Why is ionization energy of nitrogen higher than oxygen?

Nitrogen has stable half-filled 2p3 configuration. Oxygen 2p4 has one paired electron that is easier to remove, restoring the stable half-filled state. Hence IE of N is greater than IE of O.

Why does chlorine have higher electron affinity than fluorine?

Fluorine is very small so the incoming electron feels strong inter-electron repulsion in the compact 2p subshell. Chlorine 3p is larger and more diffuse, allowing the added electron to release more energy.

Which element has highest electronegativity?

Fluorine has the highest electronegativity on the Pauling scale at 4.0 due to its small size, high effective nuclear charge, and compact 2p orbitals close to the nucleus.

Periodicity In Atomic Properties

Q: Why does atomic radius decrease across a period?

Across a period each element has one more proton and one more electron in the same shell. Effective nuclear charge increases while shielding stays roughly constant, so the nucleus pulls electrons tighter and the atom shrinks.

Q: Why is a cation smaller than its parent atom?

Cation formation increases effective nuclear charge on remaining electrons and often removes an entire outer shell. Both effects contract the electron cloud.

Q: What is the diagonal relationship?

Diagonal relationship is the chemical similarity between second and third period elements one group apart: Li-Mg, Be-Al, and B-Si. It arises from comparable ionic size, electronegativity, and polarizing power along the diagonal.

ChemistryClassification Of Elements And Periodicity In PropertiesFor NEET aspirants

Periodicity in atomic properties refers to the regular, repeating variation of properties like atomic radius, ionization enthalpy, electron gain enthalpy, and electronegativity as we move across a period or down a group of the modern periodic table. Across a period, the effective nuclear charge increases while the shell number stays constant - atoms shrink and hold their electrons more tightly. Down a group, new shells are added - atoms grow and electrons are held more loosely. Understanding these five master trends predicts almost every chemical behaviour: metallic character, oxide acidity, reducing power, and even bond types.

Key Formulas & Trends - Quick Reference

Covalent radius: $r_{A} = d_{A - A} /2$ (half the internuclear distance in a homo-diatomic molecule).
Radius order (same element): $r_{vdW} > r_{metallic} > r_{covalent}$ .
Ionic radius: cation $<$ parent atom $<$ anion.
Ionization energy order: $I E_{1} < I E_{2} < I E_{3} < \dots$ (each successive removal is harder).
Effective nuclear charge: $Z_{eff} = Z - σ$ , where $σ$ is the Slater shielding constant.
Electron affinity ∝ $1/ r$ , i.e. small atoms release more energy on gaining an electron.
Pauling scale: $0.208 Δ_{A B} = x_{A} - x_{B}$ (for $x_{A} > x_{B}$ ).
% ionic character: $16 Δ x + 3.5 (Δ x)^{2}$ . A $Δ x = 1.7$ gives $\approx 50%$ ionic.
Master trend: across period → radius $↓$ ; IE, EA, EN $↑$ . Down group → reverse.

1. Atomic Radius

The atomic radius is the distance between the nucleus of an atom and the outermost electron shell. Since electron clouds have no sharp boundaries, atomic radius cannot be measured directly on a single free atom. Instead, we define three practical types depending on the environment of the atom.

Figure: Three types of atomic radii - covalent radius (overlapping atoms), metallic radius (just-touching atoms), and van der Waals radius (non-bonded atoms with a gap). Order: vdW

>

metallic

>

covalent.

1.1 Covalent Radius

One-half of the distance between the nuclei of two identical atoms joined by a single covalent bond in a homo-diatomic molecule.

r_{A} = \frac{d _{A - A}}{2}

Or equivalently, in a hetero-nuclear bond A-B:

d_{A - B} = r_{A} + r_{B}

1.2 Metallic Radius

In a metallic crystal, atoms are treated as closely packed spheres. The metallic radius is half the distance between the nuclei of two adjacent metal atoms.

Metallic radius $>$ covalent radius. For potassium: metallic radius $= 2.30 \overset{˚}{A}$ , covalent radius $= 2.03 \overset{˚}{A}$ .

1.3 van der Waals Radius (Collision Radius)

For non-metal atoms in the solid state, half the distance between the nuclei of two non-bonded adjacent atoms held together only by van der Waals forces.

