The hybrid orbitals used by central atoms in | vedclass

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(B) To determine the hybridisation,we calculate the number of electron pairs around the central atom using the formula: $	ext{Number of hybrid orbita

Vedclass · Accepted Answer

$sp$,$sp^2$ and $sp^3$

Vedclass · Answer

(B) To determine the hybridisation,we calculate the number of electron pairs around the central atom using the formula: $	ext{Number of hybrid orbitals} = 	ext{Number of sigma bonds} + 	ext{Number of lone pairs}$.In $BeCl_2$,$Be$ has $2$ sigma bonds and $0$ lone pairs,so it is $sp$ hybridised.In $BCl_3$,$B$ has $3$ sigma bonds and $0$ lone pairs,so it is $sp^2$ hybridised.In $CCl_4$,$C$ has $4$ sigma bonds and $0$ lone pairs,so it is $sp^3$ hybridised.Therefore,the hybrid orbitals are $sp$,$sp^2$ and $sp^3$ respectively.

Molecule	Hybridisation of central atom	Shape
$(a)$ $PCl_5$	$dsp^3$	square pyramidal
$(b)$ $[Ni(CN)_4]^{2-}$	$sp^3$	tetrahedral
$(c)$ $SF_6$	$sp^3d^2$	octahedral
$(d)$ $IF_3$	$dsp^3$	pyramidal

The hybrid orbitals used by central atoms in $BeCl_2$,$BCl_3$ and $CCl_4$ molecules are respectively

Similar Questions

Identify the correct set from the following table:
Molecule Hybridisation of central atom Shape
$(a)$ $PCl_5$ $dsp^3$ square pyramidal
$(b)$ $[Ni(CN)_4]^{2-}$ $sp^3$ tetrahedral
$(c)$ $SF_6$ $sp^3d^2$ octahedral
$(d)$ $IF_3$ $dsp^3$ pyramidal

Which of the following statements is not correct?

In $HCHO$,the carbon atom has which hybridization?

What is the type of hybridization of $N$ in $NH_3$?

Which molecule has $sp^3$ hybridization of the central atom?

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