Consider i HO–S(=O)₂–O–S(=O)₂–OH; ii HO–S(=O)₂–O–O–S(=O)₂–OH; iii HO–S(=O)₂–S(=O)₂–OH; iv HO–S(=O)₂–SH; v cyclic (SO₃)₄ with alternating S/O and two terminal =O per S. Let P count entries where every

Question

Consider i HO–S(=O)₂–O–S(=O)₂–OH; ii HO–S(=O)₂–O–O–S(=O)₂–OH; iii HO–S(=O)₂–S(=O)₂–OH; iv HO–S(=O)₂–SH; v cyclic (SO₃)₄ with alternating S/O and two terminal =O per S. Let P count entries where every S is+6, Q those where every S has total formal bond order6, and R those with S–O–S but no O–O. Find PQR.

Answer:

24

Solution:

P=3 from i, ii, v. Q=4 from i, ii, iii, v: iii has two doubles plus OH and S single bonds at each S, giving 6 despite OS+5. iv fails Q at terminal S. R=2 from i, v. Product3×4×2=24.

TestHub · Accepted Answer

P=3 from i, ii, v. Q=4 from i, ii, iii, v: iii has two doubles plus OH and S single bonds at each S, giving 6 despite OS+5. iv fails Q at terminal S. R=2 from i, v. Product3×4×2=24.

Chemistry - CHEMICAL BONDING Question with Solution | TestHub

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