Unit notes and practice questions

Grade 11 Chemistry unit 1: Atomic Structure and Periodic Properties of the Elements

Everything here is for unit 1 of the Grade 11 Chemistry textbook: what the chapter covers, the short notes, practice questions with their answers, and the lesson video where one exists.

8
Cards: 10
Practice questions on unit 1: 1–66
The textbook pages for unit 1

What unit 1 covers

1.Dalton's Atomic Theory and the Modern Atomic Theory
2.Postulates of Dalton's Atomic Theory
3.Postulates of Modern Atomic Theory
4.Early Experiments to Characterize the Atom
5.The Discovery of the Electron
6.Radioactivity and the Discovery of the Nucleus
7.Discovery of the Neutron
8.Make-up of the Nucleus
9.Subatomic Particles
10.Atomic Mass and Isotopes
11.Electromagnetic Radiation and Atomic Spectra
12.Electromagnetic Radiation
13.The Quantum Theory and Photon
14.Atomic Spectra
15.The Bohr Model of the Hydrogen Atom
16.Limitations of the Bohr Model
17.The Wave-Particle Duality of Matter and Energy
18.The Quantum Mechanical Model of the Atom
19.The Heisenberg's Principle
20.Quantum Numbers

Short notes for unit 1

The cards filed under this chapter on the Grade 11 Chemistry page. Open one at a time.

de Broglie WavelengthFormula

\lambda = \frac{h}{m v}

\lambda = \frac{h}{m v}

$\lambda$: de Broglie wavelength $\text{m}$
$h$: Planck's constant $\text{J}\,\text{s}$
$m$: Mass of the particle $\text{kg}$
$v$: Velocity of the particle $\text{m}\,\text{s}^{-1}$

When you use it

Use when finding the wavelength associated with a moving particle such as an electron or atom.

Watch out

Mass must always be in kilograms, not grams or atomic mass units, to cancel units with Planck's constant properly.

Drafted from Grade 11 Chemistry, pages 2-66, then checked twice before it went up

Hydrogen Electron Transition EnergyFormula

\Delta E = -2.18 \times 10^{-18} \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)

\Delta E = -2.18 \times 10^{-18} \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)

$\Delta E$: Energy change of transition $\text{J}$
$n_i$: Initial principal quantum number $\text{dimensionless}$
$n_f$: Final principal quantum number $\text{dimensionless}$

When you use it

Use to calculate the energy absorbed or emitted when an electron in a hydrogen atom jumps between two orbits.

Watch out

A negative change in energy indicates emission of a photon, while a positive change indicates absorption.

Drafted from Grade 11 Chemistry, pages 2-66, then checked twice before it went up

Hydrogen Emission SeriesReference tableLyman · Balmer · Paschen · +1

Series	Final level ($n_f$)	Initial level ($n_i$)	Spectral region
Lyman	$1$	$2, 3, 4, \dots$	Ultraviolet
Balmer	$2$	$3, 4, 5, \dots$	Visible and ultraviolet
Paschen	$3$	$4, 5, 6, \dots$	Infrared
Brackett	$4$	$5, 6, 7, \dots$	Infrared

When you use it

Use to identify the spectral series and wavelength region for electron transitions in atomic hydrogen.

Watch out

Only transitions ending at n = 2 produce visible lines. Transitions ending at n = 1 are entirely ultraviolet.

Drafted from Grade 11 Chemistry, pages 2-66, then checked twice before it went up

Photoelectric Effect EquationFormula

\text{KE}_e = h\nu - h\nu_0

\text{KE}_e = h\nu - h\nu_0

$\text{KE}_e$: Kinetic energy of the ejected electron $\text{J}$
$h$: Planck's constant $\text{J}\,\text{s}$
$\nu$: Frequency of incident radiation $\text{s}^{-1}$
$\nu_0$: Threshold frequency of the metal $\text{s}^{-1}$

When you use it

Use when finding the kinetic energy or speed of electrons ejected from a metal illuminated by light above the threshold frequency.

Watch out

No electrons are emitted if the incident frequency is less than the threshold frequency, regardless of light intensity.

