The elements in copper's group will fill the d orbital at the expense of having a half filled s orbital. This is because this configuration is more stable than a full s orbital and d orbital with 9/10 spots filled.

The periodic table is designed to organize elements by their orbitals. Electrons only enter the d orbital in the fourth period (row) at the beginning of the transition metals. Scandium is the first element to have an electron enter the d orbital. Calcium is the last element on the table that has up to p orbital electrons, and empty d orbitals.

Calcium:

Scandium:

is on the third period and eighth group of the periodic table. In this problem, you have to consider the ionic state of . This means that is deficient of electrons. The complete electron configuration of at ground state is . Notice that the electrons are subtracted from first. When electrons are removed from an atom, they are removed from the outermost shell of the atom. In this case, the outermost shell would be from the 4s orbital. However, there are only 2 electrons in this orbital so the remaining one electron would be taken from next highest energy orbital which would be the 3d orbital.

The aluminum atom has its furthest electrons in the 3p shell. We know this because aluminum is in the third row of the periodic table, and is in the p-block. The possible quantum numbers for any valence electron are:

6 more electrons are present in chromium than argon. It's more stable for d orbitals to be half-filled. Thus, one electron will fill each of the 5 d orbitals.

For the 3d subshell, n (principal quantum number) = 3 and l (azimuthal quantum number) = 2, so n + l = 5. For the 4s subshell, n = 4 and l = 0, so n+l = 4. From this information, we can see that the 4s subshell has lower energy and will fill with electrons first.

A fully filled orbital is generally more stable than a half-filled orbital. The splitting of the energy levels is also dependent on the geometry of the compound that one is analyzing. The 4s electrons are of higher energy than the 3d electrons, thus are lost first, leaving half filled d orbitals, which is the most stable configuration.

The azimuthal quantum number is the second quantum number, designated by the letter l. It gives the shape and number of subshells in a principal energy level (shell).

Atomic sulfur has 16 electrons, yielding 3p orbitals with 2, 1, and 1 electrons as seen below:

orbital diagram:

1s22s23s23p4

We consider the 3p subshell independently as the lower subshells are filled completely. There are three 3p orbitals, each of which may hold two electrons. Hund's rule tells us that each orbital of a given subshell must be singly filled before any orbital in the subshell is doubly filled.

Following this rule, each of the three orbitals are given one of the four remaining electrons, and one orbital is given a second. The 3p subshell has two orbitals occupied by unpaired electrons in atomic sulfur.

However, has two additional electrons. As a result, the 3p orbitals are filled completely and the correct answer is 0 unpaired electrons.

orbital diagram:

1s22s23s23p6

Note: this is identical to the electron configuration of Argon.

Carbon is on the second period and group of the periodic table. Each time we jump from left to right, we add an electron. Period 1 corresponds to 1s, period 2 corresponds to 2s, 2p and so forth. The first energy shell can hold a maximum of two electrons, and each subsequent shell can hold eight.