Which of the following pairs have the same ty | vedclass

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Question

(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

Vedclass · Accepted Answer

$H_2O, NH_3$

Vedclass · Answer

(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).

List-$I$	List-$II$
$(a)$ $SF_6$	$(i)$ $sp^3d^2$
$(b)$ $PCl_5$	(ii) $sp^3d$
$(c)$ $XeF_4$	(iii) $sp^3d^3$
$(d)$ $IF_7$	(iv) $sp^3d^2$

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

Similar Questions

In which of the following molecular shapes must the $d_{z^2}$ orbital not be involved in bonding?

Match the following:
List-$I$ List-$II$
$(a)$ $SF_6$ $(i)$ $sp^3d^2$
$(b)$ $PCl_5$ (ii) $sp^3d$
$(c)$ $XeF_4$ (iii) $sp^3d^3$
$(d)$ $IF_7$ (iv) $sp^3d^2$

The correct answer is:

Which one of the following is a planar molecule?

$sp^3d^2$ hybrid orbitals result in which geometry?

$H - C \equiv C - C \equiv C - CH_3$
Compare the bond lengths $a$ and $b$.

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