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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?

Similar Questions

Which hybrid orbitals are used by carbon atoms in the following molecules?
$(a)$ $CH_3-CH_3$
$(b)$ $CH_3-CH=CH_2$
$(c)$ $CH_3-CH_2-OH$
$(d)$ $CH_3-CHO$
$(e)$ $CH_3COOH$

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