MHT-CET Chemistry · Formula sheet

Structure of Atom formulas

19 formulas, 9 reference tables and 57 common traps for MHT-CET Chemistry Structure of Atom, grouped by subtopic.

Subatomic Particles, Isotopes, Isobars and Isoelectronic Species

Counting protons, neutrons and electrons in a species

Electron count of a species

e^- = Z - (\text{charge}) \qquad N = A - Z

$Z$ atomic number (protons)
$A$ mass number (nucleons)
$N$ number of neutrons

Identifying isoelectronic species by counting electrons

Isoelectronic test

e^-_1 = Z_1 - q_1 \;\overset{?}{=}\; Z_2 - q_2 = e^-_2

$Z$ atomic number of the species
$q$ charge (with sign; subtract it)
$e^-$ resulting electron count

Average atomic mass and isotope abundance ratio

Weighted average atomic mass

\bar{m} = \frac{m_1 x + m_2 (100 - x)}{100}

$\bar{m}$ average atomic mass
$m_1, m_2$ the two isotope masses
$x$ percentage abundance of isotope 1

The three subatomic particles and the nuclide notation

Particle	Charge	Relative mass	Location
Proton	$+1$	$\approx 1\ \text{u}$	Nucleus
Neutron	$0$ (neutral)	$\approx 1\ \text{u}$	Nucleus
Electron	$-1$	$\approx \tfrac{1}{1836}$ of a proton	Shells outside the nucleus Electrons are so light that the mass number counts only protons and neutrons — never electrons.

Nucleons (protons + neutrons) carry the mass; the atomic number

Z

fixes the element.

Isotopes, isobars, isotones and isoelectronic species

Term	What is the same	What differs	Example
Isotopes	Protons $Z$ (same element)	Neutrons / mass number	$^{35}\text{Cl}$ , $^{37}\text{Cl}$ Isotopes do NOT have equal neutrons — that is the false statement the bank plants.
Isobars	Mass number $A$	Element ( $Z$ )	$^{40}\text{Ar}$ , $^{40}\text{Ca}$
Isotones	Number of neutrons	$Z$ and $A$	$^{12}_{6}\text{C}$ , $^{11}_{5}\text{B}$
Isoelectronic	Number of electrons	Element and charge	$\text{Na}^{+}$ , $\text{F}^{-}$ , $\text{O}^{2-}$ , $\text{Ne}$

Ask 'what count is held fixed?' — protons, mass number, neutrons, or electrons.

Isotope counts, hydrogen-like species and radioactivity

Fact	Answer	Watch out for
Natural isotopes of nitrogen	$2$ ( $^{14}\text{N}$ , $^{15}\text{N}$ )	Hydrogen has $3$ , not nitrogenQ
Hydrogen-like species	One electron only	Neutral $\text{He}$ has $2$ e $^-$ , so it is excludedQ
Not radioactive	$\text{Ar}$ (argon)	$\text{At}$ , $\text{Po}$ , $\text{Rn}$ are all radioactiveQ

Three independent recall items — learn the exception in each.

Common traps

Mass number counts nucleons, not electrons

The mass number is protons plus neutrons only. Electrons are about

1/1836

of a nucleon, so they add nothing to the mass number — never include them.

Read the notation the right way up

^{A}_{Z}\text{X}

the top number is the mass number

A

and the bottom is the atomic number

Z

. Swapping them gives the wrong neutron count

A-Z

A neutral atom of Ca has 20 electrons, but Ca-based ions do not

The 20-electron species is neutral calcium, because electrons

= Z = 20

\text{Ca}^{2+}

would have only

18

. Check the charge before you count.

Add for negative, subtract for positive

For

\text{O}^{2-}

you add

2

electrons (

8 + 2 = 10

); for

\text{Na}^{+}

you subtract

1

(

11 - 1 = 10

). Getting the sign backwards is the single most common slip.

'Isotopes have equal neutrons' is FALSE

Isotopes share protons, not neutrons — indeed they must differ in neutrons (that is what gives them different mass numbers). The statement 'they have equal number of neutrons' is the false one to spot.

Isotones vs isobars vs isotopes

Same neutrons -> isotones; same mass number -> isobars; same protons -> isotopes.

^{14}_{6}\text{C}

and

^{16}_{8}\text{O}

share neutrons, so they are isotones, not isobars.

