The oxidation number of Chromium in Na$_2$Cr$_2$O$_7$ is

Analyze the question You need the oxidation number of chromium (Cr) in sodium dichromate, Na₂Cr₂O₇.

Concept In a neutral compound, the oxidation numbers of all atoms add up to zero. The fixed values to use are:

• Na (Group 1) is always +1
• O is usually -2 (except in peroxides)

Solution Let the oxidation number of Cr be x.

Na₂Cr₂O₇:

• 2 Na atoms: 2(+1) = +2
• 2 Cr atoms: 2(x) = 2x
• 7 O atoms: 7(-2) = -14

Sum = 0:

2 + 2x - 14 = 0 2x - 12 = 0 2x = 12 x = +6

Each Cr atom has an oxidation number of +6. Option D (+12) is the total for the two Cr atoms, not the value for one atom, which is a common trap.

Answer The correct option is **B.

Tip Always find the oxidation number per atom. If the subscript is 2, divide your total by 2. Also remember that dichromate (Cr₂O₇²⁻) and chromate (CrO₄²⁻) both have Cr in the +6 state. This comes up often in redox questions.