Question #e1818

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Question #e1818

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Answer 1 · 2016-10-02T13:45:28

Summing up the number of $σ$ $sigma$ -bond formed by the desired atom (here N) and the number of lone pair on it we can easily know the hybridization of it.

$If the sum is 2 \to hybridization - s p$ $"If the sum is 2 "->"hybridization"-sp$

$If the sum is 3 \to hybridization - s p^{2}$ $"If the sum is 3 "->"hybridization"-sp^2$

$If the sum is 4 \to hybridization - s p^{3}$ $"If the sum is 4 "->"hybridization"-sp^3$

$If the sum is 5 \to hybridization - s p^{3} d$ $"If the sum is 5 "->"hybridization"-sp^3d$

$If the sum is 6 \to hybridization - s p^{3} d^{2}$ $"If the sum is 6 "->"hybridization"-sp^3d^2$

Inspecting the Lewis structure of $H - C \equiv N :$ $H-CequivN:$ we see that N-atom has one $σ$ $sigma$ -bond and one lone pair so sum is 2.
Hence the hybridization is sp.

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Question #e1818

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