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The iron in the compound has a charge of [tex]Fe^{2+}[/tex]. What is the formula of iron(II) chromate?

Chromate ion: [tex]CrO_4^{2-}[/tex]

\begin{tabular}{cc}
A. [tex]Fe(CrO_4)_2[/tex] & B. [tex]FeCr_2O_7[/tex] \\
C. [tex]Fe_2CrO_4[/tex] & D. [tex]FeCrO_4[/tex] \\
\end{tabular}

Sagot :

GinnyAnswer
To determine the formula of iron(II) chromate, we need to understand the charges of the ions involved and ensure that the compound is electrically neutral.

1. Identify the charges of the ions:
- Iron (II) ion has a charge of [tex]\( Fe^{2+} \)[/tex].
- Chromate ion has a charge of [tex]\( CrO_4^{2-} \)[/tex].

2. Balancing the charges:
- Since the iron (II) ion has a charge of [tex]\( +2 \)[/tex] and the chromate ion has a charge of [tex]\( -2 \)[/tex], we can see that the charges are equal in magnitude but opposite in sign. Therefore, they can combine in a 1:1 ratio to form a neutral compound.

3. Writing the formula:
- Given that one [tex]\( Fe^{2+} \)[/tex] ion combines with one [tex]\( CrO_4^{2-} \)[/tex] ion to balance the charges, the resulting compound will have the formula [tex]\( FeCrO_4 \)[/tex].

Thus, the formula of iron(II) chromate is [tex]\( FeCrO_4 \)[/tex].
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