# Question #b7814

Jan 1, 2015

The $H C l$ molecule is formed between a neutral hydrogen atom and a neutral chlorine atom.

Hydrogen has 1 electron in its 1s-orbital, but it needs 1 more to form a stable configuration and complete its 1s-orbital (essentially reaching the configuration of $H e$). Chlorine, being in group 17, has 7 electrons in its outermost shell, and needs one more to form a stable octet.

A covalent bond thus forms between $H$ and $C l$, each receiving the extra electron it needed (for more on $H C l$'s covalent bond read about electronegativity).

Now, since $H C l$ is a strong acid, it dissociates completely in aqueous solution into ${H}^{+}$(which form ${H}_{3} {O}^{+}$) and $C {l}^{-}$ ions as a result of its reaction with water, which acts as a base.

Any further reactions between $H C l$ and say, s strong base, will take place between the molecules' ions, like this:

$H C {l}_{\left(a q\right)} + N a O {H}_{\left(a q\right)} r i g h t \le f t h a r p \infty n s N a C {l}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)}$

${H}_{3} {O}_{\left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-} + N {a}_{\left(a q\right)}^{-} + O {H}_{\left(a q\right)}^{-} r i g h t \le f t h a r p \infty n s N {a}_{\left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-} + 2 {H}_{2} {O}_{\left(l\right)}$

The net ionic equation, which is determined by eliminating spectator ions (ions present on both the reactants' and on the products' side) will be:

${H}_{3} {O}_{\left(a q\right)}^{+} + O {H}_{\left(a q\right)}^{-} r i g h t \le f t h a r p \infty n s 2 {H}_{2} {O}_{\left(l\right)}$