Question

When the hybridization state of Carbon atom changes from $s{p^3}$ to $s{p^2}$ and finally to sp, the angle between the hybridized orbitals?
A.Decreases gradually
B.Decreases considerably
C.Is not affected
D.Increases progressively

Answer 1

Hint: When the hybridization state of Carbon atom changes from $s{p^3}$ to $s{p^2}$ and finally to sp, the angle between the hybridized orbitals increases progressively since the angles in the $s{p^3}$ to $s{p^2}$ and sp hybridization is , and \[180^\circ \] respectively.

Complete step by step answer:
In $s{p^3}$ hybridization, the four $s{p^3}$ hybrid orbitals will arrange themselves in three-dimensional space to get as far apart as possible (to minimize repulsion). So, the geometry that achieves this is tetrahedral geometry, where the bond angle is . Example is methane $(C{H_4})$ .
In $s{p^2}$ hybridization, the three $s{p^2}$ hybrid orbitals will arrange themselves in three-dimensional space to get as far apart as possible. So, the geometry that achieves this is trigonal planar geometry, where the bond angle is \[120^\circ \] . Example is ethane $({C_2}{H_2})$ .
And in sp hybridization, two sp hybrid orbitals arrange themselves in three-dimensional space to get as far apart as possible. So, the geometry which achieves is linear geometry where the bond angle is maximum i.e. \[180^\circ \] . Example is ethyne $({C_2}{H_2})$ .
So, we can say that the bond angle of \[sp > s{p^2} > s{p^3}\]
So, when the hybridization state of Carbon atom changes from $s{p^3}$ to $s{p^2}$ and finally to sp, the angle between the hybridized orbitals increases progressively

Therefore, the correct answer is option (D).

Note: When the hybridization changes, a Pi electron cloud is formed between the \[C - C\] . This electron cloud repels the \[C - H\] bond. Thus, the angle between \[C - C\] and \[C - H\] increases and the angle between \[C - H\] and \[C - H\] decreases.