Question

Na+ ion is said to be a cation.why?
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Answer 1

Explanation:

Cation Formation

Cations are the positive ions formed by the loss of one or more electrons. The most commonly formed cations of the representative elements are those that involve the loss of all of the valence electrons. Consider the alkali metal sodium (Na). It has one valence electron in the third principal energy level. Upon losing that electron, the sodium ion now has an octet of electrons from the second principal energy level. The equation below illustrates this process.

& text{Na} qquad qquad rightarrow quad text{Na}^+ quad + quad text{e}^-\& 1s^22s^22p^63s^1 qquad 1s^22s^22p^6 (text{octet})

The electron configuration of the sodium ion is now the same as that of the noble gas neon. The term isoelectronic refers to an atom and an ion of a different atom (or two different ions) that have the same electron configuration. The sodium ion is isoelectronic with the neon atom. Consider a similar process with magnesium and with aluminum:

& text{Mg} qquad qquad rightarrow quad text{Mg}^{2+} quad + quad 2text{e}^-\& 1s^22s^22p^63s^2 qquad quad 1s^22s^22p^6 (text{octet})\& text{Al} qquad qquad quad rightarrow quad text{Al}^{3+} quad + quad 3text{e}^-\& 1s^22s^22p^63s^23p^1 quad 1s^22s^22p^6 (text{octet})

In this case, the magnesium atom loses its two valence electrons in order to achieve the same noble-gas configuration. The aluminum atom loses its three valence electrons. The Mg 2+ ion, the Al3+ ion, the Na + ion, and the Ne atom are all isoelectronic. For representative elements under typical conditions, three electrons is the maximum number that will be lost.

We can also show the loss of valence electron(s) with an electron dot diagram.

text{Na} cdot & rightarrow text{Na}^++text{e}^-\cdot text{Mg} cdot & rightarrow text{Mg}^{2+}+2text{e}^-

I hope it will help you.