NCERT Plus One (Class 11) Chemistry • Chapter 3

Classification of Elements and Periodicity in Properties

Q: Explain why (a) the first ionization enthalpy of Beryllium is higher than that of Boron, and (b) the first ionization enthalpy of Nitrogen is higher than that of Oxygen.

(a) Be > B: Be has configuration 1s 2 2s 2 and B has 1s 2 2s 2 2p 1 . A 2s-electron penetrates closer to the nucleus than a 2p-electron; thus the 2p-electron of Boron is more effectively shielded by inner 2s electrons and requires less energy to remove. (b) N > O: Nitrogen (2p x 1 2p y 1 2p z 1 ) has a stable half-filled 2p subshell with one electron in each 2p orbital (Hund's rule). Oxygen (2p 4 ) has two electrons paired in the same 2p orbital, causing increased inter-electronic repulsion , so removing the 4 th 2p electron is easier.

Q: Chlorine has a more negative electron gain enthalpy than Fluorine. Why?

In Fluorine (n = 2), the incoming electron enters a very small, compact 2p subshell and suffers significant inter-electronic repulsion from the other electrons already crowded there. In Chlorine (n = 3), the added electron enters a larger, roomier 3p subshell where electron-electron repulsion is much lower. Hence Cl releases more energy (Δ eg H = -349 kJ mol -1 ) than F (-328 kJ mol -1 ).

Q: Write the IUPAC systematic name and symbol for the element with atomic number (a) 120 and (b) 117.

(a) Z = 120 → 1 (un) + 2 (bi) + 0 (nil) + ium = Unbinilium , Symbol: Ubn . (b) Z = 117 → 1 (un) + 1 (un) + 7 (sept) + ium = Ununseptium , Symbol: Uus (official IUPAC name: Tennessine, Ts).

Q: What are isoelectronic species? Arrange N 3- , O 2- , F - , Na + , Mg 2+ , and Al 3+ in increasing order of ionic radii.

Isoelectronic species: Atoms or ions containing the same number of electrons (here, 10 electrons each). Increasing ionic radius (larger nuclear charge Z ⇒ smaller radius): Al 3+ < Mg 2+ < Na + < F - < O 2- < N 3- .

Q: State the Modern Periodic Law. How did Henry Moseley prove that atomic number is more fundamental than atomic mass?

Modern Periodic Law: "The physical and chemical properties of the elements are periodic functions of their atomic numbers." Moseley's Proof (1913): When Moseley studied characteristic X-ray spectra of elements, a plot of √ν (square root of X-ray frequency) against atomic number (Z) gave a straight line, whereas plotting √ν against atomic mass did not.

Q: What is meant by diagonal relationship in the periodic table? Give two pairs and mention two reasons for the anomalous behaviour of second-period elements.

Diagonal Relationship: Similarity in chemical properties shown by a second-period element with the diagonally adjacent third-period element of the next group. Pairs: Li & Mg , Be & Al . Reasons for anomalous behaviour of Period-2 elements: (1) Very small atomic/ionic size and high charge/radius ratio; (2) High electronegativity and absence of vacant d-orbitals (only 4 valence orbitals: 2s and 2p, capping maximum covalency at 4).

Q: Why is a cation always smaller and an anion always larger than its parent atom? Also explain why the second ionization enthalpy (Δ i H 2 ) of Na is much higher than that of Mg.

Cation Parent Atom: Gain of electron(s) increases inter-electronic repulsion and lowers effective nuclear charge, expanding the cloud (e.g., F 64 pm → F - 136 pm). Why Δ i H 2 (Na) >> Δ i H 2 (Mg): After losing 1 e - , Na + attains a highly stable, closed noble gas configuration [Ne] (1s 2 2s 2 2p 6 ) , requiring huge energy to remove a 2 nd electron. Mg + ([Ne]3s 1 ) still has a single outer 3s electron whose removal gives stable Mg 2+ ([Ne]).

Q: Define Electron Gain Enthalpy. Why is the second electron gain enthalpy of oxygen positive, and why do noble gases have large positive electron gain enthalpies?

Electron Gain Enthalpy (Δ eg H): The enthalpy change accompanying the addition of an electron to a neutral isolated gaseous atom: X(g) + e - → X - (g). Why Δ eg H 2 of O is positive: Adding a second electron to an already negatively charged O - (g) ion faces strong electrostatic repulsion , so external energy must be supplied (endothermic, +780 kJ mol -1 ). Noble Gases: Their valence shells are completely filled (ns 2 np 6 ), so an added electron must enter the next higher principal energy level , forming a very unstable state.