Drafted from Grade 11 Chemistry, pages 2-66, then checked twice before it went up

Photon Energy and WavelengthFormula

E = \frac{h c}{\lambda}

E = \frac{h c}{\lambda}

$E$: Energy of a photon $\text{J}$
$h$: Planck's constant $\text{J}\,\text{s}$
$c$: Speed of light $\text{m}\,\text{s}^{-1}$
$\lambda$: Wavelength $\text{m}$

When you use it

Use when converting between the wavelength of electromagnetic radiation and the energy carried by a single photon.

Watch out

Always convert wavelength from nanometers or micrometers to meters before calculating energy in joules.

Drafted from Grade 11 Chemistry, pages 2-66, then checked twice before it went up

The order the subshells fillRule

\mathrm{1s\;2s\;2p\;3s\;3p\;4s\;3d\;4p\;5s\;4d\;5p\;6s\;4f\;5d\;6p\;7s\;5f\;6d\;7p}

\mathrm{1s\;2s\;2p\;3s\;3p\;4s\;3d\;4p\;5s\;4d\;5p\;6s\;4f\;5d\;6p\;7s\;5f\;6d\;7p}

$\mathrm{s}$: one orbital, so the subshell holds 2 electrons
$\mathrm{p}$: three orbitals, so the subshell holds 6 electrons
$\mathrm{d}$: five orbitals, so the subshell holds 10 electrons
$\mathrm{f}$: seven orbitals, so the subshell holds 14 electrons
$2(2l+1)$: how many electrons a subshell holds, counted from its own quantum number

When you use it

Use whenever you write a ground state electron configuration. Electrons enter the lowest energy subshell first (Aufbau), an orbital takes at most two electrons and those two spin opposite ways (Pauli), and among orbitals of equal energy each takes one electron before any takes a second (Hund).

Watch out

After $\mathrm{3p}$ the next electron goes into $\mathrm{4s}$ and not $\mathrm{3d}$ , so potassium is $1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{6}\,4s^{1}$ . The trap is that the same textbook then prints scandium as $[\mathrm{Ar}]\,3d^{1}\,4s^{2}$ , because a finished configuration is written by shell number rather than by the order it filled. Fill in the order above, write the answer in shell order, and when the atom is ionised take the electrons out of $\mathrm{4s}$ first: $\mathrm{Fe^{2+}}$ is $[\mathrm{Ar}]\,3d^{6}$ and never $[\mathrm{Ar}]\,3d^{4}\,4s^{2}$ .

Chromium, copper and the half filled d subshellCommon mistakeChromium, Z = 24 · Copper, Z = 29 · Molybdenum, Z = 42 · +2

Element	Ground state	What to notice
Chromium, $Z = 24$	$[\mathrm{Ar}]\,3d^{5}\,4s^{1}$	the filling order predicts $[\mathrm{Ar}]\,3d^{4}\,4s^{2}$
Copper, $Z = 29$	$[\mathrm{Ar}]\,3d^{10}\,4s^{1}$	the filling order predicts $[\mathrm{Ar}]\,3d^{9}\,4s^{2}$
Molybdenum, $Z = 42$	$[\mathrm{Kr}]\,4d^{5}\,5s^{1}$	the same pattern as chromium, one row above it
Silver, $Z = 47$	$[\mathrm{Kr}]\,4d^{10}\,5s^{1}$	the same pattern as copper, one row above it
Niobium, $Z = 41$	$[\mathrm{Kr}]\,4d^{4}\,5s^{1}$	neither subshell ends up half filled or full

When you use it

Use when a d-block element's ground state refuses to match the plain filling order: one s electron shifts into the d subshell, because a half filled set ( $d^{5}$ ) or a completely filled set ( $d^{10}$ ) is the more stable arrangement. Check any configuration you write by adding its electrons up, since they have to total the atomic number.

