Be in BeCl_2 undergoes | vedclass

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Question

(A) The atomic number of $Be$ is $4$. The ground state electronic configuration is $1s^2, 2s^2$.In the excited state,one electron from the $2s$ orbita

Vedclass · Accepted Answer

Diagonal hybridization

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(A) The atomic number of $Be$ is $4$. The ground state electronic configuration is $1s^2, 2s^2$.In the excited state,one electron from the $2s$ orbital is promoted to the $2p$ orbital,resulting in the configuration $1s^2, 2s^1, 2p^1$.These two orbitals ($2s$ and $2p$) hybridize to form two equivalent $sp$ hybrid orbitals.This type of hybridization is known as diagonal or linear hybridization,which results in a linear geometry for the $BeCl_2$ molecule.

$Be$ in $BeCl_2$ undergoes

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