Find out the correct hybridisation of the cen | vedclass

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(B) To determine the hybridisation,we use the formula: $H = \frac{1}{2} (V + M - C + A)$,where $V$ is the number of valence electrons of the central a

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$(BCl_3: sp^2); (PCl_5: sp^3d); (NH_3: sp^3)$ and $(SF_6: sp^3d^2)$

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(B) To determine the hybridisation,we use the formula: $H = \frac{1}{2} (V + M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$1$. For $BCl_3$: $V=3, M=3$. $H = \frac{1}{2}(3+3) = 3$,which corresponds to $sp^2$ hybridisation.$2$. For $PCl_5$: $V=5, M=5$. $H = \frac{1}{2}(5+5) = 5$,which corresponds to $sp^3d$ hybridisation.$3$. For $NH_3$: $V=5, M=3$. $H = \frac{1}{2}(5+3) = 4$,which corresponds to $sp^3$ hybridisation.$4$. For $SF_6$: $V=6, M=6$. $H = \frac{1}{2}(6+6) = 6$,which corresponds to $sp^3d^2$ hybridisation.Thus,the correct hybridisation is $(BCl_3: sp^2); (PCl_5: sp^3d); (NH_3: sp^3)$ and $(SF_6: sp^3d^2)$.

Find out the correct hybridisation of the central atom in $BCl_3$,$PCl_5$,$NH_3$ and $SF_6$.

Similar Questions

Consider the following reactions:
$(i)$ $PCl_3 + 3H_2O \to H_3PO_3 + 3HCl$
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$(iii)$ $BCl_3 + 3H_2O \to H_3BO_3 + 3HCl$
$(iv)$ $XeF_6 + 3H_2O \to XeO_3 + 6HF$
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