NEET Corner — Structure of Atom the NEET Way

How NEET Tests This Chapter — and the 40-Second Mindset

NEET wants something different here than JEE does: no Bohr derivation, no ratios of time periods, no Schrödinger algebra. The chapter gives 2 to 3 questions out of 45 in Chemistry, and nearly every one is a direct formula or definition in a plain sentence.

What shows up:

Question style	What it looks like	Time you should spend
Discovery / model recall	"The neutron was discovered by…" / "Which model could not explain the stability of the atom?"	15 seconds
Quantum-number validity	"Which set of quantum numbers is not possible?"	25 seconds
Electronic configuration	"The configuration of $\mathrm{Cr}$ / $\mathrm{Cu^+}$ / $\mathrm{Fe^{3+}}$ is…"	25 seconds
Unpaired electrons and magnetism	"The number of unpaired electrons in $\mathrm{Mn^{2+}}$ is…"	30 seconds
One-formula numerical	"The energy of the second Bohr orbit of hydrogen is…" / "Wavelength of an electron accelerated through 100 V is…"	40 seconds
Spectral series	"Which series of hydrogen lies in the visible region?" / "Number of lines when the electron falls from $n = 5$ to $n = 1$"	30 seconds
Nodes and orbitals	"Number of radial nodes in a 3p orbital is…"	20 seconds
Assertion-Reason / statements	"Assertion: 4s fills before 3d. Reason: 4s has lower $(n + l)$…"	40 seconds

All eight can be finished without a rough page once the tables and shortcuts below are in your head.

Favourite topics, ranked

1. Quantum numbers — allowed ranges, "which set is impossible", how many orbitals or electrons a set describes.
2. Electronic configuration of transition-metal ions — $\mathrm{Cr}$, $\mathrm{Cu}$, $\mathrm{Fe^{2+}}$, $\mathrm{Fe^{3+}}$, $\mathrm{Mn^{2+}}$, $\mathrm{Cu^+}$, $\mathrm{Zn^{2+}}$, and the unpaired-electron count.
3. Bohr numbers — $E_n$, $r_n$, $v_n$ for hydrogen-like species and the number of spectral lines.
4. de Broglie and Heisenberg — one-line plug-ins.
5. Photoelectric effect — threshold frequency, work function, and what does and does not depend on intensity.

The 40-second mindset

NEET is 180 questions in 200 minutes, a little over a minute each. This chapter is where you bank time for Biology, so the target is not "solve it" but "solve it in 40 seconds and move on".

1. Read the last line first. It names what is wanted — energy, radius, wavelength, number of lines, unpaired electrons.
2. Spot the species and the shell. $\mathrm{H}$, $\mathrm{He^+}$ or $\mathrm{Li^{2+}}$? $n = 2$ or $n = 3$? The $Z$ and the $n$ are the whole Bohr question.
3. Write one line. $E_n = -13.6\,Z^2/n^2$ eV; $r_n = 0.529\,n^2/Z$ Å; $E = 1240/\lambda$ eV.
4. Round hard. $13.6/4 = 3.4$, $13.6/9 \approx 1.5$, $0.529 \times 4 \approx 2.1$. NEET options are usually spaced far enough apart that this rounding does not flip the answer.
5. Match to the options. If your number is 3.4 and the options are $-3.4$, $-1.51$, $-13.6$ and $-0.85$ eV, bubble and go.

Key Point: NEET rewards knowing which formula and which configuration, not the longest calculation. Treat this as a recall-plus-one-line chapter.

[NEET] Negative marking ($-1$) makes a 40-second wrong answer worse than a 40-second skip. If two lines in you are not converging on an option, mark for review and come back.

What NEET does not ask from this chapter

• Deriving Bohr's radius or energy from first principles.
• Time period, frequency of revolution, revolutions per second — rare, and only as a plug-in.
• The Schrödinger equation, wavefunction algebra, radial distribution curves in detail.
• Photoelectric problems needing three conversions — at most one: $E = 1240/\lambda$ eV, then subtract $W_0$.

More than 90 seconds on a question means either the wrong tool or the odd question of the year. Either way, move.

The Verbatim-Recall Tables — Discoveries, Models and the Definitions NEET Asks Word for Word

"Who did what, when" and "define this term" carry the same $+4$ as a hard numerical. Learn these two tables as flash-cards.

