10 (a) Step 1: Determine the number of electrons for a neutral Magnesium atom. Magnesium (Mg) has an atomic number (Z) of 12, so a neutral atom has 12 electrons. Step 2: Write the electron configuration by filling orbitals in order. $$1s^2 2s^2 2p^6 3s^2$$ The s, p configuration for a neutral Magnesium atom is $\boxed{\text{1s}^2 \text{2s}^2 \text{2p}^6 \text{3s}^2}$. 10 (b) Step 1: Understand the formation of the $\text{Mg}^{2+}$ ion. A $\text{Mg}^{2+}$ ion is formed when a neutral Magnesium atom loses 2 electrons. Electrons are lost from the outermost shell first. Step 2: Identify the outermost sub-shell. The outermost sub-shell in the neutral Magnesium atom (configuration $1s^2 2s^2 2p^6 3s^2$) is the $3s$ sub-shell. When Magnesium becomes an ion $\text{Mg}^{2+}$, the $\boxed{\text{3s}}$ sub-shell is emptied. 10 (c) Step 1: Start with the neutral Magnesium configuration and remove 2 electrons. Neutral Mg: $1s^2 2s^2 2p^6 3s^2$ Removing 2 electrons from the $3s$ sub-shell results in: $$1s^2 2s^2 2p^6$$ The resulting configuration of the $\text{Mg}^{2+}$ ion is $\boxed{\text{1s}^2 \text{2s}^2 \text{2p}^6}$. 11 (a) Step 1: Determine the charge of the Calcium ion in $\text{Ca(ClO)}_2$. Calcium (Ca) is in Group 2, so it forms a $\text{Ca}^{2+}$ ion, meaning its valency is +2. Step 2: Determine the valency of the Hypochlorite radical ($\text{ClO}$). Since the compound $\text{Ca(ClO)}_2$ is neutral, the total negative charge must balance the +2 charge from Calcium. There are two $\text{ClO}$ radicals, so each $\text{ClO}$ radical must have a valency of -1. The valency of the Hypochlorite radical ($\text{ClO}$) is $\boxed{\text{-1}}$. 11 (b) (i) Step 1: Determine the number of electrons for a neutral Chlorine atom. Chlorine (Cl) has an atomic number (Z) of 17, so a neutral atom has 17 electrons. Step 2: Write the electron configuration for a neutral Chlorine atom. $$1s^2 2s^2 2p^6 3s^2 3p^5$$ Step 3: Determine the number of electrons for a $\text{Cl}^{-}$ ion. A $\text{Cl}^{-}$ ion is formed when a neutral Chlorine atom gains 1 electron. So, it has $17 + 1 = 18$ electrons. Step 4: Write the electron configuration for the $\text{Cl}^{-}$ ion. The gained electron will fill the $3p$ sub-shell. $$1s^2 2s^2 2p^6 3s^2 3p^6$$ The s, p configuration for the $\text{Cl}^{-}$ ion is $\boxed{\text{1s}^2 \text{2s}^2 \text{2p}^6 \text{3s}^2 \text{3p}^6}$. 11 (b) (ii) Step 1: Identify the reactants and product with their chemical formulas and states. Reactants: Calcium ($\text{Ca(s)}$) and Chlorine gas ($\text{Cl}_2\text{(g)}$). Product: Calcium Chloride ($\text{CaCl}_2\text{(s)}$). Step 2: Write the unbalanced chemical equation. $$\text{Ca(s)} + \text{Cl}_2\text{(g)} \rightarrow \text{CaCl}_2\text{(s)}$$ Step 3: Balance the equation. The equation is already balanced with one Ca atom and two Cl atoms on both sides. The balanced equation with state symbols is: $$\boxed{\text{Ca(s)} + \text{Cl}_2\text{(g)} \rightarrow \text{CaCl}_2\text{(s)}}$$ Send me the next one 📸