Chapter 02: Atomic Structure2

Chapter 2 : Atomic Structure

A-Level H2 Chemistry: Chapter 2

Atomic Structure

Commonly Overlooked Concepts

1. Behaviour of Fundamental Particles in Electric Field

Although this only takes up a very small portion of this chapter, many students tend to forgot this while revising. This concept is commonly tested in Paper 1 (MCQ) and Paper 2 (Short Answer Questions), and is the easiest to score.

Note: Deflection occurs when the particle just entered the electric field (after the red dotted line)
When drawing, – Take note of the direction the particle is travelling in and its angle of deflection.
– Particles with a lighter mass and/or higher charge will have a larger angle of deflection and vice versa. – The angle of deflection for an electron is greater as explained by the _charge / ^mass ratio. The smaller the mass of the electron, the larger its angle of deflection. Angle of Deflection ∝ _charge / ^mass

Quick Example: [SAJC 2019 H2 Chem Paper 1 Qn 4]

In an experiment, a sample of gaseous ⁸⁷Sr²⁺ was passed through an electric field. The angle of deflection for ⁸⁷Sr²⁺ was observed to be 2°.

The experiment was repeated with a gaseous sample of particle Q.
Which of the following could be Q? A   ⁷⁴As³⁻
B   ¹⁹F⁻
C   ⁷⁹Se²⁻
D   ¹²⁷Te²⁻ Solution:
Using the above proportion formula, we can calculate the angle of deflection of each given option. Angle of Deflection = (k) _charge / ^mass , where k is the constant 2°= k ₂ / ⁸⁷ 1°= k ₁ / ⁸⁷

k = 87

Tip: Just include only the magnitude of the particle’s charge in the ratio

Angle of deflection of ⁷⁴As³⁻ = 87 x ₃ / ⁷⁴ = 3.5 Angle of deflection of ¹⁹F⁻ = 87 x ₁ / ¹⁹ = 4.6 Angle of deflection of ⁷⁹Se²⁻ = 87 x ₂ / ⁷⁹ = 2.2 Angle of deflection of ¹²⁷Te²⁻ = 87 x ₂ / ¹²⁷ = 1.4

Angle of deflection of Q must be smaller than 2° according to the diagram, thus the final answer would be D.

s orbitals
every shell only has one s orbital all s orbitals are spherical, symmetrical and non-directional

p orbitals
p orbitals are dumb-bell shaped and are directional p orbitals in the same subshell have the same energy (degenerates)

d orbitals
d orbitals are generally shaped like a four-bladed propeller and are directional

s orbitals	p orbitals	d orbitals
every shell only has one s orbital all s orbitals are spherical, symmetrical and non-directional	p orbitals are dumb-bell shaped and are directional p orbitals in the same subshell have the same energy (degenerates)	d orbitals are generally shaped like a four-bladed propeller and are directional

Electronic configuration:

Commonly Tested – anomalous electronic configurations.

It is key that students familiarise themselves with the following abnormalities as well as the reasons for such abnormalities in order to secure giveaway marks during examinations.

A trick to remember would be:

1. Chromium (₂₄Cr)
– Configuration : 1s²2s²2p⁶3s²3p⁶3d_xy¹3d_xz¹3d_yz¹3d_x² – _y²  ¹3d_z² ¹4s¹
– Reason : the energy difference between the 4s and 3d orbital is minimal, hence
electrons occupy the 3d orbitals singly to minimise electronic repulsion 2. Copper (₂₉Cu)
– Configuration: 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹
– Reason : the fully-filled 3d subshells are unusually stable due to symmetrical distribution of charge around the nucleus IMPORTANT REMINDERS
– 4s orbitals have lower energy than 3d orbitals
– In writing electronic configurations, 3d should still be written before 4s even though
the 4s subshell is filled before the 3d

Periodic trends:

This section of the chapter covers fundamental concepts that can be easily grasped through sheer memorisation work. Hence, the following table would provide a concise summary to aid students in remembering the periodic trends.

Note:
Students should also memorise definitions of important terms such as
– Atomic radius
– Ionic radius
– Electronegativity
– Ionisation energy

A favourite question among examination paper setters is the following deductive question on electronic configurations.

The following is a graph of the successive ionisation energies of element Y. By using the graph, predict the electronic configuration of Y.

Daunting at first, this question is manageable with this ABC method and some practice.

Step 1
A: Take note of the horizontal and vertical Axis to make sense of the data presented to you.
In this question, the horizontal axis represents the number of electrons removed while the vertical axis represents the ionisation energies of electrons in element Y.
The graph shows us that as successive electrons are removed from the outer to inner shell electrons, ionisation energy required increases. This aligns with what we know about ionisation energies of electrons. Step 2
B: Look out for Big changes in ionisation energies. From what we have learnt, ionisation energy of an electron from an inner shell is significantly larger than ionisation energy of an electron of an outer shell. Hence, we can deduce this: Step 3
C: Count the number of electrons in each electron shell. The outermost shell has 2, which makes it likely to be in a s orbital.
The next inner shell has 8 electrons, hence it should be in the next inner quantum shell.
Since this quantum shell has 8 electrons, it likely consists of s and p subshells.
We do not know if the innermost electron shell shown in the graph has 2 or more electrons, but since we know the outer two quantum shells, the answer 1s²2s²2p⁶3s² or 1s²2s²2p⁶3s²3p⁶4s² is acceptable.

Exceptions in trends of ionisation energies

Chemistry is unfortunately full of exceptions. Examiners enjoy testing students on exceptions to assess if students’ comprehension of chemistry concepts are thorough. Generally, as proton number increases across a period, you would expect the first ionisation energy of the elements to increase. But there are two exceptions to this rule: