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Question: In \[{[Fe{\left( {CN} \right)_5}(N{O^ + })]^{2 - }}\], \[Fe\] has \[ + 3\] state. It can be decided ...

In [Fe(CN)5(NO+)]2−{[Fe{\left( {CN} \right)_5}(N{O^ + })]^{2 - }}, FeFe has +3 + 3 state. It can be decided by:
A) Magnetic measurement
B) Colligative property
C) Colour
D) Hybridization

Explanation

Solution

Oxidation state of an element refers to the number of electrons that are lost, gained or are appeared to be used when joining with another atom in compounds by an atom. It can also be used to determine the ability of an atom to oxidize (which is to lose electrons) or to reduce (to gain electrons) other elements or compounds.

Complete step-by-step answer:
Transition metals have special properties which are different from that of the other metals present in the periodic table. They are known to have multiple potential oxidation states. Reduction leads to a decrease in the oxidation state while oxidation leads to increase in oxidation states.

If an atom is reduced, it has a higher number of valence shell electrons, and therefore a higher oxidation state, and is a strong oxidant. For example, oxygen (O) and fluorine (F) are very strong oxidants. On the other hand, lithium (Li) and sodium (Na) are incredibly strong reducing agents (likes to be oxidized), meaning that they easily lose 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.
Highest Oxidation State for a Transition metal = Number of Unpaired d−electrons + Two s−orbital electrons{\mathbf{Highest}}{\text{ }}{\mathbf{Oxidation}}{\text{ }}{\mathbf{State}}{\text{ }}{\mathbf{for}}{\text{ }}{\mathbf{a}}{\text{ }}{\mathbf{Transition}}{\text{ }}{\mathbf{metal}}{\text{ }} = {\text{ }}{\mathbf{Number}}{\text{ }}{\mathbf{of}}{\text{ }}{\mathbf{Unpaired}}{\text{ }}{\mathbf{d}} - {\mathbf{electrons}}{\text{ }} + {\text{ }}{\mathbf{Two}}{\text{ }}{\mathbf{s}} - {\mathbf{orbital}}{\text{ }}{\mathbf{electrons}}

Also, the magnetic moment can be found by= n(n+2) = {\text{ }}n(n + 2), when n is the number of unpaired electrons.
[Fe(CN)5(NO+)]2−{[Fe{\left( {CN} \right)_5}(N{O^ + })]^{2 - }}.
The number of unpaired electrons in Fe, it can be used to calculate its oxidation state.

Thus the correct answer is option ‘A’.

Note: In transition meals when the number of unpaired valence electrons increases, the d-orbital increases and thus the highest oxidation state increases. This happens as these valence electrons are unstable and are ready to bond with other chemical species. This means that as there is an availability of the highest number of unpaired valence electrons, the oxidation states would be highest in the middle of transition metal periods.