This from ionic compounds escape room

Fe2O3 - S

PbO2 - D

PbO - I

Fe2O - U

Fe(OH)3 - H

FeO2 - T

FeO - R

Pb2O - G

Pb(OH)2 - O

FeOH - N

Pb(OH)3 - A

Explanation:

In writing the formula of ionic compounds we consider the valency or oxidation state of each ion.

For instance, given the compound iron II oxide. The oxidation states of both iron and oxygen are +2 and -2 respectively. Ignoring the charges, this cancels out and we have FeO as the correct formula of the compound.

For Iron III oxide, the oxidation states of iron and oxygen are +3 and -2 respectively, the both atoms exchange charges. If we ignore the signs and write the exchanged numbers as subscripts we obtain the formula Fe2O3.

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.