JEE Corner — Effective Nuclear Charge, Successive Ionization Enthalpies, Scales of Electronegativity and the Ordering Questions JEE Loves

The Part of Periodicity Your Textbook Left Out

Sections 1 to 8 cover what the Board paper asks. JEE Main asks something else: the effective nuclear charge felt by a 3d electron in zinc, which element has successive ionization enthalpies of 578, 1817, 2745 and 11577 kJ/mol, why the second electron gain enthalpy of oxygen is $+780$ kJ/mol when the first is $-141$, why hafnium is the same size as zirconium, and above all arrange the following in increasing order of radius, ionization enthalpy, electron gain enthalpy, electronegativity, metallic character, oxidizing power, basicity of oxides.

This section builds that toolkit: Slater's rules, the three kinds of radius and the ionic-radius rules, successive ionization enthalpies as a code that identifies an element, the sign conventions of electron affinity and electron gain enthalpy, the Pauling, Mulliken and Allred-Rochow scales with one worked conversion each, and a map of oxides and hydroxides. The last two blocks are the ordering-question bank and the trap list.

What "beyond the textbook" means here

Textbook gives you	JEE Main also wants	JEE Advanced adds
"inner electrons shield the outer ones"	$Z_{\text{eff}} = Z - \sigma$, why $Z_{\text{eff}}$ rises by about 0.65 per element across a period	Slater's rules with actual numbers
covalent and metallic radius	van der Waals radius, why noble gases look big, the three-radius inequality	radius-ratio rules, lanthanoid contraction pairs
$\Delta_i H_2 > \Delta_i H_1$	reading a successive-IE table to find the group and the stable valence	d-block and inert-pair irregularities in IE
$\Delta_{eg} H$ of F, Cl, O, S	electron affinity sign convention, second electron gain enthalpy of O and S	the $-\tfrac{5}{2}RT$ correction
Pauling scale values	Mulliken $\chi = (IE + EA)/2$ and its conversion to Pauling units	Allred-Rochow, hybridisation and oxidation-state dependence, percent ionic character
basic, amphoteric, acidic oxides	the electronegativity rule and the oxidation-state rule for oxide character	hydride and hydroxide acid-base orders

Every rule here is stated once, boxed, illustrated with real numbers, then used in the worked questions.

Key Point: Almost every periodic trend is a fight between two quantities: the effective nuclear charge $Z_{\text{eff}}$ pulling the outer electron in, and the principal quantum number $n$ of that electron pushing it out. Across a period $Z_{\text{eff}}$ wins; down a group $n$ wins. The anomalies (Be/B, N/O, O/S, F/Cl, Ga/Al, Tl/In) are the places where a third effect — subshell type, electron pairing, poor d or f shielding — tips the balance. If you can name which of the three is operating, you can answer the question.

Screening, Effective Nuclear Charge and Slater's Rules

Why the outer electron does not feel the whole nucleus

An electron in a sodium atom is attracted by a nucleus of charge $+11$ but repelled by the other ten electrons, which sit between it and the nucleus most of the time. They screen or shield it, and the effect is bundled into the shielding constant $\sigma$:

$$\boxed{Z_{\text{eff}} = Z - \sigma}$$

$Z_{\text{eff}}$ is the charge the electron behaves as if it sees, and it sets the radius, the ionization enthalpy and the electronegativity. Three qualitative facts come first:

• Electrons in the same shell shield each other poorly, being at roughly the same distance from the nucleus. Across a period each added proton is only partly cancelled by the added electron, so $Z_{\text{eff}}$ rises — by about 0.65 per element on Slater's recipe.
• Electrons in inner shells shield well. Down a group a whole inner shell is added, and the $Z_{\text{eff}}$ felt by the outermost electron stays nearly constant (Na 3s and K 4s both feel 2.20). What changes down a group is $n$, and hence the size.
• Penetration order $s > p > d > f$: an s electron spends more time close to the nucleus than a p electron of the same shell, so it is shielded less and held more tightly. Hence 2s lies below 2p, B has a lower ionization enthalpy than Be, and d and f electrons are poor shields for the electrons outside them.

Slater's rules — the recipe

Slater (1930) estimated $\sigma$ like this. Step 1: write the configuration in these groups, in this order:

$$(1s)\ (2s\,2p)\ (3s\,3p)\ (3d)\ (4s\,4p)\ (4d)\ (4f)\ (5s\,5p)\ (5d)\ \ldots$$

$ns$ and $np$ are lumped together; $nd$ and $nf$ stand alone. Step 2: pick the electron you care about and add up the contributions of every other electron:

Electron of interest	Other electrons in the same group	Electrons in shell $n - 1$	Electrons in shell $n - 2$ or lower
$ns$ or $np$	0.35 each (0.30 if the group is $1s$)	0.85 each	1.00 each
$nd$ or $nf$	0.35 each	1.00 each (all groups to the left)	1.00 each

Electrons in groups to the right of the chosen electron contribute nothing. The d/f row says a d or f electron is buried so deep that everything inside it shields fully.

Four worked values

Sodium, 3s electron ($Z = 11$): $(1s^2)(2s^2\,2p^6)(3s^1)$. Same group: none. Shell 2: 8 at 0.85 $= 6.80$. Shell 1: 2 at 1.00 $= 2.00$. $$\sigma = 8.80, \qquad Z_{\text{eff}} = 11 - 8.80 = 2.20$$

Chlorine, 3p electron ($Z = 17$): $(1s^2)(2s^2\,2p^6)(3s^2\,3p^5)$. Same group: 6 others at 0.35 $= 2.10$. Shell 2: $8 \times 0.85 = 6.80$. Shell 1: $2 \times 1.00 = 2.00$. $$\sigma = 10.90, \qquad Z_{\text{eff}} = 17 - 10.90 = 6.10$$

Zinc, 4s electron ($Z = 30$): $(1s^2)(2s^2\,2p^6)(3s^2\,3p^6)(3d^{10})(4s^2)$. Same group: 1 other at 0.35. Shell 3 ($3s$, $3p$ and $3d$ — 18 electrons) at 0.85 $= 15.30$. Shells 1 and 2: 10 at 1.00. $$\sigma = 0.35 + 15.30 + 10.00 = 25.65, \qquad Z_{\text{eff}} = 30 - 25.65 = 4.35$$

Zinc, 3d electron: same group, 9 others at 0.35 $= 3.15$; everything to the left ($1s$, $2s2p$, $3s3p$: 18 electrons) at 1.00; the two 4s electrons are to the right and count zero. $$\sigma = 3.15 + 18.00 = 21.15, \qquad Z_{\text{eff}} = 30 - 21.15 = 8.85$$

A 3d electron in zinc feels twice the nuclear charge a 4s electron feels — the quantitative reason the 4s electrons are ionised first from every 3d metal, even though 4s fills before 3d.

