Question

Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.

(a)\({H_3}P{O_4}\)

(b)\(Al{\left( {OH} \right)_3}\)

(c)\(Se{O_2}\)

(d)\(KN{O_2}\)

(e)\(I{n_2}{S_3}\)

(f)\({P_4}{O_6}\).

Short answer

The oxidation states of the elements in the compound are

1. \(H\left( { + 1} \right),P\left( { + 5} \right),O\left( { - 2} \right)\)
2. \(Al\left( { + 3} \right),O\left( { - 2} \right),H\left( { + 1} \right)\)
3. \(Se\left( { + 4} \right),O\left( { - 2} \right)\)
4. \(K\left( { + 1} \right),N\left( { + 3} \right),O\left( { - 2} \right)\)
5. \(In\left( { + 3} \right),S\left( { - 2} \right)\)
6. \(P\left( { + 3} \right),O\left( { - 2} \right)\)

Step-by-step solution

Determine the oxidation states of elements in part (a)

Consider theoxidation numbers ofoxygen (-2) and hydrogen(+1) in (a).

Hence the oxidation state of phosphorous is +5 as indicated below:

\(\begin{aligned}{}3 \times \left( { + 1} \right) + x + 4 \times \left( { - 2} \right) = 0\\x = + 5\end{aligned}\)

Determine the oxidation states of elements in part (b)

Consider the oxidation numbers of oxygen (-2) and hydrogen (+1) in (b).

Therefore, the oxidation state of aluminium is +3 as shown below:

\(\begin{aligned}{}x + 3 \times \left( { - 2} \right) + 3 \times \left( { + 1} \right) = 0\\x = + 3\end{aligned}\)

Determine the oxidation states of elements in part (c)

Consider theoxidation number of oxygen (-2) in (c).

Therefore,theoxidation state of selenium is +4 as shown below:

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

Determine theoxidation states of elements in part (d)

Consider theoxidation numbers of oxygen (-2) and potassium (+1) in (d).

Therefore,theoxidation state of nitrogen is +3 as given below:

\(\begin{aligned}{}1 \times 1 + x + 2 \times \left( { - 2} \right) = 0\\x = + 3\end{aligned}\)

Determine the oxidation states of elements in part (e)

Consider theoxidation number of sulfur (-2) in (e).

Therefore,theoxidation state of indium is +3 as given below:

\(\begin{aligned}{}2x + 3 \times \left( { - 2} \right) = 0\\x = + 3\end{aligned}\)

Determine the oxidation states of elements in part (f)

Consider theoxidation number of oxygen(-2) in (f).

Therefore,theoxidation state of phosphorous is +3 as given below:

\(\begin{aligned}{}4x + 6 \times \left( { - 2} \right) = 0\\x = + 3\end{aligned}\)