Atomic Structure — Diagnostic Tests

Unit Tests

UT-1: Subatomic Particles and Atomic Models

Question:

(a) Describe the three subatomic particles found in an atom, stating their relative masses and charges.

(b) An atom of sodium has an atomic number of 11 and a mass number of 23. How many protons, neutrons, and electrons does it contain?

(c) Explain the difference between the plum pudding model and the nuclear model of the atom. What experiment led to the change in model?

(d) Define the term “isotope” and give an example of two isotopes of the same element, explaining how they differ.

Solution:

(a) The three subatomic particles are:

• Protons: found in the nucleus, relative mass $= 1$, charge $= +1$.
• Neutrons: found in the nucleus, relative mass $= 1$, charge $= 0$ (neutral).
• Electrons: found in electron shells (energy levels) surrounding the nucleus, relative mass $\approx 0$ (negligible compared to protons and neutrons), charge $= -1$.

(b) In a neutral sodium atom ($^{23}_{11}\text{Na}$):

• Number of protons $=$ atomic number $= 11$
• Number of neutrons $=$ mass number $-$ atomic number $= 23 - 11 = 12$
• Number of electrons $=$ number of protons $= 11$ (since the atom is neutral)

(c) The plum pudding model (proposed by J.J. Thomson) suggested that the atom was a sphere of positive charge with electrons embedded within it, like plums in a pudding. The nuclear model (proposed by Ernest Rutherford) described a tiny, dense, positively charged nucleus at the centre with electrons orbiting around it. The change was prompted by the alpha particle scattering experiment (Geiger-Marsden, 1909), where most alpha particles passed straight through gold foil but some were deflected at large angles, suggesting a concentrated positive nucleus.

(d) Isotopes are atoms of the same element (same number of protons) but with different numbers of neutrons, and therefore different mass numbers. For example, carbon-12 ($^{12}_{6}\text{C}$) has 6 protons and 6 neutrons, while carbon-14 ($^{14}_{6}\text{C}$) has 6 protons and 8 neutrons. Isotopes have the same chemical properties but different physical properties (e.g., different masses).

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UT-2: Electron Configuration and the Periodic Table

Question:

(a) State the maximum number of electrons that can occupy the first three electron shells.

(b) Write the electron configuration for the following atoms: potassium ($Z = 19$), calcium ($Z = 20$), and argon ($Z = 18$).

(c) Explain why the noble gases (Group 0) are unreactive in terms of their electron configuration.

(d) Describe the trend in atomic radius across Period 2 (from lithium to neon) and explain why this trend occurs.

Solution:

(a) The maximum number of electrons per shell is $2n^2$, where $n$ is the shell number:

• Shell 1 ($n = 1$): maximum $= 2(1)^2 = 2$ electrons
• Shell 2 ($n = 2$): maximum $= 2(2)^2 = 8$ electrons
• Shell 3 ($n = 3$): maximum $= 2(3)^2 = 18$ electrons

(b)

• Potassium ($Z = 19$): $2, 8, 8, 1$
• Calcium ($Z = 20$): $2, 8, 8, 2$
• Argon ($Z = 18$): $2, 8, 8$

Note: Potassium and calcium both have electrons in the fourth shell because the third shell only fully fills at argon (with 8 electrons at GCSE level); the remaining electrons begin filling the fourth shell.

(c) Noble gases have a full outer electron shell. For example, neon has the configuration $2, 8$, meaning its outer shell is complete with 8 electrons. Atoms tend to gain, lose, or share electrons to achieve a full outer shell. Since noble gases already have a stable, full outer shell, they have no tendency to react.

(d) Across Period 2 (from lithium to neon), the atomic radius decreases. As you move from left to right, each successive element has one more proton in the nucleus. The number of electron shells stays the same (all elements in Period 2 have two occupied shells). The increased nuclear charge pulls the outer electrons closer to the nucleus with greater force, resulting in a smaller atomic radius.

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UT-3: Ions and Relative Atomic Mass

Question:

(a) Explain the difference between an atom and an ion.

(b) Write the formula and give the electron configuration of the following ions: $\text{O}^{2-}$, $\text{Mg}^{2+}$, $\text{Al}^{3+}$.

(c) Chlorine has two isotopes: chlorine-35 (75.8% abundance) and chlorine-37 (24.2% abundance). Calculate the relative atomic mass ($A_r$) of chlorine. Give your answer to one decimal place.

(d) Explain why the relative atomic mass of an element is not usually a whole number.

Solution:

(a) An atom is electrically neutral because the number of protons equals the number of electrons. An ion is a charged particle formed when an atom gains or loses electrons. If an atom loses electrons, it becomes a positively charged ion (cation); if it gains electrons, it becomes a negatively charged ion (anion).

(b)

• $\text{O}^{2-}$: Oxygen ($Z = 8$, configuration $2, 6$) gains 2 electrons to achieve a full outer shell. Configuration becomes $2, 8$.
• $\text{Mg}^{2+}$: Magnesium ($Z = 12$, configuration $2, 8, 2$) loses its 2 outer electrons. Configuration becomes $2, 8$.
• $\text{Al}^{3+}$: Aluminium ($Z = 13$, configuration $2, 8, 3$) loses its 3 outer electrons. Configuration becomes $2, 8$.

