Worked Examples: Lists

Worked Examples: Lists#

These worked solutions correspond to the exercises on the Lists page.

How to use this notebook:

  • Try each exercise yourself first before looking at the solution

  • The code cells show both the code and its output

  • Download this notebook if you want to run and experiment with the code yourself

  • Your solution might look different - that’s fine as long as it gives the correct answer!

Setup#

We’ll import the math module here for use throughout the exercises.

import math

Exercise 1: List Slicing with Halogens#

Problem: Given the list of halogens, use list slicing to:

  • a) Extract the first three elements

  • b) Extract the last three elements

  • c) Extract every second element

  • d) Determine what using a negative number following the second colon (e.g., [::-1]) gives

First, let’s define the list:

halogens = ['fluorine', 'chlorine', 'bromine', 'iodine', 'astatine']
print(halogens)

['fluorine', 'chlorine', 'bromine', 'iodine', 'astatine']

Part (a): First three elements#

Goal: Extract ['fluorine', 'chlorine', 'bromine']

print(halogens[:3])

['fluorine', 'chlorine', 'bromine']

Explanation:

The slice [:3] means “from the start up to (but not including) index 3”.

  • Index 0: 'fluorine'

  • Index 1: 'chlorine'

  • Index 2: 'bromine'

  • Index 3: 'iodine' ← stop here (not included)

General pattern: [:n] gives you the first n elements.

Part (b): Last three elements#

Goal: Extract ['bromine', 'iodine', 'astatine']

print(halogens[-3:])

['bromine', 'iodine', 'astatine']

Explanation:

The slice [-3:] means “from the 3rd element from the end to the end”.

Negative indices count backwards from the end:

  • Index -1: 'astatine' (last element)

  • Index -2: 'iodine' (second-to-last)

  • Index -3: 'bromine' (third-to-last) ← start here

  • Index -4: 'chlorine'

  • Index -5: 'fluorine'

Alternative solution: You could also use [2:] (from index 2 onwards), but [-3:] is more flexible - it works regardless of list length.

Part (c): Every second element#

Goal: Extract ['fluorine', 'bromine', 'astatine']

print(halogens[::2])

['fluorine', 'bromine', 'astatine']

Explanation:

The slice [::2] means “from start to end, stepping by 2”.

The full slice syntax is [start:stop:step]:

  • start: where to begin (default: beginning)

  • stop: where to end (default: end)

  • step: how many indices to jump (default: 1)

So [::2] takes:

  • Index 0: 'fluorine'

  • Index 2: 'bromine'

  • Index 4: 'astatine'

Other examples:

print(halogens[1::2])  # Every second element starting from index 1
print(halogens[::3])   # Every third element

['chlorine', 'iodine']
['fluorine', 'iodine']

Part (d): Using a negative step#

Goal: Discover what [::-1] does

print(halogens[::-1])

['astatine', 'iodine', 'bromine', 'chlorine', 'fluorine']

Explanation:

Using -1 as the step value reverses the list!

[::-1] means:

  • Start: default (end of list when step is negative)

  • Stop: default (beginning of list when step is negative)

  • Step: -1 (move backwards by 1)

This is a very common Python idiom for reversing sequences.

Other negative steps:

print(halogens[::-2])  # Every second element, reversed

['astatine', 'bromine', 'fluorine']

Exercise 2: Calculating Interatomic Distances#

Problem: For each molecule, calculate the intramolecular distances between all three atoms using the formula:

\[ r = \sqrt{(x_i - x_j)^2 + (y_i - y_j)^2 + (z_i - z_j)^2} \]

Then comment on the shape of each molecule.

Molecule 1#

Given coordinates:

  • Atom 1: (0.1, 0.5, 3.2)

  • Atom 2: (0.4, 0.5, 2.3)

  • Atom 3: (-0.3, 0.3, 1.7)

First, define the atomic positions:

atom_1 = [0.1, 0.5, 3.2]
atom_2 = [0.4, 0.5, 2.3]
atom_3 = [-0.3, 0.3, 1.7]

Now calculate all three pairwise distances:

# Distance between atoms 1 and 2
r_12 = sqrt((atom_1[0] - atom_2[0])**2 + (atom_1[1] - atom_2[1])**2 + (atom_1[2] - atom_2[2])**2)

# Distance between atoms 1 and 3
r_13 = sqrt((atom_1[0] - atom_3[0])**2 + (atom_1[1] - atom_3[1])**2 + (atom_1[2] - atom_3[2])**2)