Table 1: Discoveries and models — who, when, what

Year	Scientist	What they found or proposed	The one line NEET asks
1895	Röntgen	X-rays from the anode of a discharge tube	X-rays are EM radiation, not deflected by fields
1896	Becquerel	Radioactivity ($\alpha$, $\beta$, $\gamma$ named later by Rutherford)	Spontaneous emission from uranium salts
1897	J. J. Thomson	Electron (cathode rays); measured $e/m_e = 1.758820 \times 10^{11}\ \mathrm{C\ kg^{-1}}$	Cathode rays are the same whatever gas or metal is used
1898	J. J. Thomson	Plum-pudding (watermelon) model — positive sphere with electrons embedded; mass uniformly distributed	Explained overall neutrality, failed the $\alpha$-scattering test
1909	Millikan	Charge on the electron, oil-drop method: $e = -1.602 \times 10^{-19}$ C; hence $m_e = 9.1094 \times 10^{-31}$ kg	Charge on any drop is an integral multiple of $e$
1909	Rutherford (with Geiger and Marsden)	$\alpha$-particle scattering on gold foil	Most pass undeflected, a few deflect, ~1 in 20,000 bounce back
1911	Rutherford	Nuclear model — tiny dense positive nucleus ($\sim 10^{-15}$ m) with electrons in orbits; atom $\sim 10^{-10}$ m	Could not explain stability of the atom or line spectra
1913	Bohr	Quantised orbits, $mvr = nh/2\pi$, $\Delta E = h\nu$	Explained the hydrogen spectrum, failed for multi-electron atoms
1919	Rutherford	Proton (positive particle from hydrogen)	Mass $1.67262 \times 10^{-27}$ kg, charge $+1.602 \times 10^{-19}$ C
1924	de Broglie	Matter waves, $\lambda = h/mv$	Every moving particle has an associated wavelength
1925	Uhlenbeck and Goudsmit	Electron spin — the fourth quantum number $m_s$	Spin has only two values, $+\tfrac{1}{2}$ and $-\tfrac{1}{2}$
1926	Schrödinger	Wave equation $\hat{H}\psi = E\psi$; quantum mechanics of the atom	Solutions give orbitals, quantised energies, $\psi^2$ as probability density
1926	Pauli	Exclusion principle	No two electrons in an atom can have the same set of four quantum numbers
1927	Heisenberg	Uncertainty principle, $\Delta x \cdot \Delta p \geq h/4\pi$	Position and momentum cannot both be known exactly
1932	Chadwick	Neutron ($\alpha$ on beryllium)	Neutral, mass $1.67493 \times 10^{-27}$ kg, slightly heavier than the proton

The three model one-liners that repeat: Thomson's model failed on Rutherford's scattering; Rutherford's failed on the stability of the atom (an accelerating electron should radiate and spiral in) and on the line spectrum; Bohr's failed on multi-electron spectra, the Zeeman and Stark effects, fine structure, and chemical bonding.

Also: Planck (1900) — energy is quantised, $E = h\nu$; Einstein (1905) — photoelectric effect via photons; Hertz (1887) — first observed the photoelectric effect; Moseley — atomic number from X-ray frequencies.

Table 2: The definitions NEET asks verbatim

Term	Definition as NEET wants it	Instant example
Atomic number $Z$	Number of protons in the nucleus $=$ number of electrons in a neutral atom	$\mathrm{Na}$: $Z = 11$
Mass number $A$	Number of protons $+$ number of neutrons	$^{23}_{11}\mathrm{Na}$: 11 p, 12 n
Isotopes	Atoms with the same atomic number but different mass number (same element, different neutrons)	$^{1}_{1}\mathrm{H}$, $^{2}_{1}\mathrm{H}$, $^{3}_{1}\mathrm{H}$; $^{35}\mathrm{Cl}$, $^{37}\mathrm{Cl}$
Isobars	Atoms with the same mass number but different atomic number	$^{14}_{6}\mathrm{C}$ and $^{14}_{7}\mathrm{N}$; $^{40}_{18}\mathrm{Ar}$ and $^{40}_{20}\mathrm{Ca}$
Isotones (extra)	Atoms with the same number of neutrons	$^{14}_{6}\mathrm{C}$ and $^{16}_{8}\mathrm{O}$ (8 neutrons each)
Isoelectronic species	Ions or atoms with the same number of electrons	$\mathrm{Na^+}$, $\mathrm{Mg^{2+}}$, $\mathrm{F^-}$, $\mathrm{Ne}$ (10 electrons)
Threshold frequency $\nu_0$	The minimum frequency of light below which no photoelectrons are emitted, however intense the light	Characteristic of the metal
Work function $W_0$	The minimum energy needed to eject an electron from the metal surface; $W_0 = h\nu_0$	Cs: 1.9 eV; Na: 2.3 eV; K: 2.25 eV
Photoelectric equation	$h\nu = h\nu_0 + \tfrac{1}{2}m_e v^2$	KE $= h(\nu - \nu_0)$
Black body	A perfect absorber and emitter of radiation at all frequencies	Emission depends only on temperature
Quantum	The smallest packet of energy that can be absorbed or emitted, $E = h\nu$	Called a photon for light
Orbit	A fixed circular path (Bohr) at a definite distance from the nucleus; exact position and momentum known — violates Heisenberg	$r_1 = 52.9$ pm for H
Orbital	The three-dimensional region around the nucleus where the probability of finding the electron is maximum (~90%); a one-electron wavefunction $\psi$	$1s$, $2p_x$, $3d_{z^2}$
Principal quantum number $n$	Shell; size and energy of the orbital; $n = 1, 2, 3, \ldots$	K, L, M, N
Azimuthal quantum number $l$	Subshell; shape of the orbital; $l = 0$ to $n - 1$	$s = 0$, $p = 1$, $d = 2$, $f = 3$
Magnetic quantum number $m_l$	Orientation of the orbital; $m_l = -l$ to $+l$, i.e. $2l + 1$ values	$p$: $-1, 0, +1$
Spin quantum number $m_s$	Intrinsic spin of the electron; $+\tfrac{1}{2}$ or $-\tfrac{1}{2}$	Two electrons per orbital
Aufbau principle	Electrons occupy orbitals in order of increasing energy; lower $(n + l)$ first, and for equal $(n + l)$, lower $n$ first	$4s$ before $3d$
Pauli exclusion principle	No two electrons in an atom can have the same set of four quantum numbers; hence an orbital holds at most two electrons of opposite spin	$1s^2$, not $1s^3$
Hund's rule of maximum multiplicity	Pairing in degenerate orbitals does not begin until each orbital is singly occupied, with parallel spins	N: $2p_x^1 2p_y^1 2p_z^1$
Degenerate orbitals	Orbitals of the same energy	$2p_x$, $2p_y$, $2p_z$; in hydrogen, all of $3s$, $3p$, $3d$
Effective nuclear charge $Z_{\text{eff}}$	The net positive charge felt by an electron after shielding by inner electrons; $Z_{\text{eff}} = Z - \sigma$	Reason $2s$ lies below $2p$ in multi-electron atoms
Shielding	Repulsion by inner electrons that reduces the attraction of the nucleus on the outer electron	$s > p > d > f$ in penetration

