Question

. Which of the following values is closest to the third ionization energy of Berilium? Its first Z2 ionization energy is 9.3 eV. E13.6 in eV] n- 72 a) 4.2 eV b) 9.3 eV c) 18.2 eV d) 153 eV

Answer 1

4. ans:- option (d) 153 ev

Explanation:-

The given formula of energy only valid for single electron species such as H ,He⁺ and Li²⁺etc.

we can solve this problem by comparing the third ionization energy of Li²⁺( 1 electron ) to the third ionization energy of Be²⁺ ( 2 electron ) .

Li²⁺ ( Z = 3 ) electronic configuration is : 1s¹ ( less value of third I.E) and

Be²⁺ ( Z = 4 ) electronic configuration is : 1s² ( more value of third I. E )

since s- orbital of Be²⁺ is paired and Li²⁺ is unpaired , therefore third ionization enthalpy of Be²⁺ must be greater than third ionization enthalpy of Li²⁺ .

Now , let us find the value of third ionization energy of Li²⁺( single electron species ) by using the formula:

E = -13.6 Z² / n²

here, n= shell number = 1 for Li²⁺ .

therefore , E = - 13.6 x ( 3 )² / (1)²

E = - 13.6 x 9 / 1

E = 122.4 eV

Now because third ionization energy of Be²⁺ > third ionization energy of Li²⁺

therefore , correct closest third ionization energy of Be will be (d) 153 eV .