The period-2 and period-3 tables you should be able to regenerate

For a $2p$ (or $2s$) electron, $\sigma = 0.35 \times (\text{others in } 2s2p) + 0.85 \times 2$, so each step across the period adds one proton and 0.35 of shielding — a net gain of 0.65 in $Z_{\text{eff}}$.

Element	Li	Be	B	C	N	O	F	Ne
$Z_{\text{eff}}$ (outer electron)	1.30	1.95	2.60	3.25	3.90	4.55	5.20	5.85

Element	Na	Mg	Al	Si	P	S	Cl	Ar
$Z_{\text{eff}}$ (outer electron)	2.20	2.85	3.50	4.15	4.80	5.45	6.10	6.75

Down group 1: Li 1.30, Na 2.20, K 2.20 (for K, $8 \times 0.85 + 10 \times 1.00 = 16.80$, so $19 - 16.80 = 2.20$). The near-constancy from Na onward is the point.

How $Z_{\text{eff}}$ explains the trends

Trend	In terms of $Z_{\text{eff}}$ and $n$
Radius falls across a period	same $n$, $Z_{\text{eff}}$ up by 0.65 per step: the cloud is pulled in
Radius rises down a group	$Z_{\text{eff}}$ roughly constant, $n$ up by one: a new, bigger shell
Ionization enthalpy rises across, falls down	roughly $\Delta_i H \propto Z_{\text{eff}}^2/n^2$ for the outer electron
Electronegativity rises across, falls down	Allred-Rochow: $\chi \propto Z_{\text{eff}}/r^2$
Ga is not much bigger than Al, and has a slightly higher IE	ten 3d electrons shield at only 0.85 (for 4p) — Ga 4p feels 5.00 against Al 3p's 3.50 — the d-block contraction
Hf is the same size as Zr	fourteen 4f electrons shield 5d and 6s poorly — the lanthanoid contraction
Second-period anomaly Be/B	B's 2p electron is less penetrating than Be's 2s (Slater's numbers do not see this; penetration does)

[JEE Main] Memorise the grouping, the triple 0.35 / 0.85 / 1.00, the 0.30 exception for 1s, and the "everything inside counts 1.00 for d and f" rule. Slater's rules are a model, not a law — they give the same $Z_{\text{eff}}$ for 2s and 2p, so they cannot explain the Be/B or N/O anomalies. For those, argue from penetration and pairing, not from Slater.

The Three Radii, the Ionic-Radius Rules and the Lanthanoid Contraction

Covalent, metallic and van der Waals radius — and their order

An atom has no edge, so "radius" always means half of some measured internuclear distance. There are three such distances.

Radius	Half of…	Example	Typical for
Covalent radius $r_{\text{cov}}$	the bond length between two like atoms joined by a single covalent bond	$\mathrm{Cl-Cl}$ in $\mathrm{Cl_2}$: 198 pm, so $r_{\text{cov}}(\mathrm{Cl}) = 99$ pm	non-metals
Metallic radius $r_{\text{met}}$	the distance between adjacent nuclei in the metal crystal	$\mathrm{Cu-Cu}$ 256 pm, so $r_{\text{met}}(\mathrm{Cu}) = 128$ pm; Na 186 pm	metals
van der Waals radius $r_{\text{vdW}}$	the closest distance between nuclei of two non-bonded atoms in neighbouring molecules (solid state)	Cl: 180 pm; Ne: about 160 pm; Ar: about 190 pm	noble gases, and the "size" of any atom that is not bonded

$$\boxed{r_{\text{covalent}} < r_{\text{metallic}} < r_{\text{van der Waals}}}$$

The covalent radius is smallest because bonding pulls the nuclei together and the clouds overlap. The metallic radius is larger because metallic bonding is weaker and less directional (sodium: covalent about 154 pm, metallic 186 pm). The van der Waals radius is largest because there is no overlap at all. Chlorine's pair, 99 pm (covalent) and 180 pm (van der Waals), is worth remembering.

Key Point (Why noble gases "look big"): Noble gases do not form molecules, so no covalent radius can be measured for them; only the van der Waals radius is available. A table listing Ne at 160 pm next to F at 64 pm (covalent) compares a van der Waals radius with a covalent one. That is why noble gases appear to be the largest atoms of their period, and why a correct radius comparison across a period stops at the halogen. If a question forces the noble gas in, it is the largest — but say why.

Ionic radius: the rules

A cation is smaller than its atom (a whole shell is often gone, and the remaining electrons feel a higher $Z_{\text{eff}}$); an anion is larger (extra repulsion in the same shell, lower $Z_{\text{eff}}$). Na 186 pm to $\mathrm{Na^+}$ 95 pm; F 64 pm to $\mathrm{F^-}$ 136 pm. Three rules cover every comparison JEE sets.

Rule 1 — Isoelectronic species: higher nuclear charge, smaller ion. $$\mathrm{N^{3-}}\ (171) > \mathrm{O^{2-}}\ (140) > \mathrm{F^-}\ (136) > \mathrm{Ne}\ > \mathrm{Na^+}\ (95) > \mathrm{Mg^{2+}}\ (72) > \mathrm{Al^{3+}}\ (54\ \text{pm})$$ (all ten-electron species). Same for the 18-electron set $\mathrm{S^{2-}} > \mathrm{Cl^-} > \mathrm{K^+} > \mathrm{Ca^{2+}} > \mathrm{Sc^{3+}}$. Equivalently: in an isoelectronic series, size falls as $Z/e$ (nuclear charge to electron number) rises.

Rule 2 — Same element, different charge: the more positive, the smaller. Each electron removed raises $Z_{\text{eff}}$ on those left behind. $$\mathrm{Fe}\ (126\ \text{pm}) > \mathrm{Fe^{2+}}\ (76) > \mathrm{Fe^{3+}}\ (64); \qquad \mathrm{Cl^-} > \mathrm{Cl} > \mathrm{Cl^+}; \qquad \mathrm{H^-} > \mathrm{H} > \mathrm{H^+}$$ $\mathrm{H^+}$ is extreme: a bare proton, about $10^{-5}$ of the atom's radius, which is why it never exists free and is always attached to something ($\mathrm{H_3O^+}$, $\mathrm{NH_4^+}$).

Rule 3 — Different elements, not isoelectronic: use group and period first, then charge. $\mathrm{K^+} > \mathrm{Na^+} > \mathrm{Li^+}$ (down a group); $\mathrm{Na^+}$ (95) $> \mathrm{Mg^{2+}}$ (72) (across, and more charge). Watch for diagonal near-equalities: $\mathrm{Li^+}$ (76) and $\mathrm{Mg^{2+}}$ (72) are almost the same size, one root of the Li/Mg diagonal relationship.

