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I understand that Sulfur can have an expanded orbital, thus it has a d orbital. But if you do the electron configuration for Sulfur it is [Ne]3s^2 3p^4. I don't understand how it has a d orbital. Can someone please explain?
It has access to its d orbital because it is in the 3p block, which is close enough to the 3d sub shell to use when making bonds. However, in its ground state and when it is not making any bonds, its 3d sub shell is not in use.
As mentioned above, all elements in period 3 have access to their d-orbitals because their principal quantum number n=3. With n=3, the angular quantum number l can be 0, 1, and 2, which corresponds to s, p, and d-orbitals respectively. The d-orbital of sulfur simply isn't in use during its ground state, but can still be accessed to form an expanded octet.
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