Question

Which one of the elements with the following outer orbital configurations may exhibit the largest number of oxidation states?
(1) 3d54s1 (2) 3d54s2 (3) 3d24s2 (4) 3d34s2

Answer 1

Accd. to me options (4) is correct.
Because it will show +5 oxidation state.
Option 1 will show +1 oxidation state.
Option 2 will show + 2 Oxidation state.
Option 3 will show + 4 Oxidation State

Answer 2

Answer: The correct answer is Option 4.

Explanation:

Oxidation state is defined as the number that is given to an atom when it looses or gains electron. It is always written as a superscript.

When an atom looses electron, it will attain a positive oxidation state and when an atom gains electron, it will attain a negative oxidation state.

All the given configurations are of the 1st transition metals. So, these elements will always exhibit in positive oxidation state.

For the given options:

Option 1:  $3d^54s^1$

This configuration belongs to chromium element and its common oxidation states are +6, +3 and +2.

Option 2:  $3d^54s^2$

This configuration belongs to manganese element and its common oxidation states are +7, +4 and +2.

Option 3:  $3d^24s^2$

This configuration belongs to titanium element and its common oxidation states are +4 and +3.

Option 4:  $3d^34s^2$

This configuration belongs to vanadium element and its common oxidation states are +5, +4, +3 and +2.

Hence, the correct answer is Option 4.