Question

The Lewis structure of allene is



Which of the following statements correctly gives answers for all 3 parts:

(a) Is the molecule planar?

(b) Does 1, 3 - dichloro propadiene show geometrical isomerism?

(c) Is the molecule 1, 3 - dichloro propadiene polar?

• A. (i) Non-planar
  
  (ii) No geometrical isomerism
  
  (iii) Polar
• B. (i) Planar
  
  (ii) No geometrical isomerism
  
  (iii) Polar
• C. (i) Planar
  
  (ii) Yes geometrical isomerism
  
  (iii) Polar
• D. (i) Non Planar
  
  (ii) No geometrical isomerism
  
  (iii) Not Polar

Answer 1

Answer: (A)

Hybridization of central carbon is $\text{sp}$ and two of its unhybridized p -orbitals are involved in $\pi $ - bonding. Also p - orbitals at an atom are orthogonal (at $90^{\text{ο}}$ angle), the two $\pi $ - bonds at central carbon are also orthogonal. Let these p - orbitals, involved in $\pi $ - bonding, at central carbon be $\text{p}_{\text{y}}$ and $\text{p}_{\text{z}}$ , then



Hybridization at the terminal carbons are both $\text{sp}^{2}$ . The two p - orbitals involved in $\text{sp}^{2}$ hybridization are $\text{p}_{\text{x}}$ and $\text{p}_{\text{y}}$ at left terminal and $\text{p}_{\text{x}}$ and $\text{p}_{\text{z}}$ at right terminals. Therefore, the three $\text{s} \text{p}^{2}$ hybrid orbitals at left terminal carbon are in $\text{x} \text{y}$ plane, while the same at right terminal carbon are in $\text{x} \text{z}$

plane. As a result, the two triangular planes at terminals are at right angles as shown below:



The two $\text{H - C - H}$ planes at terminals are perpendicular to each other. ( $\pi $ - bonds are also perpendicular to each other.)

Hence, (a) Molecule is non - planar. (b) It doesn't show geometrical isomerism because two H and chlorine are in different planes. (c) polar.