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← JEE Advanced 2018 · Paper 2
ChemistryNumerical+3Question 25 of 54

JEE Advanced 2018 · Paper 2, Chemistry question 7

The total number of compounds having at least one bridging oxo group among the molecules given below is ____.

N2O3\mathrm{N_2O_3}, N2O5\mathrm{N_2O_5}, P4O6\mathrm{P_4O_6}, P4O7\mathrm{P_4O_7}, H4P2O5\mathrm{H_4P_2O_5}, H5P3O10\mathrm{H_5P_3O_{10}}, H2S2O3\mathrm{H_2S_2O_3}, H2S2O5\mathrm{H_2S_2O_5}

Answer and worked solution

Official answer: 5

Check each structure for an X-O-X bridge:

Molecule Structure Bridging O?
NX2OX3\ce{N2O3} O=N−NOX2\ce{O=N-NO2} (N-N bond) No
NX2OX5\ce{N2O5} OX2N−O−NOX2\ce{O2N-O-NO2} Yes
PX4OX6\ce{P4O6} six P-O-P bridges Yes
PX4OX7\ce{P4O7} six P-O-P bridges Yes
HX4PX2OX5\ce{H4P2O5} HP(O)(OH)−O−P(O)(OH)H\ce{HP(O)(OH)-O-P(O)(OH)H} Yes
HX5PX3OX10\ce{H5P3O10} two P-O-P Yes
HX2SX2OX3\ce{H2S2O3} S=S(O)(OH)X2\ce{S=S(O)(OH)2} No
HX2SX2OX5\ce{H2S2O5} HO−S(O)−S(O)X2−OH\ce{HO-S(O)-S(O)2-OH} (S-S bond) No

Total: 5. The key also accepts 6, which counts NX2OX3\ce{N2O3} in its symmetric form O=N−O−N=O\ce{O=N-O-N=O}.

Answer: 5 (key: 5 or 6)