Modern Periodic Law and Nomenclature of Elements
Modern Periodic Law
- Core definition & intuition: The physical and chemical properties of the elements are periodic functions of their atomic numbers. This replaced Mendeleev’s law which was based on atomic mass.
- Crucial detail/Trap: Atomic number (Z), not atomic mass, is the fundamental property governing periodicity because it equals the nuclear charge and determines electronic configuration.
IUPAC Nomenclature for Elements Z > 100
- Core definition & intuition: Temporary IUPAC names for undiscovered or unnamed elements are derived from their atomic number using numerical roots: 0=nil, 1=un, 2=bi, 3=tri, 4=quad, 5=pent, 6=hex, 7=sept, 8=oct, 9=enn, followed by “ium”.
- Crucial detail/Trap: Element 119 has not been officially named yet; its temporary IUPAC name is Ununennium (from un-un-enn-ium = 1-1-9).
PYQ Problem 1:
The IUPAC name of an element with atomic number 119 is. (2022) NEET 2022
(1) Ununcotium (2) Ununennium (3) Unmilenaium (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 → 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.
Electronic Configurations and Types of Elements (s, p, d, f-Blocks)
General Electronic Configuration and Block Identification
- 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: ns1–2 (Groups 1–2)
p-block: ns2np1–6 (Groups 13–18)
d-block: (n−1)d1–10ns0–2 (Groups 3–12)
f-block: (n−2)f1–14(n−1)d0–1ns2 (Lanthanoids & Actinoids) - Crucial detail/Trap: Helium (1s2) is placed in the p-block (Group 18) despite its s2 configuration due to its noble gas chemical behavior. Palladium (Pd) has an anomalous configuration [Kr]4d10 (5s0).
PYQ Problem 2:
The element Z=114 has been discovered recently. It will belong to which of the following family group and electronic configuration? (2017-Delhi) NEET 2017
(1) Nitrogen family, [Rn] 5f46d17s27p3
(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. The nearest noble gas is Radon (Rn, Z=86). The remaining 28 electrons fill the 5f, 6d, and 7s subshells before entering 7p. The configuration is [Rn] 5f14 6d10 7s2 7p2.
Step 3 - Application: The outermost shell is n=7. It has 2 (from 7s) + 2 (from 7p) = 4 valence electrons. Group = 10 + 4 = 14. Group 14 is the Carbon family.
Conclusion: Option (3) matches both group and config and is correct.
PYQ Problem 3:
Which of the following configuration is correct for iron? (1999) AIPMT 1999
(1) 1s2, 2s22p6, 3s23p63d8
(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 1s2 2s2 2p6 3s2 3p6 4s2 3d6.
Step 3 - Application: Rearranging strictly by principal quantum number (n) gives: 1s2 2s2 2p6 3s2 3p6 3d6 4s2.
Conclusion: Option (4) is the correct standard representation.
PYQ Problem 4:
The electronic configuration of an element is 1s22s22p63s23p1. What is the atomic number of the element, which is just below the above element in the periodic table? (1995) 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: The given configuration ends in 3p1. Total electrons = 13 (Aluminium, Group 13). *Note: Standard AIPMT archives indicate a known typo in this specific year's paper where the intended configuration was 3p3 (Phosphorus, Z=15).*
Step 3 - Application: If we follow the standard intended question (Z=15, Group 15), the element just below it in Period 4 is Arsenic (As). The atomic number of As is 33. (Even if we strictly use Al, Ga is 31, which isn't an option, confirming the 3p3 typo theory).
Conclusion: Based on standard NEET/AIPMT question banks, Option (1) 33 is correct.
PYQ Problem 5:
If the atomic number of an element is 33, it will be placed in the periodic table in the: (1993) 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. The electronic configuration is [Ar] 3d10 4s2 4p3.
Step 3 - Application: The valence electrons are in the 4s and 4p subshells. Total valence electrons = 2 (from 4s) + 3 (from 4p) = 5. Group number = 10 + 5 = 15. In older notation, Group 15 is the Fifth group (Group VA).