Radius ratio — what the ion sizes decide

The ratio $r_+/r_-$ decides how many anions fit round a cation in a crystal (its coordination number). This is Class 12 solid-state material, but JEE Advanced sometimes links it to periodicity, so keep the bands:

$r_+/r_-$	Coordination number	Geometry	Example
below 0.155	2	linear	—
0.155 to 0.225	3	trigonal planar	$\mathrm{B_2O_3}$
0.225 to 0.414	4	tetrahedral	ZnS
0.414 to 0.732	6	octahedral	NaCl
0.732 to 1.0	8	cubic	CsCl

Down group 1 the cation grows, the ratio with $\mathrm{Cl^-}$ rises, and the structure shifts from the NaCl type (6:6) to the CsCl type (8:8).

Lanthanoid contraction and the pairs that come out equal

Across the lanthanoids (Ce to Lu, $Z = 58$ to 71) electrons enter the deeply buried 4f subshell, which shields the outer electrons very poorly. Atomic and ionic size shrink steadily across fourteen elements: $\mathrm{La^{3+}}$ 103 pm down to $\mathrm{Lu^{3+}}$ 86 pm. This lanthanoid contraction almost exactly cancels the size increase expected from period 5 to period 6, so each 5d element is nearly the same size as the 4d element above it:

4d element	Radius	5d element	Radius	Consequence
Zr	160 pm	Hf	159 pm	Zr and Hf occur together and are hard to separate
Nb	146 pm	Ta	146 pm	near-identical chemistry, similar ionization enthalpies
Mo	139 pm	W	139 pm	same
Ag	144 pm	Au	144 pm	but Au has much higher IE — relativistic effects add to the contraction

Ti 147 to Zr 160 pm is a 13 pm jump; Zr to Hf is 1 pm. Two exam consequences: the 4d/5d pairs have similar radii, similar ionization enthalpies and very similar chemistry; and the 5d elements are much denser than the 4d ones (same volume, nearly double the mass — W 19.3 g/cm³ against Mo 10.2).

The same logic one row up is the d-block contraction (or scandide contraction): the ten 3d electrons before gallium shield the 4p electron poorly, so Ga (135 pm) is smaller than Al (143 pm) even with one more shell, and Ga's first ionization enthalpy (579 kJ/mol) is marginally higher than Al's (577). Similarly Ge is close to Si in size and As to P.

[JEE Main] Three radius facts come as one-liners: the van der Waals radius of chlorine (180 pm) against its covalent radius (99 pm); the isoelectronic order $\mathrm{O^{2-}}$, $\mathrm{F^-}$, $\mathrm{Na^+}$, $\mathrm{Mg^{2+}}$ (reverse order of $Z$); and Zr $\approx$ Hf from the lanthanoid contraction.

Successive Ionization Enthalpies — Reading the Code

The general pattern

Each ionization is harder than the one before: the electron is pulled away from an ion of higher positive charge, and every electron that leaves raises $Z_{\text{eff}}$ on those left behind. $$\Delta_i H_1 < \Delta_i H_2 < \Delta_i H_3 < \ldots$$ The rise is not uniform. Within a shell it grows gently; when the next electron must come from an inner, noble-gas-like core, the value jumps by a factor of 3 to 6. Count the values before the jump and you have the number of valence electrons, hence the group and the stable valence.

Key Point: The number of "small" successive ionization enthalpies before the first big jump $=$ the number of valence electrons $=$ the group number (for s-block) or group number minus 10 (for p-block) $=$ the highest stable positive oxidation state.

The three elements JEE uses most (kJ/mol)

	$\Delta_i H_1$	$\Delta_i H_2$	$\Delta_i H_3$	$\Delta_i H_4$	$\Delta_i H_5$	Valence electrons	Reading
Na	496	4562	6910	9543	13354	1	jump after 1st ($\times 9$): $\mathrm{Na^+}$ is neon-like; NaCl not $\mathrm{NaCl_2}$
Mg	737	1451	7733	10543	13630	2	jump after 2nd ($\times 5$): $\mathrm{MgCl_2}$, never $\mathrm{MgCl_3}$
Al	577	1817	2745	11577	14842	3	jump after 3rd ($\times 4$): $\mathrm{AlCl_3}$; $\mathrm{Al^{3+}}$ is neon-like

Bold numbers mark the first big jump. Compare the columns, not just the rows: $\Delta_i H_1$ runs Na (496) $<$ Al (577) $<$ Mg (737); $\Delta_i H_2$ runs Mg (1451) $<$ Al (1817) $\ll$ Na (4562); $\Delta_i H_3$ runs Al (2745) $\ll$ Na (6910) $<$ Mg (7733). Which element has the highest $n$-th ionization enthalpy depends entirely on whether that electron is a valence or a core electron.

A few more rows for the identification game

Element	$\Delta_i H_1$	$\Delta_i H_2$	$\Delta_i H_3$	$\Delta_i H_4$	$\Delta_i H_5$	$\Delta_i H_6$	Jump after
Li	520	7298	11815				1st
Be	899	1757	14849	21007			2nd
B	801	2427	3660	25026	32827		3rd
C	1086	2353	4621	6223	37831		4th
Si	786	1577	3232	4356	16091	19805	4th
P	1012	1903	2912	4957	6274	21269	5th
K	419	3052	4420				1st
Ca	590	1145	4912	6491			2nd

How to read an unknown set. Say the values are 1012, 1903, 2912, 4957, 6274, 21269. Ratios of neighbours: 1.9, 1.5, 1.7, 1.3, 3.4. The only big ratio is between the 5th and 6th values, so 5 valence electrons: group 15, $ns^2\,np^3$, stable oxidation states $+3$ and $+5$, chloride $\mathrm{MCl_3}$ (or $\mathrm{MCl_5}$). The slightly larger step from the 3rd to the 4th value (2912 to 4957, ratio 1.7) is the 4th electron coming from the $ns^2$ pair — a sub-jump inside the valence shell. Sub-jumps of about 1.5 to 2 mark the s/p boundary; jumps of 3 to 9 mark the core.

The irregularities JEE builds questions on

1. Group 13: Ga $\approx$ Al, Tl $>$ In. First ionization enthalpies: B 801, Al 577, Ga 579, In 558, Tl 589 kJ/mol. The expected steady fall breaks twice. Ga's ten 3d electrons shield its 4p electron poorly (d-block contraction), so $Z_{\text{eff}}$ on that electron is high and the value does not fall below Al's. Tl sits after the 4f and 5d series: fourteen 4f plus ten 5d electrons shield poorly (lanthanoid contraction), and relativistic contraction of the 6s orbital tightens it further. The same reasoning gives Pb (716) $>$ Sn (709) in group 14 and explains the inert pair effect: the $6s^2$ pair in Tl, Pb, Bi is so tightly held that the lower oxidation states $+1$, $+2$, $+3$ are the stable ones.

2. 4s versus 3d in the transition metals. The first two electrons removed from any 3d metal are the 4s electrons, which feel the lower $Z_{\text{eff}}$ (Zn 4s 4.35, 3d 8.85). $\Delta_i H_1$ across Sc to Zn rises only gently (Sc 631 to Zn 906 kJ/mol), because the 3d electrons added alongside shield the 4s fairly well; the third ionization, which takes a 3d electron, is much larger.

