General Chemistry 202-SN1-RE • CEGEP, Montreal

Revision sheet: atomic structure and electron configuration (202-SN1-RE)

Atomic structure is the most abstract chapter of 202-SN1-RE (formerly 202-NYA), and yet the one where the errors are the most mechanical: a wavelength in nanometres, an energy per atom compared with an energy per mole, the sign of a transition. The ideas are understood; the marks go elsewhere.

This sheet separates what belongs to the course from what belongs to the exam routine: the four filling rules, the two energy formulas and their units, the periodic trends and their anomalies, and the list of the conversions that cost the most.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

This whole chapter rests on a single physical idea: an electron has only ALLOWED energies. Quantum numbers, configurations, spectra and periodicity are just four ways of reading that same constraint.

This chapter is part of 202-SN1-RE, General Chemistry

The essentials

The four quantum numbers and their bounds

• $n$ , a positive integer: the shell. It contains $n^{2}$ orbitals and therefore at most $2n^{2}$ electrons.
• $l$ , an integer from $0$ to $n-1$ : the subshell, written s, p, d, f. An $l$ equal to or greater than $n$ is FORBIDDEN.
• $m_{l}$ : all the integers from $-l$ to $+l$ , hence $2l+1$ orbitals per subshell.
• $m_{s}=+\tfrac{1}{2}$ or $-\tfrac{1}{2}$ : the spin, only two values.
• Three filling rules: Aufbau, in order of increasing energy $1s\,2s\,2p\,3s\,3p\,4s\,3d\,4p$ ; Pauli, at most two electrons per orbital and with opposite spins; Hund, one electron in each orbital before pairing up.

Each shell

n

has

n^{2}

orbitals and its subshells stop at

l=n-1

: the

2d

box does not exist, and any set that calls for it is impossible.

A set of quantum numbers is checked from left to right: first $l<n$ , then $|m_{l}|\le l$ , then $m_{s}=\pm\tfrac{1}{2}$ . A single violation makes the set impossible, and it is a course question on every exam.

The two energy formulas, and their units

• Photon: $E=h\nu=\dfrac{hc}{\lambda}$ , with $h=6.626\times 10^{-34}$ J·s, $c=3.00\times 10^{8}$ m/s and $\lambda$ IN METRES.
• Hydrogen levels: $E_{n}=-\dfrac{2.18\times 10^{-18}}{n^{2}}$ J, and for an ion with a single electron $E_{n}=-\dfrac{2.18\times 10^{-18}Z^{2}}{n^{2}}$ J.
• Transition: $\Delta E=E_{final}-E_{initial}$ . Negative when the atom EMITS, positive when it ABSORBS. The photon itself always has a positive energy $|\Delta E|$ .
• Photoelectric effect: $E_{k,max}=h\nu-W_{0}$ . Below the threshold nothing is ejected, whatever the intensity.
• Per atom or per mole: multiply by $N_{A}=6.022\times 10^{23}$ to go from joules per atom to joules per mole.

The levels crowd together toward the top: this is why transitions that fall back to

n=1

release much more energy, hence ultraviolet, than those that fall back to

n=2

, which give visible light.

Two conversions alone account for half the marks lost on this chapter: nanometres to metres ( $1$ nm $=10^{-9}$ m) and joules per atom to kilojoules per mole.

Periodicity, and the two rules that explain it

• The effective nuclear charge $Z_{eff}$ INCREASES from left to right across a period, because protons are added without adding a shell.
• Atomic radius: DECREASES from left to right (stronger attraction), INCREASES from top to bottom (one more shell).
• Ionization energy and electronegativity follow the OPPOSITE trend to the radius: the smaller the atom, the harder it is to ionize.
• A CATION is always smaller than its atom, an ANION always larger: the first often loses a whole shell, the second adds repulsion without adding a proton.
• Anomalies to know: boron below beryllium and oxygen below nitrogen for ionization energy, due respectively to the move into a p subshell and to the first paired electrons.

Along a period the nucleus pulls harder and the atom SHRINKS; from one period to the next, one more shell makes it GROW.