Key Point (Definition): Four photoelectric observations: (i) electrons come out instantly, no time lag; (ii) no emission below $\nu_0$, whatever the intensity; (iii) number of photoelectrons $\propto$ intensity; (iv) kinetic energy of photoelectrons $\propto$ frequency, independent of intensity.

[NEET] Two stock statements: isotopes have the same chemical properties because chemical properties depend on the number of electrons; isobars belong to different elements.

The Formula Shortcuts Card — Every Number NEET Needs in One Line Each

Each formula is in the form that fits a 40-second solve — eV, Å, nm, metres per second, no SI conversions unless the options force them.

Light and energy

Want	Formula	Plug-and-play values
Photon energy from wavelength	$E\,(\mathrm{eV}) = \dfrac{1240}{\lambda\,(\mathrm{nm})}$	620 nm $\to$ 2.0 eV; 400 nm $\to$ 3.1 eV; 310 nm $\to$ 4.0 eV; 248 nm $\to$ 5.0 eV
Same thing in joules	$E = h\nu = \dfrac{hc}{\lambda}$, $hc = 1.986 \times 10^{-25}$ J m	1 eV $= 1.602 \times 10^{-19}$ J
Wavenumber	$\bar{\nu} = \dfrac{1}{\lambda}$, unit $\mathrm{cm^{-1}}$ or $\mathrm{m^{-1}}$	500 nm $= 5 \times 10^{-5}$ cm $\to$ 20,000 $\mathrm{cm^{-1}}$
Photoelectric KE	$\mathrm{KE} = h\nu - W_0 = \dfrac{1240}{\lambda\,(\mathrm{nm})} - W_0\,(\mathrm{eV})$	400 nm on Na (2.3 eV): $3.1 - 2.3 = 0.8$ eV
Threshold wavelength	$\lambda_0\,(\mathrm{nm}) = \dfrac{1240}{W_0\,(\mathrm{eV})}$	Cs (1.9 eV): 653 nm
Number of photons	$N = \dfrac{\text{power} \times \text{time}}{h\nu} = \dfrac{P t \lambda}{hc}$	100 W bulb, 1 s, 500 nm: $2.5 \times 10^{20}$

Bohr model for hydrogen-like species ($\mathrm{H}$, $\mathrm{He^+}$, $\mathrm{Li^{2+}}$, $\mathrm{Be^{3+}}$)

Want	Formula	Plug-and-play values
Energy of $n$th orbit	$E_n = -13.6\,\dfrac{Z^2}{n^2}$ eV $= -2.18 \times 10^{-18}\,\dfrac{Z^2}{n^2}$ J	H: $-13.6, -3.4, -1.51, -0.85, -0.54$ eV for $n = 1$ to 5
Ground state of ions	$E_1 = -13.6\,Z^2$ eV	$\mathrm{He^+}$: $-54.4$ eV; $\mathrm{Li^{2+}}$: $-122.4$ eV
Ionisation energy	$\mathrm{IE} = +13.6\,Z^2$ eV (from $n = 1$); from $n$: $13.6 Z^2/n^2$	H from $n = 2$: 3.4 eV
Radius of $n$th orbit	$r_n = 0.529\,\dfrac{n^2}{Z}$ Å $= 52.9\,\dfrac{n^2}{Z}$ pm	H $n = 2$: 2.116 Å; $\mathrm{He^+}$ $n = 2$: 1.058 Å
Velocity in $n$th orbit	$v_n = 2.18 \times 10^6\,\dfrac{Z}{n}\ \mathrm{m\ s^{-1}}$	H $n = 1$: $2.18 \times 10^6$; $n = 2$: $1.09 \times 10^6$
Angular momentum	$mvr = \dfrac{nh}{2\pi}$	$n = 2$: $h/\pi$; $n = 3$: $3h/2\pi$
Transition energy	$\Delta E = 13.6\,Z^2\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$ eV	H, $2 \to 1$: 10.2 eV; $3 \to 2$: 1.89 eV
Wavenumber of a line	$\bar{\nu} = R_H\,Z^2\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$, $R_H = 109{,}677\ \mathrm{cm^{-1}}$	Balmer first line: $\tfrac{5}{36}R_H = 15{,}233\ \mathrm{cm^{-1}}$
Number of spectral lines, $n \to 1$	$\dfrac{n(n - 1)}{2}$	$n = 4$: 6; $n = 5$: 10; $n = 6$: 15
Number of lines, $n_2 \to n_1$	$\dfrac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$	$6 \to 2$: 10; $5 \to 3$: 3

[NEET] Ratios beat numbers. $E \propto Z^2/n^2$, $r \propto n^2/Z$, $v \propto Z/n$. "Ratio of radii of the second orbit of $\mathrm{He^+}$ and the first orbit of H" is $\dfrac{4/2}{1/1} = 2$ — no 0.529 needed.

