The state of hybridisation of B in BCl_3 is | vedclass

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(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

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$sp^2$

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(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 $sp^2$.

List-$I$ (Hybridisation)	List-$II$ (Shape)
$A$. $dsp^2$	$I$. Square planar
$B$. $sp^3$	$II$. Tetrahedral
$C$. $d^2sp^3$	$III$. Octahedral
$D$. $sp^3d$	$IV$. Trigonal bipyramidal

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

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