Successive ionization energies give the group: the sharp JUMP occurs as soon as the electron core is attacked, hence right after the last valence electron.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Leaving a wavelength in nanometres in the energy formula

3 marks, and a factor of

10^{9}

on the result

What not to write

“ $\lambda=657$ nm, so $E=\dfrac{hc}{\lambda}=\dfrac{(6.626\times 10^{-34})(3.00\times 10^{8})}{657}=3.03\times 10^{-28}$ J.”

What to write

“ $\lambda=657$ nm $=6.57\times 10^{-7}$ m, so $E=3.03\times 10^{-19}$ J.”

Why: The constants $h$ and $c$ are given in SI units. Every length that enters the same formula must be in SI units too, otherwise the result is off by exactly one billion.

2. Getting the sign wrong between emission and absorption

2 marks, and a spectral series misidentified

What not to write

“The atom goes from $n=3$ to $n=2$ , so it absorbs a photon of energy $\Delta E=+3.03\times 10^{-19}$ J.”

What to write

“ $\Delta E=E_{2}-E_{3}=-3.03\times 10^{-19}$ J: the atom GOES DOWN, so it EMITS. The photon carries off $3.03\times 10^{-19}$ J, a positive value.”

The sign of

\Delta E

follows the atom: it goes up when absorbing, down when emitting. The photon always has the same positive energy

|\Delta E|

Why: The sign of $\Delta E$ describes the ATOM, the photon energy describes the light. Going down releases energy, going up costs energy: that is the only direction to remember, and it decides whether we are talking about emission or absorption.

3. Removing the electrons from the d subshell to form a transition metal cation

the whole question, 4 marks, and the magnetism wrong as a consequence

What not to write

“Iron is $[Ar]4s^{2}3d^{6}$ , so $Fe^{2+}$ is $[Ar]4s^{2}3d^{4}$ .”

What to write

“We first remove the electrons from the shell with the largest $n$ : $Fe^{2+}$ is $[Ar]3d^{6}$ , with no $4s$ electron at all.”

We fill

4s

before

3d

, but we empty

4s

first:

Fe^{2+}

has no

4s

electron left and keeps its six

3d

electrons.

Why: The order of FILLING ( $4s$ before $3d$ ) is not the order of EMPTYING. Once $3d$ is occupied, it drops below $4s$ in energy, which becomes the outer shell and leaves first.

4. Believing that raising the light intensity will eventually eject an electron

the whole reasoning question, 4 marks

What not to write

“Red light ejects nothing from this metal, but with a lamp a hundred times more powerful it will work.”

What to write

“Ejection depends on the FREQUENCY of each photon, not on the number of photons. Below the threshold $\nu_{0}=\dfrac{W_{0}}{h}$ , nothing is ejected, whatever the intensity.”

To the left of the threshold the curve is FLAT AT ZERO: no frequency that is too low ejects anything. Intensity multiplies the number of electrons, never their energy.

Why: An electron absorbs ONE photon at a time. A hundred photons that are too weak remain a hundred photons that are too weak: this is exactly the experiment that forced the notion of the quantum.

5. Calculating an average atomic mass without weighting by the abundances

3 marks

What not to write

“Chlorine has two isotopes, $34.97$ and $36.97$ : its atomic mass is $\dfrac{34.97+36.97}{2}=35.97$ u.”

What to write

“With $75.77\%$ and $24.23\%$ : $M=0.7577(34.97)+0.2423(36.97)=35.45$ u, which the periodic table confirms.”

Chlorine 35 weighs three times as much in the mixture: the true average falls near 35, at

35.45

, not midway between the two isotopes.

Why: The mass in the table is a WEIGHTED average. Comparing your result with the periodic table value is, incidentally, the quickest check in the chapter.

6. Believing that an anion is smaller than its atom

3 marks, and a radius ranking that is entirely wrong

What not to write

“ $Cl^{-}$ has gained an electron, so it is more compact than $Cl$ .”

What to write

“ $Cl^{-}$ is LARGER than $Cl$ : one more electron for the same number of protons, hence more repulsion and a lower effective nuclear charge per electron.”

Without its outer shell the cation SHRINKS a lot; the anion, with one more electron for the same number of protons, GROWS.

Why: What decides the size is the ratio between the number of protons and the number of electrons. Adding an electron without adding a proton expands the cloud; removing one contracts it, and the cation often loses a whole shell.

