Explain the shape of $sp$,$sp^2$,and $sp^3$ hybrid orbitals.
(N/A) $sp$ orbital: The two $sp$ hybrid orbitals are oriented in opposite directions along a straight line with an angle of $180^{\circ}$ between them.
$sp^2$ orbital: These three $sp^2$ hybrid orbitals are oriented in a trigonal planar arrangement,symmetrically at an angle of $120^{\circ}$ between them.
$sp^3$ hybridization: The four $sp^3$ hybrid orbitals are directed towards the four corners of a regular tetrahedron. The angle between any two $sp^3$ hybrid orbitals is $109^{\circ} 28^{\prime}$.
$sp^2$ orbital: These three $sp^2$ hybrid orbitals are oriented in a trigonal planar arrangement,symmetrically at an angle of $120^{\circ}$ between them.
$sp^3$ hybridization: The four $sp^3$ hybrid orbitals are directed towards the four corners of a regular tetrahedron. The angle between any two $sp^3$ hybrid orbitals is $109^{\circ} 28^{\prime}$.
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Similar Questions
Match the compounds given in $Column\, I$ with the hybridisation and shape given in $Column\, II$ and mark the correct option.
$Column\, I$ $Column\, II$ $(A)\, XeF_6$ $(i)\, \text{Distorted octahedral}$ $(B)\, XeO_3$ $(ii)\, \text{Square planar}$ $(C)\, XeOF_4$ $(iii)\, \text{Pyramidal}$ $(D)\, XeF_4$ $(iv)\, \text{Square pyramidal}$
| $Column\, I$ | $Column\, II$ |
|---|---|
| $(A)\, XeF_6$ | $(i)\, \text{Distorted octahedral}$ |
| $(B)\, XeO_3$ | $(ii)\, \text{Square planar}$ |
| $(C)\, XeOF_4$ | $(iii)\, \text{Pyramidal}$ |
| $(D)\, XeF_4$ | $(iv)\, \text{Square pyramidal}$ |
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View SolutionTwo $p$-block elements $X$ and $Y$ form fluorides of the type $EF_3$. The fluoride compound $XF_3$ is a Lewis acid and $YF_3$ is a Lewis base. The hybridization of the central atoms of $XF_3$ and $YF_3$ respectively are
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