Dual nature and uncertainty

Want	Formula	Plug-and-play values
de Broglie wavelength	$\lambda = \dfrac{h}{mv} = \dfrac{h}{p} = \dfrac{h}{\sqrt{2mK}}$	Electron at $10^6$ m/s: $7.28 \times 10^{-10}$ m $= 7.28$ Å
Electron accelerated through $V$ volt	$\lambda = \dfrac{12.27}{\sqrt{V}}$ Å	100 V: 1.227 Å; 10,000 V: 0.1227 Å
Heisenberg	$\Delta x \cdot \Delta p \geq \dfrac{h}{4\pi}$, i.e. $\Delta x \cdot \Delta v \geq \dfrac{h}{4\pi m}$	For an electron, $\dfrac{h}{4\pi m_e} = 5.79 \times 10^{-5}\ \mathrm{m^2\ s^{-1}}$
Bohr's condition from de Broglie	$2\pi r = n\lambda$	Orbit is an integral number of wavelengths

Orbitals, shells and counting

Want	Formula	Values to memorise
Orbitals in a shell	$n^2$	1, 4, 9, 16 for $n = 1$ to 4
Electrons in a shell	$2n^2$	2, 8, 18, 32
Orbitals in a subshell	$2l + 1$	$s$: 1, $p$: 3, $d$: 5, $f$: 7
Electrons in a subshell	$2(2l + 1)$	$s$: 2, $p$: 6, $d$: 10, $f$: 14
Subshells in shell $n$	$n$	$n = 3$: $3s, 3p, 3d$
Radial nodes	$n - l - 1$	$3s$: 2; $3p$: 1; $3d$: 0; $4d$: 1
Angular nodes	$l$	$s$: 0; $p$: 1; $d$: 2
Total nodes	$n - 1$	$4f$: 3 (0 radial, 3 angular)
Orbital angular momentum	$\sqrt{l(l+1)}\,\dfrac{h}{2\pi}$	$p$: $\sqrt{2}\,h/2\pi$; $s$: 0
Spin-only magnetic moment	$\mu = \sqrt{n(n + 2)}$ BM, $n =$ unpaired electrons	1: 1.73; 2: 2.83; 3: 3.87; 4: 4.90; 5: 5.92

Key Point: Three numbers open most NEET numericals here: 1240 (eV nm), 13.6 (eV) and 0.529 (Å). Add 12.27 (Å, for an electron through $V$ volt) and $\sqrt{n(n+2)}$.

The unit guard

• Energy in joules: multiply the eV answer by $1.602 \times 10^{-19}$. In kJ/mol: multiply by 96.5.
• Wavelength in metres: 1 nm $= 10^{-9}$ m; 1 Å $= 10^{-10}$ m; 1 Å $= 0.1$ nm $= 100$ pm.
• Wavenumber in $\mathrm{m^{-1}}$ is 100 times the $\mathrm{cm^{-1}}$ value: $R_H = 1.09677 \times 10^7\ \mathrm{m^{-1}} = 109{,}677\ \mathrm{cm^{-1}}$.

The Series-Region Table — Hydrogen Spectrum Facts NEET Expects on Sight

One table and two habits answer every spectral-series question.

The five series of atomic hydrogen

Series	Electron falls to $n_1$	From $n_2$	Region	First line ($n_2 = n_1 + 1$)	Series limit ($n_2 = \infty$)
Lyman	1	2, 3, 4, …	Ultraviolet	$\tfrac{3}{4}R_H$; $\lambda = 121.6$ nm	$R_H$; $\lambda = 91.2$ nm
Balmer	2	3, 4, 5, …	Visible	$\tfrac{5}{36}R_H$; $\lambda = 656.5$ nm ($\mathrm{H_\alpha}$, red)	$\tfrac{1}{4}R_H$; $\lambda = 364.7$ nm
Paschen	3	4, 5, 6, …	Infrared	$\tfrac{7}{144}R_H$; $\lambda = 1876$ nm	$\tfrac{1}{9}R_H$; $\lambda = 820.6$ nm
Brackett	4	5, 6, 7, …	Infrared	$\tfrac{9}{400}R_H$; $\lambda = 4052$ nm	$\tfrac{1}{16}R_H$; $\lambda = 1459$ nm
Pfund	5	6, 7, 8, …	Far infrared	$\tfrac{11}{900}R_H$; $\lambda = 7460$ nm	$\tfrac{1}{25}R_H$; $\lambda = 2279$ nm

(Wavelengths computed with $R_H = 109{,}677\ \mathrm{cm^{-1}}$; options often round to 121.5 or 122 nm, 656 nm, 365 nm.)