Watch out

Students turn "half filled is stable" into a general licence and promote an s electron wherever the d subshell looks close to five or ten. Manganese is $[\mathrm{Ar}]\,3d^{5}\,4s^{2}$ and iron is $[\mathrm{Ar}]\,3d^{6}\,4s^{2}$ : count the electrons and the invented version comes out one short of the atomic number. The second half of the mistake shows up with ions. Once the atom ionises the $4s$ electron leaves first, so $\mathrm{Cu^{2+}}$ is $[\mathrm{Ar}]\,3d^{9}$ and $\mathrm{Cr^{3+}}$ is $[\mathrm{Ar}]\,3d^{3}$ , never $[\mathrm{Ar}]\,3d^{8}\,4s^{1}$ or $[\mathrm{Ar}]\,3d^{2}\,4s^{1}$ .

The s, p, d and f blocksReference tables · p · d · +1

Block	Outer configuration	Where it sits, and what it is like
s	$ns^{1-2}$	Groups 1 and 2, plus hydrogen and helium. All metals apart from those two, strongly electropositive with low ionization energy, and mostly ionic compounds; beryllium is the exception, since $\mathrm{BeCl_2}$ is covalent and $\mathrm{BeO}$ is amphoteric.
p	$ns^{2}\,np^{1-6}$	Groups 13 to 18. The only block that holds metals, non-metals and metalloids together, with 3 to 8 valence electrons and variable oxidation states; non-metallic character rises across a period and falls down a group.
d	$(n-1)d^{1-10}\,ns^{0-2}$	Groups 3 to 12. All metals, mostly hard and high melting, with variable oxidation states, coloured compounds and many catalysts; zinc, cadmium and mercury sit in the block but are not transition elements, because neither the atom nor its usual ion has a partly filled d subshell.
f	$(n-2)f^{1-14}\,(n-1)d^{0-1}\,ns^{2}$	The two rows printed below the main table: lanthanoids from La to Lu and actinoids from Ac to Lr, fifteen each. All metals, dense and high melting; every actinoid is radioactive, while among the lanthanoids only promethium is.

When you use it

Use when you are asked which block an element belongs to, or to write its outer configuration from its place in the periodic table. A block is named after the subshell that receives the last electron added, so read it off the configuration rather than off where the printed table puts the element.

Watch out

Students read the n in $(n-1)d$ and $(n-2)f$ as the number of the d or f subshell itself. It is the outermost shell, which is the period number. Iron sits in Period 4, so $n = 4$ and its configuration is $(n-1)d^{6}\,ns^{2} = 3d^{6}\,4s^{2}$ , never $4d^{6}\,4s^{2}$ ; cerium sits in Period 6, so $(n-2)f$ is $4f$ . Ask which period the element is in first, then subtract. The second trap is helium: it is printed above neon in Group 18 and gets written down as p-block, but $1s^{2}$ has no p electron at all, so helium is an s-block element.

Practice questions on unit 1

10 multiple-choice questions written against this chapter, each with its answer and the reason. Try one before you open the answer.

Question 1

The Greek philosopher Democritus called his tiny particles atomos. What does this word mean?

A.Indivisible
B.Empty
C.Heavy
D.Continuous

Show the answer and why

Answer: A Indivisible

Why

The word atomos means "indivisible". Democritus believed that if you divided matter into smaller and smaller pieces, you would eventually reach tiny particles that could not be divided any further.

Question 2

In Thomson's cathode ray tube, the beam of cathode rays bends away from one plate and toward the other. What charge do the cathode rays carry?

A.Negative
B.Positive
C.No charge
D.Both positive and negative

Show the answer and why

Answer: A Negative

Why

The beam bends away from the negatively charged plate and toward the positively charged plate. So the rays are negatively charged. Thomson concluded that a cathode ray is a beam of negatively charged particles, which we now call electrons.

Question 3

Which type of radioactive emission has no electric charge and is not affected by an electric or magnetic field?

A.Alpha particles
B.Beta particles
C.Gamma rays
D.Cathode rays

Show the answer and why

Answer: C Gamma rays

Why

Gamma (γ) rays are high-energy rays with no charge. They are not affected by an external electric or magnetic field. Alpha particles and beta particles are both deflected by charged plates.

Question 4

Carbon has an atomic number of 6. Naturally occurring carbon-13 (¹³C) has a mass number of 13. How many neutrons are in the nucleus of one carbon-13 atom?

