Question

The most electropositive element is:
(A) Cs
(B) C
(C) Cl
(D) K

Answer 1

Hint: Take a look at the periodic table. Think about the properties of electropositive elements present in the periodic table. Recollect the definition of electropositivity and the factors affecting it to get the answer.


Complete step by step solution:
- Electropositivity is the property of elements to donate electrons readily. Electropositivity is the tendency of an atom to easily donate electrons in order to gain stability.
- Ionization enthalpy is the amount of energy required by an atom to remove one electron from its outermost orbit.
- More easily the atom donates electrons, more is the electropositivity.
- If the ionization energy is more, electropositivity is less.
- Now, let us take a look at the factors affecting electropositivity:
- As we move from left to right in the periodic table, electropositivity decreases. This is due to reduction of atomic size because electrons are added in the same shell and so more protons are also added in the nucleus and the electrons are attracted more towards the nucleus. Therefore, we can conclude that, s-block of the periodic table has the electropositive elements and the first group is the most electropositive one in the entire periodic table.
- As we move from top to bottom in the periodic table, atomic size increases and electropositivity increases. This is because there is a shielding effect due to the addition of a greater number of shells. Due to the addition of a greater number of shells, the electron in the outermost shell is well shielded from the nuclear attraction and can be easily lost without consuming more energy. So, ionization energy is less and thus, the element will be more electropositive.
- Caesium, Cs is the most electropositive element in the periodic table. It belongs to the first group and sixth period in the periodic table. It can easily donate its one valence electron to attain noble gas configuration. The elements in the seventh period are radioactive so they aren’t considered here.
- Caesium has the electronic configuration, $\left[ Xe \right]6{{s}^{1}}$. Therefore, due to greater shielding of the electron present in the 6s orbital, because of completely filled 4d orbitals, caesium has the least ionization enthalpy and large atomic size.
- Therefore, the most electropositive element in the periodic table is Cs.

- Therefore, the correct answer is option (A).



Note: Remember caesium is an alkali metal and is the most electropositive element in the periodic table. Fluorine is the most electronegative element. As the atomic size increases, ionization enthalpy decreases and electropositivity increases.