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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

$CH_3 - C \equiv C - CH_3$

Vedclass · Answer

(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

Hybridisation involves

Choose the correct code of characteristics for the given order of hybrid orbitals of the same atom,$sp < sp^2 < sp^3$:
$(i)$ Electronegativity
$(ii)$ Bond angle between same hybrid orbitals
$(iii)$ Size
$(iv)$ Energy level

What is the type of hybridization of $N$ in $NH_3$?

The number of $sp^2 - sp^2$ sigma bonds in the compound given below is

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