Question

Q4 describe the change of atomic radio for the elements in period 2, from lithium to neon

Answer 1

Answer:

Explanation:

Atomic radii of second period elements:

Li = [He] 2s¹   atomic radii = 123 pm

Be =  [He] 2s²  atomic radii = 111 pm

B =  [He] 2s² 2p¹   atomic radii = 86 pm

C =  [He] 2s² 2p²   atomic radii = 77 pm

N =  [He] 2s² 2p³   atomic radii = 74 pm

O =  [He] 2s² 2p⁴   atomic radii = 73 pm

F =  [He] 2s² 2p⁵   atomic radii = 72 pm

Ne = [He] 2s² 2p⁶   atomic radii = 154 pm

In case of neon weak van dar waals forces are present, there is less nuclear attraction that's why its atomic radii is greater.

Definition of atomic radii:

The atomic radius is the distance between center of two bonded atoms.

Trend along period:

As we move from left to right across the periodic table the number of valance electrons in an atom increase.The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases.

Trend along group:

In group by addition of electron atomic radii increase from top to bottom due to increase in atomic number and addition of extra shell.