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

What is the bond angle around an atom that undergoes $sp^2$ hybridization?

Which of the following pairs have the same type of hybridization but different shapes?

Match the following:
| List-$I$ (Molecule/ion) | List-$II$ (Hybridisation) |
| :--- | :--- |
| $(A)$ $ICl_4^-$ | $(P)$ $sp^3$ |
| $(B)$ $NO_3^-$ | $(Q)$ $sp^3d$ |
| $(C)$ $PCl_4^+$ | $(R)$ $sp^3d^2$ |
| $(D)$ $SiF_6^{2-}$ | $(S)$ $sp^2$ |
| | $(T)$ $sp$ |

The hybridizations of the central atom in the molecules $BF_3, BeF_2, BrF_3$ are respectively

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