Question

The correct order of decreasing second ionization energy of Li, Be, Ne, C, B
A. Ne > B > Li > C > Be
B. Li > Ne > C > B > Be
C. Ne > C > B > Be > Li
D. Li > Ne > B > C > Be

Answer 1

Hint: The amount of energy required to remove the valence electron which is loosely bound to the atom is called ionization energy. Second ionization energy means the amount of energy required to remove second electrons from an atom.

Complete Solution :
- In the question it is given to find the correct order of decreasing second ionization energy of Li, Be, Ne, C, B.
- To know about the second ionization energy of the given atoms we are supposed to know the electronic configuration of the given atoms.
- The atomic number of lithium is 3 and the electronic configuration of the lithium is $1{{s}^{2}}2{{s}^{1}}$ .
- The atomic number of beryllium is 4 and the electronic configuration of the beryllium is $1{{s}^{2}}2{{s}^{2}}$ .
- The atomic number of boron is 5 and the electronic configuration of the boron is $1{{s}^{2}}2{{s}^{2}}2{{p}^{1}}$
- The atomic number of carbon is 6 and the electronic configuration of the carbon is $1{{s}^{2}}2{{s}^{2}}2{{p}^{2}}$ .
- The atomic number of neon is 10 and the electronic configuration of the neon is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$ .

- After removing one electron from the given elements the electronic configuration of the given elements will be as follows.
- Electronic configuration of the lithium (+1) is $1{{s}^{2}}2{{s}^{0}}$ .
- Electronic configuration of the beryllium (+1) is $1{{s}^{2}}2{{s}^{1}}$.
- Electronic configuration of the boron (+1) is $1{{s}^{2}}2{{s}^{2}}2{{p}^{0}}$.
- Electronic configuration of the carbon (+1) is $1{{s}^{2}}2{{s}^{2}}2{{p}^{1}}$
- Electronic configuration of the neon (+1) is $1{{s}^{2}}2{{s}^{2}}2{{p}^{5}}$
- The second ionization energy required to remove the second electron from the above atoms in decreasing order is as follows.
Li > Ne > B > C > Be
So, the correct answer is “Option D”.

Note: The second ionization energy for lithium is very high because the electronic configuration of Li (+1) resembles the electronic configuration of the stable noble gas. Boron also has filled 2s orbital after losing one electron. So, boron has high second ionization energy.