Classification of Elements and Periodicity in Properties

Classification of Elements and Periodicity in Properties NEET Chemistry Notes | High-Yield PYQs

Genesis of Periodic Classification

Early Attempts

Core definition & intuition: Early scientists attempted to group elements based on atomic weights and chemical similarities to predict properties and simplify the study of chemistry.
Dobereiner’s Triads (1829): Elements grouped in threes. The middle element's atomic weight is approximately the average of the other two (e.g., Li, Na, K).
Newlands’ Law of Octaves (1865): Every eighth element exhibits similar properties when arranged by increasing atomic weight.
Crucial detail/Trap: Newlands' law was only valid up to Calcium. It failed for heavier elements and did not account for undiscovered elements.
Mendeleev’s Periodic Law (1869): Properties of elements are a periodic function of their atomic weights.
Key Achievement: Left gaps for undiscovered elements and predicted their properties accurately. Eka-aluminium became Gallium (Ga), and Eka-silicon became Germanium (Ge).

Modern Periodic Law & Nomenclature

The Modern Periodic Table

Core definition & intuition: Henry Moseley (1913) discovered that a plot of X-ray frequency (√v) versus atomic number (Z) yields a straight line, proving atomic number is a more fundamental property than atomic mass.
Mathematical formulation: Modern Periodic Law: The physical and chemical properties of elements are periodic functions of their atomic numbers.
Structure: Long form consists of 7 horizontal periods (corresponding to principal quantum number n) and 18 vertical groups.

IUPAC Nomenclature for Z > 100

Core definition & intuition: Temporary IUPAC names for undiscovered or unnamed elements are derived from their atomic number using specific numerical roots.
Mathematical formulation: Roots: 0=nil, 1=un, 2=bi, 3=tri, 4=quad, 5=pent, 6=hex, 7=sept, 8=oct, 9=enn. Suffix: "-ium" is appended to the roots.

PYQ Problem 1:
The IUPAC name of an element with atomic number 119 is: NEET 2022
(1) Ununoctium
(2) Ununennium
(3) Unnilennium
(4) Unununnium

Step 1 - Concept: Temporary IUPAC names are formed by mapping the digits of the atomic number to their specific Latin/Greek numerical roots and appending the suffix "-ium".
Step 2 - Identify Data: For Z = 119, the digits are 1, 1, 9. The corresponding roots are: un (1), un (1), enn (9).
Step 3 - Application: Combine the roots sequentially and add the suffix: un + un + enn + ium = ununennium.
Conclusion: Option (2) is correct.

PYQ Problem 2:
Identify the incorrect match: NEET 2020
(A) Unnilunium - (i) Mendelevium
(B) Unniltrium - (ii) Lawrencium
(C) Unnilhexium - (iii) Seaborgium
(D) Unununnium - (iv) Darmstadtium
(1) (B), (ii)
(2) (C), (iii)
(3) (D), (iv)
(4) (A), (i)

Step 1 - Concept: We must decode the IUPAC temporary names to find their atomic numbers and match them with their official names.
Step 2 - Identify Data: Unununnium corresponds to Z = 111 (un=1, un=1, un=1).
Step 3 - Application: Element 111 is officially named Roentgenium (Rg), not Darmstadtium. Darmstadtium is Z = 110 (Ununnillium). Therefore, the match (D) - (iv) is incorrect.
Conclusion: Option (3) is correct.

Electronic Configurations & Element Types

s, p, d, and f-Block Elements

Core definition & intuition: The periodic table is divided into blocks based on the subshell in which the differentiating (last) electron enters.
Mathematical formulation:
s-Block (Groups 1 & 2): ns^1-2. Highly reactive metals; reactivity increases down the group. Form ionic compounds (except Li, Be).
p-Block (Groups 13-18): ns²np^1-6. Includes metals, non-metals, and metalloids. Group 18 (Noble gases) have stable ns²np⁶ configurations.
d-Block (Groups 3-12): (n-1)d^1-10ns^0-2. Transition elements bridging highly reactive s-block and less reactive p-block. Characterized by variable oxidation states and colored ions.
f-Block: (n-2)f^1-14(n-1)d^0-1ns². Lanthanoids (Z=58-71) and Actinoids (Z=90-103). All actinoids are radioactive. Elements after Uranium (Z > 92) are transuranium elements.
Crucial detail/Trap: Zn, Cd, and Hg ((n-1)d¹⁰ns²) do not exhibit typical transition metal properties because their d-orbitals are completely filled in both ground and oxidized states.

PYQ Problem 3:
The element Z=114 has been discovered recently. It will belong to which of the following family group and electronic configuration? NEET 2017
(1) Nitrogen family, [Rn] 5f¹⁴6d¹⁰7s²7p³
(2) Halogen family, [Rn] 5f¹⁴6d¹⁰7s²7p⁵
(3) Carbon family, [Rn] 5f¹⁴6d¹⁰7s²7p²
(4) Oxygen family, [Rn] 5f¹⁴6d¹⁰7s²7p⁴

Step 1 - Concept: To find the group of a p-block element, we look at its position relative to the noble gas core. The group number is 10 + number of electrons in the outermost s and p subshells.
Step 2 - Identify Data: Z = 114 is after Z = 86 (Rn, period 6). It falls in period 7. Group = 114 - 118(Og) + 18 = 14 (Carbon family).
Step 3 - Application: The outer config for Group 14 is ns²np². Here n=7, so 7s²7p². The full configuration is [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p².
Conclusion: Option (3) is correct.

