The sum of all sp3,sp2 and sp units present in solid PCl5 , solid PBr5 and solid N2O5 are:
A
4
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B
3
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C
2
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D
5
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Solution
The correct option is A 4 Solid PCl5→[PCl4]+↓sp3[PCl6]−↓sp3d2
Solid PBr5→[PBr4]+↓sp3[Br]−
Solid N2O5→[NO+2]↓sp[NO−3]↓sp2
Hence, the sum of all sp3,sp2 and sp units present in solid PCl5 , solid PBr5 and solid N2O5=1+1+1+1=4