Question

The order of the oxidation state of the phosphorus atom in \(\mathrm{H}_{3} \mathrm{PO}_{2}, \mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{3} \mathrm{PO}_{3}\), and \(\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}\) is [A] \(\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{3} \mathrm{PO}_{2}>\mathrm{H}_{3} \mathrm{PO}_{4}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}\) [B] \(\mathrm{H}_{3} \mathrm{PO}_{4}>\mathrm{H}_{3} \mathrm{PO}_{2}>\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}\) [C] \(\mathrm{H}_{3} \mathrm{PO}_{4}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}>\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{3} \mathrm{PO}_{2}\) [D] \(\mathrm{H}_{3} \mathrm{PO}_{2}>\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}>\mathrm{H}_{3} \mathrm{PO}_{4}\)

Short answer

\(\mathrm{H}_{3}\mathrm{PO}_{4}>(+5) > \mathrm{H}_{4}\mathrm{P}_{2}\mathrm{O}_{6}(+4) > \mathrm{H}_{3}\mathrm{PO}_{3}(+3) >\mathrm{H}_{3}\mathrm{PO}_{2}(+1)\)

Step-by-step solution

Understanding the Concept of Oxidation States

Identify the fact that oxidation state is the charge that an atom would have if all bonds to atoms of different elements were completely ionic. Hydrogen typically has an oxidation state of +1, and oxygen has an oxidation state of -2.

Calculating the Oxidation State of Phosphorus in \(\mathrm{H}_{3}\mathrm{PO}_{2}\)

For the compound \(\mathrm{H}_{3}\mathrm{PO}_{2}\), assuming hydrogen has an oxidation state of +1 and oxygen has an oxidation state of -2, and letting the oxidation state of phosphorus be x, the equation is:\(3(+1) + x + 2(-2) = 0\). Solving for x gives the oxidation state of phosphorus as +1.

Calculating the Oxidation State of Phosphorus in \(\mathrm{H}_{3}\mathrm{PO}_{3}\)

For \(\mathrm{H}_{3}\mathrm{PO}_{3}\), the equation is \(3(+1) + x + 3(-2) = 0\). Solving for x, the oxidation state of phosphorus is +3.

Calculating the Oxidation State of Phosphorus in \(\mathrm{H}_{3}\mathrm{PO}_{4}\)

For \(\mathrm{H}_{3}\mathrm{PO}_{4}\), the equation is \(3(+1) + x + 4(-2) = 0\). Solving for x, the oxidation state of phosphorus is +5.

Calculating the Oxidation State of Phosphorus in \(\mathrm{H}_{4}\mathrm{P}_{2}\mathrm{O}_{6}\)

For \(\mathrm{H}_{4}\mathrm{P}_{2}\mathrm{O}_{6}\), there are two phosphorus atoms, so the equation is\(4(+1) + 2x + 6(-2) = 0\). Solving for x, the oxidation state of each phosphorus is +4.

Determining the Order of Oxidation States

List the oxidation states of phosphorus in the given compounds. Compare them to determine the order: \(\mathrm{H}_{3}\mathrm{PO}_{4}\) has +5, \(\mathrm{H}_{4}\mathrm{P}_{2}\mathrm{O}_{6}\) has +4, \(\mathrm{H}_{3}\mathrm{PO}_{3}\) has +3, and \(\mathrm{H}_{3}\mathrm{PO}_{2}\) has +1. Arrange them from highest to lowest to get the order.

Key concepts

Oxidation State Calculation

Understanding how to calculate oxidation states is crucial for mastering chemistry, especially when it comes to tackling redox reactions. In compounds, the oxidation state represents the hypothetical charge an atom would have if all bonds were ionic. Take phosphorus compounds as an example — by following a simple set of rules, one can determine the oxidation state of phosphorus.

Firstly, we assign known oxidation states to elements with standard values: hydrogen, which usually has an oxidation state of +1, except in metal hydrides where it's -1, and oxygen, which typically has an oxidation state of -2, except in peroxides and F2O. Phosphorus, being more electronegative than hydrogen but less than oxygen, will accept electrons from hydrogen and donate to oxygen in these compounds.

When calculating the oxidation state for phosphorus (\(x\)), set up an algebraic equation based on the sum of oxidation states in a neutral compound equaling zero. For instance, in \(\mathrm{H}_{3}\mathrm{PO}_{2}\), hydrogen contributes +3 (\(3\times+1\)), and oxygen contributes -4 (\(2\times-2\)). Balancing the equation, phosphorus's oxidation state (x) is found to equal +1 as in \(3(+1) + x + 2(-2) = 0\).

Similarly, carry out this process for other phosphorus compounds. Upon solving, one can rank the oxidation states to understand the compound's behavior in redox reactions, which is particularly useful in advanced chemistry exams like JEE Advanced.

JEE Advanced Chemistry

Competitive exams like the JEE Advanced place a great emphasis on a candidate's ability to understand and apply chemical concepts, such as oxidation states. In these exams, a deep knowledge of oxidation state trends and the ability to quickly calculate them in complex molecules is indispensable.

Oxidation states not only explain the electronic distribution but also provide insights into the possible reactions that the compound can undergo. For JEE Advanced chemistry, one must be comfortable with several types of compounds and their properties. The question on the order of oxidation states of phosphorus in different compounds, like in the provided example, tests the conceptual underpinning and the problem-solving skills of the students.

As illustrated, by knowing how to assign oxidation numbers and use them to predict compound behavior, students can solve such problems efficiently. Building a strong foundation in these calculations will aid students in tackling synthesis, analysis, and redox reaction questions, which are frequently encountered in JEE Advanced chemistry.

Redox Reactions

Redox reactions are chemical processes involving the transfer of electrons between two species. The concepts of oxidation state are integral to understanding redox processes. Recognizing the change in oxidation states can help identify the oxidizing and reducing agents in a reaction.

In the context of phosphorus compounds, knowing the oxidation states can indicate how these compounds may act in redox reactions. A higher oxidation state suggests that the element is more likely to be reduced (gain electrons), while a lower state suggests a tendency to be oxidized (lose electrons).

For example, \(\mathrm{H}_{3}\mathrm{PO}_{4}\), with phosphorus at a +5 oxidation state, is more likely to act as an oxidizing agent as compared to \(\mathrm{H}_{3}\mathrm{PO}_{2}\) where phosphorus has a +1 oxidation state. Teaching students to recognize these patterns in oxidation states not only helps in predicting the role of compounds in redox reactions but also serves as a fundamental skill in the synthesis and analysis of chemical reactions, which is especially relevant for chemistry students preparing for exams like the JEE Advanced.