PYQ Problem 4:
The number of d-electrons in Fe²⁺ (Z=26) is not equal to the number of electrons in which one of the following? AIPMT 2015
(1) p-electrons in Cl (Z=17)
(2) d-electrons in Fe (Z=26)
(3) p-electrons in Ne (Z=10)
(4) s-electrons in Mg (Z=12)

Step 1 - Concept: We must carefully count the specific subshell electrons for each neutral atom or ion provided in the options and compare them to the d-electron count of Fe²⁺.
Step 2 - Identify Data: Fe (Z=26) is [Ar] 4s² 3d⁶. For Fe²⁺, we remove the two 4s electrons first, leaving [Ar] 3d⁶. Thus, Fe²⁺ has 6 d-electrons.
Step 3 - Application: Let's check the options: Cl p-electrons = 11 (2p⁶, 3p⁵). Fe d-electrons = 6. Ne p-electrons = 6. Mg s-electrons = 6 (1s², 2s², 3s²). Only Cl has a count (11) that is not equal to 6.
Conclusion: Option (1) is correct.

PYQ Problem 5:
A nucleus of an alkaline earth metal undergoes radioactive decay by emission of three α-particles in succession. The group of the periodic table to which the resulting daughter element would belong is: AIPMT 2005
(1) Group 14
(2) Group 16
(3) Group 4
(4) Group 6

Step 1 - Concept: Emission of an α-particle decreases the atomic number (Z) by 2. We must track the shift in the periodic table groups.
Step 2 - Identify Data: Alkaline earth metal = Group 2. Three α-particles = decrease in Z by 3 × 2 = 6.
Step 3 - Application: Shifting 6 places back across the periodic table from Group 2 (or effectively moving left by 6 groups): Group 2 → Group 18 → Group 16 → Group 14.
Conclusion: Option (1) is correct.

PYQ Problem 6:
Which of the following configuration is correct for iron? AIPMT 1999
(1) 1s², 2s²2p⁶, 3s²3p⁶3d⁵
(2) 1s², 2s²2p⁶, 3s²3p⁶, 4s²3d⁵
(3) 1s², 2s²2p⁶, 3s²3p⁶, 4s²3d⁷
(4) 1s², 2s²2p⁶, 3s²3p⁶, 3d⁶, 4s²

Step 1 - Concept: According to the Aufbau principle, electrons fill lower energy orbitals first (4s fills before 3d). However, when writing the final configuration, it is standard convention to group subshells by their principal quantum number (n).
Step 2 - Identify Data: Iron (Fe) has Z = 26. The filling order is [Ar] 3d⁶ 4s².
Step 3 - Application: Rearranging strictly by principal quantum number (n) gives: 1s², 2s²2p⁶, 3s²3p⁶, 3d⁶, 4s².
Conclusion: Option (4) is correct.

PYQ Problem 7:
Which of the following has more unpaired d-electrons? AIPMT 1999
(1) Zn⁺
(2) Fe²⁺
(3) Ni³⁺
(4) Cu⁺

Step 1 - Concept: Unpaired electrons are determined by applying Hund's Rule to the d-subshell configuration. We must first find the correct ion configuration.
Step 2 - Identify Data: Zn⁺ is 3d¹⁰4s¹ (0 unpaired d). Fe²⁺ is 3d⁶ (4 unpaired d). Ni³⁺ is 3d⁷ (3 unpaired d). Cu⁺ is 3d¹⁰ (0 unpaired d).
Step 3 - Application: Comparing the unpaired d-electron counts, Fe²⁺ has the maximum with 4.
Conclusion: Option (2) is correct.

PYQ Problem 8:
The electronic configuration of an element is 1s²2s²2p⁶3s²3p³. What is the atomic number of the element, which is just below the above element in the periodic table? AIPMT 1995
(1) 33
(2) 34
(3) 36
(4) 49

Step 1 - Concept: Elements in the same group have the same valence shell configuration. Moving "just below" an element means moving to the next period in the same group.
Step 2 - Identify Data: Z = 2+2+6+2+3 = 15 (Phosphorus, Group 15, Period 3).
Step 3 - Application: Add 18 to shift down one period in the p-block. 15 + 18 = 33 (Arsenic).
Conclusion: Option (1) is correct.

PYQ Problem 9:
If the atomic number of an element is 33, it will be placed in the periodic table in the: AIPMT 1993
(1) First group
(2) Third group
(3) Fifth group
(4) Seventh group

Step 1 - Concept: For p-block elements, the group number is calculated as 10 + (number of ns + np electrons).
Step 2 - Identify Data: Z=33 is [Ar] 3d¹⁰ 4s² 4p³. Valence shell has 5 electrons (4s²4p³).
Step 3 - Application: Group number = 10 + 5 = 15. In older notation, Group 15 is the Fifth group (Group VA).
Conclusion: Option (3) is correct.

PYQ Problem 10:
The electronic configuration of four elements are given below. Which elements does not belong to the same family as others? AIPMT 1989
(1) [Xe] 4f¹⁴5d¹⁰6s²
(2) [Kr] 4d¹⁰5s²
(3) [Ne] 3s²3p⁵
(4) [Ar] 3d¹⁰4s²

Step 1 - Concept: Elements in the same family (group) have identical valence shell configurations.
Step 2 - Identify Data: Options (1), (2), and (4) all have an ns² valence shell (Group 12).
Step 3 - Application: Option (3) has an ns²np⁵ valence shell, which belongs to Group 17 (Halogen family). It is the outlier.
Conclusion: Option (3) is correct.