Habit 1: "first line" means the smallest jump, "limit" means the biggest

• First line ($\alpha$) of any series: $n_2 = n_1 + 1$. Longest wavelength, lowest energy and frequency in that series.
• Series limit: $n_2 \to \infty$. Shortest wavelength, highest energy; wavenumber $R_H/n_1^2$.
• Second line ($\beta$): $n_2 = n_1 + 2$. Third ($\gamma$): $n_2 = n_1 + 3$.

So the longest wavelength in Balmer is $3 \to 2$, 656 nm; the shortest in Lyman is the series limit, 91.2 nm; the longest in Lyman is $2 \to 1$, 121.6 nm — the two ends of one table row.

Habit 2: energy ordering of the series

Lyman lines are the most energetic (they end at $n = 1$, the deepest level), then Balmer, Paschen, Brackett, Pfund:

$$\text{Energy or frequency:}\quad \text{Lyman} > \text{Balmer} > \text{Paschen} > \text{Brackett} > \text{Pfund}$$

$$\text{Wavelength:}\quad \text{Lyman} < \text{Balmer} < \text{Paschen} < \text{Brackett} < \text{Pfund}$$

Cross-check: the Lyman limit (91.2 nm) is shorter than every Balmer line, the Balmer limit (364.7 nm) shorter than every Paschen line. Lyman, Balmer and Paschen occupy separate windows; the infrared series from Paschen onward overlap.

Ratio questions that need no constants

Question	Set-up	Answer
Ratio of wavenumbers, Lyman first line : Balmer first line	$\tfrac{3}{4} : \tfrac{5}{36}$	$27 : 5$
Ratio of wavelengths, Balmer first : Lyman first	$\tfrac{36}{5} : \tfrac{4}{3}$	$27 : 5$ (about 5.4)
Ratio of the series limits, Lyman : Balmer (wavenumber)	$1 : \tfrac{1}{4}$	$4 : 1$
Same line in $\mathrm{He^+}$ versus H	$Z^2 = 4$	Wavenumber $\times 4$, wavelength $\div 4$
Which $\mathrm{He^+}$ transition matches the H $2 \to 1$ line?	Need $Z^2(1/n_1^2 - 1/n_2^2) = 3/4$	$4 \to 2$ in $\mathrm{He^+}$ (the $\mathrm{He^+}$ $n = 2, 4, 6, \ldots$ ladder maps to H)

Counting lines in 10 seconds

An electron in level $n$ dropping to the ground state gives $\dfrac{n(n - 1)}{2}$ lines (every pair of levels from 1 to $n$). From $n_2$ only as far as $n_1$: $\dfrac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$.

Excited level	Total lines	Of which Lyman	Balmer	Paschen
$n = 3$	3	2	1	0
$n = 4$	6	3	2	1
$n = 5$	10	4	3	2
$n = 6$	15	5	4	3

Key Point: From level $n$, lines ending at $n = 1$ number $(n - 1)$; lines ending at $n = 2$ number $(n - 2)$. The total $n(n-1)/2$ is just $(n-1) + (n-2) + \cdots + 1$.

[NEET] "Hydrogen atoms are excited to $n = 4$; the number of lines in the visible region is…" — visible means Balmer, ending at $n = 2$: $4 \to 2$ and $3 \to 2$, so 2 lines. The total (6) is the distractor.

The absorption side

Hydrogen's absorption lines at room temperature are only Lyman: essentially all atoms sit in $n = 1$, so only $1 \to n$ jumps can absorb. Emission shows all series because excited atoms occupy many levels. Dark lines on a bright background is absorption, bright on dark is emission, and the two are exact negatives.

Configuration Quick-Checks — Cr, Cu, the Ions, Isoelectronic Sets and Quantum-Number Validity

Configuration questions are the most reliable $+4$ here: a handful of memorised answers plus two rules.

The two rules that decide every ion

1. Filling: follow $(n + l)$ — $1s\ 2s\ 2p\ 3s\ 3p\ 4s\ 3d\ 4p\ 5s\ 4d\ 5p\ 6s\ 4f\ 5d\ 6p\ 7s\ 5f\ 6d\ 7p$. So $4s$ fills before $3d$.
2. Removing (making a cation): electrons leave the highest $n$ first — $4s$ empties before $3d$. Write the neutral atom, then strip $4s$, then $3d$.

Two exceptions: chromium and copper. A half-filled or completely filled $d$ subshell is extra stable (symmetry plus exchange energy), so one $4s$ electron moves into $3d$.

The atoms

Element ($Z$)	Configuration	Valence picture	Unpaired electrons
Sc (21)	$[\mathrm{Ar}]\,3d^1 4s^2$		1
Ti (22)	$[\mathrm{Ar}]\,3d^2 4s^2$		2
V (23)	$[\mathrm{Ar}]\,3d^3 4s^2$		3
Cr (24)	$[\mathrm{Ar}]\,3d^5 4s^1$ — not $3d^4 4s^2$	half-filled $d$ and $s$	6
Mn (25)	$[\mathrm{Ar}]\,3d^5 4s^2$		5
Fe (26)	$[\mathrm{Ar}]\,3d^6 4s^2$		4
Co (27)	$[\mathrm{Ar}]\,3d^7 4s^2$		3
Ni (28)	$[\mathrm{Ar}]\,3d^8 4s^2$		2
Cu (29)	$[\mathrm{Ar}]\,3d^{10} 4s^1$ — not $3d^9 4s^2$	full $d$, half $s$	1
Zn (30)	$[\mathrm{Ar}]\,3d^{10} 4s^2$		0

Same story one row down: Mo (42) is $[\mathrm{Kr}]\,4d^5 5s^1$ and Ag (47) is $[\mathrm{Kr}]\,4d^{10} 5s^1$; Au (79) ends $5d^{10} 6s^1$.