7. Writing the configuration of chromium or copper without the exception

2 marks, and the number of unpaired electrons wrong

What not to write

“Chromium, $Z=24$ : $[Ar]4s^{2}3d^{4}$ .”

What to write

“ $[Ar]4s^{1}3d^{5}$ : a half-filled $d$ subshell is more stable. Likewise copper is $[Ar]4s^{1}3d^{10}$ , never $4s^{2}3d^{9}$ .”

Chromium and copper move one

4s

electron into

3d

to get a half-filled or full

d

subshell: the number of unpaired electrons changes with it.

Why: These two exceptions are the only ones required at the CEGEP level, and they are paid for twice: first on the configuration, then on paramagnetism, which counts the unpaired electrons.

8. Confusing an energy per atom with an energy per mole

3 marks

What not to write

“The ionization of hydrogen costs $2.18\times 10^{-18}$ kJ/mol.”

What to write

“ $2.18\times 10^{-18}$ J per ATOM, that is $2.18\times 10^{-18}\times 6.022\times 10^{23}=1.31\times 10^{6}$ J/mol, or $1310$ kJ/mol.”

Why: Tables of ionization energies are in kilojoules per mole, quantum formulas in joules per atom. Going from one to the other takes $N_{A}$ and then a division by a thousand.

9. Proposing an impossible set of quantum numbers

2 marks per set, and the question is often asked five times

What not to write

“ $n=2$ , $l=2$ , $m_{l}=0$ , $m_{s}=+\tfrac{1}{2}$ describes a $2d$ electron.”

What to write

“ $l$ can only go up to $n-1$ , so $l=2$ is forbidden for $n=2$ : the $2d$ subshell does not exist.”

Why: The bounds are not conventions, they come out of solving the wave equation. This is also why there are only two electrons in the first shell and eight in the second.

10. Reading the wrong position in the successive ionization energies

3 marks

What not to write

“The jump occurs after the third ionization, so the element is in period 3.”

What to write

“The jump after the third ionization says there are THREE valence electrons, so the element is in GROUP 13, not period 3.”

The jump occurs after the THIRD ionization: three valence electrons, hence column 13. The position of the jump gives the group, never the period.

Why: The jump marks the passage into the electron core. It therefore counts valence electrons, which gives the column, never the row of the periodic table.

Which method to choose

Which energy formula to use

The quantity given in the problem

If the problem gives a wavelength or a frequency → $E=\dfrac{hc}{\lambda}$ or $E=h\nu$ , after converting to metres and hertz

Example: $657$ nm gives $3.03\times 10^{-19}$ J
If the problem gives two levels $n$ of hydrogen → $\Delta E=-2.18\times 10^{-18}\left(\dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}}\right)$ then $\lambda=\dfrac{hc}{|\Delta E|}$

Example: from $n=3$ to $n=2$ , red line at $657$ nm
If the problem is about an ion with ONE electron only → same formula multiplied by $Z^{2}$

Example: $He^{+}$ has levels four times deeper than hydrogen
If the problem is about an illuminated metal and ejected electrons → $E_{k,max}=h\nu-W_{0}$ , with the threshold $\nu_{0}=\dfrac{W_{0}}{h}$

Example: a work function of $2.0$ eV gives a threshold at $4.84\times 10^{14}$ Hz
If the problem gives a mass and a speed and talks about wavelength → de Broglie, $\lambda=\dfrac{h}{mv}$ , never the photon formula

a macroscopic object gives a wavelength of the order of $10^{-34}$ m, hence unobservable

Before any calculation, write down the target units: joules per atom, kilojoules per mole, electronvolts. The factor $N_{A}$ and the factor $1.602\times 10^{-19}$ are never forgotten once the target is written in advance.

