What is the oxidation state of sulfur in the following compounds? (a) \(\mathrm{SF}_{4} ;\) (b) \(\mathrm{S}_{2} \mathrm{F}_{10} ;\) (c) \(\mathrm{H}_{2} \mathrm{S} ;\) (d) \(\mathrm{CaSO}_{3}\).

Short Answer

Expert verified
The oxidation states of sulfur in the given compounds are: (a) +4, (b) +5, (c) -2, and (d) +4.

Step by step solution

01

Determining oxidation states in SF4

Let's denote the oxidation state of sulfur as \(x\). Fluorine usually has an oxidation state of \(-1\) in its compounds. As there are four fluorine atoms, the total charge due to fluorine is \(-4\). The molecule is neutral, so the sum of oxidation states must be equal to zero, thus \(x - 4 = 0\). By solving this equation, we find \(x = +4\).
02

Determining oxidation states in S2F10

Since we have two atoms of sulfur, we will denote their total charge as \(2x\). Same as in step one, fluorine contributes with \(-10\), and the total charge of the molecule is zero. We therefore have \(2x - 10 = 0\). Solving this equation gives \(x = +5\).
03

Determining oxidation states in H2S

In this molecule, hydrogen has its usual oxidation state of \(+1\), contributing with \(+2\). Hence, \(2 + x = 0\), thus \(x = -2\).
04

Determining oxidation states in CaSO3

In this compound, calcium has an oxidation state of \(+2\) and oxygen has an usual oxidation state of \(-2\), contributing with \(-6\) (we have three oxygen atoms in the molecule). The total charge is zero, so we have \(2 + x - 6 = 0\). This gives us the oxidation state of sulfur as \(x = +4\).

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