Question

The number of $s{{p}^{2}}-s{{p}^{2}}$ sigma bonds in the compound given below is:

A. 1
B. 3
C. 4
D. 5

Answer 1

To solve this question, identify the hybridisation of all the carbon atoms present in the structure by taking into account the $\sigma$- bonds and the $\pi$ - bonds.

Complete answer:
We know that $s{{p}^{3}}$ hybridized carbon atoms are tetrahedral carbons in this compound. They are not planar. The $s{{p}^{2}}$ hybridized carbon atoms are planar carbon atoms. Also, they have vacant p-orbitals.
Carbon atoms are usually $s{{p}^{3}}$ hybridized when all the bonds around them are single $\sigma$-bonds, $s{{p}^{2}}$ hybridized when 1 double bond or 1 $\pi$ - bond along with 3 $\sigma$- bonds is present and $sp$ hybridized when 1 triple bond or 2 $\pi$ - bonds along with 2 $\sigma$- bonds are present.
So here, all the carbons having a double bond will be $s{{p}^{2}}$ hybridized.
In this given compound, the carbon atoms that are joining the six membered ring and four membered rings are $s{{p}^{3}}$hybridized. They are not planar. The rest of the carbon atoms in this given compound is $s{{p}^{2}}$ hybridized. There are 6 $s{{p}^{2}}$ hybridized carbon atoms. Now we want to look at the $\sigma$ - bonds.
The $\sigma$- bonds are the single bonds connected with these carbon atoms. One of the bonds in the double bonds will be a $\sigma$- bond and the other will be a $\pi$- bond.
There are 4 $s{{p}^{2}}-s{{p}^{2}}$ sigma bonds in this compound. 3 in the six-membered ring and 1 in the 4-membered ring.
Therefore, the correct option is ‘C. 4’

Note: Note that in this problem, there are 3 $\sigma$- bonds that occur along with the $\pi$ - bonds. The fourth sigma bond is the one that is present between 2 $s{{p}^{2}}$ hybridized carbons that are joined with double bonds to 2 other carbon atoms. Thus, this $\sigma$- bond is solitary. Do not forget to take this into account.