How to write an electron configuration

The requested species: neutral atom, cation, anion

If neutral atom of the s or p block → direct Aufbau, stopping at the right number of electrons

Example: $S$ , $Z=16$ : $1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}$
If neutral atom of the d block → Aufbau, THEN check whether it is chromium or copper

Example: $Cr$ : $[Ar]4s^{1}3d^{5}$ , not $4s^{2}3d^{4}$
If cation of a transition metal → start from the neutral atom and remove first the electrons with the largest $n$ , hence the $4s$

Example: $Fe^{2+}$ : $[Ar]3d^{6}$
If anion → add the electrons following Aufbau, which often gives the configuration of the next noble gas

Example: $Cl^{-}$ : $[Ne]3s^{2}3p^{6}$ , isoelectronic with argon
If the problem asks about paramagnetism → draw the orbital boxes of the LAST subshell and apply Hund's rule

a single unpaired electron is enough to make the species paramagnetic

Always check the total: the sum of the exponents must equal $Z$ for a neutral atom, $Z$ minus the charge for a cation, $Z$ plus the charge for an anion. This check takes three seconds and catches almost every error.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Calculating the wavelength of an electronic transition

When to use it: The problem gives two levels of hydrogen or of a hydrogen-like ion and asks for the colour, the wavelength or the series.

1 Write the two energies separately with their sign: $E_{i}$ and $E_{f}$ , in joules.
2 Calculate $\Delta E=E_{f}-E_{i}$ and state explicitly whether it is an emission or an absorption according to its sign.
3 Take the ABSOLUTE VALUE for the photon energy, then $\lambda=\dfrac{hc}{|\Delta E|}$ with all quantities in SI units.
4 Convert to nanometres and name the region: ultraviolet below $400$ nm, visible between $400$ and $700$ nm, infrared beyond.
5 Identify the series from the FINAL level: Lyman to $n=1$ , Balmer to $n=2$ , Paschen to $n=3$ .

Concluding sentence

“The transition from $n=3$ to $n=2$ gives $\Delta E=-3.03\times 10^{-19}$ J: the atom emits a photon of $3.03\times 10^{-19}$ J, that is $\lambda=657$ nm, a red line of the Balmer series.”

The trap: Naming the series after the starting level. It is always the FINAL level that gives the name, because it is the one that sets the order of magnitude of the energy released.

Marking: Typically 1 mark for the two energies, 1 for the interpreted sign, 2 for the wavelength, 1 for the region and the series.

Justifying a periodic ranking

When to use it: The problem asks you to rank species by radius, by ionization energy or by electronegativity.

1 Locate each species in the table: period and group, writing its configuration if needed.
2 Apply the period rule, explicitly invoking $Z_{eff}$ , which grows to the right.
3 Apply the group rule, explicitly invoking the number of shells, which grows going down.
4 Treat ions separately: cation smaller, anion larger, and for an isoelectronic series, more protons means smaller.
5 Conclude with a complete ranking using the inequality symbol, and point out the anomaly if there is one.

Concluding sentence

“Within a period, $Z_{eff}$ grows from left to right, so the radius decreases; within a group, a shell is added, so the radius grows. We get $r(Na^{+})<r(Cl)<r(Na)<r(Cl^{-})$ .”

The trap: Comparing two species that are neither in the same period nor in the same group without going through a third. The table only gives trends along a row or a column.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The sum of the exponents against $Z$

Mentally add all the exponents of a configuration: the total must equal $Z$ for the neutral atom, adjusted by the charge for an ion.

An iron configuration whose exponents total $25$ instead of $26$ : an electron was dropped somewhere.
The atomic mass compared with the periodic table

Any average atomic mass you calculate must land on the table value, to within the second decimal place. It is an instant and infallible check.

A $35.97$ for chlorine, when the table shows $35.45$ : the weighting was not done.
The spectral region against the transition

A transition to $n=1$ must fall in the ultraviolet, a transition to $n=2$ in the visible. A result that contradicts this signals an error in the levels.

A wavelength of $1876$ nm announced in the Balmer series: $1876$ nm is infrared, it is Paschen.
The exponent of the power of ten

A visible photon has an energy of about $3\times 10^{-19}$ J, an ionization energy is about $10^{3}$ kJ/mol. Two landmarks that frame almost every result in the chapter.

A $3\times 10^{-28}$ J for a red photon: the nanometres were not converted.
The order of the radii reread on the table

Place the species on a sketch of the periodic table and follow the two arrows. A ranking that goes up one period AND down one group at the same time cannot be justified.

A radius for sulfur announced as larger than that of sodium: they are in the same period, so it is the other way round.

The typical problem, taken apart

The hydrogen lamp, from the red line to the ionization energy

A discharge lamp contains atomic hydrogen. The line corresponding to the transition from $n=3$ to $n=2$ is observed.

