Which of the following pairs have the same ty | vedclass

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(A) $H_2O$: Central atom $O$ is $sp^3$ hybridized with $2$ lone pairs,resulting in a bent ($V$-shape) geometry.$NH_3$: Central atom $N$ is $sp^3$ hybr

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$H_2O, NH_3$

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(A) $H_2O$: Central atom $O$ is $sp^3$ hybridized with $2$ lone pairs,resulting in a bent ($V$-shape) geometry.$NH_3$: Central atom $N$ is $sp^3$ hybridized with $1$ lone pair,resulting in a trigonal pyramidal geometry.Since both $H_2O$ and $NH_3$ are $sp^3$ hybridized but have different shapes (bent vs pyramidal),option $A$ is correct.$Ni(CO)_4$ is $sp^3$ (tetrahedral) and $[Ni(CN)_4]^{4-}$ is $dsp^2$ (square planar).$XeF_4$ is $sp^3d^2$ (square planar) and $[Fe(CO)_4]^{2-}$ is $sp^3$ (tetrahedral).$SF_4$ is $sp^3d$ (see-saw) and $CF_4$ is $sp^3$ (tetrahedral).

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

Similar Questions

In the formation of the $CH_4$ molecule,carbon makes use of:

Describe the change in hybridization (if any) of the $Al$ atom in the following reaction:
$AlCl_3 + Cl^{-} \to AlCl_4^-$

In $CO_2$ molecule,the hybridizations of carbon and oxygen atoms are respectively:

Which set of hybridisation is correct for the following compounds:
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In which of the following does the central atom $NOT$ use $sp^3$ hybrid orbitals in its bonding?

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