Determine the oxidation state of the elements in the following compounds:

(a)\(NaI\)

(b)\(GdC{l_3}\)

(c)\(LiN{O_3}\)

(d)\({H_2}Se\)

(e)\(M{g_2}Si\)

(f)\(Rb{O_2}\), rubidium superoxide

(g)\(HF\)

Consider the iodine oxidation number (-1) in (a).

Hence the oxidation state of sodium \(\left( x \right)\)is +1.

\(\begin{aligned}{}x - 1 &= 0\\x &= + 1\end{aligned}\)

Consider the chlorine oxidation number(-1) in (b).

The oxidation state of gadolinium\(\left( x \right)\) is calculated as below:

\(\begin{aligned}{}3 \times \left( { - 1} \right) + x& = 0\\x &= + 3\end{aligned}\)

Hence the oxidation state of gadolinium is +3.

Consider theoxidation number of oxygen (-2) and lithium (+1) in (c).

The oxidation state of nitrogen\(\left( x \right)\) is as calculated below:

\(\begin{aligned}{l}1 \times ( + 1) + 3 \times ( - 2) + x &= 0\\x &= + 5\end{aligned}\)

Hence the oxidation state of nitrogen is +5.

Consider the oxidation number of hydrogen (+1) in compound (d).

The oxidation number of selenium \(\left( x \right)\)is therefore -2.

\(\begin{aligned}{}2 \times ( + 1) + x &= 0\\x &= - 2\end{aligned}\)

Consider the oxidation number of magnesium (+2) in (e).

The oxidation number of silicon\(\left( x \right)\)is therefore - 4.

\(\begin{aligned}{}2 \times ( + 2) + x &= 0\\x &= - 4\end{aligned}\)

Consider the oxidation number of oxygen (-1/2) for superoxide in(f).

The oxidation number of rubidium \(\left( x \right)\)is therefore +1.

\(\begin{aligned}{}2 \times ( - \frac{1}{2}) + x & = 0\\x &= + 1\end{aligned}\)

Consider the oxidation number of fluorine, which is always -1.

Therefore, the oxidation number of hydrogen\(\left( x \right)\)has to be +1.

\(\begin{aligned}{}1 \times ( - 1) + x &= 0\\x &= + 1\end{aligned}\)