Question

metalloids having how many electrons in the last orbit?

Answer 1

There are at least 3 effects that go into determining the energy levels of orbitals: Coulomb's law, shielding, and penetration. The first of these, Coulomb's law, is the usual inverse square relationship for the electric force between charged objects. The closer they are, the greater the force. For oppositely charged objects, this means the more strongly they are attracted, and the lower their potential energy. In a single-electron atom, this is mostly the only thing at play, and as a result, the orbitals with the same llquantum number all have the same energy, while those of different nn quantum numbers have different energies.

In a multi-electron atom, though, there are the other two effects I mentioned: electrons in other orbitals can "shield" those in outer orbitals from some of the nuclear charge -- that is, they create a "screen" of negative charge around the positive charge of the nucleus, cancelling some of it out. As a result, the outer orbital rises in energy, as it is a more loosely bound state since there is "effectively" less charge to bind it. Penetration is where one orbital has a part that is inside another, so it experiences less of the shielding effect that that other orbital would otherwise provide, allowing it to be attracted more strongly and lowering its energy.

The effects can be determined by examining the radial distribution functions of the orbitals. The radial distribution function gives the probability that the electron will be found within a certain distance of the nucleus. Formally, it is the integral of the orbital wave function over a sphere of the given radius centered at the nucleus, to get a probability level.

The RDF for the 3d and 4s orbitals looks like

Note that 4s contains several peaks within the space of the 3d orbital. As a result, it experiences penetration of the shielding the 3d would otherwise provide, and is pulled closer and thus its energy level is lower.