3. Cu versus Zn. $\Delta_i H_1$: Cu 745 $<$ Zn 906 — Zn's 4s electron comes from a filled $4s^2$ pair with a higher nuclear charge. But $\Delta_i H_2$: Cu 1958 $>$ Zn 1733 — the second electron from $\mathrm{Cu^+}$ must break the stable $3d^{10}$ shell, whereas $\mathrm{Zn^+}$ merely gives up its remaining 4s electron. Same reversal for Cr/Mn: $\Delta_i H_1$ Cr 653 $<$ Mn 717, but $\Delta_i H_2$ Cr 1591 $>$ Mn 1509 because $\mathrm{Cr^+}$ is $3d^5$ (half-filled).

4. Second-period second ionization enthalpies. $\Delta_i H_2$: Li 7298 $\gg$ Be 1757; N 2856, O 3388, F 3374. $\Delta_i H_2$ of O is marginally higher than F's: $\mathrm{O^+}$ is $2p^3$ (half-filled) while $\mathrm{F^+}$ is $2p^4$ and loses a paired electron. The anomaly that puts O below N in $\Delta_i H_1$ puts F below O in $\Delta_i H_2$ — the pattern shifts one place right per electron already removed. (A tendency, not a law: by the third ionization the extra proton wins and Ne's value edges above F's again.)

Ionization enthalpy of ions — the same numbers, relabelled

$$\Delta_i H_2(\mathrm{Na}) = \Delta_i H_1(\mathrm{Na^+}) = 4562\ \mathrm{kJ/mol}, \qquad \Delta_i H_3(\mathrm{Al}) = \Delta_i H_1(\mathrm{Al^{2+}}) = 2745\ \mathrm{kJ/mol}$$ "The first ionization enthalpy of $\mathrm{Mg^+}$" is the second ionization enthalpy of Mg, 1451 kJ/mol. For a one-electron ion you can compute it: for $\mathrm{He^+}$, $13.6 \times 2^2 = 54.4$ eV $= 5250$ kJ/mol, which is $\Delta_i H_2$ of helium (the first is 2372 kJ/mol).

Ionization enthalpy, reducing power and the reactivity of metals

Low ionization enthalpy means the atom gives up its electron easily — a good reducing agent and a reactive metal. Down group 1 the ionization enthalpy falls (Li 520, Na 496, K 419, Rb 403, Cs 376 kJ/mol), so gas-phase reducing power rises Li $<$ Na $<$ K $<$ Rb $<$ Cs and caesium is the most reactive metal. Two cautions: in aqueous solution lithium is the strongest reducing agent (its tiny ion has a huge hydration enthalpy), so "strongest reducing agent among alkali metals" has two answers depending on whether the question says "in solution"; and non-metals run the other way, high ionization enthalpy and highly negative electron gain enthalpy meaning strong oxidizing power (F $>$ Cl $>$ O $>$ N, and $\mathrm{F_2} > \mathrm{Cl_2} > \mathrm{Br_2} > \mathrm{I_2}$).

[JEE Main] When a table of successive ionization enthalpies is given, do not match numbers to memory. Compute the ratios of neighbouring values, find the first ratio above 3, and count. It works even for made-up data.

Electron Affinity versus Electron Gain Enthalpy, and the Second Electron

Two names, opposite signs

$$\mathrm{X(g)} + e^- \longrightarrow \mathrm{X^-(g)}, \qquad \Delta_{eg} H$$

Electron gain enthalpy $\Delta_{eg} H$ follows the thermodynamic convention: negative when energy is released (Cl: $-349$ kJ/mol), positive when energy must be supplied (Ne: $+116$ kJ/mol). Electron affinity $A_e$ is the older quantity, defined as the energy released on adding the electron, so it carries the opposite sign: $A_e(\mathrm{Cl}) = +349$ kJ/mol (or 3.62 eV), $A_e(\mathrm{Ne}) = -116$ kJ/mol. Strictly, electron affinity is defined at 0 K while enthalpy is measured at temperature $T$; allowing for the heat capacities of the gaseous species gives $$\boxed{\Delta_{eg} H = -A_e - \tfrac{5}{2}RT}$$ At 298 K the correction $\tfrac{5}{2}RT = 2.5 \times 8.314 \times 298 = 6.2$ kJ/mol — small, and ignored in ordering questions, but JEE Advanced has tested the statement that the electron gain enthalpy is slightly more negative than the negative of the electron affinity.

The data with the anomalies (kJ/mol)

Group 1		Group 16		Group 17		Group 18
H	$-73$	O	$-141$	F	$-328$	He	$+48$
Li	$-60$	S	$-200$	Cl	$-349$	Ne	$+116$
Na	$-53$	Se	$-195$	Br	$-325$	Ar	$+96$
K	$-48$	Te	$-190$	I	$-295$	Kr	$+96$
Rb	$-47$	Po	$-174$	At	$-270$	Xe	$+77$
Cs	$-46$					Rn	$+68$

Key Point (The orders):

• Halogens: $\mathbf{Cl > F > Br > I}$ in magnitude ($-349, -328, -325, -295$). Fluorine is second, not first: its 2p shell is so small that the incoming electron feels strong repulsion from the seven electrons already there.
• Chalcogens: $\mathbf{S > Se > Te > Po > O}$ ($-200, -195, -190, -174, -141$). Oxygen is last for the same reason. (Do not write S $>$ O $>$ Se: selenium at $-195$ is far more negative than oxygen at $-141$.)
• Overall, chlorine has the most negative electron gain enthalpy of all elements.
• The alkali metals do release energy on gaining an electron ($\mathrm{H^-}$, $\mathrm{Na^-}$ exist in the gas phase): $-73$ for H down to $-46$ for Cs, becoming less negative down the group as the atom grows.

Positive and near-zero values, and why

Species	$\Delta_{eg} H$	Reason
N	about 0 (tables give $-7$ to $+7$)	the electron must pair up in the half-filled $2p^3$ shell; the exchange stability is lost
Be, Mg	positive	the electron would enter the higher-energy $np$ subshell above a filled $ns^2$
Noble gases	positive (He $+48$ to Ne $+116$)	the electron must start a new shell outside a closed octet; Ne is the most positive of all
Mn, Zn	near zero or positive	half-filled $3d^5\,4s^2$ and full $3d^{10}\,4s^2$ respectively

Among the group-18 values the order is He $<$ Ne $>$ Ar $=$ Kr $>$ Xe $>$ Rn: neon is the hardest to force an electron on, and helium is anomalous because it is so small that the added electron goes into 2s at fairly short range.

