Identify elements P, Q, R, S, T, U based on their ion configuration and charge. Then, arrange them by increasing atomic size, explaining the trend.

Part 1: Analysis of Elements P, Q, R, S, T, U

First, let's identify the elements based on the given information:

• P: Ion config 2.8, Divalent cation. Neutral atom: 2.8.2 (12 electrons). Element: Magnesium ($\text{Mg}$).
• Q: Ion config 2.8.8, Divalent anion. Neutral atom: 2.8.6 (16 electrons). Element: Sulfur ($\text{S}$).
• R: Ion config 2.8, Monovalent cation. Neutral atom: 2.8.1 (11 electrons). Element: Sodium ($\text{Na}$).
• S: Ion config 2.8.8, Monovalent anion. Neutral atom: 2.8.7 (17 electrons). Element: Chlorine ($\text{Cl}$).
• T: Ion config 2.8, Trivalent cation. Neutral atom: 2.8.3 (13 electrons). Element: Aluminium ($\text{Al}$).
• U: Ion config 2.8.8, Trivalent anion. Neutral atom: 2.8.5 (15 electrons). Element: Phosphorus ($\text{P}$).

i) Arrange the elements PQRST and U in terms of increasing atomic sizes.
All these elements are in Period 3 of the periodic table. Atomic size decreases across a period from left to right due to increasing nuclear charge pulling the electrons closer to the nucleus.
The order of elements in Period 3 from left to right is: $\text{Na (R), Mg (P), Al (T), P (U), S (Q), Cl (S)}$.
Therefore, in terms of increasing atomic size (smallest to largest):
$\text{Cl (S)} < \text{S (Q)} < \text{P (U)} < \text{Al (T)} < \text{Mg (P)} < \text{Na (R)}$

• Answer: $\boxed{\text{S} < \text{Q} < \text{U} < \text{T} < \text{P} < \text{R}}$

ii) Name the period of the periodic table to which these elements belong?
All the ions have either 2 or 3 electron shells (e.g., 2.8 or 2.8.8). This indicates that the neutral atoms have electrons in 3 main energy levels.

• Answer: $\boxed{\text{Period 3}}$

iii) Write an equation for the reaction between element R and S.
Element R is Sodium ($\text{Na}$). Element S is Chlorine ($\text{Cl}$). Sodium is a Group 1 metal, and Chlorine is a Group 17 non-metal. They react to form an ionic compound, sodium chloride.

$$2\text{Na(s)} + \text{Cl}_2\text{(g)} \to 2\text{NaCl(s)}$$

• Answer: $\boxed{2\text{Na(s)} + \text{Cl}_2\text{(g)} \to 2\text{NaCl(s)}}$

iv) Compare the electrical conductivity of elements P and T. Explain.
Element P is Magnesium ($\text{Mg}$). Element T is Aluminium ($\text{Al}$). Both are metals and conduct electricity.

• Comparison: Aluminium (T) has higher electrical conductivity than Magnesium (P).
• Explanation: Electrical conductivity in metals depends on the number of delocalized valence electrons per atom. Magnesium (P) has 2 valence electrons per atom, while Aluminium (T) has 3 valence electrons per atom. A greater number of delocalized electrons per unit volume allows for more efficient charge transport, hence higher electrical conductivity.
• Answer: $\boxed{\text{Aluminium (T) has higher electrical conductivity than Magnesium (P). Aluminium has 3 delocalized valence electrons per atom, while Magnesium has 2, leading to better charge transport in Aluminium.}}$

v) Compare the first ionization energies of elements P and R. Explain.
Element P is Magnesium ($\text{Mg}$). Element R is Sodium ($\text{Na}$). Both are in Period 3. Sodium is in Group 1, and Magnesium is in Group 2.

• Comparison: Magnesium (P) has a higher first ionization energy than Sodium (R).
• Explanation: First ionization energy generally increases across a period from left to right. Magnesium has a greater nuclear charge (+12) than Sodium (+11). Although both have the same number of electron shells, the stronger attraction from the nucleus in Magnesium pulls the valence electrons closer, requiring more energy to remove the first electron.
• Answer: $\boxed{\text{Magnesium (P) has a higher first ionization energy than Sodium (R). Magnesium has a greater nuclear charge and a smaller atomic radius than Sodium, resulting in a stronger attraction for its valence electrons.}}$

vi) Which element is the strongest oxidizing agent?
An oxidizing agent is a substance that gains electrons (gets reduced). Non-metals are generally good oxidizing agents, and their strength increases across a period and up a group.
The non-metals among the given elements are U (Phosphorus), Q (Sulfur), and S (Chlorine).
Chlorine (S) is in Group 17 (halogens) and is the most electronegative among these Period 3 non-metals. It has the strongest tendency to gain an electron to achieve a stable octet.

• Answer: $\boxed{\text{S (Chlorine)}}$

Part 2: Aluminium Chloride ($\text{AlCl}_3$) Analysis

i) State the type of bond and structure in $\text{AlCl}_3$.
From the table, $\text{AlCl}_3$ has a melting point of $-70 \text{ }^\circ\text{C}$. This is a very low melting point, characteristic of simple molecular substances, not typical ionic compounds (e.g., $\text{NaCl}$ at $801 \text{ }^\circ\text{C}$). This indicates that the bonding is predominantly covalent. In the solid and gaseous states, $\text{AlCl}_3$ exists as a dimer, $\text{Al}_2\text{Cl}_6$.

• Bond type: $\boxed{\text{Covalent}}$
• Structure: $\boxed{\text{Simple molecular (dimeric }\text{Al}_2\text{Cl}_6\text{)}}$

ii) What type of bond would $\text{AlCl}_3$ be expected to have? Give a reason.

• Expected bond type: $\boxed{\text{Ionic}}$
• Reason: Aluminium is a metal (Group 13) and Chlorine is a non-metal (Group 17). Typically, metals react with non-metals by transferring electrons to form ionic bonds. Aluminium would be expected to lose 3 electrons to form an $\text{Al}^{3+}$ ion, and Chlorine would gain 1 electron to form a $\text{Cl}^-$ ion.
• Answer: $\boxed{\text{Ionic. Aluminium is a metal and Chlorine is a non-metal; metals are expected to transfer electrons to non-metals to form ionic bonds.}}$