Genesis of Periodic Table, Modern Periodic Law, IUPAC Z>100 Naming, Blocks & Periodic Trends

📄 Official NCERT Textbook PDF

Topic 3.1 – 3.2: Genesis of Periodic Classification: Dobereiner, Newlands & Mendeleev

Walk into a giant supermarket with 118 different spices scattered randomly on the floor—finding turmeric or pepper would take hours! In 1800, chemists knew only 31 elements; by 1865, that jumped to 63, and today we have 118. Just like arranging books by genre or groceries by aisle, chemists hunted for a hidden pattern. First came triplets, then musical octaves, and finally Dmitri Mendeleev—who famously arranged element cards like a game of solitaire and boldly left blank spaces for elements nobody had even discovered yet!

Classification organizes the properties of elements and their innumerable compounds systematically, rationalizing known facts and predicting undiscovered elements.

1. Early Pioneers: Dobereiner & Newlands

Dobereiner's Law of Triads (1829): Johann Dobereiner grouped elements in threes (triads) with similar properties where the middle element's atomic weight was roughly the average of the other two:
- Li (7), Na (23), K (39) → (7 + 39) / 2 = 23
- Ca (40), Sr (88), Ba (137) → (40 + 137) / 2 = 88.5
- Cl (35.5), Br (80), I (127) → (35.5 + 127) / 2 = 81.25
Limitation: Worked only for a few elements and was dismissed as coincidence. (In 1862, A.E.B. de Chancourtois also arranged elements on a cylindrical helix.)
Newlands' Law of Octaves (1865): John Alexander Newlands arranged elements by increasing atomic weight and noticed every eighth element had properties similar to the first—just like the 8th note in a musical octave (Sa, Re, Ga, Ma, Pa, Dha, Ni, Sa). Limitation: Held true only up to Calcium (Ca)! (Awarded the Davy Medal in 1887.)

2. Lothar Meyer & Mendeleev's Periodic Law (1869)

Independently in 1869, German chemist Lothar Meyer (who plotted atomic volume, melting point & boiling point vs atomic weight) and Russian chemist Dmitri Mendeleev discovered periodic repetition. Mendeleev first published the Periodic Law:

Mendeleev's Periodic Law: "The properties of the elements are a periodic function of their atomic weights."

Mendeleev prioritized similar chemical properties (especially empirical formulas of oxides and hydrides) over strict atomic weight order. Thus he placed Tellurium (Te, 127.6) before Iodine (I, 126.9) so Iodine sat with F, Cl, Br in Group VII. Most famously, he left gaps for undiscovered elements and predicted their properties accurately:

Property	Eka-aluminium (Predicted)	Gallium (Found)	Eka-silicon (Predicted)	Germanium (Found)
Atomic weight	68	70	72	72.6
Density (g cm^-3)	5.9	5.94	5.5	5.36
Melting point	Low	302.93 K	High	1231 K
Oxide formula	E₂O₃	Ga₂O₃	EO₂	GeO₂
Chloride formula	ECl₃	GaCl₃	ECl₄	GeCl₄

Exams love asking: "Did Mendeleev strictly stick to increasing atomic weight?" NO! He inverted pairs like Te and I so elements with matching chemical properties stayed in the same group. Also memorize: Eka-aluminium = Gallium (Ga), Eka-silicon = Germanium (Ge), and Newlands' Law of Octaves works only up to Calcium (Ca).

Topic 3.3: Modern Periodic Law (Moseley) & Structure of the Long Form Periodic Table

Mendeleev arranged elements by weight, yet nagging puzzles remained—why did Argon (39.9 u) sit before Potassium (39.1 u), and Cobalt before Nickel? In 1913, a brilliant 25-year-old British physicist, Henry Moseley, fired electrons at metal targets and measured their X-rays. When he plotted the square root of X-ray frequency against Atomic Number (Z), every kink vanished into a ruler-straight line—proving that an atom's true ID card is its proton count, not its weight!

1. Moseley's Experiment & Modern Periodic Law (1913)

Henry Moseley observed regularities in characteristic X-ray spectra of elements. Plotting √ν (square root of frequency of X-rays emitted) against atomic number (Z) yielded a straight line, whereas plotting √ν against atomic mass did not. This established atomic number (Z = number of protons = nuclear charge) as the fundamental property.

Modern Periodic Law: "The physical and chemical properties of the elements are periodic functions of their atomic numbers."

Periodicity arises from the periodic repetition of similar outermost electronic configurations at regular intervals of atomic number.

2. The Long Form of the Periodic Table (Fig. 3.2)

18 Vertical Columns (Groups / Families): Numbered 1 to 18 by IUPAC (replacing old IA–VIIA, VIII, IB–VIIB and 0). Elements in a group share similar valence electronic configurations and chemical properties.
7 Horizontal Rows (Periods): Period number equals the highest principal quantum number (n) of the valence shell.
Bottom Panels (Glenn T. Seaborg): 14 Lanthanoids (4f series, Period 6) and 14 Actinoids (5f series, Period 7) placed separately below to keep the table compact.

