Question

What are the limitations of Octet Rule?

Answer 1

It's not applicable on the third period/row and on. I wouldn't call it any sort of rule because there are so many "exceptions". Some include sulfur, phosphorus, chlorine, many transition metals, etc.

The octet "rule" arises from how many allowed quantum number combinations there are for each given period of the periodic table and is in accordance with the Pauli Exclusion Principle.

The Pauli Exclusion Principle states that no two electrons may share EXACTLY the same quantum numbers.

Remember that. For the second period, which follows the octet rule:

2S ORBITALS

• The `2s` orbital has `(n,l,m_l) = (2,0,0)`.
• If two electrons are in the same orbital, the only quantum number that can be different is `m_s`, which is `pm"1/2"`.
• Therefore, one `2s` orbital multiplied by two electron spins is equal to two maximum electrons in a `2s` subshell.

2P ORBITALS

• The `2p` orbital has `(n,l,m_l) = (2,1,[-1,0,+1])`.
• If two electrons are in the same orbital, the only quantum number that can be different is `m_s`, which is `pm"1/2"`.
• Therefore, three `2p` orbitals multiplied by two electron spins is equal to six maximum electrons in a `2p` subshell.

As a result, `2 + 6 = \mathbf(8)` possible valence electrons allowed for the second period elements, which gives us the octet "rule".

But clearly this does not hold true for period 3. Let's do the same for that. Wait a minute... `3d` comes into play, right?

3S ORBITALS

• The `3s` orbital has `(n,l,m_l) = (3,0,0)`.
• If two electrons are in the same orbital, the only quantum number that can be different is `m_s`, which is `pm"1/2"`.
• Therefore, one `3s` orbital multiplied by two electron spins is equal to two maximum electrons in a `3s` subshell.

3P ORBITALS

• The `3p` orbital has `(n,l,m_l) = (3,1,[-1,0,+1])`.
• If two electrons are in the same orbital, the only quantum number that can be different is `m_s`, which is `pm"1/2"`.
• Therefore, three `3p` orbitals multiplied by two electron spins is equal to six maximum electrons in a `3p` subshell.

3D ORBITALS

• The `3d` orbital has `(n,l,m_l) = (3,2,[-2,-1,0,+1,+2])`.
• If two electrons are in the same orbital, the only quantum number that can be different is `m_s`, which is `pm"1/2"`.
• Therefore, five `3d` orbitals multiplied by two electron spins is equal to ten maximum electrons in a `3d` subshell.

You should get `2 + 6 + 10 = \mathbf(18)` possible valence electrons that can exist in period 3.

That is why chromium (`"Cr"`) can utilize `12` valence electrons, for instance.