Question #b7814

The #HCl# 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 #He#). 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 #Cl#, each receiving the extra electron it needed (for more on #HCl#'s covalent bond read about electronegativity).

Now, since #HCl# is a strong acid, it dissociates completely in aqueous solution into #H^(+)#(which form #H_3O^(+)#) and #Cl^(-)# ions as a result of its reaction with water, which acts as a base.

Any further reactions between #HCl# and say, s strong base, will take place between the molecules' ions, like this:

#HCl_((aq)) + NaOH_((aq)) rightleftharpoons NaCl_((aq)) + H_2O_((l))#

#H_3O_((aq))^(+) + Cl_((aq))^(-) + Na_((aq))^(-) + OH_((aq))^(-) rightleftharpoons Na_((aq))^(+) + Cl_((aq))^(-) + 2H_2O_((l))#

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_3O_((aq))^(+) + OH_((aq))^(-) rightleftharpoons 2H_2O_((l))#