The neutral atom hidden among its ions

When the options are

\text{Ne}

\text{O}^{2-}

\text{Na}^{+}

and

\text{Na}

, the odd one out is the neutral

\text{Na}

(

11

electrons). Always apply the charge before comparing.

Same electrons, not same protons

Isoelectronic is about electron count, so

\text{Cl}^{-}

and

\text{Ca}

(

18

20

) are not a pair even though both are 'near argon'. Recount

Z - q

each time — do not eyeball by element.

Weight by abundance, don't just average

The plain mean of

35

and

37

36

, but chlorine's true

35.5

is lower because the lighter

^{35}\text{Cl}

is far more abundant (

75\%

). Always multiply each mass by its fraction first.

Match the ratio order to the isotopes

For chlorine the

3 : 1

^{35}\text{Cl} : {}^{37}\text{Cl}

— the lighter, more abundant isotope first. A distractor offers

1 : 3

; keep the order aligned with the masses.

Hydrogen-like means one electron, not 'near hydrogen'

The test is a single electron, so

\text{He}^{+}

\text{Li}^{2+}

and

\text{Be}^{3+}

qualify but neutral

\text{He}

(

2

^-

) does not. The neutral noble gas is the planted wrong answer.

Argon is the stable one

Among

\text{At}

\text{Po}

\text{Rn}

and

\text{Ar}

, only argon is a stable, non-radioactive noble gas; the other three are radioactive heavy elements.

Electromagnetic Radiation and Wave Properties

Learn this subtopic in the notes

Wave characteristics — wavelength, frequency, wavenumber, amplitude

Wavenumber

\bar{\nu} = \dfrac{1}{\lambda}

$\bar{\nu}$ wavenumber (m^-1 or cm^-1)
$\lambda$ wavelength (m or cm, matching the wavenumber unit)

Speed of light relation, c = nu*lambda

Speed of light relation

c = \nu\lambda \qquad \Rightarrow \qquad \nu = \dfrac{c}{\lambda}

$c$ speed of light, 3 x 10^8 m/s (same for all EM radiation)
$\nu$ frequency (Hz)
$\lambda$ wavelength (m)

Planck's quantum theory and photon energy, E = h*nu = hc/lambda

Photon energy (Planck)

E = h\nu = \dfrac{hc}{\lambda} \qquad (\text{per mole: } E \times N_A)

$E$ energy of one photon (J)
$h$ Planck's constant, 6.626 x 10^-34 J s
$\nu$ frequency (Hz)
$c$ speed of light, 3 x 10^8 m/s
$\lambda$ wavelength (m)
$N_A$ Avogadro number, 6.022 x 10^23 mol^-1

The electromagnetic spectrum — order by frequency and energy

Radiation (low to high energy)	Wavelength / frequency	Energy
Radio waves	Longest wavelength, lowest frequency	Lowest energy MHT-CET — of radio waves, microwaves, IR and UV, radio waves have the LOWEST energy.
Microwaves	Long wavelength, low frequency	Very low
Infrared (IR)	Longer than visible	Low (felt as heat)
Visible light (VIBGYOR)	400–700 nm; red longest, violet shortest	Red lowest, violet highest Within visible light, VIOLET has the highest energy and RED the lowest (energy increases R->V).
Ultraviolet (UV)	Shorter than visible	Higher than visible
X-rays	Very short wavelength	High, penetrating
Gamma rays	Shortest wavelength, highest frequency	Highest energy

Energy increases from radio waves to gamma rays: E proportional to frequency proportional to 1/wavelength.

Common traps

Frequency vs wavelength — read the wording

'Number of waves passing a point per second' is frequency

\nu

. Wavelength is the length of ONE wave (a distance), and wavenumber

\bar{\nu} = 1/\lambda

is waves per unit LENGTH — do not confuse the two 'number of waves' phrasings.

Match the wavenumber unit to lambda

\bar{\nu} = 1/\lambda

gives

\text{cm}^{-1}

only if

\lambda

is in cm, and

\text{m}^{-1}

only if

\lambda

is in metres. Convert

\mu\text{m}

or nm to the required base unit FIRST (

0.25\ \mu\text{m} = 0.25 \times 10^{-6}\ \text{m}

Convert nm to metres before dividing

For

\nu = c/\lambda

in Hz,

\lambda

must be in metres. Forgetting the

\times 10^{-9}

on a nm wavelength shifts the answer by nine orders of magnitude.

