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Question

(C) For $BeCl_2$: The central atom $Be$ has $2$ valence electrons and forms $2$ bonds with $Cl$. It undergoes $sp$-hybridization and has a linear shap

Vedclass · Accepted Answer

$BCl_3$,$sp^2$,triangular planar

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(C) For $BeCl_2$: The central atom $Be$ has $2$ valence electrons and forms $2$ bonds with $Cl$. It undergoes $sp$-hybridization and has a linear shape.For $BCl_3$: The central atom $B$ has $3$ valence electrons and forms $3$ bonds with $Cl$. It undergoes $sp^2$-hybridization and has a trigonal planar (triangular planar) shape.Therefore,the correct set is $BCl_3$,$sp^2$,triangular planar.

Which one of the following is a correct set with respect to molecule,hybridisation,and shape?

Similar Questions

Consider the following reactions:
$(i)$ $PCl_3 + 3H_2O \to H_3PO_3 + 3HCl$
$(ii)$ $SF_4 + 3H_2O \to H_2SO_3 + 4HF$
$(iii)$ $BCl_3 + 3H_2O \to H_3BO_3 + 3HCl$
$(iv)$ $XeF_6 + 3H_2O \to XeO_3 + 6HF$
Then according to given information the incorrect statement is

Out of the following hybrid orbitals,the one which forms the bond at an angle of $120^o$ is:

When the hybridisation state of a carbon atom changes from $sp^3$ to $sp^2$ to $sp$,the angle between the hybridised orbitals

Explain the directional properties of bonds using $VB$ theory.

The hybridization in $PF_3$ is

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