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Atomic Structure - Bohr Models and Electronic Configurations

Grade 9IB

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

🔑Concepts

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The Bohr Model describes the atom as a small, positively charged nucleus surrounded by electrons that travel in circular orbits (energy levels) around the nucleus.

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Energy levels or shells are denoted by the principal quantum number n=1,2,3,4,…n = 1, 2, 3, 4, \dots or the letters K,L,M,N,…K, L, M, N, \dots.

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The maximum number of electrons that can be accommodated in a shell is given by the formula 2n22n^2, where nn is the shell number.

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Electrons fill the lower energy levels (closer to the nucleus) first before moving to higher energy levels. This is known as the Bohr-Bury scheme.

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The outermost shell of an atom is called the valence shell, and the electrons present in it are called valence electrons. For the first 20 elements, the valence shell can hold a maximum of 88 electrons (except for Helium, which holds 22).

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Atomic number (ZZ) represents the number of protons in the nucleus. 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, where NN is the number of neutrons.

📐Formulae

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

A=Z+NA = Z + N

Z=p=e− (for neutral atoms)Z = p = e^- \text{ (for neutral atoms)}

Valence=8−valence electrons (if valence electrons >4)\text{Valence} = 8 - \text{valence electrons (if valence electrons } > 4\text{)}

💡Examples

Problem 1:

Write the electronic configuration and draw the Bohr model distribution for a Phosphorus atom (Z=15Z = 15).

Solution:

  1. Total electrons = 1515.
  2. First shell (K,n=1K, n=1): 2(1)2=22(1)^2 = 2 electrons.
  3. Second shell (L,n=2L, n=2): 2(2)2=82(2)^2 = 8 electrons.
  4. Remaining electrons = 15−(2+8)=515 - (2 + 8) = 5.
  5. Third shell (M,n=3M, n=3): 55 electrons. Electronic Configuration: 2,8,52, 8, 5.

Explanation:

Electrons fill shells starting from n=1n=1. Since 1515 electrons are present, we fill 22 in the first, 88 in the second, and the remaining 55 in the third.

Problem 2:

Calculate the number of neutrons in an isotope of Potassium with mass number A=39A = 39 and atomic number Z=19Z = 19.

Solution:

Using the formula: A=Z+NA = Z + N 39=19+N39 = 19 + N N=39−19N = 39 - 19 N=20N = 20

Explanation:

The number of neutrons is found by subtracting the atomic number (protons) from the mass number (protons + neutrons).

Problem 3:

An element has an electronic configuration of 2,8,22, 8, 2. Identify the element and its valency.

Solution:

  1. Total electrons = 2+8+2=122 + 8 + 2 = 12.
  2. Atomic number Z=12Z = 12, which corresponds to Magnesium (MgMg).
  3. Valence electrons = 22.
  4. Valency = 22.

Explanation:

The sum of electrons gives the atomic number. Since it has 22 electrons in its outermost shell, it tends to lose them to achieve a stable octet, giving it a valency of 22.