what are oxidation numbers of (NH4)2CrO4?

for each element

4 Answers

  • N -3

    H +1

    Cr +6

    O -2

    The best way to find answers to such problems is to remember

    i] the sum of the oxidation numbers in a compound equals zero

    ii] the sum of the oxidation numbers in an ions equals the charge on the ion

    iii] in normal chemistry oxygen has fixed oxidation number of -2 and hydrogen +1 [except in hydrides]

    The normal way is to set up a smple algebra expression for each ion

    CrO4 2- Cr + (4 x -2) = -2 hence Cr = +6

    NH4+ N + 4 = +1 hence N = -3

  • H is almost always 1+, and O is almost always 2-. It’s easiest if you break the formula into its component ions, which are NH4^+ and and CrO4^2-. For the N in ammonia, there are four H+’s, and the overall charge is +1, so the N must be -3. Similarly, in chromate ion, there are 4 O’s, so a total of 8-. The overall charge is 2-, so the Cr must be +6.

  • Nh4 2cro4

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