Calculate the energy of the emitted photon and its wavelength, then identify the series and the spectral region. Then determine the ionization energy of hydrogen from its ground state, in joules per atom and in kilojoules per mole, as well as the longest wavelength able to ionize it.

Data: $E_{n}=-\dfrac{2.18\times 10^{-18}}{n^{2}}$ J, $h=6.626\times 10^{-34}$ J·s, $c=3.00\times 10^{8}$ m/s, $N_{A}=6.022\times 10^{23}$ /mol.

Step 1

$E_{3}=-\dfrac{2.18\times 10^{-18}}{9}=-2.422\times 10^{-19}$ J and $E_{2}=-\dfrac{2.18\times 10^{-18}}{4}=-5.450\times 10^{-19}$ J.

Why

Writing the two energies separately, with their negative sign, avoids the sign error on the next line. Level $2$ is LOWER than level $3$ , which is immediately visible on these two numbers.

Step 2

$\Delta E=E_{2}-E_{3}=-5.450\times 10^{-19}+2.422\times 10^{-19}=-3.028\times 10^{-19}$ J. The sign is negative, so the atom EMITS a photon of energy $3.028\times 10^{-19}$ J.

Why

The interpretation sentence is worth a mark on its own in most marking schemes, and it stops you from carrying a negative sign into the wavelength formula, where it would make no sense.

Step 3

$\lambda=\dfrac{hc}{|\Delta E|}=\dfrac{(6.626\times 10^{-34})(3.00\times 10^{8})}{3.028\times 10^{-19}}=6.566\times 10^{-7}$ m, that is $657$ nm.

Why

The formula gives the result in metres, and only at the end do we convert to nanometres to make it readable. The reverse, converting first, is exactly the error of the first trap.

Step 4

$657$ nm lies between $400$ and $700$ nm: it is VISIBLE, in the red. The final level is $n=2$ , so the line belongs to the Balmer series.

Why

Naming the series after the final level rather than the starting one is the one point to remember here, and it turns up on every exam in one form or another.

Step 5

Ionization from $n=1$ : $\Delta E=E_{\infty}-E_{1}=0-(-2.18\times 10^{-18})=2.18\times 10^{-18}$ J per atom, a POSITIVE value since energy must be supplied.

Why

Ionization is defined as the move to level $n=\infty$ , where the energy is zero by convention. This zero is not arbitrary here: it is what makes all the other energies negative.

Step 6

Per mole: $2.18\times 10^{-18}\times 6.022\times 10^{23}=1.313\times 10^{6}$ J/mol, that is $1313$ kJ/mol.

Why

This value is tabulated at $1312$ kJ/mol: recovering it validates in one go the constant used, the conversion by $N_{A}$ and the division by a thousand.

Step 7

Longest ionizing wavelength: $\lambda_{max}=\dfrac{hc}{2.18\times 10^{-18}}=9.12\times 10^{-8}$ m, that is $91.2$ nm.

Why

The word LONGEST often surprises: the greater the wavelength, the less energetic the photon, so it is indeed a maximum of wavelength that corresponds to the minimum energy required.

Step 8

Check: $91.2$ nm is far below $400$ nm, so in the far ultraviolet, and the energy of this photon is well above that of the red line, in a ratio $\dfrac{2.18}{0.3028}=7.2$ .

Why

This ratio checks the two results against each other. An ionizing line less energetic than a visible line would be absurd, and that absurdity can be seen without recalculating.

The conclusion, written out

“The transition from $n=3$ to $n=2$ emits a photon of $3.03\times 10^{-19}$ J, that is $657$ nm, a red line of the Balmer series; ionization from the ground state requires $2.18\times 10^{-18}$ J per atom, that is $1313$ kJ/mol, which corresponds to a photon of at most $91.2$ nm.”

The classic mistake on this problem: Calculating $\lambda$ while leaving $\Delta E$ negative, or leaving $\lambda$ in nanometres in $E=\dfrac{hc}{\lambda}$ . The first gives a negative wavelength, the second an energy a billion times too small: in both cases, the order of magnitude is enough to spot the error.