The second electron gain enthalpy: always positive

Adding an electron to a negative ion means pushing a negative charge onto something that already repels it, so the second electron gain enthalpy is always endothermic: $$\mathrm{O(g)} + e^- \to \mathrm{O^-(g)}, \quad \Delta_{eg} H_1 = -141\ \mathrm{kJ/mol}; \qquad \mathrm{O^-(g)} + e^- \to \mathrm{O^{2-}(g)}, \quad \Delta_{eg} H_2 = +780\ \mathrm{kJ/mol}$$ Net for $\mathrm{O(g)} \to \mathrm{O^{2-}(g)}$: $+639$ kJ/mol. For sulfur: $-200$ and about $+590$ kJ/mol, net about $+390$. Forming $\mathrm{O^{2-}}$ or $\mathrm{S^{2-}}$ in the gas phase is uphill — the oxide and sulfide ions exist only because the lattice enthalpy of the solid (MgO about $-3800$ kJ/mol, CaO about $-3400$) pays the bill many times over. The same logic says there is no isolated $\mathrm{N^{3-}}$ ion in the gas phase.

[JEE Main] Sign discipline settles most of the marks here. "Highest electron affinity" and "most negative electron gain enthalpy" are the same element (chlorine). "Positive electron gain enthalpy" is the noble gases, Be, Mg, N (roughly zero) — and the second electron of anybody. If the options mix signs, re-read the stem for which quantity it names.

Three Electronegativity Scales, and What Electronegativity Depends On

Electronegativity is the tendency of a bonded atom to pull the shared electrons toward itself. It is not measurable directly, so every scale is a recipe built on something measurable.

Pauling's scale — from bond energies

A bond A-B is always stronger than the average of A-A and B-B, and the excess grows with the bond's polarity. Pauling defined the excess with the geometric mean: $$\Delta = E(\mathrm{A-B}) - \sqrt{E(\mathrm{A-A})\,E(\mathrm{B-B})}$$ (some texts use the arithmetic mean $\tfrac{1}{2}[E(\mathrm{A-A}) + E(\mathrm{B-B})]$ instead) and then set $$\boxed{\chi_{\mathrm{A}} - \chi_{\mathrm{B}} = 0.208\sqrt{\Delta}} \quad (\Delta \text{ in kcal/mol}); \qquad \chi_{\mathrm{A}} - \chi_{\mathrm{B}} = 0.102\sqrt{\Delta} \quad (\Delta \text{ in kJ/mol})$$ The scale gives only differences, so one value is fixed by hand: fluorine $= 4.0$ (hydrogen was later set at 2.1). The ladder is Li 1.0, Be 1.5, B 2.0, C 2.5, N 3.0, O 3.5, F 4.0; Na 0.9, Mg 1.2, Al 1.5, Si 1.8, P 2.1, S 2.5, Cl 3.0; K 0.8, Rb 0.8, Cs 0.7; Br 2.8, I 2.5, At 2.2. Worked: for HCl with $E(\mathrm{H-H}) = 104.2$, $E(\mathrm{Cl-Cl}) = 58.0$, $E(\mathrm{H-Cl}) = 103.2$ kcal/mol, $\sqrt{104.2 \times 58.0} = 77.7$, $\Delta = 25.5$, $0.208 \times \sqrt{25.5} = 1.05$, so $\chi_{\mathrm{Cl}} = 2.1 + 1.05 = 3.15$, which rounds to the tabulated 3.0 (or 3.16 on the modern revised scale).

Mulliken's scale — from IE and EA

An atom that holds its own electron hard (high ionization enthalpy) and grabs an extra one readily (high electron affinity) should be strongly electronegative. Mulliken averaged the two: $$\boxed{\chi_{\mathrm{M}} = \frac{IE + EA}{2}\ (\text{in eV per atom})} \qquad \text{and} \qquad \chi_{\mathrm{Pauling}} \approx \frac{\chi_{\mathrm{M}}}{2.8}$$ With energies in kJ/mol, $1\ \mathrm{eV} = 96.5$ kJ/mol, so $\chi_{\mathrm{Pauling}} \approx \dfrac{IE + EA}{2 \times 2.8 \times 96.5} = \dfrac{IE + EA}{540}$ (kJ/mol). $EA$ here is the electron affinity — a positive number for chlorine — not the electron gain enthalpy with its negative sign. A slightly better fit is $\chi_{\mathrm{P}} = 0.336\,(\chi_{\mathrm{M}} - 0.615)$; JEE uses the simple 2.8 divisor.

Element	IE (eV)	EA (eV)	$\chi_{\mathrm{M}}$	$\chi_{\mathrm{M}}/2.8$	Pauling
F	17.42	3.40	10.41	3.72	4.0
Cl	12.97	3.62	8.29	2.96	3.0
O	13.62	1.46	7.54	2.69	3.5
S	10.36	2.07	6.22	2.22	2.5

The conversion is approximate (oxygen comes out low), but the ordering F $>$ Cl $>$ O $>$ S is preserved. Because the Mulliken value averages two energies per atom, that scale has units (eV) while the Pauling scale is dimensionless.

Allred-Rochow — from $Z_{\text{eff}}$ and radius

Electronegativity is the electrostatic pull of the nucleus on a bonding electron at the covalent radius, so it should scale as $Z_{\text{eff}}/r^2$: $$\boxed{\chi_{\mathrm{AR}} = \frac{0.359\,Z_{\text{eff}}}{r^2} + 0.744} \qquad (r \text{ in \AA}, Z_{\text{eff}} \text{ by Slater})$$ This scale explains the trends most directly: across a period $Z_{\text{eff}}$ rises and $r$ falls, both raising $\chi$; down a group $Z_{\text{eff}}$ is flat and $r$ grows, so $\chi$ falls. It also explains why the halogens are the most electronegative — smallest radii for their $Z_{\text{eff}}$. (Two further scales you may see named: Sanderson's, from electron density, and Allen's, from configuration energies; JEE names them at most.)

Electronegativity is a property of the atom in its situation

Unlike ionization enthalpy, which belongs to the isolated atom, electronegativity changes with what the atom is doing.

Factor	Effect	Example
Oxidation state	higher positive charge, higher $\chi$	$\mathrm{Fe^{3+}} > \mathrm{Fe^{2+}} > \mathrm{Fe}$; Mn in $\mathrm{Mn_2O_7}$ far more electronegative than in MnO, which is why $\mathrm{Mn_2O_7}$ is acidic
Hybridisation (s-character)	more s-character, electron held closer, higher $\chi$: $sp > sp^2 > sp^3$	carbon: about 3.3 in ethyne ($sp$), 2.75 in ethene ($sp^2$), 2.5 in ethane ($sp^3$); ethyne's C-H is weakly acidic for this reason
Charge on the atom	anion less electronegative, cation more	$\mathrm{O^-}$ attracts less than O
Neighbours	electron-withdrawing substituents raise the effective $\chi$	C in $\mathrm{CF_3}$ against C in $\mathrm{CH_3}$