(c) The relative atomic mass is calculated using the weighted average:

$A_r = \frac{(35 \times 75.8) + (37 \times 24.2)}{100} = \frac{2653 + 895.4}{100} = \frac{3548.4}{100} = 35.484 \approx 35.5$

(d) The relative atomic mass is not usually a whole number because most elements exist as a mixture of isotopes with different masses and different abundances. The $A_r$ is a weighted average of these isotope masses, which generally does not produce a whole number.

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Integration Tests

IT-1: Periodic Table Trends and Reactivity

Question:

(a) Sodium reacts vigorously with water, but magnesium reacts only slowly. Explain this difference in reactivity in terms of electron configuration.

(b) Fluorine is more reactive than chlorine. Explain why in terms of electron configuration and atomic radius.

(c) An element $X$ is in Group 1, Period 3. Write the electron configuration of $X$. Predict the formula of the oxide it forms when burned in air, and describe the appearance of its compound with chlorine.

(d) A student is given a sample of an unknown Group 2 metal. When added to water, it reacts slowly and the solution forms a white precipitate when tested with sodium hydroxide solution. Suggest what the metal might be and explain your reasoning.

Solution:

(a) Sodium ($Z = 11$) has the electron configuration $2, 8, 1$ and loses its single outer electron very easily to form $\text{Na}^+$. Magnesium ($Z = 12$) has the configuration $2, 8, 2$ and must lose two electrons to form $\text{Mg}^{2+}$. Losing two electrons requires more energy than losing one. Additionally, sodium has a larger atomic radius than magnesium (it is in Period 3 vs Period 3 but Group 1 vs Group 2), meaning its outer electron is further from the nucleus and less strongly attracted. Therefore sodium reacts more vigorously.

(b) Fluorine and chlorine are both in Group 7 (halogens) and need to gain one electron to achieve a full outer shell. Fluorine is in Period 2 while chlorine is in Period 3. Fluorine has a smaller atomic radius, meaning its outer shell is closer to the nucleus. The incoming electron is attracted more strongly to the fluorine nucleus, making it easier for fluorine to gain an electron. Therefore fluorine is more reactive than chlorine.

(c) Group 1, Period 3 is sodium ($Z = 11$). Electron configuration: $2, 8, 1$.

Sodium burns in air to form sodium oxide, $\text{Na}_2\text{O}$ (formula derived from $\text{Na}^+$ and $\text{O}^{2-}$; balanced: two $\text{Na}^+$ ions for one $\text{O}^{2-}$).

The compound with chlorine is sodium chloride, $\text{NaCl}$, which forms white crystalline solid cubes.

(d) The unknown metal is likely calcium. Group 2 metals react with water, with reactivity increasing down the group. Beryllium does not react with water; magnesium reacts slowly. The white precipitate with sodium hydroxide suggests the metal ion forms an insoluble hydroxide. $\text{Ca}(\text{OH})_2$ is only slightly soluble, producing a white precipitate when sufficient $\text{Ca}^{2+}$ is present. (Magnesium hydroxide also gives a white precipitate, but reacts more slowly with water.)

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IT-2: Combining Atomic Structure Concepts

Question:

(a) An atom of an element has 9 protons, 10 neutrons, and 10 electrons. State the atomic number, mass number, and charge of this particle. Identify the element and explain whether it is an atom or an ion.

(b) The relative atomic mass of boron is 10.8. Boron has two isotopes: boron-10 and boron-11. Calculate the percentage abundance of each isotope. Show your working.

(c) Explain how the discovery of isotopes modified Dalton’s atomic theory. State which of Dalton’s original postulates needed to be changed.

(d) A scientist determines that an element has an $A_r$ of 28.1. Which element is this likely to be? Using your knowledge of the periodic table, give its group number, period, and electron configuration.

Solution:

(a)

• Atomic number $= 9$
• Mass number $= 9 + 10 = 19$
• Charge $= 9\text{ protons} (+) - 10\text{ electrons} (-) = -1$

This is a fluoride ion, $\text{F}^-$, which is an ion because the number of protons does not equal the number of electrons, giving it a net charge of $-1$.

(b) Let $x$ = percentage abundance of boron-10, then $(100 - x)$ = percentage abundance of boron-11.

$10.8 = \frac{10x + 11(100 - x)}{100}$

$1080 = 10x + 1100 - 11x$

$1080 = 1100 - x$

$x = 20$

Boron-10 abundance = 20%, boron-11 abundance = 80%.

(c) Dalton’s atomic theory originally stated that all atoms of a given element are identical in mass and properties. The discovery of isotopes showed that atoms of the same element can have different masses (due to different numbers of neutrons). The postulate that needed changing was: “Atoms of the same element are identical” — this was modified to recognise that isotopes of the same element differ in mass.

(d) An element with $A_r = 28.1$ is silicon (Si). Silicon is in Group 4 and Period 3 of the periodic table. Its electron configuration is $2, 8, 4$. This can be confirmed: silicon-28 ($^{28}_{14}\text{Si}$) is the most abundant isotope, consistent with the given $A_r$ of 28.1.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Common Pitfalls

• Confusing atomic number with mass number: atomic number equals the number of protons, while mass number equals protons plus neutrons.
• Assuming ions always have a negative charge; cations (positive ions) form when metals lose electrons.
• Forgetting that the number of neutrons is calculated as mass number minus atomic number, not read directly from the periodic table.
• Incorrectly filling electron shells by placing too many electrons in inner shells before moving to outer shells.
• Assuming all atoms of an element are identical without considering isotopes and their relative abundances.