Question

(a)Which has the larger third ionization energy, Be or N?
(b)Which has the larger fourth ionization energy, Ga or Ge?
A. (a)Be (b)Ge
B. (a)Be (b)Ga
C. (a)N (b)Ga
D. None of these

Answer 1

Hint: We know that ionization energy is the energy required to remove the outermost valence electron from the neutral gaseous atom to form a cation. The units of ionization energy is ${\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$.

Complete step by step answer:
Let’s discuss successive ionization energies. If a gaseous atom is to lose more than one electron, they can be removed one after the other, that is, in succession and not simultaneously. Actually when the gaseous atom loses one electron to form a monovalent cation, the number of electrons in the cation decreases by one and these are held by the nucleus of cation of greater force. Therefore, with the same energy, the second electron cannot be removed and extra energy is needed to form a divalent cation. In the same manner, energy required to remove the third electron is expected to be higher. This clearly shows that electrons are removed in succession or one after the other.
Now come to the question.
In (a), we have to compare the third ionization energy of Be and N.
Let’s first write the electronic configuration of Be and N
Be=$1{s^2}2{s^2}$
N=$1{s^2}2{s^2}2{p^3}$
Now, we write the divalent cation of Be and N to compare the third ionization energy.
${\rm{B}}{{\rm{e}}^{2 + }} = 1{s^2}$
${{\rm{N}}^{2 + }} = 1{s^2}2{s^2}2{p^1}$
${\rm{B}}{{\rm{e}}^{{\rm{2 + }}}}$ possesses the noble gas configuration (stable configuration). So, its ionization energy is greater than that of ${{\rm{N}}^{{\rm{2 + }}}}$.
In (b), we have to compare the fourth ionization of Ga or Ge.
First we write the trivalent cation of both.
${\rm{G}}{{\rm{a}}^{{\rm{3 + }}}} = \left[ {{\rm{Ar}}} \right]3{d^{10}}$
${\rm{G}}{{\rm{e}}^{{\rm{3 + }}}} = \left[ {{\rm{Ar}}} \right]3{d^{10}}4{s^1}$
In ${\rm{G}}{{\rm{e}}^{{\rm{3 + }}}}$, a unpaired electron present but in ${\rm{G}}{{\rm{a}}^{{\rm{3 + }}}}$ no unpaired electron present. So, the fourth ionization energy is ${\rm{G}}{{\rm{a}}^{{\rm{3 + }}}}$ is greater than ${\rm{G}}{{\rm{e}}^{{\rm{3 + }}}}$.
So, higher ionization energy in (a) is ${\rm{B}}{{\rm{e}}^{{\rm{2 + }}}}$ and higher ionization energy in (b) is ${\rm{G}}{{\rm{a}}^{{\rm{3 + }}}}$.

Hence, the correct choice is option B.

Note: Always remember that the ionization energy to remove the second electron is termed as second ionization energy, to remove third electron the ionization energy is called third ionization energy and so on.