PYQ Problem 11:
One of the characteristic properties of non-metals is that they: AIPMT 1993
(1) Are reducing agents
(2) Form basic oxides
(3) Form cations by electron gain
(4) Are electronegative

Step 1 - Concept: Non-metals have high ionization energies and high electron affinities, meaning they tend to accept electrons rather than lose them.
Step 2 - Identify Data: Accepting electrons makes them electronegative and good oxidizing agents. They form acidic oxides, not basic.
Step 3 - Application: Therefore, the defining characteristic among the choices is that they are electronegative.
Conclusion: Option (4) is correct.

PYQ Problem 12:
In the periodic table, with the increase in atomic number, the metallic character of an element: AIPMT 1989
(1) Decreases in a period and increases in a group
(2) Increases in a period and decreases in a group
(3) Increases both in a period and the group
(4) Decreases in a period and the group

Step 1 - Concept: Metallic character is the tendency to lose electrons. This depends on atomic size and effective nuclear charge (Z_eff).
Step 2 - Identify Data: Across a period, Z_eff increases, making electron loss harder (metallic character decreases). Down a group, size increases, making electron loss easier (metallic character increases).
Step 3 - Application: Combining these trends gives the correct variation.
Conclusion: Option (1) is correct.

Periodic Trends in Physical Properties

Atomic and Ionic Radii

Core definition & intuition: Atomic radius is measured via covalent radius (non-metals) or metallic radius (metals). Ionic radius compares the size of an atom to its charged ion.
Mathematical formulation:
Trend across a Period: Decreases from left to right. Reason: Electrons are added to the same shell. Z_eff increases, pulling the electron cloud closer.
Trend down a Group: Increases from top to bottom. Reason: Principal quantum number (n) increases. The shielding effect outweighs the increased nuclear charge.
Cation (X⁺): Smaller than parent atom. (Removal of electron increases Z_eff per remaining electron).
Anion (X^-): Larger than parent atom. (Addition of electron increases electron-electron repulsion).
Crucial detail/Trap: Isoelectronic Species: Atoms/ions with the identical number of electrons (e.g., N^3-, O^2-, F^-, Ne, Na⁺, Mg²⁺, Al³⁺). In an isoelectronic series, size decreases as atomic number (Z) increases because a higher nuclear charge exerts a stronger pull on the same number of electrons.

PYQ Problem 13:
The correct decreasing order of atomic radii (pm) of Li, Be, B and C is: NEET 2024
(1) Be > Li > B > C
(2) Li > Be > B > C
(3) C > B > Be > Li
(4) Li > C > Be > B

Step 1 - Concept: All these elements belong to the same period (Period 2). Across a period, atomic radius strictly decreases from left to right due to increasing Z_eff.
Step 2 - Identify Data: The order of atomic numbers is Li(3) < Be(4) < B(5) < C(6).
Step 3 - Application: Since radius decreases left to right, the decreasing order of size must be Li > Be > B > C.
Conclusion: Option (2) is correct.

PYQ Problem 14:
The element expected to form largest ion to achieve the nearest noble gas configuration is: NEET 2023
(1) Na
(2) O
(3) F
(4) N

Step 1 - Concept: To achieve the nearest noble gas configuration (Neon, 10 electrons), these elements will form ions. Since they are isoelectronic, the size is determined by the nuclear charge (Z).
Step 2 - Identify Data: Ions formed: Na⁺, O^2-, F^-, N^3-. All have 10 electrons. The atomic numbers are Na=11, O=8, F=9, N=7.
Step 3 - Application: Size ∝ 1/Z. N has the lowest Z (7), so N^3- holds the 10 electrons the least tightly, making it the largest.
Conclusion: Option (4) is correct.

PYQ Problem 15:
Which one of the following represents all isoelectronic species? NEET 2023
(1) Na⁺, Cl^-, O^-, NO⁺
(2) N₂, O₂, N₂O₄, NO₂^-
(3) Na⁺, Mg²⁺, O^2-, F^-
(4) Ca²⁺, Ar, K⁺, Cl^-

Step 1 - Concept: Isoelectronic species must have the exact same total number of electrons.
Step 2 - Identify Data: Let's count electrons for Option (4): Ca²⁺ (20-2=18), Ar (18), K⁺ (19-1=18), Cl^- (17+1=18).
Step 3 - Application: All species in Option (4) have exactly 18 electrons, making them an isoelectronic series.
Conclusion: Option (4) is correct.

PYQ Problem 16:
From the following pairs of ions, which one is not an isoelectronic pair? NEET 2021
(1) Na⁺, Mg²⁺
(2) Mn²⁺, Fe³⁺
(3) Fe²⁺, Mn²⁺
(4) O^2-, F^-

Step 1 - Concept: To be isoelectronic, the total number of electrons (Atomic Number - Charge) must be identical for both species in the pair.
Step 2 - Identify Data: Fe²⁺ (26-2=24) and Mn²⁺ (25-2=23).
Step 3 - Application: 24 ≠ 23. This pair is not isoelectronic. (The others all match: 10/10, 23/23, 10/10).
Conclusion: Option (3) is correct.

PYQ Problem 17:
The species Ar, K⁺ and Ca²⁺ contain the same number of electrons. In which order do their radii increase? AIPMT 2015
(1) Ca²⁺ < Ar < K⁺
(2) Ca²⁺ < K⁺ < Ar
(3) K⁺ < Ar < Ca²⁺
(4) Ar < K⁺ < Ca²⁺

Step 1 - Concept: For isoelectronic species, ionic radius is inversely proportional to the nuclear charge (Z). More protons = stronger pull = smaller radius.
Step 2 - Identify Data: All have 18 electrons. Z values: Ar (18), K (19), Ca (20).
Step 3 - Application: Ca has the highest Z (smallest radius), Ar has the lowest Z (largest radius). Increasing order: Ca²⁺ < K⁺ < Ar.
Conclusion: Option (2) is correct.

