Question

Please explain to me how to answer no.2 questions

Answer 1

Answer: -

IE 1 for X = 801

Here X is told to be in the third period.

So n = 3 for X.

For 1st ionization energy the expression is

IE1 = 13.6 x Z ^2 / n^2

Where Z =atomic number.

Thus Z =( n^2 x IE 1 / 13.6)^(1/2)

Z = ( 3^2 x 801 / 13.6 )^ (1/2)

= 23

Number of electrons = Z = 23

Nearest noble gas = Argon

Argon atomic number = 18

Number of extra electrons = 23 – 18 = 5

a) Electronic Configuration= [Ar] 3d34s2

We know that more the value of atomic radii, lower the force of attraction on the electrons by the nucleus and thus lower the first ionization energy.

So more the first ionization energy, less is the atomic radius.

X has more IE1 than Y.

b) So the atomic radius of X is lesser than that of Y.

c) After the first ionization, the atom is no longer electrically neutral. There is an extra proton in the atom.

Due to this the remaining electrons are more strongly pulled inside than before ionization. Hence after ionization, the radii of Y decreases.