Name: Energy to Take Electrons Out: Ionisation Enthalpy
Uploaded: 2019-07-19
Duration: 3 min 6 s
Description: Ionisation enthalpy of an element can be the amount of energy required to remove an electron from an isolated gaseous atom. What does this mean? Learn about...

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 $2^{nd}$ .

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.

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?

Important Questions on Electrons in Atoms

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Ionization	$1 s^{0}$	$2^{nd}$	$3^{rd}$	$4^{th}$	$5^{th}$
Ionization energy/ $2 s^{1}$	$799$	$2420$	$3660$	$25000$	$32800$

The successive ionization energies of boron are shown in the table below : Ionization 1st 2nd 3rd 4th 5th Ionization energy/KJ mol-1 799 2420 3660 25000 32800 Why is there a large increase between the third and fourth ionization energies?

Important Questions on Electrons in Atoms

Learn and Prepare for any exam you want !

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?