Period (n)	Subshells Filled	Total Orbitals	No. of Elements	Range (Z)
1	1s	1	2 (Very short)	1 (H) – 2 (He)
2	2s, 2p	1 + 3 = 4	8 (Short)	3 (Li) – 10 (Ne)
3	3s, 3p	1 + 3 = 4	8 (Short)	11 (Na) – 18 (Ar)
4	4s, 3d, 4p	1 + 5 + 3 = 9	18 (Long)	19 (K) – 36 (Kr)
5	5s, 4d, 5p	1 + 5 + 3 = 9	18 (Long)	37 (Rb) – 54 (Xe)
6	6s, 4f, 5d, 6p	1 + 7 + 5 + 3 = 16	32 (Very long)	55 (Cs) – 86 (Rn)
7	7s, 5f, 6d, 7p	1 + 7 + 5 + 3 = 16	32 (Complete up to Og)	87 (Fr) – 118 (Og)

NCERT Problem 3.2: How would you justify the presence of 18 elements in the 5^th period?
Solution: For n = 5, energy order of available orbitals is 5s < 4d < 5p. Total orbitals = 1 (5s) + 5 (4d) + 3 (5p) = 9 orbitals. Each orbital holds at most 2 electrons, accommodating 9 × 2 = 18 electrons, so the 5^th period has 18 elements.

Two classic MCQ traps: (1) Moseley got a straight line by plotting √ν vs Z (square root of frequency vs atomic number), NOT ν vs Z! (2) Why does the 5^th period have 18 elements instead of 32? Because the 5f subshell has higher energy than 6s and does not fill until the 7^th period!

Topic 3.4: IUPAC Systematic Nomenclature for Elements with Atomic Numbers > 100

In the 1960s, laboratories in both the USA and the Soviet Union synthesized tiny, fleeting atoms of element 104—lasting fractions of a second! The Americans named it Rutherfordium, while the Soviets insisted on Kurchatovium. To stop Cold War naming fights before discoveries were independently verified, IUPAC invented an unbiased digital code: convert each digit of Z into Latin/Greek roots and tack on -ium!

Superheavy elements (Z > 100) are unstable and synthesized in minute quantities, causing competing priority claims. Until IUPAC ratifies an official permanent name, every element with Z > 100 receives a temporary systematic name directly from its atomic number digits:

Digit	0	1	2	3	4	5	6	7	8	9
Root	nil	un	bi	tri	quad	pent	hex	sept	oct	enn
Symbol letter	n	u	b	t	q	p	h	s	o	e

Naming Rule: Write digit roots in order + suffix "-ium". Three-letter symbol takes the first letter of each root (First letter Capital, next two lowercase).

NCERT Problem 3.1: What would be the IUPAC name and symbol for the element with atomic number 120?
Solution: Digits are 1 (un), 2 (bi), 0 (nil).
Name = Unbinilium | Symbol = Ubn.

Selected Key Elements from NCERT Table 3.5

Z	Systematic IUPAC Name	Sys. Symbol	Official Permanent IUPAC Name	Official Symbol
101	Unnilunium	Unu	Mendelevium	Md
104	Unnilquadium	Unq	Rutherfordium	Rf
106	Unnilhexium	Unh	Seaborgium (named after Glenn Seaborg)	Sg
109	Unnilennium	Une	Meitnerium	Mt
114	Ununquadium	Uuq	Flerovium (Period 7, Group 14)	Fl
117	Ununseptium	Uus	Tennessine (Period 7, Group 17)	Ts
118	Ununoctium	Uuo	Oganesson (Period 7, Group 18)	Og

Watch three spelling & symbol traps! (1) 0 = nil (n) and 9 = enn (e)—never write "non" or "zero". (2) When bi (2) or tri (3) precedes -ium, drop one 'i' so you get Unnilbium (102) and Unniltrium (103), NOT "biium" or "triium"! (3) In the 3-letter symbol, ONLY the first letter is capital (Ubn, not UBN).

Topic 3.5 – 3.6: Electronic Configurations, s, p, d, f Blocks & Metals, Non-metals, Metalloids

Look at the shape of the Modern Periodic Table—why does it look like a castle with two tall towers on the left, a wide six-column apartment on the right, a ten-column bridge in the middle, and two detached blocks floating below? That architecture wasn't drawn by an artist—it is literally the blueprint of quantum orbitals (s, p, d, f) holding 2, 6, 10, and 14 electrons!

