a) Step 1: Determine the atomic number of Beryllium. Beryllium (Be) has an atomic number (Z) of 4, meaning it has 4 electrons. Step 2: Write the electron configuration. 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) Step 1: Determine the number of electrons in a $\text{Be}^{2+}$ ion. 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. Step 2: Determine the electron configuration of $\text{Be}^{2+}$. The electron configuration for $\text{Be}^{2+}$ is $1s^2$. Step 3: Identify the Noble Gas with the same electron configuration. 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) Step 1: Determine the number of protons and electrons for the Aluminum ion. 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. Step 2: Draw the structure of the Aluminum ion. 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). $$ \text{Nucleus: } \begin{cases} 13 \text{ protons} \\ 14 \text{ neutrons (assuming mass number 27)} \end{cases} \\ \text{Electron Shells: } \\ \text{1st shell: } 2 \text{ electrons} \\ \text{2nd shell: } 8 \text{ electrons} $$ 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) Step 1: Determine the charges of the ions involved. 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 $-3$. Step 2: Balance the charges to form a neutral compound. 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. Step 3: Write the correct formula. 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. That's 2 down. 3 left today — send the next one.