Question

An ionic compound forms when lithium ( Z = 3) reacts with oxygen ( Z = 8). If a sample of the compound contains 8.4x 1021 lithium ions, how many oxide ions does it contain?

Answer 1

Answer: The number of oxide ions that must be present is
`4.2* 10^(21)`

Step-by-step explanation:

Lithium is the 3rd element of the periodic table having electronic configuration of
`1s^22s^1`

This element will loose 1 electron to form
`Li^+` ion.

Oxygen is the 8th element of the periodic table having electronic configuration of
`1s^22s^22p^4`

This element will gain 2 electrons to form
`O^(2-)` 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
`Li_2O`

According to mole concept:

1 mole of an ionic compound contains
`6.022* 10^(23)` number of ions.

In the given ionic compound, the number of lithium ions present are
`2* 6.022* 10^(23)` and number of oxide ions present are
`6.022* 10^(23)`

So, in the given ionic compound:

If
`2* 6.022* 10^(23)` number of lithium ions are present, then
`6.022* 10^(23)` number of oxide ions will be present.

So,
`8.4* 10^(21)` number of lithium ions are present, then
`(2* 6.022* 10^(23))/(6.022* 10^(23))* 8.4* 10^(21)=4.2* 10^(21)` number of oxide ions will be present.

Hence, the number of oxide ions that must be present is
`4.2* 10^(21)`