Question

The successive ionization energies of boron are shown in the table below :

Ionization	$1^{st}$	$2^{nd}$	$3^{rd}$	$4^{th}$	$5^{th}$
Ionization energy/ $KJ {mol}^{- 1}$	$799$	$2420$	$3660$	$25000$	$32800$

Why is there a large increase between the third and fourth ionization energies?

Accepted Answer

The atomic number of Boron is $5 .$ The symbol used for Boron is $B$ .
The electronic configuration of Boran is $1 s^{2} 2 s^{2} 2 p^{1}$ .

The first ionization energy of Borane. i.e; $B \to B^{+} + e^{-} = 799 {kJmol}^{- 1}$ and the electronic configuration becomes $1 s^{2} 2 s^{2} 2 p^{0}$ .

The second ionization energy of Boron. i.e; $B^{+} \to B^{2 +} + e^{-} = 2420 {kJmol}^{- 1}$ and the electronic configuration becomes $1 s^{2} 2 s^{1} 2 p^{0}$ .

The third ionization energy of Boron. i.e; $B^{2 +} \to B^{3 +} + e^{-} = 3660 {kJmol}^{- 1}$ and the electronic configuration becomes $1 s^{2} 2 s^{0} 2 p^{0}$ .

The fourth ionization energy of Boron. i.e; $B^{3 +} \to B^{4 +} + e^{-} = 25000 {kJmol}^{- 1}$ and the electronic configuration becomes $1 s^{1} 2 s^{0} 2 p^{0}$ .

The fifth ionization energy of Boron. i.e; $B^{4 +} \to B^{5 +} + e^{-} = 32800 {kJmol}^{- 1}$ and the electronic configuration becomes $1 s^{0}$ .

The large energy gap is seen between third and fourth ionization energy because of the follwoing four factors listed below:

within the same shell the penetration effect decreases in the order : $s > p > d > f$ .

$1 s^{2}$ orbital is much more stable than $2 s^{1}$ orbital.

The nuclear charge becomes higher,so the ionization energy increases.

Lawrie Ryan and Roger Norris Solutions for Chapter: Electrons in Atoms, Exercise 5: Question

Lawrie Ryan Chemistry Solutions for Exercise - Lawrie Ryan and Roger Norris Solutions for Chapter: Electrons in Atoms, Exercise 5: Question

Questions from Lawrie Ryan and Roger Norris Solutions for Chapter: Electrons in Atoms, Exercise 5: Question with Hints & Solutions

Learn and Prepare for any exam you want !