400\ \text{nm} = 400 \times 10^{-9}\ \text{m} = 4 \times 10^{-7}\ \text{m}

c is the same for every EM radiation

Radio waves, visible light and gamma rays all travel at

3 \times 10^{8}\ \text{m/s}

in vacuum. What differs between them is

\lambda

and

\nu

, never

c

Energy goes as 1/lambda, not lambda

Because

E = hc/\lambda

, a shorter wavelength means a larger energy. Do not assume the longest-wavelength radiation is the most energetic — it is the least energetic.

Per photon vs per mole

E = hc/\lambda

gives the energy of a SINGLE photon (

\sim 10^{-19}\ \text{J}

). If the question asks 'per mole', you must multiply by

N_A = 6.022 \times 10^{23}

to reach the

\sim 10^{5}\ \text{J}

range.

Amplitude does not set energy

A photon's energy depends only on frequency/wavelength, not on amplitude. Amplitude controls intensity (number of photons / brightness), which is a different quantity.

Long wavelength = LOW energy

Radio waves have the LONGEST wavelength, so by

E = hc/\lambda

they carry the least energy — a common trap is to pick them as 'highest'. Highest energy always goes to the shortest-wavelength radiation (gamma rays; violet among the visible colours).

VIBGYOR direction

Reading VIBGYOR, violet is at the high-frequency (high-energy) end and red at the low-energy end. Energy rises from Red to Violet, so 'highest energy colour' is violet, not red.

Bohr's Atomic Model

Learn this subtopic in the notes

Postulates and quantized angular momentum

Quantized angular momentum

L = mvr = \dfrac{nh}{2\pi}

$L$ angular momentum of the electron in the nth orbit
$m$ mass of the electron
$v$ speed of the electron
$r$ radius of the orbit
$n$ principal quantum number (orbit number), 1, 2, 3, ...
$h$ Planck's constant, 6.626e-34 J s

Radius of the nth orbit

Radius of nth orbit

r_n = 0.529\,\dfrac{n^2}{Z}\ \text{\AA}

$r_n$ radius of the nth orbit
$n$ orbit number (principal quantum number)
$Z$ atomic number (nuclear charge)
$0.529 \text{\AA}$ Bohr radius a_0 (= 52.9 pm)

Energy of the nth orbit

Energy of nth orbit

E_n = -2.18 \times 10^{-18}\,\dfrac{Z^2}{n^2}\ \text{J} = -13.6\,\dfrac{Z^2}{n^2}\ \text{eV}

$E_n$ energy of the electron in the nth orbit (negative)
$Z$ atomic number
$n$ orbit number
$R_H$ 2.18e-18 J = 13.6 eV, the hydrogen ground-state magnitude

Velocity of the electron in the nth orbit

Velocity of electron in nth orbit

v_n = 2.18 \times 10^{6}\,\dfrac{Z}{n}\ \text{m s}^{-1}

$v_n$ speed of the electron in the nth orbit
$Z$ atomic number
$n$ orbit number
$2.18 \times 10^{6}$ first-orbit speed in hydrogen (m/s)

Energy difference between levels and ionization energy

Energy gap between two orbits

\Delta E = 13.6\,Z^2\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)\ \text{eV}

$\Delta E$ energy absorbed (n1 to n2, up) or emitted (down)
$n_1$ lower orbit number
$n_2$ higher orbit number
$Z$ atomic number

Hydrogen-like species and limitations of the model

Test for a hydrogen-like species

\text{electrons} = Z - (\text{charge}) = 1

$Z$ atomic number (number of protons)
$\text{charge}$ the positive charge on the ion
$= 1$ hydrogen-like requires exactly one remaining electron

Rutherford's nuclear model and its drawbacks

Aspect	Rutherford's model	The problem
Structure	Tiny dense positive nucleus; electrons revolve around it; atom is mostly empty space.	This part is correct — established by alpha-particle scattering.
Stability of the atom	Electrons move in circular paths around the nucleus.	A revolving (accelerating) electron must radiate energy continuously and spiral into the nucleus, so the atom should collapse.
Atomic spectrum	Does not restrict the electron's energy.	Predicts a continuous spectrum, but hydrogen actually shows a discrete line spectrum.