The ions — the rows NEET actually asks

Ion	Electrons	Configuration	Unpaired $e^-$	$\mu$ (BM)	Magnetic behaviour
$\mathrm{Sc^{3+}}$	18	$[\mathrm{Ar}]\,3d^0$	0	0	Diamagnetic
$\mathrm{Ti^{3+}}$	19	$[\mathrm{Ar}]\,3d^1$	1	1.73	Paramagnetic
$\mathrm{V^{3+}}$	20	$[\mathrm{Ar}]\,3d^2$	2	2.83	Paramagnetic
$\mathrm{Cr^{3+}}$	21	$[\mathrm{Ar}]\,3d^3$	3	3.87	Paramagnetic
$\mathrm{Cr^{2+}}$	22	$[\mathrm{Ar}]\,3d^4$	4	4.90	Paramagnetic
$\mathrm{Mn^{2+}}$	23	$[\mathrm{Ar}]\,3d^5$	5	5.92	Paramagnetic (maximum for 3d)
$\mathrm{Fe^{3+}}$	23	$[\mathrm{Ar}]\,3d^5$	5	5.92	Paramagnetic
$\mathrm{Fe^{2+}}$	24	$[\mathrm{Ar}]\,3d^6$	4	4.90	Paramagnetic
$\mathrm{Co^{2+}}$	25	$[\mathrm{Ar}]\,3d^7$	3	3.87	Paramagnetic
$\mathrm{Ni^{2+}}$	26	$[\mathrm{Ar}]\,3d^8$	2	2.83	Paramagnetic
$\mathrm{Cu^{2+}}$	27	$[\mathrm{Ar}]\,3d^9$	1	1.73	Paramagnetic (coloured)
$\mathrm{Cu^{+}}$	28	$[\mathrm{Ar}]\,3d^{10}$	0	0	Diamagnetic (colourless)
$\mathrm{Zn^{2+}}$	28	$[\mathrm{Ar}]\,3d^{10}$	0	0	Diamagnetic

[NEET] $\mathrm{Mn^{2+}}$ and $\mathrm{Fe^{3+}}$ are both $3d^5$ — isoelectronic, five unpaired, $\mu = 5.92$ BM, the highest among common $3d$ ions. $\mathrm{Cu^+}$ and $\mathrm{Zn^{2+}}$ are both $3d^{10}$ — zero unpaired, diamagnetic. "Which ion is colourless or diamagnetic?" means find the $d^0$ or $d^{10}$ option.

Unpaired electrons in main-group species — the box-diagram habit

Draw the boxes for the outermost $p$ subshell and apply Hund's rule:

Species	Valence configuration	Unpaired
C, Si	$ns^2\,np^2$	2
N, P	$ns^2\,np^3$	3 (maximum for a $p$ block)
O, S	$ns^2\,np^4$	2
F, Cl	$ns^2\,np^5$	1
Ne, Ar, $\mathrm{O^{2-}}$, $\mathrm{N^{3-}}$	$ns^2\,np^6$	0
$\mathrm{O_2}$ molecule	(from molecular orbital theory, Class 11 Bonding)	2

Isoelectronic sets — recognise them by electron count

Electron count	Set	Size order (radius)
2	$\mathrm{H^-}$, He, $\mathrm{Li^+}$, $\mathrm{Be^{2+}}$	$\mathrm{H^-} > \mathrm{He} > \mathrm{Li^+} > \mathrm{Be^{2+}}$
10	$\mathrm{N^{3-}}$, $\mathrm{O^{2-}}$, $\mathrm{F^-}$, Ne, $\mathrm{Na^+}$, $\mathrm{Mg^{2+}}$, $\mathrm{Al^{3+}}$ (also $\mathrm{H_2O}$, $\mathrm{NH_3}$, $\mathrm{CH_4}$, HF, $\mathrm{NH_4^+}$, $\mathrm{H_3O^+}$)	$\mathrm{N^{3-}} > \mathrm{O^{2-}} > \mathrm{F^-} > \mathrm{Ne} > \mathrm{Na^+} > \mathrm{Mg^{2+}} > \mathrm{Al^{3+}}$
18	$\mathrm{P^{3-}}$, $\mathrm{S^{2-}}$, $\mathrm{Cl^-}$, Ar, $\mathrm{K^+}$, $\mathrm{Ca^{2+}}$, $\mathrm{Sc^{3+}}$, $\mathrm{Ti^{4+}}$ (also $\mathrm{H_2S}$, HCl, $\mathrm{PH_3}$)	$\mathrm{P^{3-}} > \mathrm{S^{2-}} > \mathrm{Cl^-} > \mathrm{Ar} > \mathrm{K^+} > \mathrm{Ca^{2+}}$
36	$\mathrm{Br^-}$, Kr, $\mathrm{Rb^+}$, $\mathrm{Sr^{2+}}$	anion $>$ atom $>$ cation

Key Point: In an isoelectronic series more protons pull the same electrons closer, so the radius decreases as $Z$ increases: the most negative ion is largest, the most positive cation smallest. $\mathrm{Cu^+}$ and $\mathrm{Zn^{2+}}$ have 28 electrons, not 18 — isoelectronic with each other, not with $\mathrm{K^+}$.

