Question

Explain the bond formation in sodium chloride.

Answer 1

Bond formation in Sodium Chloride $\left(NaCl\right)$

$NaCl$ is an ionic solid.

The general electronic configurations are...

$Na-{1S}^2{2S}^2{2P}^6{3S}^1$

$Cr-{1S}^2{2S}^2{2P}^6{3S}^2{3P}^5$

$Na$ is in excess of one electron to attain octet configuration.

$Cl$ has one electron less to attain octet configuration.

So, the excess one electron is transferred to a chlorine atom from $Na\left(sodium\right)$ atom and thus an ionic bond is formed due to the transfer of electrons. Chloride being a mole electronegative atom than sodium tends to attract the shared electron pair towards itself.

Thus $NaCl$ has ${Na}^{+}$ and ${Cl}^{-}$ atoms.

${Na}^{+}-{Cl}^{-}$

Chloride attains octet configuration by gaining on the electron.

Sodium attains octet configuration by losing the electron.

Thus Ionic solid $NaCl$ is stable.