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Chemistry - Atomic Structure and the Periodic Table

Grade 10IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The atom consists of a central nucleus containing protons (p+p^+) and neutrons (n0n^0), surrounded by electrons (e−e^-) in specific energy levels or shells.

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Atomic Number (ZZ) represents the number of protons in the nucleus, which defines the element. In a neutral atom, Z=number of electronsZ = \text{number of electrons}.

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Mass Number (AA) is the total number of protons and neutrons in the nucleus: A=Z+NA = Z + N.

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Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons (e.g., 1735Cl^{35}_{17}Cl and 1737Cl^{37}_{17}Cl).

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Relative Atomic Mass (ArA_r) is the weighted average mass of the isotopes of an element relative to 112\frac{1}{12} of the mass of a 12C^{12}C atom.

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The Periodic Table is arranged by increasing atomic number. Groups (vertical columns) contain elements with the same number of valence electrons, while Periods (horizontal rows) represent the number of occupied electron shells.

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Periodicity: Atomic radius decreases across a period due to increasing effective nuclear charge (ZeffZ_{eff}), and increases down a group as new electron shells are added.

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Electronegativity and Ionization Energy generally increase across a period and decrease down a group.

📐Formulae

A=Z+nA = Z + n

Ar=∑(isotopic mass×relative abundance)100A_r = \frac{\sum (\text{isotopic mass} \times \text{relative abundance})}{100}

Maximum electrons in shell n=2n2\text{Maximum electrons in shell } n = 2n^2

X(g)→X+(g)+e−(First Ionization Energy)X(g) \rightarrow X^+(g) + e^- \quad (\text{First Ionization Energy})

💡Examples

Problem 1:

Calculate the relative atomic mass (ArA_r) of Chlorine given that it has two isotopes: 35Cl^{35}Cl with an abundance of 75%75\% and 37Cl^{37}Cl with an abundance of 25%25\%.

Solution:

Ar=(35×75)+(37×25)100=2625+925100=35.5A_r = \frac{(35 \times 75) + (37 \times 25)}{100} = \frac{2625 + 925}{100} = 35.5

Explanation:

The relative atomic mass is the weighted average of the isotopic masses based on their natural abundance.

Problem 2:

Determine the number of protons, neutrons, and electrons in a Magnesium ion, 1224Mg2+^{24}_{12}Mg^{2+}.

Solution:

Protons: 1212, Neutrons: 1212, Electrons: 1010.

Explanation:

The atomic number Z=12Z=12 gives the number of protons. The number of neutrons is A−Z=24−12=12A - Z = 24 - 12 = 12. Because the ion has a 2+2+ charge, it has lost 22 electrons from its neutral state (12−2=1012 - 2 = 10).

Problem 3:

Predict and explain the trend in atomic radius when moving from Sodium (NaNa) to Argon (ArAr) across Period 3.

Solution:

The atomic radius decreases from NaNa to ArAr.

Explanation:

Across a period, the number of protons increases, which increases the positive nuclear charge. Since the electrons are being added to the same outer shell, the shielding effect remains constant, resulting in a stronger electrostatic attraction that pulls the electrons closer to the nucleus.