Question

The unit of molecular polarizability is
$
  A.{C^2}{N^{ - 1}}m \\
  B.N{m^2}{C^{ - 1}} \\
  C.{N^{ - 1}}{m^{ - 2}}{C^2} \\
  D.{C^{ - 1}}{m^2}V \\
$

Answer 1

- Hint: To get the unit of molecular polarizability you need to use the formula of it and then use the units of rest of the terms to get the unit of the asked term. Doing this will solve your problem and will give you the right unit.

Complete step-by-step solution -
When a light wave of electrical strength E hits a molecule in the medium, it creates an optical moment in the molecule. The molecule is said to have been polarized. The moment caused by the unit incident field is called the polarization of the molecule. The relationship between polarization and electron density factors, atomic radii, and molecular orientation.
The formula of polarizability is given by $\alpha = \dfrac{p}{E}$.
Where, p is the dipole moment and E is the electric field.
The unit of dipole moment is Coulomb x distance = Cm.
The unit of electric field is Newton per coulomb = N/C.
So, on considering the formula of polarizability and the above mentioned units we get the equation as $\alpha = \dfrac{{Cm}}{{\dfrac{N}{C}}} = \dfrac{{{C^2}m}}{N} = {C^2}m{N^{ - 1}}$.
Hence, the unit is ${C^2}m{N^{ - 1}}$.
So, the correct option is A.

Note – When you face such problems you need to use the formula and consider the units.The more electrons there are, the less power the nuclear charge has on the distribution of the charge, and hence the greater polarization of the atom. Polarizability is the ease with which an electron can be released. Higher electronegativity means the electrons are held together more closely. So polarization and electronegativity are inversely related. Knowing these things will help you further and will take you to the right answer.