Question

Determine the oxidation number for the indicated element in each of the following substances: (a) \(\sin \mathrm{SO}_{2},(\mathbf{b}) \mathrm{Cin} \mathrm{COCl}_{2}\) , (c) \(\mathrm{Mn}\) in \(\mathrm{KMnO}_{4},(\mathbf{d}) \mathrm{Brin} \mathrm{HBrO},(\mathbf{e}) \mathrm{P}\) in \(\mathrm{PF}_{3},(\mathbf{f}) \mathrm{O}\) in \(\mathrm{K}_{2} \mathrm{O}_{2}\)

Short answer

The oxidation numbers for the indicated elements in the given substances are as follows: (a) \(S\) in \(SO_2\) is +4. (b) \(C\) in \(COCl_2\) is +4. (c) \(Mn\) in \(KMnO_4\) is +7. (d) \(Br\) in \(HBrO\) is +1. (e) \(P\) in \(PF_3\) is +3. (f) \(O\) in \(K_2O_2\) is -1.

Step-by-step solution

(Rule of Oxidation Numbers)

Rules to find the oxidation number are: 1. The oxidation number of an atom in its elemental form is always 0. 2. The oxidation number of a monatomic ion is equal to its charge. 3. The sum of the oxidation numbers of the elements in a compound must equal the overall charge of the compound. 4. Halogens usually have an oxidation number of -1. 5. The oxidation number of oxygen is generally -2, and hydrogen is generally +1. Now we apply these rules to each substance:

(a) Oxidation number of \(S\) in \(SO_2\))

Sulfur is bonded to two oxygen atoms, which typically have an oxidation number of -2 each. As there are no charges in \(SO_2\), the sum of all oxidation numbers should be zero. Let oxidation number of S = x. So, x + 2(-2) = 0 x - 4 = 0 x = 4 The oxidation number of \(S\) in \(SO_2\) is +4.

(b) Oxidation number of \(C\) in \(COCl_2\))

Carbon is bonded to an oxygen atom (-2) and two chlorine atoms (-1 each). Again, the sum of the oxidation numbers must equal zero. Let oxidation number of C = x. So, x - 2 - 1 - 1 = 0 x - 4 = 0 x = 4 The oxidation number of \(C\) in \(COCl_2\) is +4.

(c) Oxidation number of \(Mn\) in \(KMnO_4\))

Manganese is bonded to four oxygen atoms (-2 each) and a potassium atom (+1). The sum of the oxidation numbers equals zero. Let oxidation number of Mn = x. So, x + 4(-2) + 1 = 0 x - 8 + 1 = 0 x - 7 = 0 x = 7 The oxidation number of \(Mn\) in \(KMnO_4\) is +7.

(d) Oxidation number of \(Br\) in \(HBrO\))

Bromine is bonded to one oxygen atom (-2) and one hydrogen atom (+1). The sum of the oxidation numbers equals zero. Let oxidation number of Br = x. So, x - 2 + 1 = 0 x - 1= 0 x = 1 The oxidation number of \(Br\) in \(HBrO\) is +1.

(e) Oxidation number of \(P\) in \(PF_3\))

Phosphorus is bonded to three fluorine atoms (-1 each). The sum of the oxidation numbers must equal zero. Let oxidation number of P = x. So, x - 1 - 1 - 1 = 0 x - 3 = 0 x = 3 The oxidation number of \(P\) in \(PF_3\) is +3.

(f) Oxidation number of \(O\) in \(K_2O_2\))

Oxygen atom is bonded to two potassium atoms (+1 each). The sum of the oxidation numbers must equal zero. Since there are two oxygen atoms, let oxidation number of each O be x. 2x + 2(1) = 0 2x + 2 = 0 2x = -2 x = -1 The oxidation number of \(O\) in \(K_2O_2\) is -1.

Key concepts

SO2

In the compound sulfur dioxide ((SO_2)), we need to determine the oxidation number of sulfur (S). Oxygen generally has an oxidation number of -2, based on standard oxidation rules.

• Sulfur is bonded to two oxygen atoms.
• Since there is no charge on the molecule, the sum of oxidation numbers must equal zero.
• Let the oxidation number of S be \(x\).

Equation to solve: \(x + 2(-2) = 0 \).
Solving this gives \(x - 4 = 0 \), and thus the oxidation number of sulfur is +4.

KMnO4

Potassium permanganate ((KMnO_4)) is a compound where we have to find the oxidation number of manganese (Mn). Oxygen’s oxidation number is usually -2.

• Manganese is bonded to four oxygen atoms and one potassium atom.
• Potassium, being an alkali metal, has an oxidation number of +1.
• The overall charge of KMnO_4 is zero, so the sum of oxidation numbers equals zero.

Let the oxidation number of Mn be \(x\).
Equation: \(x + 4(-2) + 1 = 0\).This solves to \(x - 7 = 0\), giving manganese an oxidation number of +7.

HBrO

In hypobromous acid ((HBrO)), the task is to find the oxidation number of bromine (Br). Using the oxidation rules:

• Oxygen usually has an oxidation number of -2.
• Hydrogen typically has an oxidation number of +1.
• Our equation should equal zero, as HBrO is a neutral compound.
• Let the oxidation number of Br be \(x\).

The equation: \(x - 2 + 1 = 0\).
This simplifies to \(x - 1 = 0\), revealing the oxidation number of bromine to be +1.

PF3

Phosphorus trifluoride ((PF_3)) is a compound where we need to find the oxidation number of phosphorus (P). Fluorine has a consistent oxidation number of -1.

• Phosphorus is bonded to three fluorine atoms.
• The sum of the oxidation numbers in this molecular compound must be zero.
• Let the oxidation number of P be \(x\).

Equation: \(x + 3(-1) = 0\).Solving, we find \(x - 3 = 0\), thus phosphorus in PF_3 has an oxidation number of +3.

Oxidation Rules

The rules of oxidation are key to determining the oxidation numbers of different elements in a chemical formula. Here's a quick rundown of these important rules:

• The oxidation number of an element in its natural state is always 0.
• For a monoatomic ion, the oxidation number is equal to its charge.
• The sum of all oxidation numbers in a compound must equal the net charge on that compound.
• Halogens generally have an oxidation number of -1 unless bonded to oxygen or other halogens.
• Oxygen usually has an oxidation number of -2, except in peroxides like (K_2O_2) where it is -1, and hydrogen is usually +1.

These rules serve as a fundamental tool for balancing chemical equations and understanding redox reactions.