Question

What is the hybridization at the two central carbon atoms of $2 - butene$?

Answer 1

Hint: We need to know the structure of $2 - butene$. Then to find out the hybridization let us understand that as atomic orbitals combine to create a new atomic orbital, it will be called its hybridization. If a carbon atom forms only sigma bonds in a hydrocarbon, its orbitals would be \[s{p^3}\]hybridized. The orbitals of a carbon atom that forms a pi bond are \[s{p^2}\]hybridized. The orbitals of a carbon atom become \[sp\]hybridized as it forms two pi bonds. Be careful that the bond must be formed between carbon and hydrogen atoms.

Complete step by step answer:
Remember that as atomic orbitals combine to create a new atomic orbital, will be called its hybridization. The new orbital will accommodate the same electron density as the previous system. The updated, hybridized orbital characteristics and energy are just the 'average' of the initial non-hybridized orbitals.
Let us look at the molecular formula of this compound $2 - butene$;
$C{H_3} - CH = CH - C{H_3}$
Now the carbon atoms we are considered about are the second and third ones when they are numbered in ascending order from left to right.
Now the rules which we require to be familiar with while handling hybridization is that;
If a carbon atom forms only sigma bonds in a hydrocarbon, its orbitals would be \[s{p^3}\]hybridized. The orbitals of a carbon atom that forms a pi bond are \[s{p^2}\]hybridized. The orbitals of a carbon atom become \[sp\]hybridized as it forms two pi bonds.
Clearly according to the molecular formula;
$C{H_3} - CH = CH - C{H_3}$
Each of the second and third carbon atoms has the same bondings since the figure is symmetric. So the hybridization of a second would be the same as the third.
Let us observe the second hydrocarbon, here there are two single bonds (to carbon and hydrogen) and one double bond (to the third hydrocarbon), this means there are three sigma bonds due to two sigma bonds from the two single bonds plus one sigma from the double bond. But the double bond also has a pi bond in it so in total there are three sigma bonds and one pi bond in the second carbon. The same will be there for the third carbon.
Now according to our previous rule, here we don’t just have sigma bonds but also have one single bond so the hybridization can only be $s{p^2}$, for both second and third carbon atoms.
$\therefore $The two central carbon atoms of $C{H_3} - CH = CH - C{H_3}$are in $s{p^2}$hybridization.

Note:
In a similar way we can also find hybridization for the outer carbon atoms (first and fourth). Taking the first carbon we know that it has four single bonds that are three to hydrogens and one to the next carbon atom. So single bonds only mean sigma bonds, therefore we can write that the hybridization for the first carbon is $s{p^3}$since only sigma bonds are present here. Due to the symmetry of the figure, the first and fourth will have the same hybridization.