Question

Write the electronic configurations of the following ions :
a.) $H^{-}$
b.) $N{a}^{+}$
c.) $O^{2-}$
d.) $F^{-}$

Answer 1

Hint: The electronic configuration is the representation of filling of electrons in the orbitals. The filling takes place as - $1s2s2p3s3p4s3d$ etc. The s-orbital can accommodate two electrons, p-orbital can accommodate six electrons and d-orbital can accommodate a total ten electrons.

Complete step by step answer:
The electronic configuration is the way of showing the number of electrons present in an orbital. The filling of electrons takes place according to Aufbau’s principle. Let us see the electronic configuration of following ions as -
a.) $H^{-}$= The hydrogen normally has one electron. The negative charge shows the presence of one more electron. Thus, the electronic configuration is - $1s^2$
b.) $N{a}^{+}$= The sodium atom normally has 11 electrons. The positive charge shows the absence of one electron. Thus, the electronic configuration is - $1s^{2}2s^{2}2p^6$
c.) $O^{2-}$= The oxygen normally has 8 electrons. The negative charge shows the presence of two more electrons. Thus, the electronic configuration is - $1s^{2}2s^{2}2p^6$
d.) $F^{-}$= The fluorine normally has nine electrons. The negative charge shows the presence of one more electron.

Thus, the electronic configuration is - $1s^{2}2s^{2}2p^6$

Note: It must be noted that the negative charge comes when any atom gains extra electrons and positive charge comes when any atom donates the electrons. Thus, if there is negative charge, the number of electrons should be added and if positive charge, then the number of electrons should be subtracted.