Question

Which of the elements listed below is most likely to exhibit an expanded octet in its compounds? o na c n s?

Answer 1

Answer: The element showing expanded octet is Sulfur (S).

Explanation:

Expanded octet is shown by few elements. This rule is applied when an element can surround more than 8 electrons around them. Some of the their period elements show expanded octet. These elements are Silicon (Si), Phosphorous (P), Sulfur (S) and Chlorine (Cl).

These elements show expanded octet because they contain vacant d-orbitals.

From the given options:

Oxygen, carbon and nitrogen atoms belong to period 2 and can only surround 8 electrons around it.

Sodium belongs to period 3 and also have d-orbitals. The last electron of sodium is present in s-orbital and the energy difference of s-orbital to d-orbital is more and this transition is not favorable.

Sulfur is the only element among the options which show expanded octet because of the presence of vacant 3d-orbital. For Example: In $SF_6$ compound, the number of electrons around sulfur are 12.

Hence, the element showing expanded octet is Sulfur (S).

Answer 2

Among the elements listed below, Sulfur (S) is the one that is most likely to exhibit an expanded octet in its compounds. It is included in the exceptions in the Octet Rule. For example, SF6, the sulfur atom has 6 bonds connected to it. It has a total of 12 electrons surrounding it rather than 8 electrons only.