Determine the oxidation number of oxygen in oxygen difluoride, $\mathrm{OF}_2$. (Note: F is the most electronegative element and is ALWAYS $-1$ in its compounds.)

1. $\mathrm{OF}_2$ is a neutral molecule, so the sum of oxidation numbers is $0$. 2. F is the most electronegative element on the periodic table, so it ALWAYS carries the negative oxidation number when bonded with anything (it cannot be oxidised further). F = $-1$. 3. Let O have oxidation number $x$. Then $x + 2(-1) = 0 \Rightarrow x = +2$. 4. This is one of the only cases where oxygen has a POSITIVE oxidation number — the standard rule "O is $-2$" only applies when O is bonded to a less electronegative element. 5. Option C ($-2$) is the usual default for O but does not apply here. Options B and D would violate the constraint that F is $-1$. _Source: NCERT Class 11 Chemistry Part 2, Ch 7, §7.3 (Oxidation Number rules, exception list for oxygen with F), p. 4–6._