Problem 36 - Chemistry in 30 days

Question

What is the oxidation number of sulphur in \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\) ? (a) \(2 / 3\) (b) \(3 / 2\) (c) \(3 / 5\) (d) \(5 / 2\)

Step-by-Step Solution

Answer

The oxidation number of sulphur in \( \mathrm{Na}_2 \mathrm{S}_4 \mathrm{O}_6 \) is \( \frac{5}{2} \), corresponding to option (d).

Step 1: Understand the Compound

The formula given is \( \mathrm{Na}_2 \mathrm{S}_4 \mathrm{O}_6 \). This is sodium tetrathionate. It consists of sodium (Na), sulphur (S), and oxygen (O).

Step 2: Assign Known Oxidation States

In compounds, sodium \( \mathrm{Na} \) typically has an oxidation state of +1, and oxygen \( \mathrm{O} \) has an oxidation state of -2.

Step 3: Set Up the Oxidation State Equation

We have 2 sodium atoms, giving a total charge of \(2 \times (+1) = +2\), and 6 oxygen atoms contributing \(6 \times (-2) = -12\). Let \(x\) be the oxidation number of sulphur. The chemical is neutral, so we have:\[ 2(+1) + 4(x) + 6(-2) = 0 \]

Step 4: Simplify the Equation

Solve the equation from the previous step:\[ 2 + 4x - 12 = 0 \]\[ 4x - 10 = 0 \]\[ 4x = 10 \]

Step 5: Solve for Sulphur's Oxidation Number

Divide both sides by 4 to solve for \(x\):\[ x = \frac{10}{4} = \frac{5}{2} \]

Step 6: Compare with Given Options

The oxidation state we found for sulphur is \( \frac{5}{2} \), which matches option (d).

Key Concepts

Sodium Tetrathionate

Sodium tetrathionate, denoted as \( \mathrm{Na}_2 \mathrm{S}_4 \mathrm{O}_6 \), is a fascinating compound made up of sodium, sulphur, and oxygen. This compound is a part of thionates, which are defined by their sulphur-oxygen linkages. It is crucial in understanding various redox reactions and often serves as a reference in the study of sulphur compounds. Sodium tetrathionate consists of:

• Two sodium ions \( \mathrm{Na}^+ \)
• Four sulphur atoms
• Six oxygen atoms

These elements combine in such a way that the overall compound remains electrically neutral. Sodium, being a metal, typically takes the role of balancing the charge contributed by the non-metals, sulphur and oxygen. Sodium tetrathionate is utilized in the chemical industry and academia for studying thiosulphate oxidation.

Sulphur Oxidation State

The oxidation state of an element within a compound reveals the number of electrons it has gained, lost, or shared. In sodium tetrathionate, the oxidation state of sulphur plays a crucial role in determining the structure and properties of the compound. Calculating the sulphur oxidation state involves understanding the typical oxidation states of the other elements.

• Sodium always has an oxidation state of \(+1\).
• Oxygen typically has an oxidation state of \(-2\).

For sulphur, which can exhibit multiple oxidation states, careful calculation is required, as it can fluctuate significantly from one compound to another. In \( \mathrm{Na}_2 \mathrm{S}_4 \mathrm{O}_6 \), we find the average oxidation number to be \( \frac{5}{2} \), indicating a fractional oxidation state, a common occurrence in complex compounds involving multiple sulphur atoms linked in series.

Chemical Neutrality

Chemical neutrality in a compound indicates that the overall charge of the compound is zero. This is a fundamental principle in chemistry, dictating that the sum of oxidation numbers in a stable compound must equal zero. Sodium tetrathionate, \( \mathrm{Na}_2 \mathrm{S}_4 \mathrm{O}_6 \), is a clear example.The neutrality is maintained by carefully balancing the total positive and negative charges: - The contribution from two sodium atoms is \(+2\).- The contribution from six oxygen atoms is \(-12\).- The sulphur atoms' contribution must therefore balance the remaining charge to zero.By setting up the oxidation state equation, chemists can solve for unknown oxidation states, maintaining the principle of neutrality and ensuring that the compound is stable and well-defined.

Oxidation State Equation

The oxidation state equation is a valuable tool for determining unknown oxidation numbers in complex compounds. In sodium tetrathionate, setting up the equation involves accounting for known oxidation states and applying the rule of chemical neutrality.In our compound \( \mathrm{Na}_2 \mathrm{S}_4 \mathrm{O}_6 \), the equation considers:- Sodium's known positive charge \(+2\).- Oxygen's charge contribution of \(-12\).Using the formula: \[ 2(+1) + 4(x) + 6(-2) = 0 \]This equation helps classify the oxidation state \(x\) of sulphur, allowing us to resolve:\[ 4x = 10 \]\[ x = \frac{5}{2} \]This fractional oxidation state not only represents the average oxidation state per sulphur atom in the compound, but also highlights the versatility of sulfur in forming varying linkages and bonds, which are significant in sulphur chemistry. This equation is instrumental for students and chemists in understanding and verifying the internal balances within compounds.