Q18E - Chemistry (OpenStax)

Question

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

(a) \({H_2}S{O_4}\)

(b) \(Ca{\left( {OH} \right)_2}\)

(c) \(BrOH\)

(d) \(ClN{O_2}\)

(e) \(TiC{l_4}\)

(f) \(NaH\)

Step-by-Step Solution

Answer

The oxidation states of the elements in the compounds given are shown below:

1. \(H\left( { + 1} \right),S\left( { + 6} \right),O\left( { - 2} \right)\)
2. \(Ca\left( { + 2} \right),O\left( { - 2} \right),H\left( { + 1} \right)\)
3. \(Br\left( { + 1} \right),O\left( { - 2} \right),H\left( { + 1} \right)\)
4. \(Cl\left( { + 1} \right),N\left( { + 3} \right),O\left( { - 2} \right)\)
5. \(Ti\left( { + 4} \right),Cl\left( { - 1} \right)\)
6. \(Na\left( { + 1} \right),H\left( { - 1} \right)\).

Step 1: Determine the oxidation states of the elements in part (a)

Consider the oxidation number of oxygen (-2) and hydrogen (+1) in \({H_2}S{O_4}\)(Sulfuric acid).

Hence the oxidation state of sulfur(\(x\)) is +5 as calculated below: 

\(\begin{aligned}{\underline{\phantom{xx}}}2 \times 1 + x + 4 \times \left( { - 2} \right) = 0\\x =  + 6\end{aligned}\)

Step 2: Determine the oxidation states of the elements in part (b)

Consider the oxidation number of oxygen (-2) and hydrogen (+1) in \(Ca{\left( {OH} \right)_2}\)(Calcium hydroxide).

Hence the oxidation state of calcium(\(x\)) is +2 as calculated below: 

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

Step 3: Determine the oxidation states of the elements in part (c)

Consider the oxidation number of oxygen (-2) and hydrogen (+1) in \(BrOH\)(Hypobromous acid).

Hence the oxidation state of bromine (\(x\)) is +1 as calculated below: 

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

Step 4: Determine the oxidation states of the elements in part (d)

Consider the oxidation number of oxygen (-2) and chlorine (+1), in presence of oxygen, in the compound formula \(ClN{O_2}\)(Chlorine nitrite).

Hence the oxidation state of nitrogen (\(x\)) is +3 as calculated below: 

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

Step 5: Determine the oxidation states of the elements in part (e)

Consider theoxidation number of chlorine (-1) in \(TiC{l_4}\)(titanium tetrachloride).

Hence the oxidation state of titanium (\(x\)) is +4 in titanium tetrachloride.

\(\begin{aligned}{\underline{\phantom{xx}}}x + 4 \times \left( { - 1} \right) = 0\\x =  + 4\end{aligned}\)

Step 6: Determine the oxidation states of the elements in part (f)

Consider the oxidation number of sodium (+1) in \(NaH\)(sodium hydride).

Hence the oxidation state of hydrogen (\(x\)) is -1 in sodium hydride.

\(\begin{aligned}{\underline{\phantom{xx}}}x + 1 \times 1 = 0\\x =  - 1\end{aligned}\)