Elements are grouped into four blocks according to the atomic orbital (subshell) receiving the last differentiating electron:

Block	Groups	General Valence Config	Salient Characteristics
s-Block	1 (Alkali) & 2 (Alkaline earth)	ns^1–2	Reactive metals, low Δ_iH, form +1 / +2 cations, predominantly ionic compounds (except Li, Be); never found pure in nature.
p-Block	13 to 18	ns² np^1–6	Together with s-block = Representative / Main Group Elements. Includes Noble gases (ns²np⁶), Halogens (Gr 17), Chalcogens (Gr 16).
d-Block (Transition)	3 to 12	(n-1)d^1–10 ns^0–2 (Pd = 4d¹⁰5s⁰)	All metals; form coloured ions, show variable valency, paramagnetism, catalytic activity. Zn, Cd, Hg (d¹⁰s²) are non-typical transition elements.
f-Block (Inner-Transition)	Bottom 2 rows	(n-2)f^1–14 (n-1)d^0–1 ns²	Lanthanoids (Ce 58–Lu 71, 4f) & Actinoids (Th 90–Lr 103, 5f, radioactive; Z > 92 = Transuranium).

Two Special Block Exceptions & Broad Categories

Helium (1s²): Strictly an s-electron element, yet placed in Group 18 (p-block column) because its valence shell is completely filled and it behaves like a noble gas.
Hydrogen (1s¹): Resembles both Group 1 (loses 1e^- → H⁺) and Group 17 (gains 1e^- → H^-), so placed separately at the top.
Metals (>78% of elements, left/centre): High m.p./b.p. (exceptions: Hg liquid; Ga 303 K & Cs 302 K low m.p.), malleable, ductile, conductors.
Non-metals (<20, top-right): Low m.p./b.p. (except B & C), brittle solids/gases, poor conductors.
Metalloids / Semi-metals: Elements bordering the zig-zag line showing both metallic & non-metallic traits: Si, Ge, As, Sb, Te.

NCERT Problem 3.3 & 3.4:
• Z = 117 → [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁵ → Group 17 (Halogens), Period 7, p-block.
• Z = 120 → [Og] 8s² → Group 2 (Alkaline earth metals), Period 8, s-block.
• Increasing metallic character of Si, Be, Mg, Na, P: P < Si < Be < Mg < Na (increases down a group, decreases L→R).

Memorize these three frequent exam traps: (1) General d-block configuration allows ns^0–2 specifically because Palladium (Pd, Z=46) is 4d¹⁰5s⁰! (2) Zn, Cd, Hg belong to the d-block, yet have completely filled d¹⁰ shells so they do NOT show typical transition metal properties. (3) The metalloid zig-zag includes Si, Ge, As, Sb, Te (not Al).

Topic 3.7.1 (a)–(b): Periodic Trends I: Atomic Radius, Ionic Radius & Isoelectronic Series

Picture holding a handful of ten identical marbles—first inside a giant balloon pulled softly from the center by 7 people (N^3-), and then inside an elastic net yanked furiously inward by 13 people (Al³⁺)! Even though both systems hold the exact same 10 electrons, the extra protons in Aluminium pull that electron cloud into less than a third of the radius. That single tug-of-war explains every atomic and ionic size trend!

Since atoms are tiny (~1.2 Å) and electron clouds have no sharp boundary, atomic size is measured from internuclear distances in combined states:

Covalent Radius (Non-metals): Half the single-bond distance in a homonuclear molecule (e.g., Cl₂ bond length = 198 pm ⇒ r_Cl = 99 pm).
Metallic Radius (Metals): Half the internuclear distance between adjacent metal cores in a crystal (e.g., Cu–Cu = 256 pm ⇒ r_Cu = 128 pm).
Noble Gases (van der Waals Radius): Monoatomic noble gases have large non-bonded van der Waals radii and are excluded from standard covalent radius period comparisons.

1. Trends in Atomic Radius

Across a Period (Left → Right): Decreases (e.g., Li 152 > Be 111 > B 88 > C 77 > N 74 > O 66 > F 64 pm). Added electrons enter the same valence shell while nuclear charge increases, boosting effective nuclear charge (Z_eff) and pulling electrons closer.
Down a Group (Top → Bottom): Increases (e.g., Li 152 < Na 186 < K 231 < Rb 244 < Cs 262 pm; F 64 < Cl 99 < Br 114 < I 133 pm). Principal quantum number n increases and inner filled shells shield the outer electrons.

2. Ionic Radius & Isoelectronic Species

Cation < Parent Atom: Losing electrons leaves fewer electrons held by the same nuclear charge (e.g., Na = 186 pm, Na⁺ = 95 pm).
Anion > Parent Atom: Gaining electrons increases inter-electronic repulsion and reduces Z_eff (e.g., F = 64 pm, F^- = 136 pm).
Isoelectronic Species: Atoms/ions having the same number of electrons (e.g., 10 e^- in N^3-, O^2-, F^-, Ne, Na⁺, Mg²⁺, Al³⁺). Size depends purely on nuclear charge (Z): higher positive nuclear charge ⇒ smaller radius!

