91Ó°ÊÓ

a) In one molecule of\(S{F_6}\)there are\(6\;{\rm{F}}\)atoms with oxidation number\( - 1\).

Since the molecule is neutral, the total charge of all of the atom should be zero. Oxidation number of\(S\)in unknown\(x\).

\(\begin{array}{l}0 = x + 6 \cdot ( - 1)\\0 = x - 6\\x = 6\end{array}\)

The oxidation number of\({\rm{S}}{{\rm{F}}_6}\)is 6.

b) Oxidation number of oxygen is\(( - 2)\)and there are two atoms of\({\rm{O}}\)in\({\rm{S}}{{\rm{O}}_2}\;{{\rm{F}}_2}\), as there are two atoms of\({\rm{F}}\)in oxidation state\(( - 1)\)and with neutral total charge:

\(\begin{array}{l}0 = x + 2 \cdot ( - 2) + 2 \cdot ( - 1)\\0 = x - 6\\x = 6.\end{array}\)

The oxidation number of \({\rm{S}}{{\rm{O}}_2}\) is 6.

c) Oxidation number of\({\rm{K}}\)and of\({\rm{H}}\)is\(( + 1)\), and there is only one atom of each in\({\rm{KHS}}\)with total charge of the molecule 0 .

\(\begin{array}{l}0 = x + 1 \cdot 1 + 1 \cdot 1\\0 = x + 2\\x = - 2.\end{array}\)

The oxidation number of \({{\rm{F}}_2}\) is \( - 2\).