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Atoms, Elements and Compounds - Atomic structure and the Periodic Table

Grade 11IGCSEChemistry

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

🔑Concepts

Atoms consist of a central nucleus containing protons (charge +1+1, relative mass 11) and neutrons (charge 00, relative mass 11), surrounded by electrons (charge 1-1, relative mass 11840\frac{1}{1840}) orbiting in shells.

The Atomic Number (ZZ) represents the number of protons in the nucleus. The Nucleon Number (Mass Number, AA) represents the total number of protons and neutrons: ZAX^A_Z X.

Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. They have the same chemical properties because they have the same number of electrons in their outer shell.

Electronic configuration describes the arrangement of electrons in shells. For the first 20 elements, shells fill in the pattern 2,8,8,22, 8, 8, 2. The number of valence (outer) electrons determines the chemical properties and the Group number.

The Periodic Table is arranged by increasing atomic number. Rows are called Periods (representing the number of electron shells) and columns are called Groups (representing the number of valence electrons).

Group I (Alkali Metals) reactivity increases down the group as the outer electron is further from the nucleus and more easily lost. Group VII (Halogens) reactivity decreases down the group as it becomes harder to attract an electron into the outer shell.

📐Formulae

A=Z+NA = Z + N where AA is the nucleon number, ZZ is the atomic number, and NN is the number of neutrons.

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

Charge of ion=number of protonsnumber of electrons\text{Charge of ion} = \text{number of protons} - \text{number of electrons}

💡Examples

Problem 1:

Calculate the number of protons, neutrons, and electrons in the ion 1531P3^{31}_{15}P^{3-}.

Solution:

Protons = 1515, Neutrons = 1616, Electrons = 1818.

Explanation:

The atomic number Z=15Z=15 defines the number of protons. The number of neutrons is AZ=3115=16A - Z = 31 - 15 = 16. Because it is a 33- ion, it has gained 33 electrons, so 15+3=1815 + 3 = 18 electrons.

Problem 2:

An element XX has two isotopes: 63X^{63}X with an abundance of 69%69\% and 65X^{65}X with an abundance of 31%31\%. Calculate the relative atomic mass (ArA_r) of XX.

Solution:

Ar=(63×69)+(65×31)100=4347+2015100=63.62A_r = \frac{(63 \times 69) + (65 \times 31)}{100} = \frac{4347 + 2015}{100} = 63.62

Explanation:

The relative atomic mass is the weighted average of the masses of the isotopes. Multiply each mass by its percentage, sum them, and divide by 100100.

Problem 3:

Predict the electronic configuration and position in the Periodic Table for an atom with 1919 protons.

Solution:

Configuration: 2,8,8,12, 8, 8, 1. Position: Period 44, Group I.

Explanation:

With 1919 electrons, the first shell takes 22, the second 88, the third 88, leaving 11 for the fourth shell. Four shells occupied means Period 44; 11 valence electron means Group I (Potassium).

Atomic structure and the Periodic Table Revision - Grade 11 Chemistry IGCSE