Isoelectronic Radius Order (10 e^-): Al³⁺ (Z=13) < Mg²⁺ (Z=12) < Na⁺ (Z=11) < F^- (Z=9) < O^2- (Z=8) < N^3- (Z=7)

NCERT Problem 3.5: Which of Mg, Mg²⁺, Al, Al³⁺ has the largest and smallest size?
Solution: Across Period 3, Mg > Al. Cations are smaller than parent atoms, and among isoelectronic 10-e^- ions Mg²⁺ > Al³⁺. Hence Largest = Mg, Smallest = Al³⁺.

Watch out for two traps! (1) In Period 2, atomic size decreases from Li to F, then suddenly jumps at Ne if you quote noble gas radii—because Neon's value is a non-bonded van der Waals radius, not covalent! (2) When sorting an isoelectronic series like N^3-, O^2-, F^-, Na⁺, Mg²⁺, Al³⁺, check whether the question asks for increasing (Al³⁺ first) or decreasing (N^3- first) order!

Topic 3.7.1 (c): Periodic Trends II: Ionization Enthalpy (ΔiH) & Period-2 Anomalies (Be > B, N > O)

Removing the first electron from an atom is like pulling a book from a shelf. As you walk from Lithium to Neon, the librarian holds every book tighter and tighter—so you expect a smooth uphill climb. Yet wait! At Beryllium to Boron and Nitrogen to Oxygen, the energy graph unexpectedly dips! Why is it easier to steal an electron from Boron than Beryllium, and from Oxygen than Nitrogen? Quantum orbital shapes and electron crowding hold the secret!

Ionization Enthalpy (Δ_iH) is the energy required to remove an electron from an isolated gaseous atom in its ground state (unit: kJ mol^-1). It is always positive (endothermic).

First (Δ_iH₁): X(g) → X⁺(g) + e^- | Second (Δ_iH₂): X⁺(g) → X²⁺(g) + e^- | Always: Δ_iH₁ < Δ_iH₂ < Δ_iH₃

1. General Periodic Trends & Shielding

Across a Period (L → R): Generally increases. Electrons enter the same principal shell; increasing nuclear charge outweighs weak inner-core shielding. Maxima occur at closed-shell Noble Gases; Minima occur at reactive Alkali Metals.
Down a Group: Decreases. Atomic radius and shielding by inner shells outweigh increasing nuclear charge.
Successive Jump: Removing an electron from a positive ion is harder than from a neutral atom. A massive jump (e.g., Na Δ_iH₁ = 496 vs Δ_iH₂ = 4562 kJ mol^-1) marks breaking into a stable noble gas core!

2. Two Famous Period-2 Anomalies (NCERT Ex 3.16!)

Anomaly 1: Be (899 kJ mol^-1) > B (801 kJ mol^-1): In Beryllium (1s²2s²), the electron is removed from a 2s orbital, which penetrates closer to the nucleus. In Boron (1s²2s²2p¹), the electron sits in a 2p orbital that is better shielded by 2s electrons and held less tightly! (Same reason: Mg > Al in Period 3!)
Anomaly 2: N (1402 kJ mol^-1) > O (1314 kJ mol^-1): Nitrogen (1s²2s²2p_x¹2p_y¹2p_z¹) has three unpaired 2p electrons in separate orbitals (stable half-filled subshell, Hund's rule). Oxygen (2p⁴) forces two electrons to pair inside the same 2p orbital, creating extra electron-electron repulsion that lowers the energy needed to knock out the 4^th electron! (Same reason: P > S in Period 3!)

Actual Period-2 Δ_iH₁ Order: Li (520) < B (801) < Be (899) < C (1086) < O (1314) < N (1402) < F (1681) < Ne (2080)

NCERT Problem 3.6: First Δ_iH of Na, Mg, Si are 496, 737, 786 kJ mol^-1. Will Al be closer to 575 or 760?
Solution: Closer to 575 kJ mol^-1 (actual: 577) because Al's outer 3p electron is effectively shielded by 3s² electrons, making Al lower than Mg (737).

NEVER rank Period-2 ionization enthalpy strictly left-to-right! Always swap the two famous pairs: B < Be and O < N. Also watch Group 13 down-the-group (NCERT Ex 3.19): Ga (579) > Al (577) slightly because Ga has poor-shielding 3d¹⁰ electrons raising its effective nuclear charge!

Topic 3.7.1 (d)–(e): Periodic Trends III: Electron Gain Enthalpy (Cl vs F) & Electronegativity

Fluorine is the smallest halogen and the undisputed bully when sharing electrons—yet when you drop a brand-new electron onto a neutral gas atom, Chlorine actually releases MORE energy than Fluorine! Why? Imagine trying to squeeze one more passenger onto a packed mini-auto (Fluorine's tiny 2p shell) versus stepping into a spacious bus (Chlorine's roomier 3p shell). Crowding pushes back!