Order:

r_{vdW} > r_{metallic} > r_{covalent}

For chlorine: van der Waals radius $= 1.80 \overset{˚}{A}$ , covalent radius $= 0.99 \overset{˚}{A}$ - almost twice as large.

1.4 Ionic Radius

When a neutral atom loses or gains electrons it becomes a cation or anion respectively, and its size changes.

Figure: Cation size decreases (Na → Na⁺) while anion size increases (Cl → Cl⁻) relative to the parent atom.

Cation is smaller than the parent atom

Effective nuclear charge on remaining electrons increases (e.g. in Na, 11 electrons share 11 protons; in Na⁺, 10 electrons share the same 11 protons).
Formation of a cation often removes the entire outermost shell, so the ion has one fewer shell.
Inter-electronic repulsion decreases, letting the electron cloud contract.

Anion is larger than the parent atom

Effective nuclear charge per electron decreases (17 protons must now attract 18 electrons in Cl⁻).
Inter-electronic repulsion increases, expanding the electron cloud.

1.5 Isoelectronic Species

Species (atoms, ions, or molecules) with the same number of electrons but different nuclear charges form an isoelectronic series.

Figure: The isoelectronic series (10 electrons each): as

Z

increases from N³⁻ to Mg²⁺, effective nuclear charge grows and the species shrinks.

Rule: In an isoelectronic series, ionic radius decreases as nuclear charge increases. Same electron cloud, stronger nuclear pull → smaller ion.

Common examples:

$N^{3 -}, O^{2 -}, F^{-}, N e, N a^{+}, M g^{2 +}, A l^{3 +}$ (all 10 electrons)
$S^{2 -}, C l^{-}, K^{+}, C a^{2 +}, S c^{3 +}$ (all 18 electrons)
$N_{2}, C O, C N^{-}$ (all 14 electrons)
$N H_{3}, H_{3} O^{+}$ (all 10 electrons)

2. Periodicity in Atomic and Ionic Radius

Figure: Master trend summary - atomic radius increases

↓

and

\leftarrow

; ionization energy, electron affinity, and electronegativity all increase

\to

and

↑

2.1 Across a Period (Normal Elements)

Moving left to right, the effective nuclear charge increases because each new electron is added to the same valence shell while a proton is added to the nucleus. The tighter pull shrinks the atom.

Element	Li	Be	B	C	N	O	F
Radius (Å)	1.23	0.89	0.80	0.77	0.74	0.74	0.72

2.2 Down a Group

The number of shells increases and although nuclear charge grows, the added inner shells screen the outer electrons effectively. Atomic size increases.

Element	Li	Na	K	Rb	Cs
Radius (Å)	1.23	1.57	2.03	2.16	2.35

Exception: The atomic radius of the inert gas at the end of a period is larger than that of the preceding halogen because for noble gases only the van der Waals radius is measurable, which is systematically larger than a covalent radius.

2.3 Transition Elements (d-block)

There are four transition series: $3 d$ (Sc-Zn), $4 d$ (Y-Cd), $5 d$ (La, Hf-Hg), and $6 d$ (Ac, Rf onwards, incomplete).

Across a transition series, atomic size first decreases, then becomes nearly constant, and finally slightly increases at the end. The addition of $(n - 1) d$ electrons screens the outer $n s$ electrons well, partly offsetting the pull of increasing nuclear charge.

Element	Sc	Ti	V	Cr	Mn	Fe	Co	Ni	Cu	Zn
Radius (Å)	1.44	1.32	1.22	1.18	1.17	1.17	1.16	1.16	1.17	1.25

2.4 Lanthanide Contraction

Across the lanthanide series (La to Lu), atomic and ionic radii decrease steadily. This is called lanthanide contraction.

Cause: the additional electrons enter the deeply buried

(n - 2) f

orbital. The mutual shielding of

f

-electrons is very poor (much weaker than

s

p

, or

d

), so the outer electrons feel almost the full increase in nuclear charge. The atom contracts slightly with each added

f

-electron.

