Question

How to understand what all oxidation states can be exhibited by d block elements??

Answer 1

Like Fe is capable of showing just two oxidation states (2 and 3)
Mn (4,3,2) and so on..

Answer 2

@cuanchi

Answer 3

@cuanchi pls help

Answer 4

Can you explain for iron?? (All the oxidation states and not just the highest)

Answer 5

Hey wait u said that the elctron would move to s orbital ..
Right ??
Bt how is it possible the s orbital is already filled with two electrons and even s orbital has a maximum occupancy of two electrons so where does it hold an extra electron of d orbital??

Answer 6

It was copied from here, http://chemwiki.ucdavis.edu/Core/Inorganic_Chemistry/Descriptive_Chemistry/Elements_Organized_by_Block/3_d-Block_Elements/1b_Properties_of_Transition_Metals/Electron_Configuration_of_Transition_Metals/Oxidation_States_of_Transition_Metals
Electrons are first lost from the 4s orbital though (because they are of higher energy), Fe(II) has lost both 4s electrons and Fe(III) has lost the 6th 3d electron.

Answer 7

@aaronq do u know a proper way to know what all oxidation states can be exhibited by d block elements...
Like say of Fe, or Mn...can you give the proper method to work these out..??

Answer 8

Not in a formal and quantitative way (which needs QM and, probably, computers) but you can provide some explanation knowing their oxidation states and knowing their initial electronic configuration. You can make \( some\) predictions about the possible oxidation states based on the fact that a full set of orbitals and half-full sets of orbitals are more stable than intermediary ones.

Electrons will first be lost first from the highest energy s orbitals available and not the d. For example, Zn only has one oxidation state 2+, because once those electrons are lost the ion is left with a full d orbital set.

Likewise, Mn has 7 electrons to lose (i.e. valence electrons), 2 from the 4s and 5 from the 3d orbitals. This explains how it can achieve a 7+ oxidation state.

Iron has 8 electrons. Fe(II) when it loses the 2 4s electrons.
Fe(III) when it loses the (III) when it loses the 6th d electron.
Further oxidation states when it loses additional electrons.

Co, likewise, loses electrons from the 4s and the 7th d electron. You may ask why Co doesn't keep losing electrons d electrons, and the answer probably due the number of protons and shielding. The Co nucleus has more protons than Fe. In Co the attraction between protons and electrons (effective nuclear charge) and the ionization energy in the environment are balanced and won't be further ionized.. unless they encounter higher energy particles or environments.

Answer 9

U mean to say that Mn can exhibit all the oxidation states like +2,+3,+4,+5,+6,+7
And Fe also has the knack to exhibit the oxidation states as
+2,+3,+4,+5..

Answer 10

yup, i forgot the word "up", in "achieve up to a 7+ oxidation state."

I mentioned that for Fe, it just keeps losing d electrons.

Answer 11

Keep losing ..
That means it can also exhibit +8 state

Answer 12

+6,+7

Answer 13

i assumed you knew (or could look up) that Fe is only observed in nature with an oxidation state of up to 6+.

Answer 14

Don't v have to count for the two paired electrons in d orbital!?

Answer 15

i dont understand what you mean by that

Answer 16

I want to say that like in Fe we have 6 electrons in d orbital out of which two are paired up ..
So in finding out the oxidation states we don't have to look for these paired electrons in d-orbital..
Is it true?