Q.iii. Which of the following element does not show oxidation state of +4 ? A. O
B. S
C. Se
D. Te

Br is in the +5 oxidation state in HOBrO$_2$ (i.e. HBrO$_3$, bromic acid).

HOBrO$_2$ is the same as HBrO$_3$ (bromic acid), containing 1 H, 1 Br and 3 O atoms in total.

Let oxidation state of Br = $x$. Using O = $-2$ and H = $+1$:

$(+1) + x + 3(-2) = 0$

$1 + x - 6 = 0 \Rightarrow x = +5$

So bromine is in the $+5$ oxidation state.

One electron short of a full octet, combined with very high electronegativity, drives halogens to gain (not lose) an electron, giving them the −1 oxidation state.

The halogens (Group 17) have the general valence-shell electronic configuration $ns^2np^5$ — one electron short of a completely filled octet ($ns^2np^6$, the stable noble-gas configuration).

Halogens are also among the most electronegative elements in the periodic table (fluorine is the most electronegative element of all) and have very high (favourable, exothermic) electron affinities. This combination means it is energetically very favourable for a halogen atom to gain one electron from another, less electronegative element, completing its octet and forming a stable halide ion, $X^-$.

This is why halogens most commonly show the $-1$ oxidation state in their compounds with less electronegative elements (metals, hydrogen, etc.) — although they can also show positive oxidation states (+1, +3, +5, +7) when combined with the even more electronegative fluorine or with oxygen.

Since fluorine is more electronegative than oxygen, oxygen is assigned a positive oxidation state in oxygen difluoride.

Oxidation number assignment gives the more electronegative element the negative oxidation state. Fluorine is the most electronegative element of all, so in $OF_2$, fluorine is $-1$ (×2 $=-2$) and oxygen must balance to $+2$ for overall neutrality. In $H_2O$, $Na_2O$, and $MgO$, oxygen (being more electronegative than H, Na, Mg) is in its usual $-2$ state.