A) Explain why S is not an ion. b) Determine the number of electrons in the neutral atoms of P, Q, R and S.

The electronic arrangement for S is $2:8:8$. This means it has $2+8+8 = 18$ electrons.

An ion is an atom or molecule that has gained or lost one or more electrons, resulting in a net electrical charge (positive or negative).

The species P$^{2-}$, Q$^{2+}$, and R$^{+}$ all have charges indicated, signifying they are ions. S has no charge indicated, meaning it is a neutral species. A neutral atom has an equal number of protons and electrons. Therefore, S represents a neutral atom, not an ion. The electronic configuration $2:8:8$ corresponds to the noble gas Argon, which exists as a neutral atom.

• P$^{2-}$ has $2:8:8$ electrons. The neutral atom P would have $18 - 2 = 16$ electrons, which is Sulfur (S).
• Q$^{2+}$ has $2:8$ electrons. The neutral atom Q would have $10 + 2 = 12$ electrons, which is Magnesium (Mg).
• R$^{+}$ has $2:8$ electrons. The neutral atom R would have $10 + 1 = 11$ electrons, which is Sodium (Na).
• S has $2:8:8$ electrons. The neutral atom S has 18 electrons, which is Argon (Ar).

All these elements are in Period 3 of the periodic table. Atomic radius generally decreases across a period from left to right due to an increasing nuclear charge pulling the electrons closer to the nucleus.

Among Sodium (Group 1), Magnesium (Group 2), Sulfur (Group 16), and Argon (Group 18), Sodium is furthest to the left in Period 3. Therefore, Sodium has the largest atomic radius.

The element with the largest atomic radius is R.
Reason: R is Sodium, which is in Group 1, Period 3. Atomic radius decreases across a period due to increasing nuclear charge. As Sodium is the furthest left among the elements in Period 3, it has the largest atomic radius.

Given density $= 1.134 \text{ gm/cm}^3 = 1.134 \text{ g/mL}$.
Volume $= 1 \text{ L} = 1000 \text{ mL}$.
Mass of solution $= \text{Density} \times \text{Volume}$

Percentage purity $= 37\%$.
Mass of pure HCl $= 37\% \text{ of } 1134 \text{ g}$

Molar mass of H $= 1.008 \text{ g/mol}$
Molar mass of Cl $= 35.453 \text{ g/mol}$
Molar mass of HCl $= 1.008 + 35.453 = 36.461 \text{ g/mol}$

Moles of HCl $= \frac{\text{Mass of pure HCl}}{\text{Molar mass of HCl}}$

Molarity $= \frac{\text{Moles of HCl}}{\text{Volume of solution in L}}$

Rounding to a reasonable number of significant figures (e.g., three, based on 37% purity):
The molarity of the solution is $\boxed{\text{11.5 M}}$.