PYQ Problem 18:
Be²⁺ is isoelectronic with which of the following ions? AIPMT 2014
(1) Li⁺
(2) Na⁺
(3) Mg²⁺
(4) H⁺

Step 1 - Concept: Isoelectronic species have the same number of electrons.
Step 2 - Identify Data: Be (Z=4), Be²⁺ has 4 - 2 = 2 electrons.
Step 3 - Application: Li⁺ (Z=3) has 3 - 1 = 2 electrons. They match.
Conclusion: Option (1) is correct.

PYQ Problem 19:
Which of the following orders of ionic radii is correctly represented? AIPMT 2014
(1) Na⁺ > F^- > O^2-
(2) O^2- > F^- > Na⁺
(3) Al³⁺ > Mg²⁺ > N^3-
(4) H^- > H⁺ > H

Step 1 - Concept: In an isoelectronic series, radius decreases as Z increases.
Step 2 - Identify Data: O^2-, F^-, Na⁺ all have 10 electrons. Z values are 8, 9, 11 respectively.
Step 3 - Application: Radius decreases as Z increases. Therefore, O^2- > F^- > Na⁺.
Conclusion: Option (2) is correct.

PYQ Problem 20:
Identify the wrong statement in the following: AIPMT 2012
(1) Atomic radius of the elements decreases as one moves across from left to right in the 2nd period.
(2) Amongst isoelectronic species, smaller the positive charge on the cation, smaller is the ionic radius.
(3) Amongst isoelectronic species, greater the negative charge on the anion, larger is the ionic radius.
(4) Atomic radius of the elements increases as one moves down the first group.

Step 1 - Concept: For isoelectronic cations, a higher positive charge means fewer electrons but the same number of protons, leading to a stronger Z_eff and a smaller radius.
Step 2 - Identify Data: Statement (2) claims "smaller positive charge → smaller radius".
Step 3 - Application: This is false. A smaller positive charge (e.g., Na⁺ vs Mg²⁺) actually results in a *larger* radius because the nuclear pull per electron is weaker.
Conclusion: Option (2) is the wrong statement.

PYQ Problem 21:
Among the elements Ca, Mg, P and Cl, the order of increasing atomic radii is: AIPMT 2010
(1) Mg < Ca < Cl < P
(2) Cl < P < Mg < Ca
(3) P < Cl < Ca < Mg
(4) Ca < Mg < P < Cl

Step 1 - Concept: Size increases down a group and decreases across a period.
Step 2 - Identify Data: P, Cl, Mg are in Period 3. Ca is in Period 4. Period 3 order: Cl < P < Mg. Period 4 is larger: Ca is largest.
Step 3 - Application: Combining these facts yields: Cl < P < Mg < Ca.
Conclusion: Option (2) is correct.

PYQ Problem 22:
The correct order of the decreasing ionic radii among the following isoelectronic species is: AIPMT 2010
(1) K⁺ > Ca²⁺ > Cl^- > S^2-
(2) Ca²⁺ > K⁺ > S^2- > Cl^-
(3) Cl^- > S^2- > Ca²⁺ > K⁺
(4) S^2- > Cl^- > K⁺ > Ca²⁺

Step 1 - Concept: Decreasing radii means ordering from largest to smallest. For isoelectronic species, lowest Z is largest.
Step 2 - Identify Data: Z values: S (16), Cl (17), K (19), Ca (20).
Step 3 - Application: Lowest Z first: S^2- > Cl^- > K⁺ > Ca²⁺.
Conclusion: Option (4) is correct.

PYQ Problem 23:
Identify the correct order of the size of the following: AIPMT 2007
(1) Ca²⁺ < K⁺ < Ar < Cl^- < S^2-
(2) Ar < Ca²⁺ < K⁺ < Cl^- < S^2-
(3) Ca²⁺ < Ar < K⁺ < Cl^- < S^2-
(4) Ca²⁺ < K⁺ < Ar < S^2- < Cl^-

Step 1 - Concept: Apply the isoelectronic rule (18 electrons each). Highest Z is smallest, lowest Z is largest.
Step 2 - Identify Data: Z values: Ca (20), K (19), Ar (18), Cl (17), S (16).
Step 3 - Application: Order from smallest to largest: Ca²⁺ < K⁺ < Ar < Cl^- < S^2-.
Conclusion: Option (1) is correct.

PYQ Problem 24:
Ionic radii are: AIPMT 2004
(1) Inversely proportional to square of effective nuclear charge
(2) Directly proportional to effective nuclear charge
(3) Directly proportional to square of effective nuclear charge
(4) Inversely proportional to effective nuclear charge

Step 1 - Concept: Stronger Z_eff pulls electrons closer, shrinking the radius.
Step 2 - Identify Data: Radius ∝ 1/Z_eff.
Step 3 - Application: This means it is inversely proportional to the effective nuclear charge.
Conclusion: Option (4) is correct.

