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B
False
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Solution
The correct option is B
False
1s orbital + 1p orbital ⇒ 2 sp hybrid orbital 1s orbital + 2p orbital ⇒ 3 sp2 hybrid orbital 1s orbital + 3p orbital ⇒ 4 sp3 orbital (hybrid) So, whenever we are getting sp2 hybrid orbitals, that means we are getting three sp2 hybrid orbitals. s - charcter = Numberofs−orbitalsinvolvedinhybridizationNumberofhybridorbitals s - charcter = 13 p - charcter = 23