Rutherford got the nucleus right but could not explain atomic stability or the line spectrum — Bohr's quantized orbits fixed both.

Common traps

Rutherford placed the electrons OUTSIDE the nucleus

The nucleus holds the protons (and neutrons) and nearly all the mass; the electrons revolve around it. A statement putting electrons inside the nucleus, or giving the nucleus a negative charge, is the false one in a 'NOT true about Rutherford's model' question.

The stability failure is a CLASSICAL-physics problem

The collapse prediction comes from classical electromagnetism (an accelerating charge radiates). Bohr did not change the nuclear picture — he quantized the orbits (fixed energy levels, no radiation while in an orbit) to rescue stability and explain the line spectrum.

It is n h / 2 pi, not 2 pi n / h

The correct form is

mvr = \dfrac{nh}{2\pi}

. Inverted look-alikes such as

mvr = \dfrac{2\pi n}{h}

mv = \dfrac{2\pi rn}{h}

are the standard wrong options — h sits on top,

2\pi

underneath.

Angular momentum ignores Z

Radius and energy scale with

Z

, but angular momentum

L = \dfrac{nh}{2\pi}

does not. The 4th orbit of

\text{He}^+

and the 4th orbit of

\text{H}

have the same angular momentum.

Divide by Z for ions

The bare formula

r_n = 0.529\,n^2\ \text{\AA}

is only for hydrogen (

Z=1

). For

\text{He}^+,\ \text{Li}^{2+},\ \text{Be}^{3+},\ \text{B}^{4+}

you must divide by

Z = 2, 3, 4, 5

— forgetting this gives an answer that is

Z

times too large.

Angstrom vs pm

Answers are often listed in pm. Convert:

0.529\ \text{\AA} = 52.9\ \text{pm}

, so multiply an Angstrom answer by 100.

r_4(\text{H}) = 8.464\ \text{\AA} = 846.4\ \text{pm}

Keep the minus sign

Orbital energy is negative because the electron is bound. Options that drop the sign (a positive energy) are wrong. The magnitude is largest for

n=1

and approaches 0 as

n \to \infty

Z is squared, n is squared

Both appear squared:

E_n \propto \dfrac{Z^2}{n^2}

. For

\text{He}^+

(

Z=2

) the energy is

2^2 = 4

times more negative than hydrogen at the same

n

— a factor of 4, not 2.

Velocity goes as 1/n, not 1/n squared

Radius scales as

n^2/Z

and energy as

Z^2/n^2

, but velocity scales as

Z/n

— a single power of each. Don't square the

n

here.

Higher Z, faster electron

A larger nuclear charge pulls the electron in tighter (smaller

r

) AND makes it move faster (larger

v

). So

\text{He}^+

is faster than

\text{H}

in the same orbit number.

Ionization energy is positive

The orbital energy

E_n

is negative, but the ionization energy you must supply is its magnitude with a plus sign:

\text{I.E.} = -E_n = +13.6\,Z^2/n^2\ \text{eV}

. For H from

n=1

it is

+13.6\ \text{eV}

, not

-13.6

Bigger n subtracted from smaller n

\Delta E = 13.6\,Z^2\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)

, the lower orbit

n_1

supplies the first (larger) term. Swapping

n_1

and

n_2

flips the sign — keep

n_1 < n_2

for a positive absorbed energy.

Bohr formulas are single-electron only

r_n,\ E_n,\ v_n

apply only to one-electron species. Do not use them for

\text{Li}^+

\text{He}

, or any neutral/multi-electron atom — count the electrons (

Z

minus the charge) first.

Bohr explains hydrogen, not the Zeeman effect

Among 'which is NOT correct about Bohr' options, the false one is usually 'it explains the Zeeman effect'. Bohr's model FAILED to explain the Zeeman effect, multi-electron spectra, fine structure, and chemical bonding.

Rutherford vs Bohr on electron energy

A 'not true about Rutherford's model' question is answered by 'it describes the energies of electrons' — Rutherford's model did not; describing electron energies is Bohr's contribution.

