Question

Under normal condition which one of the following electronic configuration is able to form di-positive ion
(A)$[Ar]4{s^1}$
(B)$[Ne]2{s^2}3{p^6}$
(C)$[Ne]3{s^2}$
(D)None of these

Answer 1

Hint: An atom or a group of atoms which carry a positive or negative electric charge because of the loss or gain of one or more number of electrons are called ions. The ions which are positively charged are called cations and the ions that are negatively charged are called cations. An ion carrying two units of positive charge is called a di-positive ion.

Complete answer:The first option has nineteen electrons, so it is the electronic configuration of the element potassium. The second option has eighteen electrons, so it is the electronic configuration of the element argon. The third option has twelve electrons, so it is the electronic configuration of the element magnesium. The one which wants to get stable will easily lose the di-positive ion. Neon is a noble gas and has stable electronic configuration. If magnesium loses two valence electrons, its electronic configuration will be the same as that of neon. Potassium cannot lose two electrons as it has only one electron in its valence shell. Argon will not lose two electrons because it will reduce its stability. So, magnesium will form a di-positive ion.
Therefore, option C is the correct answer.

Note:
If the outermost shell of an atom is complete, it means it has stable electronic configuration. An incomplete valence shell indicates that the atom has unstable electronic configuration. Half-filled and fully filled orbitals increase the stability.