Question

The ionization energy will be maximum for the process:
A)$Ba \to B{a^{2 + }}$
B) $Be \to B{e^{2 + }}$
C) $Cs \to C{s^{2 + }}$
D) $Li \to L{i^{2 + }}$

Answer 1

We know that the ionization energy of an atom is the amount of energy required to remove an electron from the gaseous form of that atom or ion.
Example:
${\text{Na}}\left( {\text{g}} \right) \to {\text{N}}{{\text{a}}^{\text{ + }}}\left( {\text{g}} \right){\text{ + }}{{\text{e}}^{\text{ - }}}$
As moving across the period the atomic radius decreases therefore the outer electrons are closer to the nucleus and more strongly held by the nucleus and thus it requires a lot of energy to remove the outermost electron.
Second Ionization energy:
The energy required to remove a second electron from a singly charged gaseous cation is called as second ionization energy.
Example:
${\text{N}}{{\text{a}}^{\text{ + }}}\left( {\text{g}} \right) \to {\text{N}}{{\text{a}}^{{\text{2 + }}}}\left( {\text{g}} \right){\text{ + }}{{\text{e}}^{\text{ - }}}$
The second Ionization energy is almost ten times greater than the first ionization energy because the number of electrons causing repulsions is reduced.

Complete answer:
The ionization energy will be greatest for the cycle; $Be \to B{e^{2 + }}$smaller is the iota, more is the energy expected to eliminate an electron, i.e., ionization energy. Additionally, expulsion of two electrons needs more energy. Atomic number of Beryllium is four. To change over into$B{e^{2 + }}$, two electrons are delivered from 2s, which are firmly connected to the core, along these lines; it requires a lot of ionization energy. Hence option B is correct.
We know that atomic size and ionization energy are inversely proportional thus barium and cesium do not have greater ionization energy. Thus option A and C are incorrect.
As we know that the atomic number of Lithium is three. To change over into$L{i^{2 + }}$, two electrons are delivered from 2s, which are firmly connected to the core, along these lines; it requires a lot of ionization energy. But its energy is lower than barium thus option D is incorrect.
Hence, according to the theory, the correct answer to the question is option B.

Note:
-We need to know that the first ionization energy of a particle (or) atom is the energy needed to eliminate one mole of electrons from one mole of gaseous molecules (or) particles.
Example:
-The first ionization energy of nitrogen molecules is greater than the nitrogen atom. This can be explained on the basis of their electronic configuration.
-The ground state electronic configuration of nitrogen molecule is,
$\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\pi 2{p_x}^2\pi 2{p_y}^2\sigma 2{p_z}^2$
-The electronic configuration of ${\text{N}}$=$1{s^2}2{s^2}2{p^3}$
-The molecules which have fully filled configurations have more ionization energy than the one with half-filled configurations. The nitrogen molecule has fully filled configuration so its first ionization energy is higher than the nitrogen atom.