Which one of the following equations is the net ionic equation for the reaction of aqueous sodium hydroxide (NaOH) with aqueous hydrochloric acid (HCl)? HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) H+(aq) + Cl‒(aq) + Na+(aq) + OH‒(aq) → Na+(aq) + Cl‒(aq) + H2O(l) H+(aq) + Cl‒(aq) + Na+(aq) + OH‒(aq) → NaCl(aq) + H2O(l) H+(aq) + OH‒(aq) → H2O(l)

H+(aq) +  OH-(aq) → H2O(l)

Explanation:

Step 1: Data given

aqueous sodium hydroxide = NaOH(aq)

aqueous hydrochloric acid =HCl(aq)

Step 2: The (un)balanced equation

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

For 1 mol HCl we need 1 mol NaOH to produce 1 mol NaCl and 1 mol H2O

This equation is already balanced.

Step 3: The net ionic equation

H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) → Na+(aq) + Cl-(aq) + H2O(l)

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

H+(aq) +  OH-(aq) → H2O(l) (The last option is correct)

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.