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State the most stable oxidation state of Cr?
A. +3
B. +4
C. +2
D. +6

Answer
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Hint: Chromium has an atomic number of 24 and the common oxidation state of chromium is +6, +3 and +2. The oxidation state of an atom is the total no. of the electron which can be lost or gained by an atom to form a bond with other molecules.

Complete answer:
- In the given question, we have to find the most stable oxidation state of chromium from the given options.
- Firstly, we have to write the electronic configuration of the chromium atom i.e.
(Ar) 3d5 4s1
- So, firstly the electron will be lost from the outermost shell of the atom i.e. 4s and then from the 3d shell.
- Now, as we know that chromium shows a wide range of oxidation states among which +3 is the most stable oxidation state.
- Whereas +1, +4 and +5 oxidation states of chromium are rare.
- When chromium loses 3 electrons then the electronic configuration becomes (Ar) 3d3 which is most stable because at the t2g level the dxy, dyz and dxzorbitals are half-filled.
- It means that one electron is present in each orbital due to which it is more stable than the d5 configuration.
- That’s why the oxidation state of +3 is the most stable oxidation state than the +4, +2 and +6 oxidation states.
- Some compounds show an oxidation state of +2 such as CrO and Cr(OH)2. Whereas the compounds such as H2Cr2O7 showing +6 oxidation state are strong oxidiser.
Therefore, the most stable oxidation state is +3 due to the stable (Ar) 3d3 configuration.

Hence option A is the correct one.

Note: Int2g, 't' means the triply degenerate orbitals and g stands for gerade which consist of three orbitals of d subshell i.e. dxy, dyz and dxzorbitals. When there is a presence of an unsymmetrical ligand, then undergo splitting of d orbitals into t2g and e.g. levels take place.


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