Problem 11 - A Complete Resource Book in Chemistry for JEE Main

Question

The oxidation state of oxygen in \(\mathrm{O}_{2} \mathrm{PtF}_{6}\) is (a) zero (b) \(-1 / 2\) (c) \(+1 / 2\) (d) \(+1\)

Step-by-Step Solution

Answer

The oxidation state of oxygen in \( O_2PtF_6 \) is (b) \(-1/2\).

Step 1: Understand the Compound

The compound \( O_2PtF_6 \) consists of \( O_2 \), which is a molecule with two oxygen atoms, bonded to \( PtF_6 \), a complex containing \( Pt \) and fluorine atoms. To determine the oxidation state of oxygen in this compound, we need to focus on the overall charges and known oxidation states in \( PtF_6 \).

Step 2: Assign Known Oxidation States

Fluorine, being a highly electronegative element, always has an oxidation state of \(-1\). In \( PtF_6 \), there are six fluorines, contributing a total of \(-6\) to the oxidation state. This implies that \( Pt \) in \( PtF_6 \) must have an oxidation state of \(+6\) to balance the charges, as \( PtF_6 \) is a neutral compound.

Step 3: Calculate Oxidation State of \( O_2 \)

Knowing that \( PtF_6 \) is in the \(+6\) oxidation state and that the entire \( O_2PtF_6 \) compound is neutral, the \( O_2 \) molecule needs to balance the \(+6\) from \( PtF_6 \). This suggests that the net oxidation state for the two oxygen atoms together needs to be \(-1\) to ensure the compound remains neutral. Therefore, the oxidation state of each oxygen atom in \( O_2 \) is \(-\frac{1}{2}\).

Step 4: Choose the Correct Answer

From the calculation, each oxygen atom in \( O_2 \) has an oxidation state of \(-\frac{1}{2}\). So the correct option is (b) \(-1/2\).

Key Concepts

Oxidation States in Compounds

The concept of oxidation states in compounds is crucial for understanding the way atoms are bonded and how they participate in chemical reactions. Oxidation states help to track the number of electrons an atom gains, loses, or shares when it forms a compound. They are useful for:

• Predicting the results of redox (reduction-oxidation) reactions.
• Clarifying how electrons are transferred between atoms.
• Determining the overall charge balance of a chemical compound.

Typically, the oxidation state of an atom in a molecule or ion reflects its ability to withdraw or donate electrons. Every atom in a compound has an oxidation state which can be determined based on established rules, such as the oxidation state of fluorine is always (-1) and oxygen is typically (-2).
In compounds, the sum of the oxidation states must equal the overall charge of the compound. For neutral compounds, like the one we're considering, the total oxidation states must sum to zero. Becoming accustomed to assigning these states makes it easier to solve complex reactions and understand the formation of compounds.

Oxygen Oxidation State

The oxidation state of oxygen is a divergent concept when compared to other elements. Generally, oxygen bears an oxidation state of (-2) in most of its compounds. This is because oxygen has a high electronegativity and usually pulls electrons towards itself. However, there are notable exceptions. For example:

• In peroxides, like (H_2O_2), each oxygen has an oxidation state of (-1).
• In superoxides, each oxygen carries an oxidation state of (-rac{1}{2}).
• In compounds with fluorine, like (OF_2), oxygen's oxidation state can be (+2).

These exceptions occur due to the presence of elements that exhibit higher electronegativity or specific bonding arrangements.
A critical takeaway is that the oxidation state in less typical compounds, like (O_2PtF_6), must be calculated based on known oxidation states of other atoms, ensuring charge neutrality is maintained.

Transition Metal Complexes

Transition metal complexes are fascinating due to their ability to form complex structures with varied oxidation states. These complexes involve a central metal atom, often a transition metal, bonded to surrounding ligands (such as halogens or organic molecules). Key characteristics of transition metal complexes include:

• The metal center, which can exhibit multiple oxidation states. This is due to the involvement of d-orbitals in forming bonds.
• Ligands, which can influence the oxidation state of the metal and the arrangement of electrons.
• Colorful appearances resulting from electron transitions between energy levels.

For instance, in (PtF_6), platinum exhibits a (+6) oxidation state. This high oxidation state balances the highly negative state of the fluorines.
Transition metals can often stabilize unusual oxidation states in neighboring atoms, such as oxygen in the compound (O_2PtF_6). Their unique electron configurations allow for the diverse interactions that stabilize such compounds. Understanding these principles is essential for predicting and explaining the behavior of these complex compounds in chemical reactions.