Consider the following reactions
NiS + HNO₃ + HCl → A + NO + S + H₂O; A + NH₄OH + H₃C-, C, =N-OH; |; H₃C-, C, =N-OH → B + NH₄Cl + H₂O
The number of protons that do not involve in hydrogen bonding in the product B is _________.

Solution

Chemistry ma’am · 4 steps

B is the red nickel dimethylglyoxime complex. Count its hydrogens, then remove the two in the O–H···O bridges.

1. Find A. Aqua regia oxidises the sulphide ion to sulphur and leaves nickel as its chloride: 3NiS + 2HNO₃ + 6HCl → 3NiCl₂ + 2NO + 3S + 4H₂O. So A is NiCl₂.
2. Find B. In ammoniacal solution, Ni²⁺ reacts with two molecules of dimethylglyoxime (dmgH₂). Each ligand loses one O–H proton, giving the square planar red complex Ni(dmgH)₂, which is B.

   NiCl₂ + 2NH₄OH + 2dmgH₂ → Ni(dmgH)₂ + 2NH₄Cl + 2H₂O

3. Count the protons. Each dmgH⁻ is C₄H₇N₂O₂⁻, so B has 2 × 7 = 14 hydrogens. Of these, 2 are the remaining O–H hydrogens. Each one bridges to an O⁻ of the other ligand, forming two intramolecular O–H···O hydrogen bonds.
4. So. The other hydrogens are the methyl hydrogens: 4 CH₃ groups × 3 = 12. C–H hydrogens do not take part in hydrogen bonding, so the answer is 12.

The common slip: 14. Counting every hydrogen in Ni(dmgH)₂ gives 14, but the two O–H hydrogens are in the O–H···O hydrogen bonds and must be left out.

The idea to keep. In Ni(dmgH)₂ each dmg loses one O–H proton. The two O–H that remain hold the ligands together through hydrogen bonds. Only H on N, O or F can hydrogen bond, so C–H hydrogens never count.

Answer from the NTA final answer key. Solution by PrepWiser.