1. Electron Gain Enthalpy (Δ_egH)

Enthalpy change when an electron is added to a neutral gaseous atom: X(g) + e^- → X^-(g). Depending on the atom, it may be exothermic (Δ_egH < 0) or endothermic (Δ_egH > 0).

Halogens (Group 17): Have highly negative Δ_egH because gaining 1 e^- completes a stable noble gas configuration.
Noble Gases (Group 18): Have large POSITIVE Δ_egH (He +48, Ne +116, Ar +96 kJ mol^-1) because the added electron must enter the next higher principal shell!
General Trends: Becomes more negative L → R across a period (smaller size, higher Z_eff) and less negative down a group.

Key Anomaly — Why is Δ_egH of Cl (-349) more negative than F (-328 kJ mol^-1)? (And S -200 more negative than O -141?)
In F (or O), the added electron enters the compact n = 2 (2p) quantum level and suffers strong inter-electronic repulsion from electrons already crowded there. In Cl (or S), the electron enters the larger n = 3 (3p) level with much less repulsion!

NCERT Problem 3.7: Among P, S, Cl, F, which has the most negative and least negative Δ_egH?
Solution: Most negative = Cl (-349 kJ mol^-1) (beats F due to lower 3p repulsion). Least negative = P (leftmost in Period 3 here, plus stable half-filled 3p³).

2. Electronegativity (EN)

A qualitative measure of the ability of an atom in a chemical compound to attract shared electrons toward itself. Unlike Δ_iH and Δ_egH, electronegativity is NOT directly measurable and varies slightly with the bonded partner.

Period 2	Li (1.0)	Be (1.5)	B (2.0)	C (2.5)	N (3.0)	O (3.5)	F (4.0)
Period 3	Na (0.9)	Mg (1.2)	Al (1.5)	Si (1.8)	P (2.1)	S (2.5)	Cl (3.0)

Pauling Scale assigns maximum 4.0 to Fluorine. Electronegativity increases across a period (correlating with rising non-metallic character) and decreases down a group (metallic character rises; e.g., Cs = 0.7).

NEVER confuse these three traps! (1) Most electronegative element = Fluorine (4.0), whereas most negative electron gain enthalpy = Chlorine (-349 kJ mol^-1)! (2) Second electron gain enthalpy (O^-(g) + e^- → O^2-(g)) is ALWAYS POSITIVE because electrostatic repulsion opposes adding an e^- to an already negative ion! (3) Note sign convention: Electron Affinity A_e = -Δ_egH (at 0 K).

Topic 3.7.2 – 3.7.3: Valency, Oxidation States, Diagonal Relationship & Reactivity of Oxides

Why do Lithium and Magnesium—sitting diagonally apart like chess knights—act almost like twins, while Lithium differs sharply from Sodium right below it? And why does sodium oxide turn red litmus blue while chlorine heptoxide turns blue litmus red? Everything connects back to atomic size, orbital availability, and how strongly elements hold or give away electrons!

1. Periodicity of Valence & Oxidation States

For representative elements, valence is usually equal to the number of outermost electrons or (8 − valence electrons):

Group	1	2	13	14	15	16	17	18
Valence e^-	1	2	3	4	5	6	7	8
Valence	1	2	3	4	3, 5	2, 6	1, 7	0, 8
Hydride / Oxide	LiH / Li₂O	CaH₂ / MgO	B₂H₆ / Al₂O₃	CH₄ / CO₂	NH₃ / N₂O₅	H₂O / SO₃	HF / Cl₂O₇	—

Oxidation state is the apparent charge assigned using electronegativity: in OF₂ (EN: F > O), F = -1 and O = +2; in Na₂O (EN: O > Na), O = -2 and Na = +1.

NCERT Problem 3.8 & 3.9:
• Compound formulas: Si (Gr 14, val 4) + Br (Gr 17, val 1) → SiBr₄; Al (val 3) + S (val 2) → Al₂S₃.
• In [AlCl(H₂O)₅]²⁺, Al has Oxidation State = +3, whereas its Covalency = 6 (bonded to 1 Cl^- and 5 H₂O ligands).

2. Anomalous Properties of Second Period Elements & Diagonal Relationship

The first member of Groups 1, 2, and 13–17 (Li to F) differs from heavier group members because of: (i) small size, (ii) large charge/radius ratio, (iii) high electronegativity, and (iv) only 4 valence orbitals (2s, 2p) (no 2d orbitals!).

