Respuesta :
Answer: The number of oxide ions that must be present is [tex]4.2\times 10^{21}[/tex]
Explanation:
Lithium is the 3rd element of the periodic table having electronic configuration of [tex]1s^22s^1[/tex]
This element will loose 1 electron to form [tex]Li^+[/tex] ion.
Oxygen is the 8th element of the periodic table having electronic configuration of [tex]1s^22s^22p^4[/tex]
This element will gain 2 electrons to form [tex]O^{2-}[/tex] ion.
By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.
So, the chemical formula of compound is [tex]Li_2O[/tex]
According to mole concept:
1 mole of an ionic compound contains [tex]6.022\times 10^{23}[/tex] number of ions.
In the given ionic compound, the number of lithium ions present are [tex]2\times 6.022\times 10^{23}[/tex] and number of oxide ions present are [tex]6.022\times 10^{23}[/tex]
So, in the given ionic compound:
If [tex]2\times 6.022\times 10^{23}[/tex] number of lithium ions are present, then [tex]6.022\times 10^{23}[/tex] number of oxide ions will be present.
So, [tex]8.4\times 10^{21}[/tex] number of lithium ions are present, then [tex]\frac{2\times 6.022\times 10^{23}}{6.022\times 10^{23}}\times 8.4\times 10^{21}=4.2\times 10^{21}[/tex] number of oxide ions will be present.
Hence, the number of oxide ions that must be present is [tex]4.2\times 10^{21}[/tex]
