Indian Forest Service (Mains), Chemistry optional · 2025 · Paper I
Q1(a). Equivalent bonds in nitrate
Question
The observed nitrogen–oxygen bond length in nitrate ion is 1.21 Å. The calculated lengths for N–O and N=O are 1.36 and 1.15 Å, respectively. Account for this with requisite structures.
Answer in brief
Nitrate is a planar, resonance-delocalised ion whose three equivalent N–O bonds have partial double-bond character rather than one permanently localised double bond.
Key points
- Each octet-obeying contributor contains one N=O bond and two N–O single bonds; nitrogen has formal charge +1 and the two singly bonded oxygens −1.
- There are three equivalent contributors, obtained by placing the double bond on each oxygen in turn; their sum represents one electronic structure.
- The simple resonance bond-order average is (2 + 1 + 1)/3 = 4/3 for every N–O bond; delocalisation spreads the negative charge over the oxygens.
- The observed length lies between the single- and double-bond values, consistent with partial double-bond character and equivalent bonds.
- The ion is approximately trigonal planar with O–N–O angles of 120°.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
Coversnitrate ion resonanceresonance structures of NO₃⁻bond order 4/3N–O bond lengthformal charge
All parts of Question 1
- 1(a) Equivalent bonds in nitrate8 marks
- 1(b) Formula from occupied crystal voids8 marks
- 1(c) Formation free energy of uranium hydride8 marks
- 1(d) Pressure dependence of solid Gibbs energy8 marks
- 1(e) Fast and slow electrode systems8 marks