Maximum Covalency = 4: Boron forms only [BF₄]^-, whereas Period-3 Aluminium (with 3s, 3p, 3d = 9 orbitals) expands its shell to form [AlF₆]^3-!
pπ–pπ Multiple Bonds: Small 2^nd-period atoms (C, N, O) readily form strong C=C, C≡C, N≡N, C=O, C=N bonds.
Diagonal Relationship: Li resembles Mg and Be resembles Al (diagonally adjacent pairs) because their ionic radii / charge-to-radius ratios and electronegativities match closely (e.g., Li and Mg form covalent halides/organometallics unlike other alkali metals).

3. Chemical Reactivity & Nature of Oxides (3.7.3)

Reactivity Extremes: Highest at left (Group 1 metals lose e^- easily; increases down group: Li < Na < K < Rb < Cs) and extreme right (Group 17 halogens gain e^- easily; decreases down group: F > Cl > Br > I); lowest in the middle.
Oxides Across Period 3 (NCERT Problem 3.10): Leftmost Basic (Na₂O + H₂O → 2NaOH) → Middle Amphoteric (Al₂O₃, As₂O₃) or Neutral (CO, NO, N₂O) → Rightmost Strongly Acidic (Cl₂O₇ + H₂O → 2HClO₄).

Watch three high-yield exam points: (1) Oxidation state and covalency are NOT always identical—in [AlCl(H₂O)₅]²⁺, oxidation state of Al is +3 while covalency is 6! (2) Second-period elements CANNOT exceed covalency 4 ([BF₆]^3- does not exist). (3) Know neutral oxides (CO, NO, N₂O) vs amphoteric oxides (Al₂O₃, As₂O₃)!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March (Repeated Favorite) • 2 Marks): Explain why (a) the first ionization enthalpy of Beryllium is higher than that of Boron, and (b) the first ionization enthalpy of Nitrogen is higher than that of Oxygen.

(a) Be > B: Be has configuration 1s²2s² and B has 1s²2s²2p¹. A 2s-electron penetrates closer to the nucleus than a 2p-electron; thus the 2p-electron of Boron is more effectively shielded by inner 2s electrons and requires less energy to remove.

(b) N > O: Nitrogen (2p_x¹2p_y¹2p_z¹) has a stable half-filled 2p subshell with one electron in each 2p orbital (Hund's rule). Oxygen (2p⁴) has two electrons paired in the same 2p orbital, causing increased inter-electronic repulsion, so removing the 4^th 2p electron is easier.

Q2 (Kerala HSE SAY / Model (High Yield) • 2 Marks): Chlorine has a more negative electron gain enthalpy than Fluorine. Why?

In Fluorine (n = 2), the incoming electron enters a very small, compact 2p subshell and suffers significant inter-electronic repulsion from the other electrons already crowded there.

In Chlorine (n = 3), the added electron enters a larger, roomier 3p subshell where electron-electron repulsion is much lower. Hence Cl releases more energy (Δ_egH = -349 kJ mol^-1) than F (-328 kJ mol^-1).

Q3 (Kerala HSE March • 1 Marks): Write the IUPAC systematic name and symbol for the element with atomic number (a) 120 and (b) 117.

(a) Z = 120 → 1 (un) + 2 (bi) + 0 (nil) + ium = Unbinilium, Symbol: Ubn.

(b) Z = 117 → 1 (un) + 1 (un) + 7 (sept) + ium = Ununseptium, Symbol: Uus (official IUPAC name: Tennessine, Ts).

Q4 (Kerala HSE Model / SAY • 1 Marks): What are isoelectronic species? Arrange N^3-, O^2-, F^-, Na⁺, Mg²⁺, and Al³⁺ in increasing order of ionic radii.

Isoelectronic species: Atoms or ions containing the same number of electrons (here, 10 electrons each).

Increasing ionic radius (larger nuclear charge Z ⇒ smaller radius):
Al³⁺ < Mg²⁺ < Na⁺ < F^- < O^2- < N^3-.

Q5 (Kerala HSE March • 2 Marks): State the Modern Periodic Law. How did Henry Moseley prove that atomic number is more fundamental than atomic mass?

Modern Periodic Law: "The physical and chemical properties of the elements are periodic functions of their atomic numbers."

Moseley's Proof (1913): When Moseley studied characteristic X-ray spectra of elements, a plot of √ν (square root of X-ray frequency) against atomic number (Z) gave a straight line, whereas plotting √ν against atomic mass did not.

Q6 (Kerala HSE Model • 2 Marks): What is meant by diagonal relationship in the periodic table? Give two pairs and mention two reasons for the anomalous behaviour of second-period elements.

Diagonal Relationship: Similarity in chemical properties shown by a second-period element with the diagonally adjacent third-period element of the next group. Pairs: Li & Mg, Be & Al.

Reasons for anomalous behaviour of Period-2 elements: (1) Very small atomic/ionic size and high charge/radius ratio; (2) High electronegativity and absence of vacant d-orbitals (only 4 valence orbitals: 2s and 2p, capping maximum covalency at 4).

