If an element can show only +2 and +4 oxidation states in its compounds, which of the below can be its electronic configuration? A) 3d44s2 B) 2s22p4 C) 2s22p2 D) 3d54s1 E) 3s23p1

The oxidation state of an element is determined by the number of electrons it can lose, gain, or share when forming compounds. In this case, we are looking for an element that can show +2 and +4 oxidation states, meaning it can lose 2 or 4 electrons.

Let's examine each option:

A) 3d44s2: This configuration can lose 2 electrons from the 4s orbital to form a +2 oxidation state. It can also lose 4 electrons (2 from the 4s orbital and 2 from the 3d orbital) to form a +4 oxidation state. So, this could be a possible answer.

B) 2s22p4: This configuration can gain 2 electrons to complete the 2p orbital, forming a -2 oxidation state. It cannot form a +2 or +4 oxidation state.

C) 2s22p2: This configuration can gain 2 electrons to complete the 2p orbital, forming a -2 oxidation state. It cannot form a +2 or +4 oxidation state.

D) 3d54s1: This configuration can lose 1 electron from the 4s orbital to form a +1 oxidation state. It can also lose 5 electrons (1 from the 4s orbital and 4 from the 3d orbital) to form a +5 oxidation state. It cannot form a +2 or +4 oxidation state.

E) 3s23p1: This configuration can lose 1 electron from the 3p orbital to form a +1 oxidation state. It can also gain 2 electrons to complete the 3p orbital, forming a -2 oxidation state. It cannot form a +2 or +4 oxidation state.

Therefore, the only electronic configuration that can show both +2 and +4 oxidation states is option A) 3d44s2.