(IIT JEE 2000)
a) MnO4-
b) Cr(CN)63-
c) NiF62-
d) CrO2Cl2
Logic:
Oxidation state (or oxidation number) indicates the formal charge on one atom when all other atoms are removed from the molecule or ion. The sum of oxidation numbers of all the atoms is equal to the charge on the molecule or ion.
Solution:
a) On MnO4-, the charge is -1 and hence sum of oxidation numbers is equal to -1.
Therefore:
Mn + 4('O') = -1
Mn + 4(-2) = -1
Mn = +7
Note: The usual oxidation number of 'O' in compounds is -2.
b) On Cr(CN)63-, the charge is -3.
Hence:
Cr + 6(CN-) = -3
Cr + 6(-1) = -3
Cr = +3
Note: The charge on cyanide ion is -1.
c) On NiF62-, the charge is -2
Thus:
Ni + 6(F) = -2
Ni + 6(-1) = -2
Ni = +4
d) The charge on CrO2Cl2 is zero.
Therefore:
Cr + 2(O) + 2(Cl) = 0
Cr + 2(-2) + 2(-1) = 0
Cr = +6
Conclusion:
The correct option is 'd'.
(AdiChemistry)
a) +8 and +8
b) +8 and +6
c) +6 and +8
d) +6 and +6
Explanation:
* Both will show maximum oxidation number - 8. However, Ru will be less stable in this oxidation state.
Conclusion:
Correct option is: 'a'.
(IIT JEE 2000)
a) [MnO4]-
b) [Co(NH3)6]3+
c) [Fe(CN)6]3-
d) [Cr(H2O)6]3+
Explanation:
| Compound | Constituent ions/molecules | Atomic number of metal | |||
| outer e.c of metal | no. of electrons removed to get metal ion | outer e.c of metal ion | |||
| [MnO4]- | Mn7+ + 4O2- | 25 | 3d5 4s2 | 7 | 3d0 4s0 |
| [Co(NH3)6]3+ | Co3+ + 6NH3 | 27 | 3d7 4s2 | 3 | 3d6 4s0 |
| [Fe(CN)6]3- | Fe3+ + 6CN- | 26 | 3d6 4s2 | 3 | 3d5 4s0 |
| [Cr(H2O)6]3+ | Cr3+ + 6H2O | 24 | 3d5 4s1 | 3 | 3d3 4s0 |
Conclusion:
Correct option is: 'a'.
(IIT JEE 2004)
a) MnO2, FeCl3
b) [MnO4]-, CrO2Cl2
c) [Fe(CN)6]3-, [Co(CN)3]
d) [NiCl4]2-, [CoCl4]-
Explanation:
| Option | Compound | Calculation of oxidation number | Comment |
| a | MnO2 | Mn + 2O = 0 Mn + 2(-2) = 0 Mn = +4 |
For Mn, the highest oxidation number is +7 |
| FeCl3 | Fe + 3Cl = 0 Fe + 3(-1) = 0 Fe = +3 |
Highest oxidation state for Fe is +6 | |
| b | [MnO4]- | Mn + 4O = -1 Mn + 4(-2) = -1 Mn = +7 |
Mn is in its highest oxidation state +7 |
| CrO2Cl2 | Cr + 2(O) + 2(Cl) = 0 Cr + 2(-2) + 2(-1) = 0 Cr = +6 |
Cr is also in its highest oxidation state +6 | |
| c | [Fe(CN)6]3- | Fe + 6CN- = -3 Fe + 6(-1) = -3 Fe = +3 |
Not in its highest O.S |
| [Co(CN)3] | Co + 3CN- = 0 Co + 3(-1) = 0 Co = +3 |
Highest oxidation number possible for Co is +4 | |
| d | [NiCl4]2- | Ni + 4Cl = -2 Ni + 4(-1) = -2 Ni = +2 |
For Ni also the highest O.S possible is +4 |
| [CoCl4]- | Co + 4Cl = -1 Co + 4(-1) = -1 Co = +3 |
Conclusion:
Correct option is: 'b'.
1) What is the common oxidation state of d-block elements?
2) Why transition elements show variable oxidation states?