`Cr_(2)O_(7)^(2-) + 14H^(+) to I_(2) + 2Cr^(3+) + 7H_(2)O`
Which ions are not in balanced position in above reaction?

Options

A. `H^(+)` and `H_(2)O`
B. `Cr_(2)O_(7)^(2-)` and `Cr^(3+)`
C. `I^(-)` and `I_(2)`
D. All are balanced

Answer

C

Explanation

Since charge is not balanced for both the sides, on solving we get:
`14H^(+) + Cr_(2)O_(7)^(2-) + 6I^(-) to 3I_(2) + 2Cr_(2)^(3+) + 7H_(2)O`
Now, charge on LHS = charge on RHS
Hence, `I^(-)` and `I_(2)` are not balanced in the given equation.