Number of hybridised orbitals present in N(SiH3)3 is x and number of pure orbitals in N(SiH3)3 is y then xy is ______

Nitrogen undergo sp2 hybridisation Silicon undergo sp3 hybridisation Hybridised orbitals = 15 Pure orbitals = 10

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Number of hybridised orbitals present in N(SiH₃)₃is x and number of pure orbitals in N(SiH₃)₃ is y then $\frac{x}{y}$ is ______

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answer is 1.5.

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Nitrogen undergo sp² hybridisation
Silicon undergo sp³ hybridisation
Hybridised orbitals = 15
Pure orbitals = 10

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