The question states that the oxygen and fluorine atoms have the same electron configuration as neon. The electron configuration for neon is , with ten total electrons and eight valence electrons. To match this configuration, both the oxygen and the fluorine atom must have eight valence electrons (two in the  orbital and six in the  orbital).

Recall that atoms in the oxygen group have six valence electrons, whereas halogens have seven valence electrons; therefore, oxygen has six and fluorine has seven valence electrons. To reach a total of eight valence electrons (octet) the oxygen atom must have gained two electrons, giving it a charge of . Similarly, fluorine must have gained one electron, giving it a charge of . Fluorine has half the of charge of oxygen because it gained half the amount of electrons to complete its octet.

Cobalt is a transition metal; therefore, it is found in the D block of the periodic table. A ground state cobalt atom has an electron configuration of . The last orbitals that gain or lose electrons must be either the  or  orbitals, since these are the orbitals with highest energy and located farthest from the nucleus.

Recall that electrons are filled from orbitals of low energy to high energy. A  orbital has a lower energy than a  orbital. This means that when you are filling electrons, the last orbital you fill is the  orbital. When you are assigning electrons to each orbital you assign two electrons to the  orbital and then the remaining seven electrons to the five  orbitals; therefore,  orbitals are filled last when gaining electrons.

When an element loses electrons, the first orbital that loses electrons is the outermost orbital. This occurs because the attractive force of the nucleus on the electron will be weakest in the outermost orbital (because it is farthest away from nucleus); therefore, it will be easy to pull the electron away from the nucleus. In cobalt, the outermost orbital is the  orbital (because it has the highest shell number). This means that electrons will be lost from the  orbital before the  orbital. 

Notice that the orbital that gains electron last is not the same orbital that loses electrons first. Gaining electrons is dependent on the energy of the orbital, and losing electrons is dependent on the location of the orbital. The highest energy orbital will gain electron last and the outermost orbital will lose electron first.

The question states that the element is a transition element and it only has one electron in its  orbital. Recall that transition metals usually have two electrons in its  orbital; however, some transition metals lose one of the electrons from their  orbital and move it to one of their  orbital. This occurs because of a phenomenon called half-shell stability. Half-shell stability states that an element is more stable when all the orbitals are half filled.

This is best exemplified by the transition metal chromium. The electron configuration of chromium, conventionally, would be . This means that chromium has two electrons in the  orbital, one electron in four  orbitals, and an empty  orbital. Half-shell stability states that chromium is more stable if all five  orbitals have an electron, rather than having an empty  orbital; therefore, an electron from the  orbital will be moved to the empty  orbital to fulfill half-shell stability. This gives one electron in the  orbital and one electron in each of the five  orbitals.

Notice that half-shell stability is also observed in molybdenum () and tungsten (). Molybdenum will have one electron in its  orbital and one electron in each of its  orbitals, and tungsten will have one electron in its  orbital and one electron in each of its  orbitals.

Valence electrons will be housed in the outer shell (highest numbered orbital) of the electron configuration.

Calcium:

Magnesium:

Zinc:

Copper:

Arsenic:

Calcium, magnesium, and zinc all have two valence electrons in their highest energy orbital. Copper has only one valence electron. Arsenic has five total valence electrons in the fourth shell.

An atom that has a charge of  has lost two electrons. The two lost are always the valence electrons, which are most easily taken from the molecule. Strontium has a normal electron configuration with 38 electrons. In shorthand notation, this would be:

Losing two valence electrons will remove the 5s electrons. This leave the configuration as:

This makes sense because krypton has a stable valence octet. Strontium is stable as an ion, meaning that it will also have an octet, allowing it to match the configuration for krypton.

Valence electrons in the d subshell can be odd because a half-filled d orbital is more stable than one with three or four electrons. While strontium, titanium, and vanadium have two electrons in their 4s orbitals, chromium has one in the 4s shell and puts five in the 3d shell.

This creates a half-filled 4s and 3d shells, which are more stable than a full 4s shell and partial 3d shell. Chromium and manganese thus have the same number of 3d electrons: five.

This particular configuration denotes a particle with ten total electrons. The sodium atom, with eleven electrons, is the only one listed that could not have this configuration. Ionized  sodium, however, symbolized as Na⁺, does apply. (Be careful to distinguish neutral atoms and ions).    

Nuclear orbitals will always fill from the innermost to the outermost subshells. Using the  rule, we can approximate the order in which these orbitals will fill. Because electrons fill starting with the centermost orbitals, the electron that is closest to the nucleus will belong to the orbital that fills first.

corresponds to the principle quantum number, the first number in the given orbital location. is the azimuthal quantum number, and dictates the shape of the orbital.

5s and 3d produce the same number from the  equation, but in the event of a tie we always pick the orbital with the lowest letter (s < p < d < f). The innermost orbital will be 5s.

When an element loses an electron it is generally taken away from the highest electron shell. The electron configuration of iron (Fe) is . The 4s orbital is farther away from the nucleus than the 3d orbital, therefore the electron configuration of Fe⁺ will be .

When determining electronic configuration, the answer is made much easier by starting with the next smallest noble gas in brackets. As a result, [Ar] is an appropriate way to incorporate every previous electron before argon.

After argon, vanadium has five other electrons to distribute, and because vanadium is a transitional element, it will fill its 3d subshells before filling the 4p subshells. The 4s subshell is filled first, and the last three electrons are placed into the 3d subshells.

[Ar]3d³4s²