Question

For bromine within it’s electron configuration the valence electrons are 4s² and 4p⁵ and not 3d¹⁰ and 4p⁵, my chemistry teacher explained this as 4s² being less shielded than the 3d orbital. Valence electrons are the electrons that get ionised correct? If the 4s² orbital experiences more Zeff then wouldn’t it be harder to ionise as there’s a greater degree of electrostatic attraction? And the 3d orbital, since it experiences more repulsive forces, would be easier to ionise?

Answer 1

The valence electrons of bromine are in the 4s and 4p orbitals, and they are ionized first due to their higher energy level, despite the initially stronger electrostatic attraction they experience from the nucleus compared to the 3d electrons.

Step-by-step explanation:

The valence electrons of bromine are those electrons in the highest occupied principal energy level, which in this case, include the 4s and 4p orbitals, not the 3d orbital.

Valence electrons are indeed the electrons that are most likely to be involved in chemical reactions and are the ones that can be ionized or shared during the formation of chemical bonds.

While it's true that the 4s orbital experiences more effective nuclear charge (Zeff) than the 3d orbital and thus electrons are more strongly attracted to the nucleus, the energy levels of orbitals are also affected by their principal quantum number.

In the context of ionization, the valence electrons are typically those that are lost first, and for transition metals and elements with similar properties, the 4s electrons are usually lost before the 3d electrons—even though the 4s orbitals are filled before the 3d orbitals when moving across the periodic table.

This seeming contradiction arises because once the 3d orbitals begin to fill, they shield the 4s electrons, and the energy level of the 4s electrons becomes higher than that of the 3d electrons. This makes the 4s electrons easier to ionize despite being more strongly attracted to the nucleus initially.