Problem 49 - Chemistry The Central Science In Si Units

Question

Determine the oxidation number for the indicated element in each of the following substances: (a) \(\mathrm{S}\) in \(\mathrm{SO}_{3},\) (b) Ti in \(\mathrm{TiCl}_{4}\) (c) \(\mathrm{P}\) in \(\mathrm{AgPF}_{6}\), (d) \(\mathrm{Nin} \mathrm{HNO}_{3},\) (e) \(\mathrm{S}\) in \(\mathrm{H}_{2} \mathrm{SO}_{3},(\mathbf{f}) \mathrm{O}\) in \(\mathrm{OF}_{2}\).

Step-by-Step Solution

Answer

(a) +6, (b) +4, (c) +5, (d) +5, (e) +4, (f) +2.

Step 1: Oxidation Number Rules

Review the rules for assigning oxidation numbers: (1) The oxidation number of an atom in its elemental form is 0. (2) The sum of oxidation numbers in a neutral compound must be zero. (3) For ions, the sum must match the ion's charge. (4) Specific elements typically have the same oxidation numbers: Oxygen is usually -2, Hydrogen is +1, Fluorine is -1, etc.

Step 2: Oxidation Number of S in SO3

In \( ext{SO}_3\), oxygen is usually \(-2\). There are three oxygen atoms, so the total contribution from oxygen is \(-6\). For the compound to be neutral, sulfur must have an oxidation number that makes the total zero: \( ext{x} + 3(-2) = 0\), solving gives \( ext{x} = +6\).

Step 3: Oxidation Number of Ti in TiCl4

In \( ext{TiCl}_4\), chlorine is usually \(-1\). There are four chlorine atoms, contributing \(-4\) total. Titanium has to balance this out to zero: \( ext{x} + 4(-1) = 0\), solving gives \( ext{x} = +4\).

Step 4: Oxidation Number of P in AgPF6

In \( ext{AgPF}_6\), fluorine is \(-1\) and there are six fluorines, giving \(-6\). Silver ( ext{Ag}) is in the \(+1\) oxidation state for a neutral compound. Hence, \( ext{x} + 6(-1) = -1\), solving gives \( ext{x} = +5\).

Step 5: Oxidation Number of N in HNO3

In \( ext{HNO}_3\), hydrogen is \(+1\), and each oxygen \(-2\) (total \(-6\) for three oxygens). To balance, the oxidation number of nitrogen must satisfy \(+1 + ext{x} + 3(-2) = 0\), giving \( ext{x} = +5\).

Step 6: Oxidation Number of S in H2SO3

In \( ext{H}_2 ext{SO}_3\), hydrogen contributes \(+2\) and oxygen \(-6\) (since it's \(-2\) per atom). For sulfur, \(+2 + ext{x} - 6 = 0\), solving gives \( ext{x} = +4\).

Step 7: Oxidation Number of O in OF2

In \( ext{OF}_2\), even though oxygen is usually \(-2\), fluorine is more electronegative and is \(-1\). Two fluorines contribute \(-2\). Oxygen then must be \(+2\) to balance: \( ext{x} + 2(-1) = 0\), so \( ext{x} = +2\).

Key Concepts

SO3

Sulfur trioxide, written chemically as \( \text{SO}_3 \), is a compound in which the sulfur atom needs to adjust its oxidation state to balance the overall charge contributed by oxygen.
Here, oxygen typically carries an oxidation number of \(-2\). Since \( \text{SO}_3 \) is a neutral molecule, the sum of the oxidation states of all the atoms inside it must equal zero. With three oxygen atoms present, their total contribution to the charge is \(-6\).
To bring the overall charge to zero, sulfur must have an oxidation number of \(+6\).

• Thus, in \( \text{SO}_3 \), sulfur has an oxidation number of \(+6\).

This process highlights the fundamentals of determining oxidation states using known oxidation numbers of elements and balancing charges.

TiCl4

Titanium tetrachloride \( \text{TiCl}_4 \) contains titanium and chlorine atoms. Chlorine, a halogen, generally possesses an oxidation number of \(-1\).
When four chlorine atoms are involved, their total contribution to the compound's charge results in \(-4\). Since \( \text{TiCl}_4 \) is neutral, the oxidation number of titanium must counterbalance this.

• Therefore, the titanium atom has an oxidation number of \(+4\).

This balance maintains the compound's neutrality and demonstrates how specific rules and known oxidation states can determine unknown quantities in chemical structures.

AgPF6

Silver hexafluorophosphate, \( \text{AgPF}_6 \), consists of silver, phosphorus, and fluorine atoms. Fluorine is highly electronegative and uniformly carries an oxidation number of \(-1\).
With six fluorine atoms, the cumulative charge becomes \(-6\). The silver atom in neutral compounds usually holds a \(+1\) oxidation state. In this compound, the sum of charges needs to reflect the overall compound's charge, which in this case is \(-1\) as it forms an anion with phosphorus.

• Solving for phosphorus: \( \text{x} + 6(-1) = -1 \) results in phosphorus having an oxidation number of \(+5\).

This example serves to illustrate how the consideration of each component's typical oxidation state affects the calculation of the unknown element's state.

HNO3

Nitric acid, or \( \text{HNO}_3 \), includes hydrogen, nitrogen, and oxygen. Oxygen usually takes \(-2\), leading the total oxidative contribution from the three oxygen atoms to \(-6\).
Hydrogen, commonly \(+1\), adds a positive charge. Using the rule that the sum must equal zero in neutral molecules, we find nitrogen's oxidation number through simple algebra:

• The equation \( +1 + \text{x} + 3(-2) = 0 \) reveals nitrogen's oxidation number to be \(+5\).

This balance shows how even when multiple elements converge, knowledge of common oxidation states and basic algebra allows determination of specific unknown values.

H2SO3

Sulfurous acid represented by \( \text{H}_2\text{SO}_3 \) contains hydrogen, sulfur, and oxygen. Two hydrogens contribute a total of \(+2\). Oxygen's presence, each \(-2\), gives a cumulative charge of \(-6\) across three atoms.
For the entire neutral molecule, these contributions must balance out.

• This leads to calculating that sulfur has an oxidation number of \(+4\), as derived from: \( +2 + \text{x} - 6 = 0 \).

Illustrating how changes in the number of oxygen or hydrogen in a chemical formula affects other elements' oxidation states provides a deeper understanding of balancing equations.

OF2

In oxygen difluoride (\( \text{OF}_2 \)), an unusual case is observed. Despite oxygen's usual state of \(-2\), fluorine, being more electronegative, has \(-1\).
For two fluorines, that reaches \(-2\) overall, compelling oxygen to step in with \(+2\) to ensure the neutrality of the compound.

• The calculation \( \text{x} + 2(-1) = 0 \) substantiates this unusual assignment where oxygen shows a \(+2\) oxidation state.

This serves as a notable exception, emphasizing that general rules might deviate under specific conditions, exhibiting the variability and adaptability required when balancing chemical equations.