Electronegativity difference, bond polarity and percent ionic character

The bigger $\Delta\chi = \lvert \chi_{\mathrm{A}} - \chi_{\mathrm{B}} \rvert$, the more polar the bond. Pauling's rough rule: $\Delta\chi \geq 1.7$ means more than 50% ionic character — treat as ionic; between 0.4 and 1.7 polar covalent; below 0.4 essentially non-polar. Two formulas quantify it:

$$\text{Pauling: } \% \text{ ionic} = 100\left[1 - e^{-(\Delta\chi)^2/4}\right]; \qquad \boxed{\text{Hannay-Smyth: } \% \text{ ionic} = 16(\Delta\chi) + 3.5(\Delta\chi)^2}$$

For HCl ($\Delta\chi = 0.9$): Hannay-Smyth gives $14.4 + 2.8 = 17.2\%$; the dipole-moment route (observed 1.03 D against $e \times 127$ pm $= 6.10$ D for a fully ionic bond) gives $16.9\%$. For HF ($\Delta\chi = 1.9$): $30.4 + 12.6 = 43\%$. For $\Delta\chi = 1.7$ Pauling's formula gives 51% — the source of the "1.7 rule". Bond polarity in the hydrogen halides runs HF $>$ HCl $>$ HBr $>$ HI, the reverse of their acid strength — a trap covered in the next block.

[JEE Main] Three numbers to keep ready: the divisor 2.8 (Mulliken to Pauling), the factor 0.208 (Pauling from kcal bond energies) and the threshold 1.7 (50% ionic). When a question gives IE and EA in kJ/mol, divide by 96.5 first or use 540 as the divisor — do not average kJ values and then divide by 2.8.

Acidic, Basic and Amphoteric — Oxides, Hydroxides and Hydrides Across the Table

The two rules that predict oxide character

Rule 1 — electronegativity of the element. If E is far less electronegative than oxygen (a metal, $\chi \lesssim 1.5$), the E-O bond is ionic and the oxide releases $\mathrm{O^{2-}}$, which takes protons from water: basic ($\mathrm{Na_2O} + \mathrm{H_2O} \to 2\,\mathrm{NaOH}$). If E is close to oxygen ($\chi \gtrsim 2$, a non-metal), the E-O bond is covalent and the oxide gives an oxoacid with water: acidic ($\mathrm{SO_3} + \mathrm{H_2O} \to \mathrm{H_2SO_4}$; $\mathrm{Cl_2O_7} + \mathrm{H_2O} \to 2\,\mathrm{HClO_4}$). In between ($\chi$ about 1.5 to 2, near the metalloid line) the oxide can act either way: amphoteric ($\mathrm{Al_2O_3}$, $\mathrm{ZnO}$, $\mathrm{BeO}$, $\mathrm{SnO}$, $\mathrm{PbO}$, $\mathrm{Ga_2O_3}$, $\mathrm{As_2O_3}$, $\mathrm{Sb_2O_3}$, $\mathrm{Cr_2O_3}$). A few low-oxidation-state non-metal oxides do not react with water at all: neutral (CO, NO, $\mathrm{N_2O}$, $\mathrm{H_2O}$).

Across period 3 the whole spectrum appears in one row:

Oxide	$\mathrm{Na_2O}$	MgO	$\mathrm{Al_2O_3}$	$\mathrm{SiO_2}$	$\mathrm{P_4O_{10}}$	$\mathrm{SO_3}$	$\mathrm{Cl_2O_7}$
Character	strongly basic	basic	amphoteric	weakly acidic	acidic	strongly acidic	very strongly acidic
Product with water	NaOH	$\mathrm{Mg(OH)_2}$	—	— ($\mathrm{H_4SiO_4}$ with alkali)	$\mathrm{H_3PO_4}$	$\mathrm{H_2SO_4}$	$\mathrm{HClO_4}$

So basicity falls and acidity rises across a period; basicity rises down a group ($\mathrm{BeO}$ amphoteric $<$ MgO $<$ CaO $<$ SrO $<$ BaO; $\mathrm{Li_2O} < \mathrm{Na_2O} < \mathrm{K_2O} < \mathrm{Rb_2O} < \mathrm{Cs_2O}$; in group 15, $\mathrm{N_2O_3}$ and $\mathrm{P_4O_6}$ acidic, $\mathrm{As_2O_3}$ and $\mathrm{Sb_2O_3}$ amphoteric, $\mathrm{Bi_2O_3}$ basic).

Rule 2 — oxidation state of the element. The same element in a higher oxidation state is smaller, more electronegative and more covalently bonded to oxygen, so its oxide is more acidic. This is the rule for transition metals:

Element	Basic	Amphoteric	Acidic
Mn	MnO ($+2$), $\mathrm{Mn_2O_3}$ ($+3$)	$\mathrm{MnO_2}$ ($+4$)	$\mathrm{MnO_3}$ ($+6$), $\mathrm{Mn_2O_7}$ ($+7$, gives $\mathrm{HMnO_4}$)
Cr	CrO ($+2$)	$\mathrm{Cr_2O_3}$ ($+3$)	$\mathrm{CrO_3}$ ($+6$, gives $\mathrm{H_2CrO_4}$)
V	VO ($+2$), $\mathrm{V_2O_3}$ ($+3$)	$\mathrm{VO_2}$ ($+4$), $\mathrm{V_2O_5}$ ($+5$, mainly acidic)	—
Pb, Sn	—	PbO, SnO ($+2$)	$\mathrm{PbO_2}$, $\mathrm{SnO_2}$ ($+4$, more acidic than $+2$)
N	—	—	$\mathrm{N_2O_3} < \mathrm{N_2O_5}$ (acidity rises with oxidation state); $\mathrm{N_2O}$, NO neutral
Cl	—	—	$\mathrm{Cl_2O} < \mathrm{ClO_2} < \mathrm{Cl_2O_7}$

Two combined orderings appear as MCQs: $\mathrm{Cl_2O_7} > \mathrm{SO_3} > \mathrm{P_4O_{10}} > \mathrm{SiO_2}$ in acidic strength (across the period), and $\mathrm{Mn_2O_7} > \mathrm{CrO_3} > \mathrm{V_2O_5}$ (higher oxidation state, further right).

Hydroxides: basicity follows the metal

$\mathrm{E-O-H}$ can break at E-O (giving $\mathrm{OH^-}$: a base) or at O-H (giving $\mathrm{H^+}$: an acid). A large, low-charge, low-electronegativity E pushes the break to E-O:

$$\mathrm{LiOH} < \mathrm{NaOH} < \mathrm{KOH} < \mathrm{RbOH} < \mathrm{CsOH}; \qquad \mathrm{Be(OH)_2} < \mathrm{Mg(OH)_2} < \mathrm{Ca(OH)_2} < \mathrm{Sr(OH)_2} < \mathrm{Ba(OH)_2}$$ $$\text{across period 3: } \mathrm{NaOH} > \mathrm{Mg(OH)_2} > \mathrm{Al(OH)_3}\ (\text{amphoteric}) > \mathrm{Si(OH)_4}\ (\text{weak acid}) > \ldots > \mathrm{HClO_4}$$

$\mathrm{Be(OH)_2}$, $\mathrm{Al(OH)_3}$, $\mathrm{Zn(OH)_2}$, $\mathrm{Sn(OH)_2}$, $\mathrm{Pb(OH)_2}$ and $\mathrm{Cr(OH)_3}$ are amphoteric — they dissolve in both acid and excess alkali ($\mathrm{Al(OH)_3} + \mathrm{OH^-} \to [\mathrm{Al(OH)_4}]^-$). Hydroxides of the same metal in higher oxidation states are more acidic, as for oxides.

