Question

Why cesium ( at.no.55) requires little energy to release its one electron present in the outermost shell

Answer 1

The atomic radius (distance between nucleus and outermost shell) of Caesium is quite bigger. Thus, the outermost electron does not experience a great force of attraction from the nucleus. Hence, it is easy for it to give the electron in the outer most shell and requires little energy to release one electron

Answer 2

Because it is large in size and on losing an electron an element attains noble gas configuration.

Explanation:

The ionization energy is the energy required to remove an electron from the valence shell of an isolated gaseous atom of an element.

The ionization energy is the energy required to remove an electron from the valence shell of an isolated gaseous atom of an element.

Factors on which ionization energy depend:

a) size of an element ,larger the size more lesser the effective nuclear charge on electron thus lower the ionization energy.

b) the electronic configuration of an element. If on losing an electron an element attains noble gas configuration or half filled stability then it will easily lose an electron.

The size increases as we move from top to bottom in a group. As, the electrons get added up in the new shell, the electron in the outermost orbital gets far away from the nucleus. And hence, the electron will be easier to remove. Thus, the ionization energy decreases from top to bottom in a group.

Cesium with atomic number of 55 has 55 electrons with noble gas configuration of $Xe6s^1$. As on losing one electron, it acquires noble gas configuration, thus can easily lose one electron.

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