Question

Chromate is a polyatomic ion with a charge of negative two. it contains chromium and oxygen in a one-to-four ratio. what is its formula?

Answer 1

$CrO_{4} ^{-2}$

Because the ratio of the chromium to oxygen should be 1 is to 4, the first formula that comes to mind is $CrO_{4} ^{-2}$, with 1 chromium atom and 4 oxygen atoms.

The most common oxidation states of chromium are +6, +3, and +2. Less common are the oxidation states +5, +4, and +1 states, which are present in a few stable compounds. 

The oxide ion has a charge of -2. With the chromate formula being $CrO_{4} ^{-2}$, the chromium oxidation state (x) is calculated to be 

                            x + 4*(-2) = -2
                                        x  = -2+8
                                        x  = +6

The chromium oxidation state is +6, which is a known common oxidation state of chromium. 

Testing other 1:4 ratios of Cr and O by multiplying the ratio with different factors would give, 
 
               $Cr_{2}O_{8} ^{-2}$
               $Cr_{3}O_{12} ^{-2}$

and so on. 

Determining the oxidation state (x) of chromium for the two possible chromate ion formulas,

1.  $Cr_{2}O_{8} ^{-2}$

                          2x + 8*(-2) = -2
                                      2x  = -2+16
                                      2x  = +14
                                        x  = +7

2.  $Cr_{3}O_{12} ^{-2}$

                        3x + 12*(-2) = -2
                                      3x  = -2+24
                                      3x  = +22
                                        x  = +7.3

As observed, using chromate ion formulas in which the 1:4 ratio is multiplied by factors greater than 1, the calculated oxidation states of chromium are not  considered to be possible. Therefore, the only possible formula for the chromate ion with a chromium to oxygen ratio of 1:4, and an ionic charge of -2 is $Cr O_{4}$. 
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