PYQ Problem 25:
The ions O^2-, F^-, Na⁺, Mg²⁺ and Al³⁺ are isoelectronic. Their ionic radii show: AIPMT 2003
(1) A significant increase from O^2- to Al³⁺
(2) A significant decrease from O^2- to Al³⁺
(3) An increase from O^2- to F^- and then decrease from Na⁺ to Al³⁺
(4) A decrease from O^2- to F^- and then increase from Na⁺ to Al³⁺

Step 1 - Concept: As atomic number increases in an isoelectronic series, the radius continuously decreases.
Step 2 - Identify Data: Z increases from O (8) to Al (13).
Step 3 - Application: Therefore, the radius shows a significant decrease from O^2- to Al³⁺.
Conclusion: Option (2) is correct.

PYQ Problem 26:
Which one of the following is correct order of the size of iodine species? AIPMT 1997
(1) I⁺ > I^- > I
(2) I^- > I > I⁺
(3) I > I^- > I⁺
(4) I > I⁺ > I^-

Step 1 - Concept: An anion is always larger than its neutral atom due to increased repulsion. A cation is always smaller than its neutral atom.
Step 2 - Identify Data: I^- (anion), I (neutral), I⁺ (cation).
Step 3 - Application: Order from largest to smallest: I^- > I > I⁺.
Conclusion: Option (2) is correct.

PYQ Problem 27:
Which of the following ion is the largest in size? AIPMT 1996
(1) K⁺
(2) Ca²⁺
(3) Cl^-
(4) S^2-

Step 1 - Concept: Apply the isoelectronic rule. All have 18 electrons.
Step 2 - Identify Data: Z values: K(19), Ca(20), Cl(17), S(16).
Step 3 - Application: S^2- has the lowest Z (16), therefore it exerts the weakest pull on the 18 electrons, making it the largest.
Conclusion: Option (4) is correct.

PYQ Problem 28:
Which of the following has the smallest size? AIPMT 1996
(1) Al³⁺
(2) F^-
(3) Na⁺
(4) Mg²⁺

Step 1 - Concept: Apply the isoelectronic rule. All have 10 electrons.
Step 2 - Identify Data: Z values: Al(13), F(9), Na(11), Mg(12).
Step 3 - Application: Al³⁺ has the highest Z (13), therefore it pulls the 10 electrons the tightest, making it the smallest.
Conclusion: Option (1) is correct.

PYQ Problem 29:
Na⁺, Mg²⁺, Al³⁺ and Si⁴⁺ are isoelectronic. The order of their ionic size is: AIPMT 1993
(1) Na⁺ > Mg²⁺ < Al³⁺ < Si⁴⁺
(2) Na⁺ < Mg²⁺ > Al³⁺ > Si⁴⁺
(3) Na⁺ > Mg²⁺ > Al³⁺ > Si⁴⁺
(4) Na⁺ < Mg²⁺ > Al³⁺ < Si⁴⁺

Step 1 - Concept: Size decreases as Z increases in an isoelectronic series.
Step 2 - Identify Data: Z values: Na (11), Mg (12), Al (13), Si (14).
Step 3 - Application: Order from largest to smallest: Na⁺ > Mg²⁺ > Al³⁺ > Si⁴⁺.
Conclusion: Option (3) is correct.

PYQ Problem 30:
In the periodic table from left to right in a period, the atomic volume: AIPMT 1993
(1) Decreases
(2) Increases
(3) Remains same
(4) First decrease then increases

Step 1 - Concept: Atomic volume depends on atomic radius. Radius initially decreases across a period due to Z_eff.
Step 2 - Identify Data: However, towards the end of the period, noble gases have large van der Waals radii, and halogens have high electron repulsion.
Step 3 - Application: Therefore, atomic volume first decreases, then increases towards the rightmost elements.
Conclusion: Option (4) is correct.

Ionization Enthalpy (Δ_iH)

Core definition & intuition: Energy required to remove an electron from an isolated gaseous atom in its ground state. Always positive (endothermic process).
Mathematical formulation:
Successive Enthalpies: IE₁ < IE₂ < IE₃. It becomes increasingly difficult to remove electrons from a positively charged core.
Trend across a Period: Generally increases from left to right (increasing Z_eff).
Trend down a Group: Decreases from top to bottom (size increases, shielding increases).
Crucial detail/Trap (Period 2 Exceptions):
Be > B: The 2s electron of Be is more deeply penetrating and stable than the 2p electron of B.
N > O: Nitrogen has a stable, exactly half-filled 2p³ configuration, requiring more energy to break compared to Oxygen's 2p⁴ configuration (where inter-electronic repulsion makes losing an electron easier).

PYQ Problem 31:
Arrange the following elements in increasing order of first ionization enthalpy: Li, Be, B, C, N NEET 2024
(1) Li < Be < C < B < N
(2) Li < Be < N < B < C
(3) Li < Be < B < C < N
(4) Li < B < Be < C < N

Step 1 - Concept: Generally IE increases across a period, but we must account for the anomaly between Group 2 and Group 13.
Step 2 - Identify Data: Be (2s²) is more stable than B (2p¹). So B < Be.
Step 3 - Application: The correct increasing order is Li < B < Be < C < N.
Conclusion: Option (4) is correct.

PYQ Problem 32:
Given below are two statements:
Assertion (A): Ionisation enthalpy increases along each series of the transition elements from left to right. However, small variations occur.
Reason (R): There is corresponding increase in nuclear charge which accompanies the filling of electrons in the inner d-orbitals. NEET 2023
(1) (A) is correct but (R) is not correct.
(2) (A) is not correct but (R) is correct.
(3) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(4) Both (A) and (R) are correct (R) is not the correct explanation of (A).

