Determine the atomic number of Beryllium.

Beryllium (Be) has an atomic number (Z) of 4, meaning it has 4 electrons.

The electron configuration for a neutral Beryllium atom is $1s^2 2s^2$.
The s, p configuration for the Beryllium atom is $\boxed{1s^2 2s^2}$.

b)

A neutral Beryllium atom has 4 electrons. When it forms a $\text{Be}^{2+}$ ion, it loses 2 electrons.
Number of electrons in $\text{Be}^{2+} = 4 - 2 = 2$ electrons.

The electron configuration for $\text{Be}^{2+}$ is $1s^2$.

The Noble Gas with 2 electrons and an electron configuration of $1s^2$ is Helium (He).
The Noble Gas is $\boxed{\text{Helium}}$.

c)
Beryllium prefers to lose 2 electrons because it requires less energy to remove the two valence electrons from its outer $2s$ shell to achieve a stable electron configuration like Helium ($1s^2$). Gaining 6 electrons would require significantly more energy due to increased electron-electron repulsion and the nucleus's inability to effectively attract and hold that many additional electrons.

a)

Aluminum (Al) has an atomic number (Z) of 13, meaning a neutral atom has 13 protons and 13 electrons.
Aluminum typically forms a $\text{Al}^{3+}$ ion by losing its 3 valence electrons.
Number of protons = 13
Number of electrons in $\text{Al}^{3+}$ = $13 - 3 = 10$ electrons.

The $\text{Al}^{3+}$ ion has 13 protons in its nucleus and 10 electrons arranged in two shells (2 electrons in the first shell, 8 electrons in the second shell).

The structure of the Aluminum ion ($\text{Al}^{3+}$) is represented as:
(A central nucleus containing 13 protons, surrounded by two electron shells. The first shell has 2 electrons, and the second shell has 8 electrons.)

b)

Aluminum forms an ion with a charge of $+3$, so it is $\text{Al}^{3+}$.
The phosphate radical is given as $\text{PO}_4^{3-}$, which has a charge of $-3$.

For an ionic compound to be electrically neutral, the total positive charge must equal the total negative charge.
One $\text{Al}^{3+}$ ion provides a $+3$ charge.
One $\text{PO}_4^{3-}$ ion provides a $-3$ charge.
Since $+3 + (-3) = 0$, one $\text{Al}^{3+}$ ion combines with one $\text{PO}_4^{3-}$ ion.

Therefore, the formula for Aluminum Phosphate is $\boxed{\text{AlPO}_4}$.
The formula is not $\text{Al}_3(\text{PO}_4)$ because that would result in a net charge of $(3 \times +3) + (1 \times -3) = +9 - 3 = +6$, which is not a neutral compound.