Question

Show the formation of $N{a}_{2}O$ and $MgO$ by transfer of electrons:

Answer 1

Atomic number of sodium - 11

Electronic configuration of sodium - $1s^{2}2s^{2}2p^{6}3s^1$

It has one valence electron, it will lose one electron to get octet configuration

Atomic number of oxygen - 8

Electronic configuration of oxygen - $1s^{2}2s^{2}2p^4$

It is two electrons short of octet configuration, thus it will gain two electrons.

During the bond formation between sodium and oxygen, two sodium atoms loses one electron each to oxygen atom,

$2Na-2e^{-}\to 2N{a}^{+}O+2e^{-}\to O^{-2}2N{a}^{+}+O^{-2}\to N{a}_{2}O$

Atomic number of Magnesium - 12

Electronic configuration of Magnesium -$1s^{2}2s^{2}2p^{6}3s^2$

It has two valence electrons , it will lose two electrons to get octet configuration

Atomic number of oxygen -8

Electronic configuration of oxygen $-1s^{2}2s^{2}2p^4$

It is two electrons short octet configuration, thus it will gain two electrons.

During the bond formation between Magnesium and Oxygen, one magnesium atom loses two electrons to oxygen atom.

$Mg-2e^{-}\to M{g}^{2+}O+2e^{-}\to O^{-2}{Mg}^{+2}+O^{-2}\to MgO$