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Atomic Structure - The nuclear atom

Grade 12IBChemistry

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

🔑Concepts

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

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Protons and neutrons are collectively known as nucleons. The mass of the atom is concentrated in the nucleus, as the mass of an electron is approximately 11836\frac{1}{1836} of the mass of a proton.

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The atomic number (ZZ) represents the number of protons in the nucleus, which defines the identity of the element. In a neutral atom, ZZ also equals the number of electrons.

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The mass number (AA) is the total number of protons and neutrons in the nucleus.

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Isotopes are atoms of the same element (same ZZ) with different numbers of neutrons (different AA). Isotopes of an element show identical chemical properties because they have the same electron configuration, but differ in physical properties such as density and boiling point.

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Relative atomic mass (ArA_r) is the weighted average mass of an atom of an element, taking into account its naturally occurring isotopes, relative to 112\frac{1}{12} of the mass of a 12C^{12}C atom.

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Mass spectrometry is used to determine the relative atomic mass of an element by measuring the m/zm/z (mass-to-charge) ratio and relative abundance of its isotopes.

📐Formulae

A=Z+N (where N is the number of neutrons)A = Z + N \text{ (where } N \text{ is the number of neutrons)}

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

Charge of ion=(number of protons)−(number of electrons)\text{Charge of ion} = (\text{number of protons}) - (\text{number of electrons})

💡Examples

Problem 1:

Determine the number of protons, neutrons, and electrons in a gallium ion 3171Ga3+^{71}_{31}Ga^{3+}.

Solution:

Protons = 3131, Neutrons = 4040, Electrons = 2828.

Explanation:

The atomic number Z=31Z = 31 indicates 3131 protons. The mass number A=71A = 71, so neutrons N=A−Z=71−31=40N = A - Z = 71 - 31 = 40. Since the ion has a 3+3+ charge, it has lost 33 electrons: 31−3=2831 - 3 = 28 electrons.

Problem 2:

A sample of neon consists of 90.48%90.48\% 20Ne^{20}Ne and 9.52%9.52\% 22Ne^{22}Ne. Calculate the relative atomic mass (ArA_r) of this neon sample to two decimal places.

Solution:

Ar=(20×90.48)+(22×9.52)100=20.19A_r = \frac{(20 \times 90.48) + (22 \times 9.52)}{100} = 20.19

Explanation:

The relative atomic mass is the weighted average of the isotopic masses. We multiply each mass by its percentage abundance, sum them up, and divide by 100100.

Problem 3:

Explain why 35Cl^{35}Cl and 37Cl^{37}Cl have the same chemical properties.

Solution:

Chemical properties are determined by the number and arrangement of electrons, specifically the valence electrons. Since both isotopes have 1717 protons, they both have 1717 electrons in a neutral state, leading to the same electron configuration: [Ne]3s23p5[Ne] 3s^2 3p^5.

Explanation:

Isotopes only differ in the number of neutrons in the nucleus, which affects mass-related physical properties but not the electronic interactions responsible for chemical bonding.

The nuclear atom - Revision Notes & Key Formulas | IB Grade 12 Chemistry