Step 1 - Concept: IE generally increases across the d-block but slowly due to inner d-electron shielding.
Step 2 - Identify Data: Assertion is True. Reason is True: added protons increase nuclear charge, but added d-electrons partially shield the 4s electrons.
Step 3 - Application: The shielding effect explains the slow, small variation increase, making R the correct explanation for A.
Conclusion: Option (3) is correct.

PYQ Problem 33:
The correct order of first ionization enthalpy for the given four elements is: NEET 2022
(1) C < F < N < O
(2) C < N < F < O
(3) C < N < O < F
(4) C < O < N < F

Step 1 - Concept: Baseline trend is C < N < O < F, but we must apply the half-filled orbital exception.
Step 2 - Identify Data: N (2p³) is half-filled and more stable than O (2p⁴). So O < N.
Step 3 - Application: Final order: C < O < N < F.
Conclusion: Option (4) is correct.

PYQ Problem 34:
If first ionization enthalpies of elements X and Y are 419 kJ mol^-1 and 590 kJ mol^-1, respectively and second ionization enthalpies of X and Y are 3069 kJ mol^-1 and 1145 kJ mol^-1, respectively. Then correct statement is: NEET 2022
(1) Both X and Y are alkaline earth metals.
(2) X is an alkali metal and Y is an alkaline earth metal.
(3) X is an alkaline earth metal and Y is an alkali metal.
(4) Both X and Y are alkali metals.

Step 1 - Concept: A massive jump between successive ionization enthalpies indicates the removal of a core electron after achieving a noble gas configuration.
Step 2 - Identify Data: For X: IE₁ = 419, IE₂ = 3069 (Massive jump → 1 valence electron → Alkali metal). For Y: IE₁ = 590, IE₂ = 1145 (Gradual increase → 2 valence electrons → Alkaline earth metal).
Step 3 - Application: X is an alkali metal, Y is an alkaline earth metal.
Conclusion: Option (2) is correct.

PYQ Problem 35:
For the second period elements, the correct increasing order of first ionisation enthalpy is: NEET 2019
(1) Li < Be < B < C < N < O < F < Ne
(2) Li < B < Be < C < O < N < F < Ne
(3) Li < B < Be < C < N < O < F < Ne
(4) Li < Be < B < C < O < N < F < Ne

Step 1 - Concept: Recall the two major exceptions in Period 2: Be > B and N > O.
Step 2 - Identify Data: Normal trend: Li < Be < B < C < N < O < F < Ne.
Step 3 - Application: Inserting the exceptions yields: Li < B < Be < C < O < N < F < Ne.
Conclusion: Option (2) is correct.

PYQ Problem 36:
Amongst the elements with following electronic configurations, which one of them may have the highest ionisation energy? AIPMT 2009
(1) [Ne] 3s²3p²
(2) [Ar] 3d¹⁰4s²4p³
(3) [Ne] 3s²3p¹
(4) [Ne] 3s²3p³

Step 1 - Concept: Half-filled orbitals provide extra stability, increasing IE. Size also plays a role (smaller size = higher IE).
Step 2 - Identify Data: (4) is P (half-filled 3p³). (2) is As (half-filled 4p³).
Step 3 - Application: P is smaller than As, so its valence electrons are closer to the nucleus. Therefore, P has a higher IE than As.
Conclusion: Option (4) is correct.

PYQ Problem 37:
With which of the following electronic configuration, an atom has the lowest ionisation enthalpy? AIPMT 2007
(1) 1s²2s²2p³
(2) 1s²2s²2p⁶3s¹
(3) 1s²2s²2p⁶
(4) 1s²2s²2p⁵

Step 1 - Concept: Alkali metals (Group 1) have the largest size in their period and a single valence electron, resulting in the lowest IE.
Step 2 - Identify Data: (1) N, (2) Na, (3) Ne, (4) F.
Step 3 - Application: Na is an alkali metal with the configuration 1s²2s²2p⁶3s¹.
Conclusion: Option (2) is correct.

PYQ Problem 38:
Correct order of 1st IP among following elements Be, B, C, N, O is: AIPMT 2001
(1) B < Be < C < O < N
(2) B < Be < C < N < O
(3) Be < B < C < N < O
(4) Be < B < C < O < N

Step 1 - Concept: Apply the exceptions: B < Be and O < N.
Step 2 - Identify Data: Base sequence is Be, B, C, N, O.
Step 3 - Application: Merging the anomalies gives: B < Be < C < O < N.
Conclusion: Option (1) is correct.

PYQ Problem 39:
The first ionization potential (in eV) of Be and B, respectively are: AIPMT 1998
(1) 8.29, 9.32
(2) 9.32, 9.32
(3) 8.29, 8.29
(4) 9.32, 8.29

Step 1 - Concept: Recall the exception: IE of Be > IE of B because 2s electrons are more penetrating than 2p.
Step 2 - Identify Data: Be must have the higher value.
Step 3 - Application: 9.32 > 8.29. Therefore, Be is 9.32 and B is 8.29.
Conclusion: Option (4) is correct.

PYQ Problem 40:
Which electronic configuration of an element has abnormally high difference between second and third ionization energy? AIPMT 1993
(1) 1s², 2s², 2p⁶, 3s¹
(2) 1s², 2s², 2p⁶, 3s²3p¹
(3) 1s², 2s², 2p⁶, 3s²3p²
(4) 1s², 2s², 2p⁶, 3s²

Step 1 - Concept: A high jump between 2nd and 3rd IE means the element has exactly 2 valence electrons (Alkaline earth metal).
Step 2 - Identify Data: Option (4) is Mg, with 2 valence electrons (3s²).
Step 3 - Application: After removing 2 electrons, it hits the stable Ne core, making the 3rd removal extremely difficult.
Conclusion: Option (4) is correct.

