Question

Which of the following represents the given mode of hybridization $s{{p}^{2}}-s{{p}^{2}}-sp-sp$ from left to right
A.
B.
C.
D.

Answer 1

Hint: The mode of hybridization is given by the number of sigma bonds ($\sigma -$bonds) attached to that particular carbon atom. To approach this question we will have to count the number of sigma bonds present attached to each carbon centre present in the given compounds.

Complete Step by Step Answer:
There is one sigma bond present in the double bond and triple bond between two carbon atoms. And all single bonds are sigma bonds. The mode of hybridization is equal to the number of sigma bonds present in that particular carbon present in the compound. This is because the sigma bonds are formed by the overlap of hybridised orbitals present on that atom but pi bonds are formed by unhybridized orbitals present.

When the carbon atom is linked with two sigma bonds the hybridization is $sp$, when linked with three sigma bonds the hybridization is $s{{p}^{2}}$ and when linked with four sigma bonds hybridization is $s{{p}^{3}}$.

Now we will have to count sigma bonds present on each carbon present in the given compound by numbering each carbon atom by $1,2,3$and $4$from left to right.
From (A), we can see there is one double bond between $1\And 2$ , one single bond between $2\And 3$, and one triple bond between $3\And 4$ present.

Thus in (A) the hybridization mode from left to right is $s{{p}^{2}}-s{{p}^{2}}-sp-sp$.


In (B) the hybridization mode from left to right is $sp-sp-sp-sp$.


The hybridization mode from left to right in (C) is $s{{p}^{2}}-sp-sp-s{{p}^{2}}$.

Finally, the hybridization mode from left to right in (D) is $s{{p}^{2}}-s{{p}^{2}}-s{{p}^{2}}-s{{p}^{2}}$.

Thus, option (A) is correct.

Note: The hybridization is only for sigma bonds this is because sigma bonds are formed by overlapping of hybrid orbitals but pi bonds are formed by sidewise overlapping of these atomic orbitals. That’s why pi-bonds do not undergo hybridization.