Question

Calculate the oxidation number of sulphur, chromium and nitrogen in \(\mathrm{H}_{2} \mathrm{SO}_{5}\), \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) and \(\mathrm{NO}_{3}^{-}\). Suggest structure of these compounds. Count for the fallacy.

Short answer

Sulphur: +6 in H₂SO₅; Chromium: +6 in Cr₂O₇²⁻; Nitrogen: +5 in NO₃⁻.

Step-by-step solution

Determine Oxidation Numbers of Sulphur in H₂SO₅

To determine the oxidation number of sulphur in \( \mathrm{H}_{2} \mathrm{SO}_{5} \): - Hydrogen (H) has an oxidation number of +1. - Oxygen (O) typically has an oxidation number of -2. - Let the oxidation number of sulphur (S) be \( x \). - The compound is neutral, so the sum of the oxidation numbers must be zero: \(2(+1) + x + 5(-2) = 0\). - Simplifying gives: \(2 + x - 10 = 0\) - Solving for \( x \), we get: \( x = 8\).- Therefore, the oxidation number of sulphur is +8. However, because +8 is improbable for sulfur, re-evaluation suggests non-standard bonding or alternatives, but conventional rules point to possible +6 with peroxide linkage.

Determine Oxidation Numbers of Chromium in Cr₂O₇²⁻

For \(\mathrm{Cr}_2\mathrm{O}_7^{2-} \):- Let the oxidation number of chromium (Cr) be \( y \).- Oxygen has an oxidation number of -2.- The overall charge of the ion is -2.- The equation considering the charge is \( 2y + 7(-2) = -2 \).- Simplifying, \( 2y - 14 = -2 \) which leads to \( 2y = 12 \) and \( y = +6 \).- Therefore, the oxidation number of chromium is +6.

Determine Oxidation Numbers of Nitrogen in NO₃⁻

For \( \mathrm{NO}_{3}^{-} \):- Let the oxidation number of nitrogen (N) be \( z \).- Oxygen has an oxidation number of -2.- The overall charge of the ion is -1.- The equation is \( z + 3(-2) = -1 \).- Simplifying leads to \( z - 6 = -1 \) which gives \( z = +5 \).- Therefore, the oxidation number of nitrogen is +5.

Suggest Structures of Compounds

For \( \mathrm{H}_2\mathrm{SO}_5 \), considering typical oxidation states, it may involve peroxy linkage with sulfur having a +6 oxidation state.For \( \mathrm{Cr}_2\mathrm{O}_7^{2-} \), the structure is a dichromate ion with two chromium atoms each having +6 oxidation states and sharing oxygens.For \( \mathrm{NO}_3^{-} \), it's nitrate ion, where nitrogen has a standard +5 oxidation state with double-bond and single-bonded oxygens resonating.

Key concepts

Sulphur Oxidation States

Sulphur is a versatile element capable of exhibiting various oxidation states in compounds. In the context of the compound \( \mathrm{H}_{2} \mathrm{SO}_{5} \), sulphur can appear to have an oxidation number of +8 by direct calculation. However, this is typically unrealistic due to electron configuration limits. Instead, this high figure indicates a need to consider specialized structural features or unconventional bonding patterns, such as peroxy linkages.

• Sulphur commonly exhibits oxidation states ranging from -2 to +6.
• The +6 state is more common and feasible, especially in sulfate (\( \mathrm{SO}_{4}^{2-} \)) and certain other oxyanions.
• In \( \mathrm{H}_{2} \mathrm{SO}_{5} \), the more realistic structure might incorporate a peroxy bond, influencing the apparent oxidation state through non-traditional bond saturation.

This adaptability in oxidation states allows sulphur to be foundational in chemistry, bridging between typical and atypical oxidation conditions.

Chromium Compounds

Chromium is another fascinating element that displays multiple oxidation states in its compounds. Within the dichromate ion \( \mathrm{Cr}_2 \mathrm{O}_7^{2-} \), each chromium atom achieves an oxidation state of +6. Chromium's ability to attain such a high oxidation state stems from its electron configuration, which allows it to lose multiple electrons readily.

• Chromium typically exhibits oxidation states of +2, +3, and +6.
• In the +6 oxidation state, chromium forms stable compounds like \( \mathrm{CrO}_4^{2-} \) (chromate) and \( \mathrm{Cr}_2 \mathrm{O}_7^{2-} \) (dichromate).
• The dichromate ion's structure is characterized by shared oxygen atoms that contribute to chromium's stability at this higher oxidation level.

This multi-faceted element fascinates chemists due to its varied roles in redox reactions and industrial applications.

Nitrogen Compounds

Nitrogen is an essential element that displays a wide range of oxidation states across its compounds. In the nitrate ion \( \mathrm{NO}_3^{-} \), nitrogen commonly exhibits an oxidation state of +5. This is achieved through the bonding nature within the ion, featuring double and single bonds between nitrogen and oxygen atoms.

• Nitrogen has oxidation states ranging from -3 in ammonia (\( \mathrm{NH}_3 \)) to +5 in nitrates.
• In the nitrate ion, nitrogen forms one double bond and two single bonds with oxygen, resulting in an overall resonance stability.
• The +5 oxidation state of nitrogen reflects its capability to lose five electrons and fill the valence shell with bonding shared electrons.

This versatility in oxidation states enriches nitrogen's role in biological systems, industrial processes, and environmental chemistry.