Q7 (Kerala HSE March • 3 Marks): Why is a cation always smaller and an anion always larger than its parent atom? Also explain why the second ionization enthalpy (Δ_iH₂) of Na is much higher than that of Mg.

Cation < Parent Atom: Loss of electron(s) leaves fewer electrons pulled by the same nuclear charge (and often removes the outer shell), shrinking the radius (e.g., Na 186 pm → Na⁺ 95 pm).

Anion > Parent Atom: Gain of electron(s) increases inter-electronic repulsion and lowers effective nuclear charge, expanding the cloud (e.g., F 64 pm → F^- 136 pm).

Why Δ_iH₂(Na) >> Δ_iH₂(Mg): After losing 1 e^-, Na⁺ attains a highly stable, closed noble gas configuration [Ne] (1s²2s²2p⁶), requiring huge energy to remove a 2^nd electron. Mg⁺ ([Ne]3s¹) still has a single outer 3s electron whose removal gives stable Mg²⁺ ([Ne]).

Q8 (Kerala HSE SAY • 3 Marks): Define Electron Gain Enthalpy. Why is the second electron gain enthalpy of oxygen positive, and why do noble gases have large positive electron gain enthalpies?

Electron Gain Enthalpy (Δ_egH): The enthalpy change accompanying the addition of an electron to a neutral isolated gaseous atom: X(g) + e^- → X^-(g).

Why Δ_egH₂ of O is positive: Adding a second electron to an already negatively charged O^-(g) ion faces strong electrostatic repulsion, so external energy must be supplied (endothermic, +780 kJ mol^-1).

Noble Gases: Their valence shells are completely filled (ns²np⁶), so an added electron must enter the next higher principal energy level, forming a very unstable state.

Q9 (Kerala HSE / NCERT • 3 Marks): Write the general outer electronic configuration of s-, p-, d-, and f-block elements. Justify why (a) Helium (1s²) is placed in Group 18 and (b) the 5^th period has 18 elements.

General Configurations: s-block: ns^1–2 | p-block: ns²np^1–6 | d-block: (n-1)d^1–10ns^0–2 | f-block: (n-2)f^1–14(n-1)d^0–1ns².

(a) Helium in Group 18: Although its last electron enters 1s, Helium has a completely filled valence shell (1s²) and exhibits chemically inert properties matching Group 18 noble gases.

(b) 18 elements in 5^th period: The orbitals filled in Period 5 follow energy order 5s < 4d < 5p, giving 1 + 5 + 3 = 9 orbitals, which hold 9 × 2 = 18 electrons.

Q10 (Kerala HSE / NCERT Problem 3.10 • 3 Marks): Classify the following oxides as acidic, basic, amphoteric, or neutral: Na₂O, Cl₂O₇, Al₂O₃, CO. Show by chemical reactions with water that Na₂O is basic and Cl₂O₇ is acidic.

Classification: Na₂O = Basic | Cl₂O₇ = Acidic | Al₂O₃ = Amphoteric | CO = Neutral.

Reactions with water:
Na₂O + H₂O → 2NaOH (forms strong base; turns red litmus blue)
Cl₂O₇ + H₂O → 2HClO₄ (forms strong perchloric acid; turns blue litmus red).

Q11 (Kerala HSE March (NCERT Ex 3.31) • 4 Marks): The Δ_iH₁, Δ_iH₂, and Δ_egH (in kJ mol^-1) of elements I–V are: I (520, 7300, -60), II (419, 3051, -48), III (1681, 3374, -328), IV (2372, 5251, +48), V (738, 1451, -40). Identify: (a) least reactive element, (b) most reactive metal, (c) most reactive non-metal, (d) metal forming stable binary halide MX₂.

(a) Least reactive element: IV — highest Δ_iH₁ (2372) and positive Δ_egH (+48), typical of a noble gas (He).

(b) Most reactive metal: II — lowest Δ_iH₁ (419) and large jump to Δ_iH₂ (3051), an alkali metal (K).

(c) Most reactive non-metal: III — very high Δ_iH₁ (1681) and highly negative Δ_egH (-328), a halogen (F).

(d) Metal forming stable MX₂: V — relatively low Δ_iH₁ (738) and Δ_iH₂ (1451) without a huge second jump, an alkaline earth metal (Mg).

Q12 (Kerala HSE Improvement • 1 Marks): Account for the fact that reactivity increases down Group 1 (Li < Na < K < Rb < Cs) whereas it decreases down Group 17 (F > Cl > Br > I).

Group 1 metals react by losing their valence electron; as atomic size increases and ionization enthalpy decreases down the group, electron loss becomes easier, so reactivity increases.

Group 17 halogens react by gaining an electron; as atomic size increases down the group, attraction for the incoming electron decreases (less negative Δ_egH and lower EN), so reactivity decreases.