If the following orbitals are hybridized, then the number of hybrid orbitals will be:
$2 s$ , $2 p_{x}$ , $2 p_{y}$ , $2 p_{z}$

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Solution

The correct option is D
4

Always remember,
Number of hybrid orbitals = Number of orbitals involved in hybridization.
Here, number of orbitals involved = 4
$2 s$ $2 p_{x}$ $2 p_{y}$ $2 p_{z}$
$∴$ Number of hybrid orbitals = 4
4 ${s p}^{3}$ hybrid orbitals

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Similar questions

2 s, 2 p_{x}, 2 p_{y}, 2 p_{z}

2 s, 2 p_{x}, 2 p_{y}, 2 p_{z}

If the following orbital's are hybridized, then the hybridization will be respectively:

(i) One s orbital, two p orbitals (ii) One s orbital, three p orbitals (iii) One s orbital, one p orbital

x -

1 s

2 s

1 s

2 p_{x}

2 p_{y}

2 p_{y}

1 s

2 s

σ (1 s) < σ^{*} (1 s) < σ (2 s) < σ^{*} (2 s) < [π (2 p_{y}) π (2 p_{z})] < σ (2 p_{x}) < [π^{*} (2 p_{y}) π^{*} (2 p_{z})] < σ^{*} (2 p_{x})

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