Question

Which among the following elements have lowest value of $I{E}_1$

• A. Pb
• B. Sn
• C. Si
• D. C

Answer 1

The correct option is B

Sn

We know the relation between atomic radius and IE

(a) Atomic radius: The values of ionization potential of an element decreases as its atomic radius increases. This is because the electrostatic force of attraction between the nucleus and the outermost electron decreases as the distance between them increases. So the energy required for the removal of the electron will comparatively be less

Ionization Potential $\alpha$ $\frac{1}{AtomicRadius}$

$\to$ Therefore, down the group from top to bottom, as atomic radius increases, therefore IE decreases

Then, the order of $I{E}_1$ should be

C > Si > Sn > Pb

But, The ionization energy of tin{Sn} is less than that of lead)Pb).It is due to the poor shielding of d- and f- electrons in Pb due to which it feels greater attraction from nucleus.Therefore,

C > Si > Pb > Sn

$\Rightarrow$ s

(True/False)