Consequence for group trends: Because of lanthanide contraction, the atomic radii of the 4d (2nd transition series) and 5d (3rd transition series) elements in the same group are almost identical (except Y-La). For example, the size of Zr $\approx$ Hf, Nb $\approx$ Ta, Mo $\approx$ W. This is why 4d and 5d elements have very similar chemistry.

Solved Example 1

Compare the sizes of Cl, Cl⁻, and Fe²⁺.

Solution:

Use the electron-to-proton ratio (higher ratio → looser hold → larger size):

For Cl: $Z / e = 17/17 = 1.00$

For Cl⁻: $Z / e = 17/18 = 0.944$

For Fe²⁺: $Z / e = 26/24 = 1.083$

Lower $Z / e$ means a bigger species, so: Cl⁻ > Cl > Fe²⁺.

Solved Example 2

The radius of argon is greater than the radius of chlorine. Explain.

Solution:

For chlorine, the reported radius is the covalent radius, which is half the inter-nuclear distance in Cl₂. For argon, only the van der Waals radius is available (Ar does not form covalent bonds). Because van der Waals radii are always significantly larger than covalent radii of the same element, argon appears larger than chlorine even though its effective nuclear charge is greater.

3. Ionization Enthalpy (Ionization Energy)

The ionization enthalpy ( $I E_{1}$ ) is the minimum energy required to remove the most loosely bound electron from one mole of isolated gaseous atoms in their ground state.

A_{(g)} + I E_{1} \to A_{(g)}^{+} + e^{-}

A_{(g)}^{+} + I E_{2} \to A_{(g)}^{2 +} + e^{-}

A_{(g)}^{2 +} + I E_{3} \to A_{(g)}^{3 +} + e^{-}

Because effective nuclear charge felt by remaining electrons increases at each step:

I E_{1} < I E_{2} < I E_{3} < \dots

Figure: Successive ionization energies of sodium - a huge jump between

I E_{1}

and

I E_{2}

because

I E_{2}

breaks into the stable [Ne] core.

3.1 Factors Affecting Ionization Enthalpy

Atomic radius: $I E \propto 1/ r$ . A bigger atom holds its outer electron more loosely.
Effective nuclear charge: $I E \propto Z_{eff}$ . A higher effective pull → higher $I E$ .
Penetration effect: For the same shell, the order of penetration is $s > p > d > f$ . An $s$ -electron is closer to the nucleus and harder to remove than a $p$ -electron of the same shell.
Shielding (screening): Inner electrons block outer electrons from feeling the nuclear pull. $I E \propto 1/ shielding$ .
Stability of half-filled and fully-filled configurations: Configurations like $p^{3}, p^{6}, d^{5}, d^{10}, f^{7}, f^{14}$ are extra-stable. Removing an electron from these is harder - a bump in $I E$ .

3.2 Periodicity in Ionization Enthalpy

Across a period (normal elements)

$I E$ generally increases because effective nuclear charge grows while the shell number is fixed.

Exceptions across a period

IIA > IIIA: $I E$ of Group 2 (e.g. Be, Mg) is greater than Group 13 (B, Al). Reason: Group 2 has fully filled $n s^{2}$ , extra stable; Group 13 has an easily-removable single $p$ -electron.
VA > VIA: $I E$ of Group 15 (N, P, As) is greater than Group 16 (O, S, Se). Reason: half-filled $p^{3}$ of Group 15 is extra stable, so P > S and As > Se. But for heavier pairs Sb-Te and Bi-Po, the normal trend takes over (Sb $<$ Te, Bi $<$ Po).

Down a group

$I E$ generally decreases as atomic size grows.

Exceptions down a group

Group 13: $I E$ of Ga ≈ $I E$ of Al (not lower as expected). Before Ga, the $3 d^{10}$ sub-shell is filled but shields poorly - so $Z_{eff}$ on the outer $4 p$ electron of Ga is larger than expected.
6th period Tl, Pb: $I E$ of Tl and Pb is larger than that of In and Sn respectively - this time due to the poor shielding from the intervening $4 f^{14}$ and $5 d^{10}$ electrons (lanthanide contraction effect).