Quantum-number validity — six one-liners

1. $l$ can never equal or exceed $n$: $(n = 1, l = 1)$, $(n = 2, l = 2)$, $(n = 3, l = 3)$ are impossible.
2. $\lvert m_l \rvert$ can never exceed $l$: $(l = 1, m_l = +2)$, $(l = 2, m_l = -3)$ are impossible.
3. $m_s$ is only $+\tfrac{1}{2}$ or $-\tfrac{1}{2}$: $m_s = 0$ or $m_s = 1$ is impossible.
4. $n = 0$ or a negative $n$ is impossible; $l = 0$ with $m_l = 0$ is always fine (an $s$ orbital).
5. A set $(n, l, m_l)$ names one orbital; $(n, l, m_l, m_s)$ names one electron; $(n, l)$ names a subshell of $2(2l + 1)$ electrons; $n$ alone names a shell of $2n^2$ electrons.
6. "How many electrons have $n = 3, l = 2, m_s = +\tfrac{1}{2}$?" — five orbitals in $3d$, one electron of that spin each: 5. "How many with $n = 4, m_s = -\tfrac{1}{2}$?" — 16 orbitals, so 16.

Set $(n, l, m_l, m_s)$	Verdict	Why
$(1, 0, 0, +\tfrac{1}{2})$	Valid	$1s$
$(2, 1, -1, -\tfrac{1}{2})$	Valid	$2p$
$(3, 2, +2, +\tfrac{1}{2})$	Valid	$3d$
$(4, 3, -3, -\tfrac{1}{2})$	Valid	$4f$
$(2, 2, 0, +\tfrac{1}{2})$	Invalid	$l = n$
$(3, 1, +2, -\tfrac{1}{2})$	Invalid	$m_l > l$
$(3, 0, 0, 0)$	Invalid	$m_s = 0$
$(1, 1, 0, +\tfrac{1}{2})$	Invalid	no $1p$

The last electron of an element: Na — $(3, 0, 0, +\tfrac{1}{2})$; Cl — $(3, 1, \pm 1 \text{ or } 0, -\tfrac{1}{2})$ depending on convention; K — $(4, 0, 0, +\tfrac{1}{2})$, not $(3, 2, \ldots)$, because $4s$ fills before $3d$.

The NEET Traps List — Where the Minus One Comes From

Trap 1: orbit versus orbital

An orbit is Bohr's fixed circular path; an orbital is a probability region from the wave picture. Orbits are well-defined paths and violate Heisenberg; orbitals have shapes ($s$ spherical, $p$ dumbbell, $d$ double dumbbell) and nodes. "The shape of an orbit" is a trick — orbits are always circular, only orbitals have $l$-dependent shapes. An orbital holds at most 2 electrons; an orbit holds $2n^2$.

Trap 2: the $l$ range versus the $m_l$ range

$l$ runs from $0$ to $n - 1$ ($n$ values); $m_l$ runs from $-l$ to $+l$ ($2l + 1$ values). Students swap them under pressure: for $n = 3$, $l$ is $0, 1, 2$ (never 3); for $l = 2$, $m_l$ is $-2, -1, 0, +1, +2$ (five values, not three). Orbitals in a subshell is $2l + 1$, not $2n + 1$.

Trap 3: 4s fills before 3d — but empties first

Filling follows $(n + l)$: $4s$ ($n + l = 4$) before $3d$ ($n + l = 5$). Once $3d$ holds electrons it lies lower than $4s$, so a cation loses $4s$ first. $\mathrm{Fe^{2+}}$ is $[\mathrm{Ar}]\,3d^6$, not $3d^4 4s^2$; $\mathrm{Cu^+}$ is $3d^{10}$, not $3d^9 4s^1$. Deleting from the right end of the written configuration (the $3d$ end) is the commonest error here.

Trap 4: energy of 3d versus 4s — hydrogen versus everyone else

In hydrogen energy depends only on $n$: $3s = 3p = 3d$, degenerate, all below $4s$. In multi-electron atoms it depends on $(n + l)$, so $4s$ lies below $3d$ for filling. "In hydrogen, which is lower, $3d$ or $4s$?" wants $3d$; the same question for potassium wants $4s$.

Trap 5: wavenumber units

$R_H = 109{,}677\ \mathrm{cm^{-1}} = 1.09677 \times 10^7\ \mathrm{m^{-1}}$. With $\bar{\nu}$ in $\mathrm{cm^{-1}}$, $\lambda = 1/\bar{\nu}$ is in cm; multiply by $10^7$ for nm. Options in metres against a number in centimetres puts you off by 100, and the option list usually carries both.

Trap 6: eV versus J

$E_1 = -13.6$ eV $= -2.18 \times 10^{-18}$ J, and both appear in the options, so read the unit in the last line. The joule figure is per atom; per mole it is $-1312$ kJ/mol (multiply by $N_A$, divide by 1000). "1312 kJ/mol" and "13.6 eV per atom" are one fact.

