Problem 16 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

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 [Adv. 2017] (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) \(\mathrm{H}_{3} \mathrm{PO}_{2}>\mathrm{H}_{3} \mathrm{PO}_{3}>\mathrm{H}_{A} \mathrm{P}_{2} \mathrm{O}_{6}>\mathrm{H}_{2} \mathrm{PO}_{4}\)

Step-by-Step Solution

Answer

(c) \( \mathrm{H}_{3} \mathrm{PO}_{4} > \mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6} > \mathrm{H}_{3} \mathrm{PO}_{3} > \mathrm{H}_{3} \mathrm{PO}_{2} \).

Step 1: Calculate Oxidation State of Phosphorus in H3PO2

Apply the formula for oxidation state calculation: Let the oxidation state of phosphorus be \( x \). In \( \mathrm{H}_{3}\mathrm{PO}_{2} \), the equation according to oxidation rules is: \( 3(+1) + x + 2(-2) = 0 \). Solve for \( x \): \( 3 + x - 4 = 0 \), which gives \( x = +1 \).

Step 2: Calculate Oxidation State of Phosphorus in H3PO4

Using a similar method, let the oxidation state of phosphorus be \( x \). The equation for \( \mathrm{H}_{3}\mathrm{PO}_{4} \) is: \( 3(+1) + x + 4(-2) = 0 \). Solve for \( x \): \( 3 + x - 8 = 0 \), which gives \( x = +5 \).

Step 3: Calculate Oxidation State of Phosphorus in H3PO3

For \( \mathrm{H}_{3}\mathrm{PO}_{3} \), again apply the oxidation state calculation: \( 3(+1) + x + 3(-2) = 0 \). Solve for \( x \): \( 3 + x - 6 = 0 \), resulting in \( x = +3 \).

Step 4: Calculate Oxidation State of Phosphorus in H4P2O6

For \( \mathrm{H}_{4}\mathrm{P}_{2}\mathrm{O}_{6} \), let \( x \) be the oxidation state for phosphorus and write the equation: \( 4(+1) + 2x + 6(-2) = 0 \). Simplify and solve for \( x \): \( 4 + 2x - 12 = 0 \), resulting in \( 2x = +8 \), thus \( x = +4 \).

Step 5: Determine the Order of Oxidation States

From calculations, the oxidation states are: \( \mathrm{H}_{3} \mathrm{PO}_{2} = +1 \), \( \mathrm{H}_{3} \mathrm{PO}_{3} = +3 \), \( \mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6} = +4 \), \( \mathrm{H}_{3} \mathrm{PO}_{4} = +5 \). So, order them from highest to lowest: \( \mathrm{H}_{3}\mathrm{PO}_{4} > \mathrm{H}_{4} \mathrm{P}_{2}\mathrm{O}_{6} > \mathrm{H}_{3}\mathrm{PO}_{3} > \mathrm{H}_{3}\mathrm{PO}_{2} \).

Key Concepts

Phosphorus Compounds

Phosphorus compounds are essential in both chemistry and biology. These compounds typically include phosphorus bonded with oxygen, hydrogen, or other elements. The different types of phosphorus compounds include phosphoric acids, which vary in the number of oxygen atoms attached to phosphorus as well as the overall structure of the molecules. For instance:

• Hypophosphorous Acid (\( \mathrm{H}_{3}\mathrm{PO}_{2}\)): This compound contains one oxygen atom bonded to phosphorus.
• Phosphorous Acid (\( \mathrm{H}_{3}\mathrm{PO}_{3}\)): Here, phosphorus is bonded with three oxygen atoms.
• Phosphoric Acid (\( \mathrm{H}_{3}\mathrm{PO}_{4}\)): This is widely known and involves four oxygen atoms around the phosphorus.
• Pyrophosphoric Acid (\( \mathrm{H}_{4}\mathrm{P}_{2}\mathrm{O}_{6}\)): Comprises two phosphorus and six oxygen atoms leading to more complex interactions.

Understanding these compounds is crucial as they often play significant roles in chemical processes, from energy transfer in the form of ATP in biological systems to industrial applications.

Oxidation Number Calculation

Calculating oxidation numbers is vital in understanding chemical compounds and reactions. It helps us determine the electron transfer between atoms within a molecule. To calculate the oxidation state of phosphorus, follow these basic rules in balancing the oxidation equation:

• Assign an oxidation state of +1 for hydrogen and -2 for oxygen.
• Assume the total charge of the entire compound is zero for neutral compounds.
• Use algebraic equations to solve for the unknown oxidation state.
  • For \( \mathrm{H}_{3}\mathrm{PO}_{2}\): Calculate \( 3(+1) + x + 2(-2) = 0\) to find \( x = +1\).
  • For \( \mathrm{H}_{3}\mathrm{PO}_{4}\): Calculate \( 3(+1) + x + 4(-2) = 0 \) to find \( x = +5\).
  • For \( \mathrm{H}_{3}\mathrm{PO}_{3}\): Calculate \( 3(+1) + x + 3(-2) = 0 \) to find \( x = +3\).
  • For \( \mathrm{H}_{4}\mathrm{P}_{2}\mathrm{O}_{6}\): Calculate \( 4(+1) + 2x + 6(-2) = 0\) to find \( x = +4\).

Applying these calculations to phosphorus compounds enables chemists to predict and understand the behavior of these elements within various reactions.

Chemical Reactions

Chemical reactions often involve changes in oxidation states. The oxidation state helps us identify how electrons are transferred during such reactions. In many reactions, especially redox reactions, understanding the oxidation states can clarify which elements are oxidized and which are reduced. Phosphorus compounds participate in many important reactions:

• In biological systems, phosphorus plays a critical role in energy transfer through the utilization of ATP, an energy-carrying molecule with phosphate groups.
• Phosphate ions from phosphoric acid reactions coordinate with metals in solutions to form metal phosphates, which are significant in material science.
• In agriculture, phosphates derived from phosphorus compounds are key components of fertilizers.

Interpreting the oxidation states and chemical reactivity of phosphorus compounds allows scientists and engineers to harness these reactions for practical uses in various industries.