3.3 Transition Series Trends

Across a transition series (e.g. $3 d$ : Sc to Zn), $I E$ increases slowly. Increase in $Z_{eff}$ is partly cancelled by the shielding of newly added $(n - 1) d$ electrons.
From Sc to Cr, $I E$ rises steadily; from Cr onwards it fluctuates due to half- and fully-filled stability.
Down a transition group: $I E$ decreases from 1st to 2nd series (size dominates) but slightly increases from 2nd to 3rd series (lanthanide contraction), except for Y → La.

Solved Example 3

Magnesium has first and second ionization potentials

7.646 eV

and

15.035 eV

respectively. Calculate the energy required to convert

24 mg

of Mg atoms into Mg²⁺ ions.

Solution:

$M g_{(g)} \to M g_{(g)}^{+} + e^{-}$ ; $I E_{1} = 7.646 eV$

$M g_{(g)}^{+} \to M g_{(g)}^{2 +} + e^{-}$ ; $I E_{2} = 15.035 eV$

Total per atom $= 7.646 + 15.035 = 22.681 eV$ .

Converting to $kJ/mol$ ( $1 eV = 96.49 kJ/mol$ ): $22.681 \times 96.49 = 2188.5 kJ/mol$ .

Moles in $24 mg$ of Mg $= 0.024/24 = 1 \times 1 0^{- 3} mol$ .

Energy required $= 2188.5 \times 1 0^{- 3} = 2.188 kJ$ .

Solved Example 4

Compare qualitatively the first and second ionization energies of Cu and Zn.

Solution:

$I E_{1}$ : Cu is $[A r] 3 d^{10} 4 s^{1}$ ; removing the single $4 s$ electron leaves an extra-stable $3 d^{10}$ core, so $I E_{1}$ of Cu is relatively low. Zn is $[A r] 3 d^{10} 4 s^{2}$ ; removing one $4 s$ electron does not leave a specially stable configuration. Hence $I E_{1} (Cu) < I E_{1} (Zn)$ .

$I E_{2}$ : Cu⁺ has $[A r] 3 d^{10}$ - the next electron must come from the very stable $d^{10}$ core, needing high energy. Zn⁺ has $[A r] 3 d^{10} 4 s^{1}$ - the next electron is again a lone $4 s$ , much easier. Hence $I E_{2} (Cu) > I E_{2} (Zn)$ .

4. Electron Affinity (Electron Gain Enthalpy)

The electron affinity is the amount of energy released when an electron is added to an isolated gaseous atom in its lowest energy state, forming a mono-negative ion.

A_{(g)} + e^{-} \to A_{(g)}^{-} + E_{1} (energy released)

A_{(g)}^{-} + e^{-} + energy supplied \to A_{(g)}^{2 -} (energy absorbed)

The second electron addition is endothermic because the incoming electron is repelled by the already-negative ion.

4.1 Factors Affecting Electron Affinity

Atomic size: smaller atoms have higher $E A$ ; $E A \propto 1/ r$ .
Effective nuclear charge: higher $Z_{eff}$ → more attraction on the incoming electron → higher $E A$ .
Shielding effect: more inner-shell shielding → lower $E A$ .
Stability of half- and fully-filled configurations: if adding an electron would break a stable configuration, $E A$ is low (or even effectively zero).

4.2 Periodicity in Electron Affinity

Across a period

$E A$ generally increases because $Z_{eff}$ increases and atomic size decreases.

Down a group

$E A$ generally decreases because atomic size increases.

4.3 Important Exceptions

Alkaline earth metals (Group IIA - Be, Mg) have $E A \approx 0$ because they have stable $n s^{2}$ configuration; adding an electron would force it into a higher-energy $p$ -orbital.
Alkali metals (Group IA) also have very low (nearly zero) $E A$ - they prefer to lose their single valence electron, not gain another.
Group 15 (N, P) has $E A$ less than Group 14 (C, Si) because $n s^{2} n p^{3}$ is a stable half-filled configuration in Group 15.
Noble gases: theoretical $E A = 0$ because their $n s^{2} n p^{6}$ octet is already complete.
Cl > F: Chlorine has the highest known electron affinity - higher than fluorine. This is because F is very small and its incoming electron faces strong electron-electron repulsion in the compact $2 p$ orbital.

Solved Example 5

Formation of Cl⁻ is exothermic but formation of O²⁻ is endothermic. Explain.