Trap 7: kinetic energy depends on frequency, not intensity

Doubling the intensity doubles the number of photoelectrons but leaves their kinetic energy unchanged. Only a higher frequency raises the KE. Below $\nu_0$, no electrons at any intensity. Assertion-Reason questions set this trap often.

Trap 8: the sign of energy and "higher" versus "lower"

$E_n$ is negative, so a larger $n$ is less negative, i.e. higher: $E_3 = -1.51$ eV is higher than $E_2 = -3.4$ eV, and $3 \to 2$ releases the positive difference, 1.89 eV. The gap between consecutive levels shrinks as $n$ grows: $E_2 - E_1 = 10.2$ eV, $E_3 - E_2 = 1.89$ eV, $E_4 - E_3 = 0.66$ eV.

Trap 9: number of lines versus lines in one series

"Lines when the electron returns from $n = 5$" is $5 \times 4/2 = 10$. Balmer lines from that level: $5 - 2 = 3$. Lyman: $5 - 1 = 4$. The question tells you which; read it.

Trap 10: Hund's rule versus Pauli's principle

Pauli says two electrons per orbital with opposite spins — a rule about the maximum. Hund says singly occupy degenerate orbitals first, parallel spins — a rule about the order. Nitrogen's $2p_x^1\,2p_y^1\,2p_z^1$ is Hund; "an orbital cannot hold three electrons" is Pauli. Aufbau is a third rule, about which subshell fills first.

Trap 11: mass number is not atomic mass

Mass number $A$ is a whole number (protons + neutrons). Atomic mass (35.5 u for chlorine) is a weighted average of isotopes. Neutrons $= A - Z$, always an integer. $^{37}\mathrm{Cl}$: 17 protons, 20 neutrons, 17 electrons; $^{37}\mathrm{Cl^-}$: 18 electrons.

Trap 12: nodes — radial, angular, total

Radial nodes $= n - l - 1$; angular nodes $= l$; total $= n - 1$. $3p$ has one radial and one angular node; $2s$ one radial and none angular; $3d$ zero radial. "Nodes" without qualification usually means total, $n - 1$.

Key Point: Most NEET losses here are reading errors, not physics errors: orbit/orbital, $l$/$m_l$, fills/empties, eV/J, cm/m, lines/lines-in-a-series. Underline the noun, then compute.

The 40-second checklist (say it before you bubble)

1. Which species ($Z$) and which level ($n$) — and did I square the right one?
2. Did I strip $4s$ before $3d$ for the cation, and remember Cr and Cu?
3. Is the answer in the unit asked for — eV or J, nm or m, $\mathrm{cm^{-1}}$ or $\mathrm{m^{-1}}$?
4. Lines in a series or total lines; radial or total nodes?
5. Does my rounded number match exactly one option?

Solved Examples

Question 1: Energy and radius of a Bohr orbit

Find the energy (in eV) and the radius (in Å) of the electron in the third orbit of a hydrogen atom.

Answer:

Here $Z = 1$ and $n = 3$.

$E_n = -13.6\,\dfrac{Z^2}{n^2}$ eV, so $E_3 = -\dfrac{13.6}{9} = -1.51$ eV.

$r_n = 0.529\,\dfrac{n^2}{Z}$ Å $= 0.529 \times 9 = 4.76$ Å.

Ans: $E_3 = -1.51$ eV; $r_3 = 4.76$ Å (476 pm). Watch out: Options in joules need $-1.51 \times 1.602 \times 10^{-19} = -2.42 \times 10^{-19}$ J.

Question 2: Ground-state energy of a hydrogen-like ion

The ionisation energy of hydrogen is 13.6 eV. What is the energy of the electron in the ground state of $\mathrm{Li^{2+}}$, and what is its ionisation energy?

Answer:

Lithium has $Z = 3$, and $\mathrm{Li^{2+}}$ has one electron, so Bohr's formula applies.

$E_1 = -13.6 \times Z^2 = -13.6 \times 9 = -122.4$ eV.

Ionisation energy takes the electron from $n = 1$ to $n = \infty$: $+122.4$ eV.

Ans: $E_1 = -122.4$ eV; ionisation energy $= 122.4$ eV. Watch out: IEs of $\mathrm{H}$, $\mathrm{He^+}$, $\mathrm{Li^{2+}}$ go as $1 : 4 : 9$ — 13.6, 54.4, 122.4 eV.

Question 3: Photon energy from wavelength

Light of wavelength 400 nm falls on a sodium surface whose work function is 2.3 eV. Find the energy of a photon and the maximum kinetic energy of the ejected electrons.

Answer:

$E = \dfrac{1240}{\lambda\,(\mathrm{nm})} = \dfrac{1240}{400} = 3.1$ eV.

$\mathrm{KE}_{\max} = h\nu - W_0 = 3.1 - 2.3 = 0.8$ eV.

In joules, if asked: $0.8 \times 1.602 \times 10^{-19} = 1.28 \times 10^{-19}$ J.

Ans: Photon energy 3.1 eV; $\mathrm{KE}_{\max} = 0.8$ eV. Watch out: Check the photon energy clears the work function. At 600 nm, $E = 2.07$ eV $< 2.3$ eV and no electrons come out.