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Chapter 02: Atomic Structure2

June 16, 2021 by Admin

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.

Orbitals:

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:

Crazy Copper

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.

	Down the group	Across the period
Trend	Increase: Atomic radius Decrease: Electronegativity Ionisation energy	Increase: Electronegativity Ionisation energy Decrease: Electronegativity Ionisation energy
Explanation	General explanation: Nuclear charge (NC) increases as the number of protons increases down the group. With the increase in the number of shells, valence electrons are further away from the nucleus. Hence shielding effect (SE) increases. Atomic radius explanation: (Add on) Increase in SE is greater than increase in NC hence valence electrons are less strongly attracted to nucleus, causing the atomic radius to increase. Electronegativity explanation: (Add on) Hence the ability of an atom to attract bonding electrons to itself decreases, causing electronegativity to decrease. Ionisation energy explanation: (Add on) The valence electrons are less strongly attracted to the nucleus and less energy is required to remove the valance electrons, causing ionisation energy to decrease.	General explanation: The number of protons increases across the period, hence nuclear charge increases. Successive electrons added to the same outermost shell hence increase in shielding effect is negligible. Increase in NC outweighs increase in SE, causing effective NC to increase. Atomic radius explanation: (Add on) Valence electrons are more strongly attracted to the nucleus, hence the atomic radius decreases. Electronegativity explanation: (Add on) Hence the ability of an atom to attract bonding electrons to itself increases, causing electronegativity to increase across the period. Ionisation energy explanation: (Add on) The valence electrons are more strongly attracted to the nucleus, thus more energy is required to remove the valence electrons, causing ionisation energy to generally increase across the period.

Down the group

Across the period

Trend

Increase:

Atomic radius

Decrease:

Electronegativity
Ionisation energy

Increase:

Electronegativity
Ionisation energy

Decrease:

Electronegativity
Ionisation energy

Explanation

General explanation: Nuclear charge (NC) increases as the number of protons increases down the group. With the increase in the number of shells, valence electrons are further away from the nucleus. Hence shielding effect (SE) increases.

Atomic radius explanation: (Add on) Increase in SE is greater than increase in NC hence valence electrons are less strongly attracted to nucleus, causing the atomic radius to increase.

Electronegativity explanation: (Add on) Hence the ability of an atom to attract bonding electrons to itself decreases, causing electronegativity to decrease.

Ionisation energy explanation: (Add on) The valence electrons are less strongly attracted to the nucleus and less energy is required to remove the valance electrons, causing ionisation energy to decrease.

General explanation: The number of protons increases across the period, hence nuclear charge increases. Successive electrons added to the same outermost shell hence increase in shielding effect is negligible. Increase in NC outweighs increase in SE, causing effective NC to increase.

Atomic radius explanation: (Add on) Valence electrons are more strongly attracted to the nucleus, hence the atomic radius decreases.

Electronegativity explanation: (Add on) Hence the ability of an atom to attract bonding electrons to itself increases, causing electronegativity to increase across the period.

Ionisation energy explanation: (Add on) The valence electrons are more strongly attracted to the nucleus, thus more energy is required to remove the valence electrons, causing ionisation energy to generally increase across the period.

Down the group
Trend Increase: Atomic radius Decrease: Electronegativity Ionisation energy
Explanation General explanation: Nuclear charge (NC) increases as the number of protons increases down the group. With the increase in the number of shells, valence electrons are further away from the nucleus. Hence shielding effect (SE) increases. Atomic radius explanation: (Add on) Increase in SE is greater than increase in NC hence valence electrons are less strongly attracted to nucleus, causing the atomic radius to increase. Electronegativity explanation: (Add on) Hence the ability of an atom to attract bonding electrons to itself decreases, causing electronegativity to decrease. Ionisation energy explanation: (Add on) The valence electrons are less strongly attracted to the nucleus and less energy is required to remove the valance electrons, causing ionisation energy to decrease.

Down the group

Trend Increase:

Atomic radius

Decrease:

Electronegativity
Ionisation energy

Explanation

Across the period
Trend Increase: Electronegativity Ionisation energy Decrease: Electronegativity Ionisation energy
Explanation General explanation: The number of protons increases across the period, hence nuclear charge increases. Successive electrons added to the same outermost shell hence increase in shielding effect is negligible. Increase in NC outweighs increase in SE, causing effective NC to increase. Atomic radius explanation: (Add on) Valence electrons are more strongly attracted to the nucleus, hence the atomic radius decreases. Electronegativity explanation: (Add on) Hence the ability of an atom to attract bonding electrons to itself increases, causing electronegativity to increase across the period. Ionisation energy explanation: (Add on) The valence electrons are more strongly attracted to the nucleus, thus more energy is required to remove the valence electrons, causing ionisation energy to generally increase across the period.

Across the period

Trend Increase:

Electronegativity
Ionisation energy

Decrease:

Electronegativity
Ionisation energy

Explanation

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:

Mg and Al	P and S
Electronic configuration of Mg = 1s²2s²2p⁶3s² Electronic configuration of Al = 1s²2s²2p⁶3s²3p¹ Due to the increased shielding experienced by the 3p electron in Al, it takes less energy to be removed than the valence electron in the 3 subshell of Mg. Thus the first ionisation energy of Al is smaller than the first ionisation energy of Mg.	Electronic configuration of P =1s²2s²2p⁶3s²3pₓ¹3pᵧ¹3p_z¹ Electronic configuration of S =1s²2s²2p⁶3s²3pₓ²3pᵧ¹3p_z¹ It takes less energy to remove the paired electron in the 3pₓ² orbital of S compared to the unpaired electron in the 3pₓ¹ of P due to inter-electron repulsion. Thus the first ionisation energy of S is smaller than the first ionisation energy of P.

Mg and Al
Electronic configuration of Mg = 1s²2s²2p⁶3s² Electronic configuration of Al = 1s²2s²2p⁶3s²3p¹ Due to the increased shielding experienced by the 3p electron in Al, it takes less energy to be removed than the valence electron in the 3 subshell of Mg. Thus the first ionisation energy of Al is smaller than the first ionisation energy of Mg.

P and S
Electronic configuration of P =1s²2s²2p⁶3s²3pₓ¹3pᵧ¹3p_z¹ Electronic configuration of S =1s²2s²2p⁶3s²3pₓ²3pᵧ¹3p_z¹ It takes less energy to remove the paired electron in the 3pₓ² orbital of S compared to the unpaired electron in the 3pₓ¹ of P due to inter-electron repulsion. Thus the first ionisation energy of S is smaller than the first ionisation energy of P.

Continue Reading on Chapter 3: Chemical Bonding