV$ Q = 28 / 7 = 4. Step 1: Tin (Sn) belongs to Group 14, where the stability of the +4 oxidation state is higher than that of +2 for the lighter members. Step 2: Because Sn4+ is highly stable, Sn2+ readily acts as a reducing agent by losing two electrons to convert into Sn4+: Sn^2+ -> Sn^4+ + 2e^-. Step 3: Therefore, the reducing nature of SnCl2 is due to Sn4+ being more stable than Sn2+, matching option (c).

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