# Distance between atoms 2 and 3
r_23 = sqrt((atom_2[0] - atom_3[0])**2 + (atom_2[1] - atom_3[1])**2 + (atom_2[2] - atom_3[2])**2)

print(f"Distance between atoms 1 and 2: {r_12:.4f}")
print(f"Distance between atoms 1 and 3: {r_13:.4f}")
print(f"Distance between atoms 2 and 3: {r_23:.4f}")

---------------------------------------------------------------------------
NameError                                 Traceback (most recent call last)
Cell In[10], line 2
      1 # Distance between atoms 1 and 2
----> 2 r_12 = sqrt((atom_1[0] - atom_2[0])**2 + (atom_1[1] - atom_2[1])**2 + (atom_1[2] - atom_2[2])**2)
      4 # Distance between atoms 1 and 3
      5 r_13 = sqrt((atom_1[0] - atom_3[0])**2 + (atom_1[1] - atom_3[1])**2 + (atom_1[2] - atom_3[2])**2)

NameError: name 'sqrt' is not defined

Explanation of the formula:

For each distance calculation:

  • atom_1[0] accesses the x-coordinate of atom 1

  • atom_1[1] accesses the y-coordinate of atom 1

  • atom_1[2] accesses the z-coordinate of atom 1

We calculate:

  1. \((x_1 - x_2)^2\) using (atom_1[0] - atom_2[0])**2

  2. \((y_1 - y_2)^2\) using (atom_1[1] - atom_2[1])**2

  3. \((z_1 - z_2)^2\) using (atom_1[2] - atom_2[2])**2

  4. Sum these three values

  5. Take the square root

Analysis of Molecule 1:

The distance \(r_{13}\) (1.565) is not equal to the sum \(r_{12} + r_{23}\) (1.892).

In fact, \(r_{13} < r_{12} + r_{23}\), which means the three atoms are not collinear (not in a straight line).

Conclusion: Molecule 1 has a bent geometry. Atom 2 sits at an angle between atoms 1 and 3.

This is similar to the structure of water (H₂O) or sulfur dioxide (SO₂), where the central atom forms an angle with the two outer atoms.

Molecule 2#

Given coordinates:

  • Atom 1: (-0.1, 0.5, 1.5)

  • Atom 2: (0.2, 0.5, 2.6)

  • Atom 3: (0.5, 0.5, 3.7)

Define the atomic positions:

atom_1 = [-0.1, 0.5, 1.5]
atom_2 = [0.2, 0.5, 2.6]
atom_3 = [0.5, 0.5, 3.7]

Calculate the distances:

r_12 = sqrt((atom_1[0] - atom_2[0])**2 + (atom_1[1] - atom_2[1])**2 + (atom_1[2] - atom_2[2])**2)
r_13 = sqrt((atom_1[0] - atom_3[0])**2 + (atom_1[1] - atom_3[1])**2 + (atom_1[2] - atom_3[2])**2)
r_23 = sqrt((atom_2[0] - atom_3[0])**2 + (atom_2[1] - atom_3[1])**2 + (atom_2[2] - atom_3[2])**2)

print(f"Distance between atoms 1 and 2: {r_12:.4f}")
print(f"Distance between atoms 1 and 3: {r_13:.4f}")
print(f"Distance between atoms 2 and 3: {r_23:.4f}")

Distance between atoms 1 and 2: 1.1402
Distance between atoms 1 and 3: 2.2804
Distance between atoms 2 and 3: 1.1402

Analysis of Molecule 2:

Check if the atoms are collinear:

The distance \(r_{13}\) (2.280) is equal to the sum \(r_{12} + r_{23}\) (also 2.280).

Notice also that \(r_{12} = r_{23}\) = 1.140 - the two bonds are the same length!

Conclusion: Molecule 2 has a linear geometry. The three atoms are arranged in a straight line, with atom 2 positioned exactly midway between atoms 1 and 3.

This is similar to the structure of carbon dioxide (CO₂) or beryllium chloride (BeCl₂), where a central atom forms 180° bonds with two identical atoms.|

Summary#

1. List Slicing#

  • [:n] - first n elements

  • [-n:] - last n elements

  • [::step] - every step-th element

  • [::-1] - reverse the list

2. Molecular Geometry from Coordinates#

For a triatomic molecule ABC:

Linear geometry: \(r_{AC} = r_{AB} + r_{BC}\)

  • Atoms arranged in a straight line

  • Bond angle = 180°

  • Example: CO₂

Bent geometry: \(r_{AC} < r_{AB} + r_{BC}\)

  • Atoms form an angle

  • Bond angle < 180°

  • Example: H₂O, SO₂

The distance formula in 3D:

\[r = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}\]

This is the 3D extension of Pythagoras’ theorem.

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