Match each item in Column I with the oxidation number in Column II (H is $+ 1$ , Ca is $+ 2$ , K is $+ 1$ , F is $- 1$ , O is $- 2$ unless the species is a superoxide). Column I: (P) C in $CaC X_{2}$ (acetylide ion $C X_{2} X^{2 -}$ ); (Q) N in $N X_{2} H X_{4}$ ; (R) O in $KO X_{2}$ ; (S) Xe in $XeO X_{3} F X_{2}$ . Column II: (1) $- 2$ ; (2) $- 1$ ; (3) $- \frac{1}{2}$ ; (4) $+ 8$ .

Answer: D

CaC X_{2}

2 + 2 x = 0

gives

- 1

N X_{2} H X_{4}

2 x + 4 = 0

gives

- 2

KO X_{2}

1 + 2 x = 0

gives

- \frac{1}{2}

XeO X_{3} F X_{2}

x - 6 - 2 = 0

gives

+ 8

. So P-2, Q-1, R-3, S-4.