Chemistry · Periodicity in Elements
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In which of the following chromium has different oxidation state?
- A
K2Cr2O7
- B
CrCl3
- C
CrO3
- D
K2CrO4
i.Oxidation state in K2Cr2O7=2+2x+(-14)=0
=2x=12;x=6.
ii.Oxidation state in CrCl3=x+(-3)=0
=x=3
iii.Oxidation state in CrO3=x+(-6)=0
=x=6.
There is no need to further find out the oxidation state of Cr in K2CrO4 because we already have a compound(CrCl3) having an odd oxidation state,so that is our answer.
The oxidation state of Cr in K2Cr2O7 is +6. Each oxygen contributes -2, and potassium contributes +1, leading to the equation: 2(1) + 2x + 7(-2) = 0, which simplifies to x = 6.
The oxidation state of Cr in CrCl3 is +3. Each chlorine contributes -1, resulting in the equation: x + 3(-1) = 0, which simplifies to x = 3.
The oxidation state of Cr in CrO3 is +6. Each oxygen contributes -2, leading to the equation: x + 3(-2) = 0, which simplifies to x = 6.
The oxidation state of Cr in K2CrO4 is +6. Each oxygen contributes -2, and potassium contributes +1, resulting in the equation: 2(1) + x + 4(-2) = 0, which simplifies to x = 6.
Tagged under Chemistry · Periodicity in Elements · 2016