A.6
B.7
C.13
D.19

Show the answer and why

Answer: B 7

Why

The mass number is the number of protons plus neutrons. Carbon has 6 protons, so neutrons = 13 − 6 = 7. Carbon-13 has seven neutrons, as the chapter states.

Question 5

Water is always 11.2% hydrogen by mass. What mass of hydrogen is in 18.0 g of water?

A.2.02 g
B.15.98 g
C.0.112 g
D.0.888 g

Show the answer and why

Answer: A 2.02 g

Why

Mass of hydrogen = 0.112 × 18.0 g = 2.02 g. Option B is the mass of oxygen, and options C and D are the answers for 1.00 g of water, not 18.0 g.

Question 6

Ethiopian National Radio, Addis Ababa station, broadcasts its AM signal at a frequency of 2400 kHz. Using c = 3.00 × 10⁸ m/s, what is the wavelength of this radio wave in metres?

A.125.0 m
B.0.125 m
C.12.5 m
D.1250 m

Show the answer and why

Answer: A 125.0 m

Why

Use λ = c / ν. The frequency is 2400 kHz = 2.4 × 10⁶ s^-1. So λ = (3.00 × 10⁸ m/s) / (2.4 × 10⁶ s^-1) = 125.0 m, as in Example 1.3.

Question 7

According to the quantum mechanical model, how many orbitals are there in the principal energy level n = 3?

A.3
B.6
C.9
D.18

Show the answer and why

Answer: C 9

Why

The number of orbitals in an energy level is n². For n = 3, that is 3² = 9. These are one 3s orbital, three 3p orbitals and five 3d orbitals (1 + 3 + 5 = 9).

Question 8

Silver has two naturally occurring isotopes. ¹⁰⁷Ag has a mass of 106.90509 amu and 51.84% abundance. ¹⁰⁹Ag has a mass of 108.90476 amu and 48.16% abundance. What is the atomic mass of silver?

A.106.91 amu
B.107.87 amu
C.108.90 amu
D.107.00 amu

Show the answer and why

Answer: B 107.87 amu

Why

Multiply each isotope mass by its fractional abundance and add the results. 106.90509 × 0.5184 = 55.42 amu. 108.90476 × 0.4816 = 52.45 amu. The total is 55.42 + 52.45 = 107.87 amu.

Question 9

Copper(I) chloride emits blue light with a wavelength of 600 nm. Using h = 6.63 × 10^-34 J·s and c = 3.0 × 10⁸ m/s, what is the energy of one quantum (photon) of this light?

A.3.32 × 10-19 J
B.3.32 × 10-34 J
C.6.63 × 10-19 J
D.5.00 × 1014 J

Show the answer and why

Answer: A 3.32 × 10-19 J

Why

First find the frequency: ν = c / λ = 3.0 × 10⁸ / 6.0 × 10^-7 = 0.50 × 10¹⁵ s^-1. Then E = hν = 6.63 × 10^-34 × 0.50 × 10¹⁵ = 3.32 × 10^-19 J, as in Example 1.4.

Question 10

In a hydrogen atom, the electron falls from n = 3 to n = 2. The energy emitted is 3.03 × 10^-19 J. What is the wavelength of the emitted light?

A.656 nm
B.486 nm
C.434 nm
D.122 nm

Show the answer and why

Answer: A 656 nm

Why

Use ∆E = hc / λ, so λ = hc / ∆E = (6.63 × 10^-34 J·s × 3.0 × 10⁸ m/s) / (3.03 × 10^-19 J) = 6.56 × 10^-7 m = 656 nm, as in Example 1.7.

The app holds 12 questions on this unit, marked as you go. These are the first 10.

The textbook pages for unit 1

Unit 1 runs from page 1 to page 66 of the Ministry of Education edition.

Ministry of Education edition, English, 342 pages

Download the PDF

All Grade 11 Chemistry short notes

Practise this unit in the app

Temari turns this chapter into timed practice and marks it as you go, with an AI tutor to explain anything that did not land.

Start free

Temari.et is also an Android app, free on Google Play.

About the Temari app