Hydrogen Spectrum and the Rydberg Equation

Learn this subtopic in the notes

Rydberg equation — wavenumber of a spectral line

Rydberg equation

\bar{\nu} = \dfrac{1}{\lambda} = R_H\, Z^2\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right) \qquad (n_1 < n_2)

$\bar{\nu}$ wavenumber of the line (cm^-1 or m^-1)
$\lambda$ wavelength of the line
$R_H$ Rydberg constant = 1.097 x 10^7 m^-1 = 109677 cm^-1
$Z$ atomic number (Z = 1 for hydrogen)
$n_1$ lower orbit (fallen TO), the larger positive fraction
$n_2$ upper orbit (fallen FROM)

Longest wavelength, series limit, and number of spectral lines

Number of spectral lines from orbit n

\text{No. of lines} = \dfrac{n(n-1)}{2}

$n$ highest orbit the electron is excited to

The spectral series of hydrogen

Series	Falls to (n1)	From (n2)	Region
Lyman	1	2, 3, 4, ...	Ultraviolet (UV)Q MHT-CET 2024 — the series for a jump from n2 = infinity to n1 = 1 is the Lyman series.
Balmer	2	3, 4, 5, ...	VisibleQ MHT-CET 2023 + 2021 — Balmer is the ONLY series in the visible region.
Paschen	3	4, 5, 6, ...	Infrared (IR)
Brackett	4	5, 6, 7, ...	Infrared (IR)
Pfund	5	6, 7, 8, ...	Infrared (IR)

Memory aid: La-Ba-Pa-Bra-Pf for n1 = 1, 2, 3, 4, 5. Only Balmer (n1 = 2) is visible; Lyman is UV; the rest are IR.

Common traps

n1 is the smaller orbit — keep the bracket positive

\bar{\nu} = R_H\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)

with

n_1 < n_2

. For a fall from

n = 5

n = 2

, use

n_1 = 2, n_2 = 5

— NOT the reverse. Swapping them flips the sign and gives a negative (wrong) wavenumber.

Wavenumber and wavelength are reciprocals

\bar{\nu} = 1/\lambda

. If a question asks for the wavelength, compute

\bar{\nu}

first, then take its reciprocal. Reporting

\bar{\nu}

as the wavelength (or vice versa) is a classic careless slip.

Balmer is visible, Lyman is not

The visible series is Balmer (falls to n = 2). Lyman (falls to n = 1) is ultraviolet. A large share of MHT-CET marks are lost by picking Lyman for the 'visible region' question.

The series is named by the LOWER orbit, not the upper one

A jump 'from n = 4 to n = 2' is a Balmer line, because it lands on

n_1 = 2

. Don't name a series by the starting (upper) orbit — always look at where the electron finishes.

Longest wavelength = smallest gap, not the biggest jump

The longest wavelength corresponds to the least energy, i.e. the closest pair of orbits (

n_2 = n_1 + 1

). Students often plug in

n_2 = \infty

— that gives the SHORTEST wavelength (the series limit), the exact opposite.

n(n-1)/2 counts every downward jump

The formula

\dfrac{n(n-1)}{2}

is the number of distinct lines when the electron can cascade down in ALL possible ways from orbit

n

. For

n = 4

that is 6, not 3.

Quantum Mechanical Model — de Broglie, Heisenberg and Quantum Numbers

Learn this subtopic in the notes

de Broglie wavelength — wave-particle duality

de Broglie wavelength and momentum

\lambda = \dfrac{h}{mv} = \dfrac{h}{p} \qquad p = \dfrac{h}{\lambda}

$\lambda$ de Broglie wavelength (m)
$h$ Planck's constant, 6.63e-34 J s
$m$ mass of the particle (kg)
$v$ velocity of the particle (m/s)
$p$ momentum, p = mv (kg m/s)

Heisenberg's uncertainty principle

Heisenberg uncertainty relation

\Delta x \cdot \Delta p \geq \dfrac{h}{4\pi} \qquad \Delta x \cdot \Delta v \geq \dfrac{h}{4\pi m}

$\Delta x$ uncertainty in position
$\Delta p$ uncertainty in momentum
$\Delta v$ uncertainty in velocity
$h$ Planck's constant

