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) \(+\frac{1}{2}\) (d) \(+1\)

Step-by-Step Solution

Answer

The oxidation state of oxygen in \( \mathrm{O}_2 \mathrm{PtF}_6 \) is \( +\frac{1}{2} \) (Option c).

Step 1: Understand the Compound

The compound in question is dioxygenyl hexafluoroplatinate, denoted as \( \mathrm{O}_{2} \mathrm{PtF}_{6} \). This is a complex ion where \( \mathrm{O}_{2}^+ \) is the cation and \( \mathrm{PtF}_{6}^{-} \) is the anion.

Step 2: Assign Oxidation State to Fluorine

In \( \mathrm{PtF}_{6}^{-} \), fluorine, being highly electronegative, always has an oxidation state of \(-1\). Since there are six fluorine atoms, the total oxidation state contributed by fluorine is \(-6\).

Step 3: Calculate Oxidation State of Platinum

As \( \mathrm{PtF}_{6}^{-} \) is a polyatomic ion with an overall charge of \(-1\), the oxidation state of platinum in \( \mathrm{PtF}_{6} \) must be \(+5\) to balance the \(-6\) contributed by fluorine, meeting the \(-1\) overall charge of the ion.

Step 4: Evaluate the Cation 50_251^+

The cation \( \mathrm{O}_2^+ \) has an overall positive charge of \(+1\). Let the oxidation state of each oxygen atom be \( x \). The equation \( 2x = +1 \) arises from the need to balance the charge, leading to \( x = +\frac{1}{2} \).

Step 5: Identify Oxidation State of Oxygen in Compound

The oxidation state of oxygen in \( \mathrm{O}_2 \mathrm{PtF}_6 \), specifically in the \( \mathrm{O}_2^+ \) ion, is determined to be \(+\frac{1}{2}\) per oxygen atom as calculated.

Key Concepts

Oxidation State

Oxidation state is an essential concept in chemistry used to describe the degree of oxidation of an atom in a chemical compound.
It tells us how many electrons an atom has gained, lost, or shared. The oxidation state can be either positive, negative, or zero.
Understanding oxidation states helps in analyzing chemical reactions, especially redox reactions. It's important to remember that:

• The oxidation state of a pure element is always zero.
• In compounds, hydrogen is usually +1, and oxygen is usually -2.
• In ions, the sum of oxidation states is equal to the ion's charge.

Determining the oxidation state involves using these rules to understand electron transfer in chemical reactions. By breaking down complex compounds into known rules, we can assign correct oxidation numbers to help visualize changes during reactions.

Diatomic Molecule

A diatomic molecule consists of two atoms bonded together. It can be homonuclear, like O_2 or N_2, where both atoms are of the same element, or heteronuclear, where they are different.
Diatomic molecules are a key concept in chemical bonding, showcasing how atoms achieve stability through sharing electrons to form covalent bonds.In the context of the compound O_2PtF_6, let's focus on the O_2^+ ion:

• It resembles the typical diatomic molecule O_2 but exists in an oxidized form.
• Each oxygen atom shares an overall positive charge due to electron removal, resulting in an oxidation state of +\(\frac{1}{2}\) per oxygen atom.

Diatomic species like O_2 often participate actively in redox chemistry, especially when they turn into charged particles like O_2^+ due to their high reactivity and electronegativity.

Oxidation Number Calculation

Calculating the oxidation number allows us to determine how electrons are distributed in different atoms within a compound. Let's look at how to calculate the oxidation number in a compound step-by-step:1. **Identify Known Oxidation States:** Start with elements like fluorine (always -1 in compounds) and metals which often have set oxidation states.2. **Overall Charge Balance:** In polyatomic ions like PtF_6^-, use the charge balance method to deduce the unknown oxidation states. The sum of oxidation states equals the ion's charge.3. **Apply Rules for Oxygen and Others:** Generally, oxygen is -2, but in O_2^+, it is +\(\frac{1}{2}\) since the sum for O_2^+ is +1.This step-by-step approach ensures accuracy when calculating oxidation numbers, which is crucial when predicting and balancing chemical equations.

Transition Metal Complex

Transition metal complexes are fascinating structures that showcase the ability of transition metals to form various bonds. These complexes often consist of a central metal atom bonded to surrounding ligands, like Pt(F_6) in the case of O_2PtF_6. **Key points:**

• Platinum here acts as the central transition metal, forming a complex with six fluorine ligands.
• The oxidation state of platinum is critical to the overall charge of this complex. In PtF_6^-, platinum has an oxidation state of +5.
• Transition metals exhibit a range of oxidation states, essential in catalytic processes, and these complexes can undergo structural and electronic changes in reactions.

By understanding these complexes, students also gain insight into advanced topics like coordination chemistry, bridging molecular and macroscopic properties.

Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions involve the transfer of electrons between two species. These reactions are fundamental to energy exchange processes in both biological and chemical systems. Basics of Redox Reactions:

• "Oxidation" refers to the loss of electrons, leading to an increase in oxidation state.
• "Reduction" is the gain of electrons, which decreases the oxidation state.
• During the reaction, one species is oxidized, and another is reduced.

In the compound O_2PtF_6, understanding the redox balance is key. O_2^+ demonstrates how diatomic molecules like oxygen undergo oxidation, gaining a positive charge. Meanwhile, the PtF_6^- part ensures charge balance through reduction, holding the interplay necessary for forming such complexes. Recognizing these exchanges is vital to understanding the complex behavior of transition metals and their role in catalysis and industrial processes.