# d-BLOCK - OXIDATION STATES MCQ IIT JEE

## 1) Amongst the following, identify the species with an atom in oxidation state +6.

(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'.

## Related questions

### 2) The maximum oxidation states shown by Ru and Os, respectively, are:

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'.

### 3) The complex ion which has no d-electron in the central metal ion is:

(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'.

### 4) The pair of compounds having metals in their highest oxidation state is:

(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'.

### Homework

1) What is the common oxidation state of d-block elements?

2) Why transition elements show variable oxidation states?

 Color of ions of transition elements (d-block) >