Shell capacity, orbital energy order and nodes

Shell capacity, subshell capacity and nodes

\text{orbitals} = n^2,\quad e^- _{\max} = 2n^2,\quad e^-_{\text{subshell}} = 2(2l+1) \qquad \text{nodes} = n-1,\ \text{radial} = n-l-1

$n$ principal quantum number (shell)
$l$ azimuthal quantum number (subshell)
$n^2$ number of orbitals in the shell
$2n^2$ maximum electrons in the shell

The four quantum numbers

Quantum number	Symbol	What it describes	Allowed values
Principal	n	Shell / main energy level and size of the orbital	1, 2, 3, ... (positive integers)
Azimuthal (subsidiary)	l	Subshell and shape of the orbital (s, p, d, f)	0 to (n-1); coded 0=s, 1=p, 2=d, 3=f l runs only from 0 up to n-1. For n=3, l can be 0, 1 or 2 — never 3.
Magnetic	m_l	Orientation of the orbital in space (which orbital)	-l to +l, i.e. (2l+1) values
Spin	m_s	Direction of the electron's spin	+1/2 or -1/2 only

l fixes the shape (s/p/d/f); the orbital label is n followed by that letter.

Orbital shapes from l

l value	Subshell	Shape	Orbitals in subshell
0	s	Spherical	1
1	p	Dumbbell (two lobes)	3
2	d	Four-lobed clover leaf (except d(z2))	5 d(z2) is the exception: two lobes along z plus a doughnut ring in the xy-plane — a different shape from the other four.
3	f	Complex multi-lobed	7

Number of orbitals in a subshell is 2l+1: s=1, p=3, d=5, f=7.

Common traps

Divide by momentum, not by mass alone

The denominator is

mv

(the momentum), not just the mass. Multiply mass by velocity first, then divide

h

by that product.

Convert Ångström to metres

1\,\text{Å} = 10^{-10}\,\text{m}

and

1\,\text{nm} = 10^{-9}\,\text{m}

. A wavelength given in Å or nm must be in metres before you divide, or the power of ten will be wrong.

It is a fundamental limit, not an instrument error

The uncertainty is built into nature — it is not due to imperfect instruments. Even a perfect measuring device cannot beat the

\dfrac{h}{4\pi}

bound.

Don't confuse it with Pauli or Aufbau

Position-and-momentum together = Heisenberg. Pauli's exclusion principle is about no two electrons having the same four quantum numbers; Aufbau is about the order of filling. The bank swaps these as distractors.

l ranges from 0 to n-1

The azimuthal quantum number cannot equal or exceed

n

. For

n = 3

the allowed

l

values are 0, 1, 2 only — there is no 3f (that would need

n \geq 4

m_l ranges from -l to +l

The magnetic quantum number runs over the

2l+1

integers from

-l

+l

, including 0. A p subshell (

l = 1

) has

m_l = -1, 0, +1

— three orbitals, not two.

d(z2) is the shape exception

When asked which d orbital has a different shape, the answer is

d_{z^2}

. The other four (

d_{xy}, d_{yz}, d_{xz}, d_{x^2-y^2}

) are all four-lobed clover leaves.

Break an (n+l) tie with the smaller n

When two orbitals share the same

n+l

(e.g. 3d and 4p both give 5), the one with the smaller n has the lower energy: 3d is below 4p. Only after comparing

n+l

do you look at

n

Degeneracy of 2s and 2p is a hydrogen-only fact

In the hydrogen atom, energy depends only on

n

, so 2s and 2p are degenerate. In any multi-electron atom the (n+l) rule splits them — 2s is below 2p. The bank's degeneracy question is specifically about hydrogen.

Electronic Configuration and Pauli/Hund Rules

Learn this subtopic in the notes

Ground-state configurations and the half-filled/fully-filled anomaly

The stable-subshell anomaly (Cr, Cu)

\text{Cr}: [\text{Ar}]\,4s^1 3d^5 \qquad \text{Cu}: [\text{Ar}]\,4s^1 3d^{10}

Counting unpaired electrons

Unpaired electrons in a subshell

\text{unpaired} = \begin{cases} k, & k \le N \\ 2N - k, & k > N \end{cases}

$k$ electrons in the subshell
$N$ number of orbitals in the subshell (p:3, d:5, f:7)

