To find the oxidation number (O.N) of Chromium in the dichromate ion, [tex]\(\text{Cr}_2\text{O}_7^{2-}\)[/tex], we can use the following steps:
1. Recall that the sum of the oxidation numbers of all atoms in a polyatomic ion must equal the overall charge of the ion.
2. The oxidation number of oxygen is typically -2.
3. Let the oxidation number of chromium be [tex]\( x \)[/tex].
Now, [tex]\( \text{Cr}_2\text{O}_7^{2-} \)[/tex] consists of 2 chromium ([tex]\(\text{Cr}\)[/tex]) atoms and 7 oxygen ([tex]\(\text{O}\)[/tex]) atoms. We can set up the equation based on the sum of the oxidation numbers:
[tex]\[ 2(\text{O.N of Cr}) + 7(\text{O.N of O}) = \text{Overall charge of the ion} \][/tex]
Substitute the known values into the equation:
[tex]\[ 2(x) + 7(-2) = -2 \][/tex]
Simplify the equation:
[tex]\[ 2x - 14 = -2 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ 2x = 12 \][/tex]
[tex]\[ x = 6 \][/tex]
Therefore, the oxidation number of chromium ([tex]\(\text{Cr}\)[/tex]) in the dichromate ion [tex]\( \text{Cr}_2\text{O}_7^{2-} \)[/tex] is [tex]\( +6 \)[/tex].