Solution:

For chlorine, only one electron is added, so:

$C l_{(g)} + e^{-} \to C l_{(g)}^{-}$ ; $Δ H = - E A_{1}$ (energy released, exothermic).

For oxygen, two electrons must be added to reach $O^{2 -}$ :

$O_{(g)} + e^{-} \to O_{(g)}^{-}$ ; $Δ H = - E A_{1}$ (exothermic).

$O_{(g)}^{-} + e^{-} \to O_{(g)}^{2 -}$ ; $Δ H = + E A_{2}$ (endothermic - the incoming electron is repelled by $O^{-}$ ).

Net: $Δ H_{net} = - E A_{1} + E A_{2}$ . Since $E A_{2} ≫ E A_{1}$ , the overall process is endothermic.

Solved Example 6

Why is lithium a more powerful reducing agent than sodium in aqueous medium?

Solution:

Reducing power in solution depends on the standard electrode potential, which involves three steps: sublimation, ionization, and hydration. Li⁺ is a very small ion with a very high charge density, so its hydration enthalpy is exceptionally large and highly exothermic. This overrides the higher ionization energy of Li and makes the overall $M \to M_{(a q)}^{+}$ change more favourable, giving Li the more negative reduction potential - hence stronger reducing power in aqueous medium.

5. Electronegativity

Electronegativity is the tendency of an atom in a covalent bond to attract the shared pair of bonding electrons toward itself. Unlike ionization energy and electron affinity (which are properties of isolated atoms), electronegativity is a property of a bonded atom.

5.1 Electronegativity Scales

Pauling's Scale (1932)

Pauling related the extra bond energy of a heteronuclear bond to the difference in electronegativity between the two atoms:

0.208 Δ_{A B} = x_{A} - x_{B} (if x_{A} > x_{B})

where

Δ_{A B} = E_{A - B} - E_{A - A} \times E_{B - B}

and all bond energies are in

kcal/mol

Mulliken's Scale

Mulliken defined electronegativity as the arithmetic mean of ionization potential and electron affinity of the atom (both in $eV$ ):

x_{M} = \frac{I E + E A}{2}

Mulliken values are

\approx 2.8

times larger than Pauling values on the raw eV scale.

Alfred-Rochow Scale

Electronegativity treated as the electrostatic force of attraction between the nucleus and a valence electron:

x = \frac{0.359 Z _{eff}}{r ^{2}} + 0.744, Z_{eff} = Z - σ

where

r

is the covalent radius in Å and

σ

is the Slater shielding constant.

Figure: Pauling electronegativity scale - F (4.0) is the most electronegative element; Cs (0.7) the least.

5.2 Factors Affecting Electronegativity

Atomic radius: $E N \propto 1/ r$ . Smaller atoms are more electronegative.
Effective nuclear charge: $E N \propto Z_{eff}$ .
Oxidation state: $E N$ of an atom increases as its positive oxidation state increases (higher positive charge pulls bonding electrons harder).
Hybridization ( $s$ -character): More $s$ -character in a hybrid orbital → higher $E N$ (because $s$ -electrons are closer to the nucleus).

Hybridization	$s$ -character	Electronegativity (C)
$s p^{3}$	25 %	2.48
$s p^{2}$	33.3 %	2.75
$s p$	50 %	3.25

5.3 Periodicity in Electronegativity

Across a period, $E N$ increases due to increasing $Z_{eff}$ .
Down a group, $E N$ decreases due to increasing atomic size.
Highest $E N$ : Fluorine (4.0) - highest in the periodic table.
Lowest $E N$ : Caesium (0.7). Fr should be less than Cs but its measured value is slightly higher due to the abnormally large jump of +32 in nuclear charge between Cs and Fr (the intervening $4 f$ and $5 d$ shells contract the outer $7 s$ orbital of Fr).
Noble gases: $E N$ is effectively zero (they don't form bonds; not defined on standard scales).

5.4 Applications of Electronegativity

Partial ionic character in a covalent bond

The ionic character of a covalent bond increases as $Δ x$ between the bonded atoms increases.

% ionic character

= 16 ∣Δ x ∣ + 3.5 (Δ x)^{2}

A $Δ x = 1.7$ gives about 50 % ionic character. Order in H-X bonds: $H F > H C l > H B r > H I$ .

