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Chemistry - Atomic Structure (Protons, Neutrons, Electrons, Valency, Isotopes)

Grade 8ICSE

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

🔑Concepts

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An atom is the smallest unit of matter, consisting of a central nucleus and surrounding shells.

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Protons (p+p^+) are positively charged particles located in the nucleus with a mass of 1 amu1 \text{ amu}.

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Neutrons (n0n^0) are neutral particles found in the nucleus with a mass of approximately 1 amu1 \text{ amu}.

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Electrons (e−e^-) are negatively charged particles that revolve around the nucleus in fixed orbits. Their mass is negligible (≈1/1837 amu\approx 1/1837 \text{ amu}).

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Atomic Number (ZZ) is defined as the number of protons in the nucleus of an atom. In a neutral atom, Z=number of protons=number of electronsZ = \text{number of protons} = \text{number of electrons}.

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Mass Number (AA) is the sum of the number of protons and neutrons in the nucleus: A=p+nA = p + n.

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Electronic Configuration describes the distribution of electrons in different shells (K,L,M,N...K, L, M, N...). The maximum number of electrons in a shell is governed by the 2n22n^2 rule.

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Valency is the combining capacity of an atom. It is determined by the number of electrons in the outermost shell (valence electrons).

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Isotopes are atoms of the same element that have the same Atomic Number (ZZ) but different Mass Numbers (AA). They have the same chemical properties but different physical properties.

📐Formulae

Z=pZ = p

A=p+nA = p + n

n=A−Zn = A - Z

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

Valency (for metals)=Number of valence electrons\text{Valency (for metals)} = \text{Number of valence electrons}

Valency (for non-metals)=8−Number of valence electrons\text{Valency (for non-metals)} = 8 - \text{Number of valence electrons}

💡Examples

Problem 1:

An element XX is represented as 1327X^{27}_{13}X. Calculate the number of protons, electrons, and neutrons present in it.

Solution:

p=13p = 13, e=13e = 13, n=14n = 14

Explanation:

From the notation ZAX^{A}_{Z}X, the Atomic Number Z=13Z = 13, so Protons (pp) = 1313. Since it is a neutral atom, Electrons (ee) = 1313. The Mass Number A=27A = 27. Number of Neutrons n=A−Z=27−13=14n = A - Z = 27 - 13 = 14.

Problem 2:

Determine the electronic configuration and valency of an atom with Atomic Number Z=16Z = 16.

Solution:

Configuration: 2,8,62, 8, 6; Valency: 22

Explanation:

With Z=16Z = 16, the electrons are distributed as: KK shell = 22, LL shell = 88, and MM shell = 66. Since there are 66 valence electrons, the atom needs 22 more to complete its octet. Therefore, Valency=8−6=2\text{Valency} = 8 - 6 = 2.

Problem 3:

Identify the isotopes of Hydrogen and write their notation.

Solution:

Protium (11H^1_1H), Deuterium (12H^2_1H), and Tritium (13H^3_1H)

Explanation:

All three have the same atomic number (Z=1Z=1) but different mass numbers (A=1,2,3A=1, 2, 3 respectively) due to differing numbers of neutrons (0,1,20, 1, 2).