Hydrides: two different rules for two directions

Across a period the acidity of the hydride tracks the electronegativity of the central atom, because a more electronegative E stabilises $\mathrm{E^-}$ better: $\mathrm{CH_4} < \mathrm{NH_3} < \mathrm{H_2O} < \mathrm{HF}$. $\mathrm{NH_3}$ is a base (lone pair on a not-very-electronegative N), $\mathrm{H_2O}$ is neutral/amphiprotic, HF is a weak acid.

Down a group the rule flips: acidity is decided by the E-H bond strength, and the bond gets longer and weaker as E grows. $$\text{acid strength: } \mathrm{HF} < \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI}; \qquad \mathrm{H_2O} < \mathrm{H_2S} < \mathrm{H_2Se} < \mathrm{H_2Te}; \qquad \text{basicity: } \mathrm{NH_3} > \mathrm{PH_3} > \mathrm{AsH_3} > \mathrm{SbH_3}$$ HI is the strongest hydrohalic acid although H-I is the least polar of them; HF is the weakest although H-F is the most polar, because the 570 kJ/mol H-F bond is very hard to break (and $\mathrm{F^-}$ is strongly hydrogen-bonded to undissociated HF). Polarity governs across a period; bond strength governs down a group.

Key Point (One-line summary): Basic character of oxides and hydroxides rises with metallic character (down a group, leftward across a period, lower oxidation state). Acidic character rises with non-metallic character and with oxidation state. Hydride acidity rises across a period with electronegativity and down a group with weakening E-H bonds.

[JEE Main] "Which oxide is amphoteric?" — the metalloid neighbourhood (Be, Al, Ga, Zn, Sn, Pb, As, Sb) or a $+3$/$+4$ transition-metal oxide ($\mathrm{Cr_2O_3}$, $\mathrm{MnO_2}$). "Most acidic oxide" — highest oxidation state of the most non-metallic element listed. "Most basic hydroxide" — the biggest, least electronegative metal in its lowest oxidation state.

The Ordering-Question Bank

Every JEE Main paper carries at least one "arrange in increasing/decreasing order" from this chapter. Below: the recipe for each property, fifteen ready-made sets with reasons, then the anomaly checklist to run before you commit.

Recipes

Property	Across a period ($\to$)	Down a group ($\downarrow$)	Then check for
Atomic radius	decreases	increases	noble gas (van der Waals, exclude); Ga $\approx$ Al; Zr $\approx$ Hf
Ionic radius	isoelectronic: falls with $Z$	increases	same element: more positive is smaller
$\Delta_i H_1$	increases	decreases	Be $>$ B; N $>$ O; Mg $>$ Al; P $>$ S; Ga $>$ Al; Tl $>$ In; Pb $>$ Sn
$\Delta_i H_2$, $\Delta_i H_3$	shift the anomaly one place right per electron removed	—	Na $\gg$ Mg for $\Delta_i H_2$; Cu $>$ Zn for $\Delta_i H_2$
$\Delta_{eg} H$ (magnitude)	increases	decreases	Cl $>$ F; S $>$ O; N, Be, Mg, noble gases positive
Electronegativity	increases	decreases	N $=$ Cl $=$ 3.0; C $=$ S $=$ I $=$ 2.5; higher oxidation state or more s-character raises it
Metallic character / reducing power	decreases	increases	in solution Li is the strongest reducing agent
Non-metallic character / oxidizing power	increases	decreases	F $>$ Cl $>$ O $>$ N as oxidizers (not the $\Delta_{eg} H$ order)
Basicity of oxides	decreases	increases	falls with oxidation state of the same element
Acidity of oxides	increases	decreases	rises with oxidation state of the same element

Fifteen ready-made sets

#	Property	Order	Why
1	Atomic radius	Na $>$ Mg $>$ Al $>$ Si $>$ P $>$ S $>$ Cl (186, 160, 143, 117, 110, 104, 99 pm)	$Z_{\text{eff}}$ up 0.65 per step, same shell
2	Atomic radius	Cs $>$ Rb $>$ K $>$ Na $>$ Li (262, 244, 231, 186, 152 pm)	new shell each step
3	Ionic radius (isoelectronic)	$\mathrm{N^{3-}} > \mathrm{O^{2-}} > \mathrm{F^-} > \mathrm{Na^+} > \mathrm{Mg^{2+}} > \mathrm{Al^{3+}}$	ten electrons, nuclear charge 7 to 13
4	Ionic radius (same element)	$\mathrm{Fe} > \mathrm{Fe^{2+}} > \mathrm{Fe^{3+}}$; $\mathrm{Cl^-} > \mathrm{Cl} > \mathrm{Cl^+}$	more positive, higher $Z_{\text{eff}}$ on the rest
5	$\Delta_i H_1$, period 2	Ne $>$ F $>$ N $>$ O $>$ C $>$ Be $>$ B $>$ Li	N/O and Be/B swaps
6	$\Delta_i H_1$, period 3	Ar $>$ Cl $>$ P $>$ S $>$ Si $>$ Mg $>$ Al $>$ Na	P/S and Mg/Al swaps
7	$\Delta_i H_1$, group 13	B $>$ Tl $>$ Ga $>$ Al $>$ In (801, 589, 579, 577, 558)	d-block and lanthanoid contraction
8	$\Delta_i H_1$, group 14	C $>$ Si $>$ Ge $>$ Pb $>$ Sn (1086, 786, 762, 716, 709)	Pb above Sn for the same reason as Tl above In
9	$\Delta_i H_2$	Na $>$ Al $>$ Mg; Li $>$ Be; O $>$ F $>$ N	second electron of Na and Li is a core electron; $\mathrm{O^+}$ is half-filled
10	$\Delta_{eg} H$ (more negative first)	Cl $>$ F $>$ Br $>$ I; S $>$ Se $>$ Te $>$ Po $>$ O	period-2 compactness penalty
11	$\Delta_{eg} H$ (most positive first)	Ne $>$ Ar $=$ Kr $>$ Xe $>$ Rn $>$ He	closed shells; He anomalous
12	Electronegativity	F $>$ O $>$ N $=$ Cl $>$ Br $>$ C $=$ S $=$ I $>$ H $\approx$ P (both 2.1); $sp > sp^2 > sp^3$ carbon	Pauling values; s-character
13	Metallic character	K $>$ Mg $>$ Al $>$ B; Cs $>$ Rb $>$ K $>$ Na $>$ Li	low $\Delta_i H$, large size
14	Oxidizing power	F $>$ Cl $>$ O $>$ N; $\mathrm{F_2} > \mathrm{Cl_2} > \mathrm{Br_2} > \mathrm{I_2}$	electronegativity plus (for $\mathrm{F_2}$) weak F-F bond and high hydration enthalpy of $\mathrm{F^-}$
15	Basicity of oxides	$\mathrm{Na_2O} > \mathrm{MgO} > \mathrm{Al_2O_3} > \mathrm{SiO_2} > \mathrm{P_4O_{10}} > \mathrm{SO_3} > \mathrm{Cl_2O_7}$; MnO $>$ $\mathrm{MnO_2}$ $>$ $\mathrm{Mn_2O_7}$	metallic character; oxidation state

