Question

The stability of ferric ion is due to
(A) half filled f-orbitals
(B) half filled d-orbitals
(C) completely filled f-orbitals
(D) completely filled d-orbitals

Answer 1

Hint: Ferric is an ionic form of iron which has valency as three plus. Ferrous is also an ionic form of iron but it has a valency as two plus. The stability of half filled and fully filled elements or ions are more than the other configurations.

Complete step by step solution:
Iron is a chemical element of group eight it is denoted by the symbol Fe. Atomic number of iron is twenty-six. It belongs to the first transition series. Iron is the fourth most common element in the earth's crust. The most common oxidation states of iron are iron (ll) and iron (lll). Iron shares many common properties of group eight and transition metals.
Iron has twenty-six electrons in one atom. So, the electronic configuration of iron is as follows:
First two electrons are present in the s orbital of the first shell.
Then we come to the second shell where there is s orbital and p orbital present. So, again another two electrons are filled in the s orbital of the second shell, that is they fill 2s. Then to fill p orbital we need six electrons, so 2p is filled completely by another six electrons.
So, in the same way we fill the 3s and 3p orbital of the third shell. Before filling the 3d orbital we need to fill the s orbital of the next shell, that is we need to completely fill the s orbital of the fourth shell.
Therefore, after filling 4s by two electrons we fill 3d by remaining electrons. As iron only has twenty-six electrons the d orbital is not filled completely. In d orbital only six orbitals are filled.
So, when we say ferric it means iron loses three electrons, that is it loses two electrons from 4s orbital and one electron from 3d orbital. So, we can say that a ferric d orbital is filled by 5 electrons, that is , half-filled and half-filled orbitals are stable. So, ferric is stable due to half-filled d orbital.
The electronic configuration of iron is $ 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^6} $ . The electronic configuration of ferric ion is $ 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^0}3{d^5} $ .
The correct answer is option B.

Note:
Iron also forms a wide range of oxidation states that is from minus two to plus seven such as ferrocene, ferrioxalate and pursian blue.