The state of hybridisation of B in BCl_3 is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The atomic number of Boron $(B)$ is $5$. Its ground state electronic configuration is $1s^2 2s^2 2p^1$.In the excited state,one electron from the

Vedclass · Accepted Answer

$sp^2$

Vedclass · Answer

(B) The atomic number of Boron $(B)$ is $5$. Its ground state electronic configuration is $1s^2 2s^2 2p^1$.In the excited state,one electron from the $2s$ orbital is promoted to the $2p$ orbital,resulting in the configuration $1s^2 2s^1 2p_x^1 2p_y^1$.These three orbitals $(2s, 2p_x, 2p_y)$ undergo $sp^2$ hybridisation to form three equivalent $sp^2$ hybrid orbitals.These three $sp^2$ hybrid orbitals overlap with the $3p$ orbitals of three $Cl$ atoms to form three $B-Cl$ bonds.Therefore,the hybridisation of $B$ in $BCl_3$ is $sp^2$.

The state of hybridisation of $B$ in $BCl_3$ is

Similar Questions

$sp^3d^2$ hybrid orbitals are formed by the mixing of which atomic orbitals?

Oxidation of carbon with concentrated sulphuric acid gives the corresponding oxides of $C, S$ and $H$. The hybridisation of the central atoms in the oxides of $C$ and $H$ are respectively.

The maximum number of $90^{\circ}$ angles between bond pair-bond pair of electrons is observed in

The central atom in a molecule is in $sp^2$ hybrid state. The shape of the molecule will be

The $PCl_5$ molecule is a result of the hybridisation of

Vedclass Test Series

Exam Paper Generator

Online Exam Module