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: 

Since this is a monovalent cation, the non-ionic electron configuration will be  (note that the electron was removed from the s shell and not the d shell, as this is a transition metal). This electron configuration corresponds to iron, so the cation must be .

First, we will write down the electronic configuration for the ground state of Cobalt:

Remember that a half-filled orbital is very stable, and when it is possible for an ion to have a electron configuration it will. In order to attain this, two electrons will be removed from the orbitals and one will be removed from the orbital, yielding:

We discover that element X is sodium, due to the masive jump after the first ionization energy. Sodium has filled 1s and 2s orbitals, and a half filled 3s orbital. We know that sodium will have the base configuation of [Ne], with one additional electron in the third shell, due to its position in the third period. This leads us to our answer of 1s²2s²2p⁶3s¹.

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.

In order to write the shorthand electronic configuration correctly, choose the last noble gas before the element of interest. The last noble gas before phosphorus is neon. Put the atomic symbol for neon in brackets and continue assigning electron configurations until phosphorus. After neon, we are in the third row, so the s and p orbitals are in the third energy level. Notice that phosphorus has 5 valence electrons since it is in group V. Also notice that in the electron configuration, there are 5 electrons in the highest electron energy level (3).

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.

 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.  

Element in the group 11 (Cu, Ag and Au) prefer to have a closed shell configuration for the  orbitals because is energetically more stable. This lives only one electron in the  orbital. The  oxidation state is easy to achieve because the formed cation exhibits a closed shell configuration for his  level. The last is known as the 18 electrons rule.

For transition metals with six valence electrons, the d orbital is half-filled and s orbital electron is taken to achieve that. The half-filled d orbital in the form [Ar]4s¹3d⁵, is more stable than in the form [Ar]4s²3d⁴. 

[Ar]4s¹4d⁵ is wrong because period 4 transition metals need to fill the 3d orbital before the 4d orbital, under the Aufbau principle.

[Ar]3d⁵ is wrong because it has an incorrect number of electrons.