Question

Given $\Delta H$ for the process $Li(g)\to L{{i}^{3+}}(g)+3{{e}^{-}}$ is 19800 kJ/mol and $I{{E}_{1}}$for Li is 520 then $I{{E}_{2}}$and$I{{E}_{3}}$ of Li are respectively (approx value)
(a)- 7505, 11775
(b)- 520, 19280
(c)- 11775, 19280
(d)- Data is insufficient.

Answer 1

Hint: The ionization enthalpy of a process is equal to the sum of all successive ionization enthalpies in the process. The second ionization enthalpy will be always greater than the first ionization enthalpy and the third ionization enthalpy will be greater than the second ionization enthalpy.

Complete step by step answer:
Ionization enthalpy is the minimum amount of energy that is used to extract the most loosely bound electron from an isolated gaseous atom so that it can be converted into its gaseous cation. It is also known as the ionization potential because it is the minimum potential difference required to remove the most loosely bound electron to form a cation.
When the first electron is removed and the energy required to remove the most loosely bound electron is called the first ionization enthalpy.
Similarly to extract the second and third electron it is called second and third ionization enthalpy respectively.
The ionization enthalpy of the overall process is equal to the sum of all successive ionization enthalpies in the process.
The $\Delta H$ for the process $Li(g)\to L{{i}^{3+}}(g)+3{{e}^{-}}$ is 19800 kJ/mol.

In this process, 3 electrons are removed.
$I{{E}_{1}}+ I{{E}_{2}} + I{{E}_{3}} = 19800$
Given the first ionization is $520\text{ }kJ/mol$
Therefore, $520 + I{{E}_{2}} + I{{E}_{3}} = 19800$
$I{{E}_{2}} + I{{E}_{3}} = 19800-520$
$I{{E}_{2}} + I{{E}_{3}} = 19280$

Let us see which option equals to 19280
(a)- $7505 + 11775 = 19280$
(b)- $520 + 19280 = 19800$
(c)- $11775 + 19280 = 31055$

Hence, the second ionization and third ionization enthalpy values are 7505 and 117755 Kj/mol respectively. So, the correct answer is “Option A”.

Note: The higher value of the ionization energy for the second and third compared to the first is due to the nuclear force of attraction which increases when an electron is removed, since the number of protons becomes more than the number of electrons, therefore more energy is required.