Chlorine has higher ionization energy
Ionization energy is defined as the minimum amount of energy required to remove the most loosely bound electron, the valence electron, of an isolated neutral gaseous atom, molecule or ion. It is quantitatively expressed in symbols as
X + energy → X+ + e−
Where X is any atom, molecule or ion capable of being ionized, X+ is that atom or molecule with an electron removed, and e− is the removed electron. This is generally an endothermic process.
The ionization energy of Sodium (alkali metal) is 496KJ/mol whereas Chlorine's first ionization energy is 1251.1 KJ/mol.
Alkali metals (IA group) have small ionization energies, especially when compared to halogens. Because as we move across the period from left to right, in general, the ionization energy increases. The atoms become smaller which causes the nucleus to have greater attraction for the valence electrons. Therefore, the electrons are more difficult to remove.
[tex]\boxed{{\text{Chlorine}}}[/tex] has higher ionization energy than sodium.
Further Explanation:
Ionization energy:
It is the amount of energy that is required to remove the most loosely bound valence electrons from the isolated neutral gaseous atom. It is denoted by IE. The value of IE is related to the ease of removing the outermost valence electrons. If these electrons are removed so easily, small ionization energy is required and vice-versa. It is inversely proportional to the size of the atom.
Ionization energy trends in the periodic table:
1. Along the period, IE increases due to the decrease in the atomic size of the succeeding members. This results in the strong attraction of electrons and hence are difficult to remove.
2. Down the group, IE decreases due to the increase in the atomic size of the succeeding members. This results in the lesser attraction of electrons and hence are easy to remove.
Sodium and chlorine are present in period 3 of the periodic table. Sodium lies to the left region of the period while chlorine lies to the right.
The atomic number of sodium atom [tex]\left({{\text{Na}}}\right)[/tex] that lies in left region of the period 3 is 11 and its electronic configuration is [tex]{\mathbf{1}}{{\mathbf{s}}^{\mathbf{2}}}{\mathbf{2}}{{\mathbf{s}}^{\mathbf{2}}}{\mathbf{2}}{{\mathbf{p}}^{\mathbf{6}}}{\mathbf{3}}{{\mathbf{s}}^{\mathbf{1}}}[/tex] . The atomic number of chlorine is 17 and its electronic configuration is [tex]{\mathbf{1}}{{\mathbf{s}}^{\mathbf{2}}}{\mathbf{2}}{{\mathbf{s}}^{\mathbf{2}}}{\mathbf{2}}{{\mathbf{p}}^{\mathbf{6}}}{\mathbf{3}}{{\mathbf{s}}^{\mathbf{2}}}{\mathbf{3}}{{\mathbf{p}}^{\mathbf{5}}}[/tex] . Sodium has only one electron in its outermost valence shell that can be removed easily in order to achieve the nearest stable noble gas configuration of He, resulting in its low ionization energy. Chlorine is one electron short of noble gas so it can gain an electron easily, but its removal requires a large amount of energy. So the ionization energy of chlorine is higher than that of sodium.
Learn more:
1. Rank the elements according to first ionization energy: https://brainly.com/question/1550767
2. Write the chemical equation for the first ionization energy of lithium: https://brainly.com/question/5880605
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Periodic classification of elements
Keywords: ionization energy, sodium, atomic number, electron, neutral, isolated, gaseous atom, IE, chlorine, group, period, higher.