Electron Gain Enthalpy (Δ_egH)

Core definition & intuition: Enthalpy change when an electron is added to an isolated neutral gaseous atom. It can be exothermic (negative) or endothermic (positive).
Mathematical formulation:
Trend across a Period: Becomes more negative from left to right (as size decreases and Z_eff increases). Halogens have the most negative values.
Trend down a Group: Becomes less negative from top to bottom (due to increasing size).
Crucial detail/Trap: Electron gain enthalpy of O or F is less negative than that of the succeeding element (S or Cl). Reason: O and F are extremely small. Adding an electron to the n=2 quantum level causes significant inter-electronic repulsion. Noble Gases have large positive electron gain enthalpies.

PYQ Problem 41:
The formation of the oxide ion, O^2-(g) from oxygen atom requires first an exothermic step and then an endothermic step as shown below:
O(g) + e^- → O^-(g); Δ_egH° = -141 kJ mol^-1
O^-(g) + e^- → O^2-(g); Δ_egH° = +780 kJ mol^-1
Thus, process of formation of O^2- in gas phase is unfavourable even though O^2- is isoelectronic with neon. It is due to the fact that, AIPMT 2015
(1) O^- ion has comparatively smaller size than oxygen atom
(2) Oxygen is more electronegative
(3) Addition of electron in oxygen results in larger size of the ion
(4) Electron repulsion outweighs the stability gained by achieving noble gas configuration

Step 1 - Concept: Adding a negative electron to an already negative ion creates massive electrostatic repulsion.
Step 2 - Identify Data: The second step involves adding an e^- to O^-.
Step 3 - Application: This electron repulsion requires a large input of energy (+780 kJ/mol), outweighing the stability of the neon configuration.
Conclusion: Option (4) is correct.

PYQ Problem 42:
What is the value of electron gain enthalpy of Na⁺ if IE₁ of Na = 5.1 eV? AIPMT 2011
(1) +2.55 eV
(2) +10.2 eV
(3) -5.1 eV
(4) -10.2 eV

Step 1 - Concept: Electron gain enthalpy of an ion is the exact reverse thermodynamic process of the ionization of its parent neutral atom.
Step 2 - Identify Data: Reaction for IE: Na → Na⁺ + e^- (ΔH = +5.1 eV). Reaction for EGE of Na⁺: Na⁺ + e^- → Na.
Step 3 - Application: Reverse the sign of the IE. ΔH = -5.1 eV.
Conclusion: Option (3) is correct.

PYQ Problem 43:
Which of the following represents the correct order of increasing electron gain enthalpy with negative sign for the elements O, S, F and Cl? AIPMT 2010
(1) O < S < F < Cl
(2) Cl < F < O < S
(3) O < S < F < Cl
(4) F < S < O < Cl

Step 1 - Concept: "Increasing with negative sign" means least negative to most negative. Halogens > Chalcogens. Period 3 > Period 2 due to repulsion anomalies.
Step 2 - Identify Data: Cl > F and S > O.
Step 3 - Application: Combining these yields the correct order: O < S < F < Cl.
Conclusion: Option (1) / (3) is correct.

PYQ Problem 44:
Which one of the following arrangements represents the correct order of electron gain enthalpy (with negative sign) of the given atomic species? AIPMT 2005
(1) Cl < F < S < O
(2) O < S < F < Cl
(3) S < O < Cl < F
(4) F < Cl < O < S

Step 1 - Concept: Apply the exact logic from the previous problem regarding anomalies and group trends.
Step 2 - Identify Data: O < S and F < Cl. Also, Halogens > Chalcogens.
Step 3 - Application: The correct sequence from least negative to most negative is O < S < F < Cl.
Conclusion: Option (2) is correct.

PYQ Problem 45:
Which of the following elements has the maximum electron affinity? AIPMT 1999
(1) I
(2) Br
(3) Cl
(4) F

Step 1 - Concept: Electron affinity generally becomes more negative across a period and less negative down a group, with the Period 2 anomaly.
Step 2 - Identify Data: Cl has higher electron affinity than F because the small size of F causes strong inter-electronic repulsions in the 2p subshell.
Step 3 - Application: Chlorine has the maximum (most negative) electron affinity among all elements.
Conclusion: Option (3) is correct.

Electronegativity

Core definition & intuition: Qualitative measure of the ability of an atom in a compound to attract shared electrons to itself.
Mathematical formulation:
Scales: Pauling scale is most common (F = 4.0, highest).
Trend across a Period: Increases (due to shrinking atomic radius and higher Z_eff).
Trend down a Group: Decreases (due to increasing atomic radius).
Crucial detail/Trap: Electronegativity is directly related to non-metallic properties and inversely related to metallic properties.

PYQ Problem 46:
Arrange the following elements in increasing order of electronegativity: N, O, F, C, Si NEET 2024
(1) O < F < N < C < Si
(2) F < O < N < C < Si
(3) Si < C < N < O < F
(4) Si < C < O < N < F

Step 1 - Concept: EN increases left to right across a period and decreases down a group.
Step 2 - Identify Data: Si (Group 14, P3), C, N, O, F (P2). EN decreases down a group: Si < C. EN increases left to right: C < N < O < F.
Step 3 - Application: Combine the trends: Si < C < N < O < F.
Conclusion: Option (3) is correct.