The anomaly checklist — run it before you answer

1. Is a noble gas in the radius list? Its radius is van der Waals; if forced, it is the largest.
2. Is Be/B, N/O, Mg/Al or P/S in an ionization-enthalpy list? Swap them (filled s or half-filled p wins).
3. Is it the second or third ionization enthalpy? Shift every anomaly one place right per electron already gone; check whether the electron now comes from the core.
4. Is F or O in an electron-gain list? Put Cl above F and S above O (and Se, Te above O too).
5. Are Ga/Al, Tl/In, Pb/Sn in an IE list? The heavier one is higher.
6. Are Zr/Hf, Nb/Ta in a radius list? Nearly equal.
7. Is the question about oxidizing power rather than $\Delta_{eg} H$? Then F beats Cl.
8. Is it reducing power in water? Then Li beats Cs.
9. Is a transition-metal oxide series given? Rank by oxidation state, not by position.
10. Are the hydrogen halides given as acids? HI strongest, HF weakest — the reverse of polarity.
11. Does the list mix electronegativity of the same element in different oxidation states or hybridisations? Higher oxidation state and more s-character win.
12. Is "electron affinity" (positive for Cl) or "electron gain enthalpy" (negative for Cl) the word used? Fix the sign before ranking.

[JEE Main] When two options differ only by the position of one pair, that pair is the anomaly the setter planted. Find the pair, name the rule that flips it, and you have the answer without checking the rest of the sequence.

The Traps JEE Sets (read before every test)

#	Trap	The fix
1	$Z_{\text{eff}}$ of Zn 3d computed with 0.85 for the inner shells	for a d or f electron every inner electron counts 1.00: Zn 3d $\sigma = 3.15 + 18 = 21.15$, $Z_{\text{eff}} = 8.85$
2	Forgetting that 3d belongs to shell $n - 1$ for a 4s electron	Zn 4s: the 18 electrons of $3s\,3p\,3d$ all count 0.85: $Z_{\text{eff}} = 4.35$
3	Using 0.35 for the 1s partner	the 1s group uses 0.30: He 1s $Z_{\text{eff}} = 1.70$
4	Comparing Ne's radius (van der Waals, 160 pm) with F's (covalent, 64 pm)	different definitions; exclude noble gases from a period's radius order
5	Writing $\mathrm{Na^+} > \mathrm{F^-}$ because Na is a bigger atom	isoelectronic: more protons, smaller; $\mathrm{F^-}$ (136) $> \mathrm{Na^+}$ (95)
6	Assuming Ga is much bigger than Al and has a lower IE	d-block contraction: Ga 135 pm $<$ Al 143 pm; IE Ga 579 $>$ Al 577
7	Expecting Hf to be bigger than Zr	lanthanoid contraction: 159 vs 160 pm — practically equal
8	Ranking $\Delta_i H_2$ like $\Delta_i H_1$	$\Delta_i H_2$: Na (4562) $\gg$ Mg (1451); Cu (1958) $>$ Zn (1733); O $>$ F
9	Identifying an element from IE numbers by memory	compute neighbour ratios; the first ratio above 3 marks the core; count values before it
10	Group 13 IE written as steadily decreasing	B $>$ Tl $>$ Ga $>$ Al $>$ In; group 14: Pb $>$ Sn
11	"First ionization enthalpy of $\mathrm{Na^+}$" answered as 496	it is $\Delta_i H_2$ of Na, 4562 kJ/mol
12	Electron affinity of Cl quoted as $-349$	electron affinity is $+349$ kJ/mol (3.62 eV); electron gain enthalpy is $-349$
13	F given the most negative $\Delta_{eg} H$	Cl ($-349$) $>$ F ($-328$); and S ($-200$) $>$ Se ($-195$) $>$ O ($-141$)
14	Second electron gain enthalpy of O taken as negative	it is $+780$ kJ/mol; every second electron gain enthalpy is positive
15	$\Delta_{eg} H$ of N taken as strongly negative	about zero (half-filled $2p^3$); Be, Mg and noble gases positive
16	Mulliken $\chi$ computed from kJ/mol and then divided by 2.8	convert to eV first (divide by 96.5), or use $(IE + EA)/540$ in kJ/mol
17	Using $\Delta_{eg} H$ with its negative sign in Mulliken's formula	use the electron affinity (positive for halogens)
18	Electronegativity treated as a fixed property of the element	rises with oxidation state ($\mathrm{Fe^{3+}} > \mathrm{Fe^{2+}}$) and with s-character ($sp > sp^2 > sp^3$)
19	HF called the strongest hydrohalic acid because H-F is the most polar	HI $>$ HBr $>$ HCl $>$ HF: bond strength decides down a group
20	$\mathrm{Mn_2O_7}$ called basic because Mn is a metal	oxidation state $+7$: strongly acidic, gives $\mathrm{HMnO_4}$; MnO is basic, $\mathrm{MnO_2}$ amphoteric
21	Oxidizing power ranked by $\Delta_{eg} H$ (Cl $>$ F)	oxidizing power F $>$ Cl $>$ Br $>$ I; the F-F bond is weak and $\mathrm{F^-}$ is strongly hydrated
22	"Strongest reducing agent among alkali metals" answered as Cs without reading the medium	gas phase / by IE: Cs; in aqueous solution: Li (hydration enthalpy)
23	Radius ratio bands mixed up	0.225 to 0.414 tetrahedral (4); 0.414 to 0.732 octahedral (6); 0.732 to 1 cubic (8)
24	Slater's rules used to explain Be/B or N/O	Slater gives the same $Z_{\text{eff}}$ to 2s and 2p; those anomalies are penetration and pairing, not screening

[JEE Main] Nearly every JEE Main question from this section is one rule wearing a costume. Strip it: "the element whose successive IEs are …" becomes "count before the jump"; "electron affinity 3.62 eV" becomes "$\Delta_{eg} H = -349$"; "Zr and Hf" becomes "lanthanoid contraction"; "$\mathrm{Mn_2O_7}$" becomes "oxidation state $+7$, acidic"; "which is most polar / which is the strongest acid" becomes "polarity across, bond strength down". Translate first, then answer.