Bond strength

Bond energy of a covalent A-B bond increases with $Δ x$ . Because H-X bond strength decreases in the order $H F > H C l > H B r > H I$ , the acid strength increases in the reverse order: $H I > H B r > H C l > H F$ .

Acidic and basic nature of oxides

Across a period, oxides go from basic (Na₂O) → amphoteric (Al₂O₃) → acidic (Cl₂O₇). As $E N$ of the element rises, the $M$ -O bond becomes more covalent and the oxide more acidic.

Metallic vs non-metallic character

Metallic character decreases as $E N$ increases. Across a period, metals give way to non-metals; down a group, elements become more metallic.

Solved Example 7

Arrange the following molecules in order of decreasing stability given electronegativities: H = 2.1, F = 4.0, Cl = 3.0, Br = 2.8, I = 2.3, N = 3.0. Molecules: HF, NCl₃, HBr, HI, HCl.

Solution:

Compute $∣Δ x ∣$ for each bond (higher $Δ x$ → more polar → stronger → more stable):

Molecule	HF	HCl	HBr	HI	NCl₃
$∣Δ x ∣$	1.9	0.9	0.7	0.2	0.0

Decreasing stability order: HF > HCl > HBr > HI > NCl₃.

Solved Example 8

Which two elements would combine in the most violent fashion?

Solution:

The most violent combination happens between the most electropositive and most electronegative stable elements. That is Cs (EN = 0.7) and F (EN = 4.0), giving an extremely exothermic ionic reaction to form CsF.

6. Some Other Periodic Properties

6.1 Atomic Volume

Atomic volume (atomic weight / density) increases down a group as more shells are added. Across a period, it varies cyclically - decreasing at first, hitting a minimum in the middle, then rising again.

6.2 Density

Density of solid elements varies periodically. Across a period, density first rises, reaches a maximum near the middle of the transition series, and then falls.

Highest density among solids: Osmium (22.6 g/cm³). Highest density among liquids at room temperature: Mercury (13.6 g/cm³).

6.3 Melting and Boiling Points

Melting points show periodic variation. In a period, they rise, reach a peak in the middle (carbon in period 2, silicon in period 3 - network solids), and then drop sharply for the non-metals.

Element	Li	Be	B	C	N	O	F	Ne
m.p. (°C)	181	1277	2030	3727	-210	-219	-220	-249

6.4 Nature of Oxides

Across a period, oxides change from basic (Na₂O) to amphoteric (Al₂O₃) to acidic (Cl₂O₇). Down a group, the basic character of oxides increases.

6.5 Nature of Oxyacids

Across a period from left to right, the strength of oxyacids of non-metals increases. Down a group, oxyacid strength decreases. E.g. HClO₄ > HBrO₄ > HIO₄.

6.6 Nature of Hydrides

Across a period, hydrides change from basic (NaH) to acidic (HCl). Down a group of non-metals, hydride acidity increases while stability decreases:

H F < H C l < H B r < H I

(acid strength increases)

H F > H C l > H B r > H I

(thermal stability decreases)

H F < H C l < H B r < H I

(reducing power increases)

7. Effective Nuclear Charge and Shielding Effect

In a multi-electron atom, the nuclear pull felt by any outer electron is less than the actual nuclear charge, because inner-shell electrons partially cancel the pull. This reduced pull is called the effective nuclear charge, $Z_{eff}$ .

Slater's rule:

Z_{eff} = Z - σ

where

σ

is the shielding (screening) constant, computed from the number and type of inner electrons.

Every periodic trend eventually reduces to a change in $Z_{eff}$ :

Across a period, $Z$ increases but $σ$ hardly changes (same shell) → $Z_{eff}$ increases → atoms shrink, $I E$ and $E N$ rise.
Down a group, $Z$ increases but so does $σ$ (added shells shield strongly) → $Z_{eff}$ is nearly constant → radius grows and $I E, E N$ fall.

8. Diagonal Relationship

Certain second-period elements - Li, Be, and B - closely resemble the third-period elements Mg, Al, and Si respectively, which are one group to the right and one period down.

