Sketch a graph for the variation of first ionization energy against atomic number for the elements across period 3 of the periodic table (Na - Ar). Explain the general trend and any dips.

SECTION B: INORGANIC CHEMISTRY

3. (a) This question concerns the elements across period 3 of the periodic table (Na - Ar).

(i) Sketch a graph for the variation of first ionization energy against atomic number for the elements.
The graph shows a general increase in first ionization energy across Period 3 due to increasing nuclear charge and decreasing atomic radius. The dips at Aluminium (Al) and Sulfur (S) are due to the removal of a $3\text{p}$ electron (Al) and a paired $3\text{p}$ electron (S), respectively, which are easier to remove than the preceding elements' electrons.

\begin{tikzpicture}[xscale=0.8, yscale=0.015] % Data points for Period 3 first ionization energies (kJ/mol) % Na: 496, Mg: 738, Al: 578, Si: 787, P: 1012, S: 1000, Cl: 1251, Ar: 1521 \def\Na{496} \def\Mg{738} \def\Al{578} \def\Si{787} \def\P{1012} \def\S{1000} \def\Cl{1251} \def\Ar{1521} % Axes \draw[->] (0,0) -- (9,0) node[right] {Atomic Number}; \draw[->] (0,0) -- (0,1600/15) node[above] {First Ionization Energy (kJ/mol)}; % X-axis labels (elements) \foreach \x/\label in {1/Na, 2/Mg, 3/Al, 4/Si, 5/P, 6/S, 7/Cl, 8/Ar} \node[below] at (\x,0) {\label}; % Y-axis ticks and labels \foreach \y in {0,250,500,750,1000,1250,1500} \draw (0,\y/15) -- (-0.1,\y/15) node[left] {\y}; % Plotting the points and connecting them \draw[blue, thick] plot[mark=o] coordinates { (1,\Na/15) (2,\Mg/15) (3,\Al/15) (4,\Si/15) (5,\P/15) (6,\S/15) (7,\Cl/15) (8,\Ar/15) }; \end{tikzpicture}

(ii) Complete the table below by giving the chemical formulae of the stable oxides of the elements.

Element	Oxide Formula
Na	$\boxed{\text{Na}_2\text{O}}$
Mg	$\boxed{\text{MgO}}$
P	$\boxed{\text{P}_4\text{O}_{10}}$
Cl	$\boxed{\text{Cl}_2\text{O}_7}$

(iii) Write a balanced equation for the reaction of silicon tetrachloride with water.

$$\boxed{\text{SiCl}_4\text{(l)} + 2\text{H}_2\text{O(l)} \to \text{SiO}_2\text{(s)} + 4\text{HCl(aq)}}$$

(iv) Write two equations to show the amphoteric nature of aluminium oxide ($\text{Al}_2\text{O}_3$).
Aluminium oxide reacts with both acids and bases.

Reaction with an acid (e.g., hydrochloric acid):

$$\boxed{\text{Al}_2\text{O}_3\text{(s)} + 6\text{HCl(aq)} \to 2\text{AlCl}_3\text{(aq)} + 3\text{H}_2\text{O(l)}}$$

Reaction with a base (e.g., sodium hydroxide):

$$\boxed{\text{Al}_2\text{O}_3\text{(s)} + 2\text{NaOH(aq)} + 3\text{H}_2\text{O(l)} \to 2\text{Na}[\text{Al(OH)}_4]\text{(aq)}}$$

(or $\text{Al}_2\text{O}_3\text{(s)} + 2\text{NaOH(aq)} \to 2\text{NaAlO}_2\text{(aq)} + \text{H}_2\text{O(l)}$)

(v) What makes sulfur dioxide gas acidic?
Sulfur dioxide ($\text{SO}_2$) is an acidic oxide because it reacts with water to form sulfurous acid ($\text{H}_2\text{SO}_3$), which then dissociates to produce hydrogen ions ($\text{H}^+$) in solution.

$$\text{SO}_2\text{(g)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_2\text{SO}_3\text{(aq)}$$ $$\text{H}_2\text{SO}_3\text{(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{HSO}_3^-\text{(aq)}$$

3. (b)

(i) Give the outermost electronic configuration for the s-block elements (group I and group II).

• Group I (Alkali metals): $\boxed{\text{ns}^1}$
• Group II (Alkaline earth metals): $\boxed{\text{ns}^2}$

(ii) For the elements Lithium and Beryllium, write equations representing:

• A. First ionization energy for Lithium $$\boxed{\text{Li(g)} \to \text{Li}^+\text{(g)} + \text{e}^-}$$
• B. Second ionization energy for Beryllium $$\boxed{\text{Be}^+\text{(g)} \to \text{Be}^{2+}\text{(g)} + \text{e}^-}$$

(iii) Which of these two elements in b(ii) will have a higher second ionization energy? Explain.

• Lithium will have a higher second ionization energy.
• Explanation:
  • Lithium has an electron configuration of $1\text{s}^2 2\text{s}^1$. After the first ionization, $\text{Li}^+$ has an electron configuration of $1\text{s}^2$, which is a stable noble gas configuration (like Helium). Removing a second electron from this very stable, fully filled inner shell requires a significantly large amount of energy.
  • Beryllium has an electron configuration of $1\text{s}^2 2\text{s}^2$. After the first ionization, $\text{Be}^+$ has an electron configuration of $1\text{s}^2 2\text{s}^1$. The second electron is removed from the $2\text{s}$ orbital, which is an outer shell electron and not from a stable noble gas core. Therefore, less energy is required to remove the second electron from $\text{Be}^+$ compared to $\text{Li}^+$.