Question

The higher oxidation states are usually exhibited by the members in the middle of a series of transition elements. explain

Answer 1

ʜɪ ᴛʜᴇʀᴇ, ᴄʜᴀɴᴅᴀɴ!

ʜᴇʀᴇ'ꜱ yᴏᴜʀ ᴀɴꜱᴡᴇʀ ᴛᴏ ᴛʜᴇ qᴜᴇꜱᴛɪᴏɴ:

Q. whч αrє thє híghєr σхídαtíσn ѕtαtєѕ uѕuαllч єхhíвítєd вч thє mєmвєrѕ ín thє míddlє σf α ѕєríєѕ σf trαnѕítíσn єlєmєntѕ?

Ans. (Extra: To clearly understand the phenomena of oxidation states of transition metals, we have to understand how the unpaired d-orbital electrons bond!
There are five orbitals in the d subshell manifold. As the number of unpaired valence electrons increases, the d-orbital increases, the highest oxidation state increases. This is because unpaired valence electrons are unstable and eager to bond with other chemical species.)
*****************************************
ᴛʜᴇ ᴇxᴀᴄᴛ ʀᴇᴀꜱᴏɴ:-
Therefore, the oxidation states would be the highest in the very middle of the transition metal periods due to the presence of the highest number of unpaired valence electrons.

To determine the oxidation state, unpaired d-orbital electrons are added to the 2s orbital electrons since the 3d orbital is located before the 4s orbital in the periodic table.
****************************************
ᴇxᴩʟᴀɴᴀᴛɪᴏɴ:-
For example: Scandium has one unpaired electron in the d-orbital. It is added to the 2 electrons of the s-orbital and therefore the oxidation state is +3. So that would mathematically look like: 1s electron + 1s electron + 1d electron = 3 total electrons = oxidation state of +3. The formula for determining oxidation states would be (with the exception of copper and chromium):

Highest Oxidation State for a Transition metal = Number of Unpaired d-electrons + Two s-orbital electrons

The number of d-electrons range from 1 (in Sc) to 10 (in Cu and Zn).

Scandium is one of the two elements in the first transition metal period which has only one oxidation state (zinc is the other, with an oxidation state of +2). All the other elements have at least two different oxidation states. Manganese, which is in the middle of the period, has the highest number of oxidation states, and indeed the highest oxidation state in the whole period since it has five unpaired electrons (see table below).

It was mentioned previously that both copper and chromium do not follow the general formula for transition metal oxidation states. This is because copper has 9 d-electrons, which would produce 4 paired d-electrons and 1 unpaired d-electron. Since copper is just 1 electron short of having a completely full d-orbital, it steals an electron from the s-orbital, allowing it to have 10 d-electrons.

Likewise, chromium has 4 d-electrons, only 1 short of having a half-filled d-orbital, so it steals an electron from the s-orbital, allowing chromium to have 5 d-electrons.

*****************************************
(In the image above, the blue-boxed area is the d block, or also known as transition metals.)
****************************************

HΩPΣ IT HΣLPS! :-)

ⓑⓔ ⓑⓡⓐⓘⓝⓛⓨ✌

Answer 2

Here is your answer!

The elements which give the greatest number of oxidation states occur in or near the middle of the series. For example, manganese exhibits all the oxidation states from +2 to +7. The lesser number of oxidation states at the extreme ends result either from too few electrons to lose or share or too many d electrons (hence fewer orbitals available in which to share electrons with others) for higher valence. For example Scandium and Titanium can lose a maximum of three and four electrons respectively, because of which they exhibit lower oxidation state. Manganese on the other hand has seven unpaired electrons, and hence it can exhibit oxidation state from +2 to +7, depending upon the number of electrons it loses. Copper and zinc, which are present on the extreme right of the table have completely filled d-orbitals, hence they do not have many orbitals to share electrons with others.

Hope it helps you.