Figure: Diagonal relationships between Li-Mg, Be-Al, and B-Si; each pair shares similar ionic size and polarizing power.

The diagonal relationship arises because moving right increases

Z_{eff}

(shrinks atoms) while moving down increases size - the two effects roughly cancel along the diagonal. The result: comparable ionic size, electronegativity, and polarizing power.

Diagonal relationships are predominant only in the lighter non-transition elements. Examples of shared behaviour:

Li and Mg: both form nitrides directly with N₂; both form normal oxides (not peroxides); both form covalent alkyl compounds.
Be and Al: both form covalent chlorides that hydrolyze; oxides and hydroxides of both are amphoteric.
B and Si: both form covalent, acidic oxides; both form volatile hydrides.

Solved Example 9

Arrange Na⁺, Mg²⁺ and Al³⁺ ions in order of increasing (a) extent of hydration, (b) hydration energy, (c) size of hydrated ions, (d) ionic mobility, (e) size of ions (bare).

Solution:

All three are isoelectronic (10 e⁻). Bare-ion size decreases with increasing charge and nuclear pull: bare size Al³⁺ $<$ Mg²⁺ $<$ Na⁺.

Smaller, more highly charged cations have higher charge density → attract more water molecules → larger hydration shell:

(a) Extent of hydration: Na⁺ < Mg²⁺ < Al³⁺

(b) Hydration energy: Na⁺ < Mg²⁺ < Al³⁺

(d) Ionic mobility (opposite of hydrated size): Al³⁺ < Mg²⁺ < Na⁺

(e) Size of bare ions: Al³⁺ < Mg²⁺ < Na⁺

Solved Example 10

The hydride of nitrogen (NH₃) is the most stable among the hydrides of Group 15 (V-group) elements. Explain.

Solution:

Down Group 15, the electropositive character of the central atom increases from N to Bi. Nitrogen has the highest electronegativity in the group. In NH₃, the H atoms act as electropositive partners, and the polar N-H bond is strong and short. As we move down (PH₃, AsH₃, SbH₃, BiH₃), the M-H bond becomes longer, weaker, and less polar, so hydride stability decreases: $N H_{3} > P H_{3} > A s H_{3} > S b H_{3} > B i H_{3}$ .

Solved Example 11

Why does BiCl₅ not exist?

Solution:

Bismuth (Group 15, 6th period) shows the strongest inert-pair effect among Group 15. The $6 s^{2}$ pair is reluctant to participate in bonding because of poor shielding by the intervening $4 f^{14}$ and $5 d^{10}$ electrons - the pair becomes energetically "inert". So Bi prefers the $+ 3$ oxidation state (using only three $6 p$ electrons) rather than the $+ 5$ state which would require the $6 s^{2}$ pair too. Hence BiCl₃ exists but BiCl₅ does not.

Common Mistakes to Avoid

Watch out

Assuming F has the highest electron affinity. Fluorine has the highest electronegativity (4.0), but chlorine has a higher electron affinity. F is so small that inter-electron repulsion in its compact $2 p$ shell reduces the energy released on adding an electron.
Confusing "ionization energy decreases across period" with "increases". Across a period, IE increases (except for the two dips at IIA→IIIA and VA→VIA). Down a group, IE decreases.
Forgetting the two IE exceptions. $I E (Be) > I E (B)$ and $I E (N) > I E (O)$ because Be has stable $2 s^{2}$ and N has stable $2 p^{3}$ half-filled. Missing these costs marks in every NEET/JEE paper.
Confusing electron affinity sign conventions. $E A$ is defined as energy released - a positive $E A$ means an exothermic addition. But $Δ_{e g} H$ (electron gain enthalpy) is negative for exothermic. They describe the same thing with opposite signs.
Applying $Z / e$ ratio without care. The ratio works for isoelectronic-style comparisons but not for atoms of very different shell counts. Always check the electron configurations first.
Treating "atomic radius" as one number. The "radius" of a noble gas is a van der Waals radius; for a metal it's metallic; for a halogen it's covalent. Comparing across types can flip the apparent trend (that's why Ar > Cl).
Ignoring lanthanide contraction in group trends. The near-equality of 4d and 5d radii (Zr ≈ Hf, Nb ≈ Ta) is not a coincidence - it's caused by the $4 f^{14}$ contraction. This also explains anomalies in Tl and Pb IE values.
Assuming diagonal relationship applies to all groups. It is meaningful only for Li-Mg, Be-Al, and B-Si - the lighter non-transition elements at the top-left of the table.

