Ammonium dichromate, $(NH X_{4}) X_{2} Cr X_{2} O X_{7}$ , contains two $NH X_{4} X^{+}$ and one $Cr X_{2} O X_{7} X^{2 -}$ (H is $+ 1$ , O is $- 2$ ). Find the sum of the oxidation numbers of all N atoms and all Cr atoms in one formula unit. Give the nearest integer.

Numerical answer type, tolerance 0.5.

Answer: 6

N in

NH X_{4} X^{+}

x + 4 = + 1

x = - 3

; two N give

- 6

. Cr in

Cr X_{2} O X_{7} X^{2 -}

2 y - 14 = - 2

y = + 6

; two Cr give

+ 12

. Sum

= - 6 + 12 = + 6