Question

First, second and third IP values are 100eV, 150eV, and 1500 eV. The element can be:
A. Be
B. B
C. F
D. Na

Answer 1

The elements present in group 2 and group 15 have completely filled or partially filled electronic configuration, therefore it requires a large amount of energy to remove an electron from the orbital with completely filled electrons.

Complete answer:
-Ionization energy is defined as the measure of the struggle in separating the electrons from the atom or ion. It is also defined as the amount of energy that a gaseous atom present in the ground state absorbs to release an electron to form a cation.
$H(g) \to {H^ + }(g) + {e^ - }$
-The energy applied is expressed in terms of kJ/mol.
$M\xrightarrow[{100}]{{I{E_1}}}{M^ + }\xrightarrow[{150}]{{I{E_2}}}{M^{2 + }}\xrightarrow[{1500}]{{I{E_3}}}{M^{3 + }}$
-The ionization energy for converting M to ${M^ + }$ is 100.
-The ionization energy for converting ${M^ + }$ to ${M^{2 + }}$ is 150.
-The ionization energy for converting ${M^{2 + }}$ to ${M^{3 + }}$ is 1500.
-As we can see that the ionization energy is very high during the conversion of ${M^{2 + }}$ to ${M^{3 + }}$. It means that the ion or atom has reached the stability. The atom must be a noble gas element or Group 2 element which contains completely filled or half-filled electrons in the valence shell.
-As, the ionization energy is very high to remove the electrons from the half-filled or completely filled electronic configuration.
-From the given element, beryllium is the alkaline earth metal which contains half-filled electrons in its valence shell.

Therefore, the correct option is (A).

Note: The ionization energy is dependent on the atomic radius. In the periodic table as we move from right to left the atomic radius increases and the ionization energy decreases. Because of the shielding effect the ionization energy decreases as we move from top to bottom of the periodic table. The ionization energy of alkali metal is considerably low as compared to halogens.