The three orbital-filling rules

Rule	Statement	Consequence
Aufbau principle	Orbitals are filled in order of increasing energy (the $(n+l)$ rule).	Filling order $1s, 2s, 2p, 3s, 3p, 4s, 3d, \dots$
Pauli's exclusion principle	No two electrons in an atom can have the same set of all four quantum numbers.	Max 2 electrons per orbital, with opposite spins.Q The bank quotes this one almost verbatim — 'no two electrons ... identical set of four quantum numbers' is always Pauli, never Heisenberg's uncertainty principle.
Hund's rule	Degenerate orbitals are singly occupied before any pairing begins.	Maximum number of parallel-spin unpaired electrons in a subshell.Q Watch the phrasing: 'pairing does not occur unless each orbital of the subshell has one electron' is Hund's rule.

Aufbau sets the order, Pauli caps each orbital at two, Hund spreads before it pairs.

Common traps

Pauli is about four quantum numbers, not position

The Pauli exclusion principle is stated in terms of the four quantum numbers

(n, l, m_l, m_s)

— two electrons in the same orbital already share

n, l, m_l

, so they must differ in spin

m_s

. Do not confuse it with the Heisenberg uncertainty principle, which is about position and momentum, a common distractor.

Hund means all singly first, then pair

For

2p^3

(nitrogen) Hund's rule gives

\uparrow\ \uparrow\ \uparrow

— three unpaired electrons — not

\uparrow\downarrow\ \uparrow\ \_\,

. Pairing in a subshell begins only after every degenerate orbital already holds one electron.

Chromium is 3d⁵4s¹, not 3d⁴4s²

The single most-tested anomaly: chromium's ground state is

[\text{Ar}]\,4s^1\,3d^5

, giving a stable half-filled d-subshell. Writing

4s^2\,3d^4

(naive Aufbau) is the classic error — and it changes the unpaired count from the correct 6 to a wrong 4.

Copper's 4s is singly occupied

Copper is

[\text{Ar}]\,4s^1\,3d^{10}

, not

4s^2\,3d^9

. Because

3d^{10}

is completely paired, copper has only 1 unpaired electron (the lone 4s), not 2 or more.

Half-filled subshells hold the most unpaired electrons

Among second-period atoms, nitrogen

(2p^3)

beats oxygen

(2p^4)

and fluorine

(2p^5)

— a half-filled

p^3

is where every orbital is singly occupied, giving the maximum 3 unpaired. More total electrons does not mean more unpaired.

Fully-filled means zero unpaired

Zinc's

3d^{10}4s^2

has all subshells complete, so it has zero unpaired electrons — and no accessible excited state changes that. A filled

d^{10}

s^2

contributes nothing to the unpaired count.

Structure of Atom formulas

Subatomic Particles, Isotopes, Isobars and Isoelectronic Species

Counting protons, neutrons and electrons in a species

Identifying isoelectronic species by counting electrons

Average atomic mass and isotope abundance ratio

The three subatomic particles and the nuclide notation

Isotopes, isobars, isotones and isoelectronic species

Isotope counts, hydrogen-like species and radioactivity

Common traps

Electromagnetic Radiation and Wave Properties

Wave characteristics — wavelength, frequency, wavenumber, amplitude

Speed of light relation, c = nu*lambda

Planck's quantum theory and photon energy, E = h*nu = hc/lambda

The electromagnetic spectrum — order by frequency and energy

Common traps

Bohr's Atomic Model

Postulates and quantized angular momentum

Radius of the nth orbit

Energy of the nth orbit

Velocity of the electron in the nth orbit

Energy difference between levels and ionization energy

Hydrogen-like species and limitations of the model

Rutherford's nuclear model and its drawbacks

Common traps

Hydrogen Spectrum and the Rydberg Equation

Rydberg equation — wavenumber of a spectral line

Longest wavelength, series limit, and number of spectral lines

The spectral series of hydrogen

Common traps

Quantum Mechanical Model — de Broglie, Heisenberg and Quantum Numbers

de Broglie wavelength — wave-particle duality

Heisenberg's uncertainty principle

Shell capacity, orbital energy order and nodes

The four quantum numbers

Orbital shapes from l

Common traps

Electronic Configuration and Pauli/Hund Rules

Ground-state configurations and the half-filled/fully-filled anomaly

Counting unpaired electrons

The three orbital-filling rules

Common traps

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