What is the oxidation number of chromium in potassium dichromate, $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$?

1. $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ is a neutral molecule, so the algebraic sum of all oxidation numbers is $0$. 2. Let Cr have oxidation number $x$. 3. Contributions: 2 K atoms $= 2(+1) = +2$. 7 O atoms $= 7(-2) = -14$. 2 Cr atoms $= 2x$. 4. Sum to zero: $(+2) + 2x + (-14) = 0 \Rightarrow 2x = +12 \Rightarrow x = +6$. 5. Option A is the Cr$^{3+}$ state seen AFTER reduction (e.g. in acidic dichromate titrations). Option C confuses Mn$^{+7}$ in KMnO$_4$. Option D is the SUM $2x$, a common slip when students forget to divide by the number of Cr atoms. _Source: syllabus Chemistry Part 2, Ch 7, §7.3 (Oxidation Number applied to dichromate), pp. 4-6._