Learn by heart

• $l$ runs from $0$ to $n-1$ ; $m_{l}$ from $-l$ to $+l$ ; $m_{s}=\pm\tfrac{1}{2}$ . A shell $n$ holds $2n^{2}$ electrons.
• $E=\dfrac{hc}{\lambda}$ with $\lambda$ IN METRES: $1$ nm $=10^{-9}$ m.
• $E_{n}=-\dfrac{2.18\times 10^{-18}Z^{2}}{n^{2}}$ J. $\Delta E<0$ means EMISSION, $\Delta E>0$ ABSORPTION.
• Series named after the FINAL level: Lyman to $1$ (UV), Balmer to $2$ (visible), Paschen to $3$ (IR).
• Photoelectric effect: $E_{k,max}=h\nu-W_{0}$ ; below the threshold, the intensity changes nothing.
• Cations: remove first the electrons with the largest $n$ , so the $4s$ before the $3d$ .
• Exceptions to know: $Cr=[Ar]4s^{1}3d^{5}$ and $Cu=[Ar]4s^{1}3d^{10}$ .
• A cation is SMALLER than its atom, an anion LARGER. The jump in successive ionizations gives the GROUP.

Frequently asked questions

Why are 4s electrons removed before 3d electrons when forming an ion?

Because the order of filling is not the order of emptying. The 4s subshell fills first because it is then the lowest in energy, but once the 3d is occupied, it drops below. The outer shell therefore becomes the fourth, and its electrons leave first. The Fe²⁺ ion is thus written with only 3d electrons.

How do I know which spectral series a line belongs to?

Look at the final level of the transition, never the starting level. A landing on the first level gives the Lyman series, in the ultraviolet. A landing on the second gives the Balmer series, in the visible. A landing on the third gives the Paschen series, in the infrared.

Why does brighter light not eject an electron more easily?

Because an electron absorbs a single photon at a time. If that photon does not have enough energy, nothing happens, and multiplying the number of photons does not change the energy of each one. Only the frequency counts for crossing the threshold. Intensity then raises the number of ejected electrons, never their energy.

Is an ion larger or smaller than the atom it comes from?

A cation is smaller, because it has lost electrons, often a whole outer shell, while the number of protons has not changed. An anion is larger, because it has gained an electron without gaining a proton: the repulsion between electrons increases and the cloud expands.

How do I find the group of an element from its ionization energies?

Look for the sharp jump between two successive ionizations. The number of electrons removed before that jump is the number of valence electrons, so it gives the column of the periodic table. A jump after the second ionization signals two valence electrons, that is, an alkaline earth metal.

Which electron configurations are exceptions at the CEGEP level?

Only two are required: chromium and copper. Chromium keeps one electron in the 4s subshell and five in the 3d subshell, rather than two and four. Copper takes one and ten rather than two and nine. In both cases, a half-filled or completely filled d subshell is more stable.

Practise it

Corrected exercises: Atomic structure and electron configuration, 202-SN1-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-chem-sn1-atomic-structure. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: atomic structure and electron configuration (202-SN1-RE)

The essentials

The four quantum numbers and their bounds

The two energy formulas, and their units

Periodicity, and the two rules that explain it

The mistakes that cost marks

1. Leaving a wavelength in nanometres in the energy formula

2. Getting the sign wrong between emission and absorption

3. Removing the electrons from the d subshell to form a transition metal cation

4. Believing that raising the light intensity will eventually eject an electron

5. Calculating an average atomic mass without weighting by the abundances

6. Believing that an anion is smaller than its atom

7. Writing the configuration of chromium or copper without the exception

8. Confusing an energy per atom with an energy per mole

9. Proposing an impossible set of quantum numbers

10. Reading the wrong position in the successive ionization energies

Which method to choose

Which energy formula to use

How to write an electron configuration

How the answer is expected to be written

Calculating the wavelength of an electronic transition

Justifying a periodic ranking

Check before you hand in

The typical problem, taken apart

The hydrogen lamp, from the red line to the ionization energy

Learn by heart

Frequently asked questions

Why are 4s electrons removed before 3d electrons when forming an ion?

How do I know which spectral series a line belongs to?

Why does brighter light not eject an electron more easily?

Is an ion larger or smaller than the atom it comes from?

How do I find the group of an element from its ionization energies?

Which electron configurations are exceptions at the CEGEP level?

Corrected exercises: Atomic structure and electron configuration, 202-SN1-RE

See also

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