Frequently Asked Questions

Q1. Why does atomic radius decrease across a period?

Across a period, each successive element has one more proton and one more electron, but the electron is added to the same principal shell. Effective nuclear charge on the outer electrons therefore increases while shielding stays roughly constant. The nucleus pulls the electron cloud tighter, so the atom shrinks.

Q2. Why is the ionization energy of nitrogen higher than that of oxygen?

Nitrogen has the configuration $2 s^{2} 2 p^{3}$ - a stable half-filled $p$ sub-shell. Oxygen has $2 s^{2} 2 p^{4}$ ; its extra electron enters a $p$ orbital that already has one electron, creating some inter-electron repulsion. Removing that electron from oxygen restores the stable half-filled state, so oxygen loses its outermost electron more easily than nitrogen. Hence $I E (N) > I E (O)$ .

Q3. Why does chlorine have a higher electron affinity than fluorine?

Fluorine is very small, so its $2 p$ subshell is highly compressed. When an additional electron is added, it experiences strong repulsion from the seven already-crowded valence electrons. In chlorine, the $3 p$ orbitals are larger and more diffuse, so the incoming electron encounters less repulsion. The net energy released is greater for chlorine, giving it a higher electron affinity than fluorine.

Q4. What is lanthanide contraction and why does it matter?

Lanthanide contraction is the steady decrease in atomic and ionic radii across the lanthanide series (La to Lu) as the $4 f$ subshell fills. Because $f$ -electrons shield poorly, the outer electrons feel almost the full extra nuclear charge and the atom shrinks with each added electron. Its main consequence is that 4d and 5d transition metals in the same group have nearly identical radii (Zr ≈ Hf, Nb ≈ Ta), giving them very similar chemistry.

Q5. Why is a cation smaller than its parent atom?

Losing electrons decreases inter-electron repulsion and, more importantly, increases the effective nuclear charge on each remaining electron. Cation formation often removes an entire outer shell, so the ion has one fewer shell than the parent atom. Both effects contract the electron cloud, making the cation significantly smaller.

Q6. Which element has the highest electronegativity and why?

Fluorine has the highest electronegativity on the Pauling scale (4.0). It is small, has a high effective nuclear charge, and its incoming electrons enter the compact $2 p$ orbital very close to the nucleus. All three factors combine to give fluorine an unmatched pull on shared electrons.

Q7. What is the diagonal relationship?

Diagonal relationship is the striking similarity in chemistry between certain second-period and third-period elements one group apart: Li-Mg, Be-Al, and B-Si. It arises because moving right (increasing $Z_{eff}$ ) and moving down (increasing size) tend to cancel along the diagonal, producing similar ionic size, electronegativity, and polarizing power.

Q8. What is effective nuclear charge?

Effective nuclear charge $Z_{eff}$ is the net positive charge experienced by a specific electron in a multi-electron atom, after accounting for shielding by other electrons. Slater's formula $Z_{eff} = Z - σ$ gives a practical estimate. Most periodic trends - atomic radius, IE, EA, EN - track $Z_{eff}$ directly.

Q9. Why does the second ionization energy of sodium jump so sharply above the first?

Sodium is $[Ne] 3 s^{1}$ . Removing the single $3 s$ electron leaves the very stable [Ne] noble-gas core. Removing a second electron requires breaking into that stable octet - a much larger energy cost. That is why $I E_{2}$ of Na is nearly ten times $I E_{1}$ . The size of the jump is a fingerprint of where the valence-to-core boundary lies.

Q10. What are isoelectronic species and how does their size vary?

Isoelectronic species have the same number of electrons but different nuclear charges. For example $N^{3 -}, O^{2 -}, F^{-}, N e, N a^{+}, M g^{2 +}$ all have 10 electrons. As the nuclear charge increases from 7 (N) to 12 (Mg), the same electron cloud is pulled more strongly, so ionic radius steadily decreases across the series.

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