Problem 152 - A Complete Resource Book in Chemistry for JEE Main

Question

In which of the following compounds the oxidation state of oxygen is other than \(-2 ?\) (a) \(\mathrm{H}_{2} \mathrm{O}_{2}\) (b) \(\mathrm{O}_{2}\) (c) \(\mathrm{O}_{2} \mathrm{~F}_{2}\) (d) \(\mathrm{H}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

Compounds with unusual oxygen states are \(\mathrm{H}_2\mathrm{O}_2\) and \(\mathrm{O}_2\mathrm{F}_2\).

Step 1: Calculating Oxygen's Oxidation State in H2O2

In hydrogen peroxide, \(\mathrm{H}_2\mathrm{O}_2\), each hydrogen has an oxidation state of +1. Let \(x\) be the oxidation state of the oxygen atom. The overall charge of the compound is 0, so \(2(+1) + 2x = 0\). This simplifies to \(2 + 2x = 0\), giving \(x = -1\).

Step 2: Calculating Oxygen's Oxidation State in O2

In molecular oxygen, \(\mathrm{O}_2\), it is in its elemental form. The oxidation state of an element in its natural form is 0, so the oxidation state of oxygen here is 0.

Step 3: Calculating Oxygen's Oxidation State in O2F2

In dioxygen difluoride, \(\mathrm{O}_2\mathrm{~F}_2\), fluorine always has an oxidation state of -1. Let \(y\) be the oxidation state of oxygen. The compound is neutral, so \(2y + 2(-1) = 0\). This simplifies to \(2y - 2 = 0\), giving \(y = +1\).

Step 4: Calculating Oxygen's Oxidation State in H2O

In water, \(\mathrm{H}_2\mathrm{O}\), each hydrogen has an oxidation state of +1. Let \(z\) be the oxidation state of the oxygen atom. The overall charge of the compound is 0, so \(2(+1) + z = 0\). This simplifies to \(2 + z = 0\), giving \(z = -2\).

Step 5: Identifying Compounds with Unusual Oxygen Oxidation States

Comparing all calculations, the oxidation states of oxygen are: \(\mathrm{H}_2\mathrm{O}_2: -1\), \(\mathrm{O}_2: 0\), \(\mathrm{O}_2\mathrm{~F}_2: +1\), and \(\mathrm{H}_2\mathrm{O}: -2\). The oxidation states of oxygen in \(\mathrm{H}_2\mathrm{O}_2\) and \(\mathrm{O}_2\mathrm{~F}_2\) are different from the usual -2 state.

Key Concepts

oxidation state of oxygen

The oxidation state, or oxidation number, is a concept in chemistry that represents the total number of electrons an atom gains or loses to form a chemical bond.
Oxygen, when part of a compound, typically exhibits an oxidation state of \(-2\). For example, in water \(\mathrm{H}_{2} \mathrm{O}\), oxygen has an oxidation state of \(-2\), balancing the \(+1\) oxidation state of the hydrogen atoms.
This usual \(-2\) state reflects oxygen's strong electronegativity, allowing it to attract electrons towards itself.

• In \(\mathrm{H}_{2} \mathrm{O}_{2}\) (hydrogen peroxide), however, oxygen takes on an unusual oxidation state of \(-1\). Each hydrogen atom still contributes \(+1\), but overall balance leads to \(-1\) for oxygen.


• In \(\mathrm{O}_{2} \mathrm{~F}_{2}\) (dioxygen difluoride), oxygen exhibits an even rarer oxidation state of \(+1\). This occurs because fluorine, being more electronegative than oxygen, pulls electron density towards itself with its \(-1\) state.

Understanding oxygen's oxidation states is crucial for predicting molecule behavior and interactions.

unusual oxidation states

Unusual oxidation states are those that differ from an atom's more common oxidation states in specific compounds. For oxygen, the typical oxidation state is \(-2\).
When oxygen is found in compounds like \(\mathrm{H}_{2}\mathrm{O}_{2}\) and \(\mathrm{O}_{2} \mathrm{~F}_{2}\), it adopts oxidation states that are out of the ordinary.

• In hydrogen peroxide, the presence of a single bond between oxygen atoms is a key reason for oxygen's \(-1\) state, as electrons are not fully transferred.


• Dioxygen difluoride's aggressive electronegativity from fluorine forces oxygen into a positive oxidation state, which is rare for a typical non-metallic element like oxygen.

Unusual oxidation states can result from such highly electronegative interactions, electron-sharing dynamics, and overall molecular geometry.
These unique states challenge typical expectations and help chemists understand more about molecular structures and reactivity.

oxidation states in compounds

In any compound, the oxidation states must add up to match the overall charge of the compound.
This balance allows chemists to determine oxidation states for various elements within complex molecules.

• When analyzing a compound such as \(\mathrm{O}_{2}\), an element in its natural form, we assume an oxidation state of \(0\) because atoms in a pure element share electrons equally:
• In \(\mathrm{H}_{2}\mathrm{O}\), which is neutral, we know that the \(+1\) from each hydrogen atom requires the oxygen to balance the total charge resulting in \(-2\) for oxygen.
• In \(\mathrm{O}_{2} \mathrm{~F}_{2}\), balancing the \(-1\) oxidation state of fluorine demands a \(+1\) state for each oxygen atom.

By applying these fundamental rules, chemists solve complex problems with greater ease.
This balance helps us understand molecular interaction and predict behavior, leading to new discoveries.

chemistry problem solving

Solving chemistry problems involves understanding key concepts, like oxidations states, and applying this understanding in a systematic way.
Systematic problem-solving in chemistry often begins by identifying the total charge and recognizing known oxidation states of elements within a compound.

• By setting up equations based on these knowns, like establishing that the sum of the oxidation states equals the charge of the molecule, we can solve for unknowns.


• For instance, if you're given the compound \(\mathrm{H}_{2}\mathrm{O}_{2}\), knowing hydrogen is typically \(+1\), you can set up an equation to find that each oxygen is \(-1\).

Some problems require acknowledging elements with fixed oxidation nature, like fluorine, which is always \(-1\), and adapting to unusual scenarios like the \(+1\) of oxygen in \(\mathrm{O}_{2}\mathrm{F}_{2}\).
Mastering these skills can lead to success in chemistry, where sometimes even minor mistakes can lead to large errors.