What is the hybridisation of Be in BeF_2 mole | vedclass

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(C) The hybridisation state of $Be$ in $BeF_2$ is determined using the formula: $H = \frac{1}{2}(V + M - C + A)$.Here,$V$ is the number of valence ele

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$sp$

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(C) The hybridisation state of $Be$ in $BeF_2$ is determined using the formula: $H = \frac{1}{2}(V + M - C + A)$.Here,$V$ is the number of valence electrons of the central atom $(Be)$ = $2$.$M$ is the number of monovalent atoms attached $(F)$ = $2$.$C$ is the cationic charge = $0$.$A$ is the anionic charge = $0$.Substituting these values: $H = \frac{1}{2}(2 + 2 - 0 + 0) = 2$.$A$ value of $2$ corresponds to $sp$ hybridisation.Thus,the hybridisation of $Be$ in $BeF_2$ is $sp$.

What is the hybridisation of $Be$ in $BeF_2$ molecule?

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