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(B) The geometry of a molecule depends on the hybridization of its carbon atoms.$sp^{3}$ hybridized carbon atoms have a tetrahedral geometry with a bo

Vedclass · Accepted Answer

$CH_3 - C \equiv C - CH_3$

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(B) The geometry of a molecule depends on the hybridization of its carbon atoms.$sp^{3}$ hybridized carbon atoms have a tetrahedral geometry with a bond angle of $109^{\circ} 28^{\prime}$.$sp^{2}$ hybridized carbon atoms have a trigonal planar geometry with a bond angle of $120^{\circ}$.$sp$ hybridized carbon atoms have a linear geometry with a bond angle of $180^{\circ}$.In $CH_3 - C \equiv C - CH_3$ (but$-2-$yne),the two central carbon atoms are $sp$ hybridized,making the central part of the molecule linear.

Considering the state of hybridization of carbon atoms,find out the molecule among the following which is linear?

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