PYQ Problem 47:
Which statement is wrong? AIPMT 1997
(1) Bond energy of F₂ > Cl₂
(2) Electronegativity of F > Cl
(3) F is more oxidising than Cl
(4) Electron affinity of Cl > F

Step 1 - Concept: Evaluate each statement based on periodic properties and the specific anomalies of the second-period elements.
Step 2 - Identify Data: F₂ bond energy is anomalously low due to severe lone-pair repulsions between the small F atoms. Cl₂ > F₂.
Step 3 - Application: Statement (1) claims F₂ > Cl₂, which is false. The other statements are factually correct.
Conclusion: Option (1) is correct.

PYQ Problem 48:
Pauling's electronegativity values for elements are useful in predicting: AIPMT 1989
(1) Polarity of the molecules
(2) Position in the E.M.F. series
(3) Coordination numbers
(4) Dipole moments

Step 1 - Concept: Electronegativity difference between two bonded atoms dictates the electron sharing distribution.
Step 2 - Identify Data: This difference predicts the ionic/covalent nature of a bond.
Step 3 - Application: Therefore, it is directly useful in predicting the polarity of the molecules.
Conclusion: Option (1) is correct.

Mixed Property Orders

PYQ Problem 49:
In which of the following options, the order of arrangement does not agree with the variation of property indicated against it? NEET 2016
(1) Li < Na < K < Rb (increasing metallic radius)
(2) Al³⁺ < Mg²⁺ < Na⁺ < F^- (increasing ionic size)
(3) B < C < N < O (increasing first ionization enthalpy)
(4) I < Br < Cl < F (increasing electron gain enthalpy)

Step 1 - Concept: We need to find the option where the stated trend contradicts established chemical principles.
Step 2 - Identify Data: Check (3): B < C < N < O is WRONG. N has a stable half-filled orbital, so N > O. Correct order is B < C < O < N. Check (4): I < Br < Cl < F is WRONG for electron gain enthalpy. Cl has the highest, so it should be I < Br < F < Cl.
Step 3 - Application: Both (3) and (4) are incorrect arrangements. (Note: standard NEET key accepted both or marked as disputed).
Conclusion: Options (3) and (4) are incorrect.

PYQ Problem 50:
Which of the following order is wrong? AIPMT 2002
(1) NH₃ < PH₃ < AsH₃ - Acidic
(2) Li < Be < B < C - 1st IP
(3) Al₂O₃ < MgO < Na₂O < K₂O - Basic
(4) Li⁺ < Na⁺ < K⁺ < Cs⁺ - Ionic radius

Step 1 - Concept: Evaluate the 1st IP trend for Period 2 elements.
Step 2 - Identify Data: Option (2) lists Li < Be < B < C.
Step 3 - Application: This is WRONG. Be (2s²) > B (2s²2p¹). Correct order is Li < B < Be < C.
Conclusion: Option (2) is correct.

Periodic Trends in Chemical Properties

Periodicity of Valence/Oxidation States

Core definition & intuition: Valence is the number of electrons in the outermost shell, or 8 minus that number. Oxidation state is the charge acquired based on electronegativity differences in a compound.
Crucial detail/Trap: In OF₂, F is -1, O is +2. In Na₂O, O is -2, Na is +1. Fluorine always shows a -1 oxidation state due to its highest electronegativity.

Anomalous Properties of 2nd Period Elements

Core definition & intuition: The first element of Groups 1, 2, and 13-17 (Li, Be, B, C, N, O, F) differs significantly from the rest of its group.
Reasons for Anomaly: Extremely small size, large charge/radius ratio, high electronegativity, and absence of d-orbitals.
Crucial detail/Trap: Absence of d-orbitals: Limits their maximum covalency to 4 (using one 2s and three 2p orbitals). They cannot expand their octet (e.g., B forms [BF₄]^- but Al forms [AlF₆]^3-). pπ-pπ Bonding: First members have a high tendency to form multiple bonds to themselves (C=C, N≡N) and other second period elements (C=O). Diagonal Relationship: Similarities between the first element of one group and the second element of the next group (Li & Mg, Be & Al, B & Si) due to similar charge/size ratios.

Chemical Reactivity & Oxides

Core definition & intuition: Reactivity is highest at the two extremes (Alkali metals via electron loss, Halogens via electron gain) and lowest in the center.
Mathematical formulation (Nature of Oxides):
Extreme Left (Metals): Strongly Basic Oxides (e.g., Na₂O).
Extreme Right (Non-metals): Strongly Acidic Oxides (e.g., Cl₂O₇).
Center Elements/Metalloids: Amphoteric Oxides (react with both acids and bases, e.g., Al₂O₃, As₂O₃, ZnO) or Neutral Oxides (CO, NO, N₂O).

PYQ Problem 51:
The correct sequence given below containing neutral, acidic, basic and amphoteric oxide each, respectively, is: NEET 2023
(1) NO, ZnO, CO₂, CaO
(2) ZnO, NO, CaO, CO₂
(3) NO, CO₂, ZnO, CaO
(4) NO, CO₂, CaO, ZnO

Step 1 - Concept: We must classify each oxide given in the options based on its acid-base character.
Step 2 - Identify Data: Neutral: NO, CO, N₂O. Acidic: Non-metal oxides like CO₂, SO₃, Cl₂O₇. Basic: Metal oxides like CaO, Na₂O. Amphoteric: ZnO, Al₂O₃.
Step 3 - Application: Matching these to the required sequence (neutral, acidic, basic, amphoteric) gives: NO, CO₂, CaO, ZnO.
Conclusion: Option (4) is correct.