Conclusion: Option (3) is correct.
PYQ Problem 6:
The number of d-electrons in Fe2+ (Z=26) is not equal to the number of electrons in which one of the following? (2015) NEET 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] 3d6 4s2. 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 (1s22s22p63s23p5): p-electrons = 6 (from 2p) + 5 (from 3p) = 11.
- Fe (neutral): 3d6 → 6 d-electrons.
- Ne (1s22s22p6): p-electrons = 6.
- Mg (1s22s22p63s2): s-electrons = 2 + 2 + 2 = 6.
Only Cl has a count (11) that is not equal to 6.
Conclusion: Option (1) is correct.
PYQ Problem 7:
Which of the following has more unpaired d-electrons? (1999) 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 by removing electrons from the highest principal quantum number (n) first.
Step 2 - Identify Data:
- Zn+ (Z=30): [Ar] 3d10 4s1 → 0 unpaired d-e⁻.
- Fe2+ (Z=26): [Ar] 3d6 → 4 unpaired d-e⁻ (↑↓ ↑ ↑ ↑ ↑).
- Ni3+ (Z=28): [Ar] 3d7 → 3 unpaired d-e⁻ (↑↓ ↑↓ ↑ ↑ ↑).
- Cu+ (Z=29): [Ar] 3d10 → 0 unpaired d-e⁻.
Step 3 - Application: Comparing the unpaired d-electron counts: Fe2+ has the maximum with 4.
Conclusion: Option (2) is correct.
PYQ Problem 8:
Which of the following has the smallest size? (1996) AIPMT 1996
(1) Al3+ (2) F- (3) Na+ (4) Mg
Step 1 - Concept: For isoelectronic species (same number of electrons), the ionic radius is inversely proportional to the nuclear charge (Z). The greater the number of protons, the tighter the electrons are pulled, resulting in a smaller size.
Step 2 - Identify Data: Al3+, F-, and Na+ all have 10 electrons. Their nuclear charges (Z) are: Al = 13, Na = 11, F = 9. Mg is a neutral atom with 12 electrons, making it significantly larger than the 10-electron ions.
Step 3 - Application: Among the isoelectronic species, Al3+ has the highest nuclear charge (13), meaning it pulls the 10 electrons the closest. Therefore, Al3+ is the smallest.
Conclusion: Option (1) is correct.
Periodic Trends in Properties of Elements
Atomic and Ionic Radii
- Core definition & intuition: Atomic radius decreases across a period (due to increasing effective nuclear charge, Zeff) and increases down a group (due to the addition of new electron shells, n).
- Crucial detail/Trap: Cations are always smaller than their parent atom (loss of outer shell/reduced shielding), and anions are larger (increased electron repulsion). Noble gas radii (van der Waals) are larger than the covalent radii of preceding halogens, so they are excluded from standard periodic trend comparisons.
PYQ Problem 9:
The correct decreasing order of atomic radii (pm) of Li, Be, B and C is (2024) 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 Zeff.
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 10:
Among the elements Ca, Mg, P and Cl the order of increasing atomic radii is: (2010 Mains) 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: We must apply both group and period trends. Size increases down a group and decreases across a period.
Step 2 - Identify Data: Ca and Mg are in Group 2; Ca is below Mg, so Ca > Mg. P and Cl are in Period 3; Cl is to the right of P, so P > Cl. Comparing metals (Mg) to non-metals (P, Cl) in the same period, metals are larger, so Mg > P > Cl.
Step 3 - Application: Combining these facts: Cl < P < Mg, and Mg < Ca. The full increasing order is Cl < P < Mg < Ca.
Conclusion: Option (2) is correct.
PYQ Problem 11:
Identify the wrong statement in the following: (2012 Pre) AIPMT 2012
(1) Atomic radius decreases left to right in 2nd period
(2) Amongst isoelectronic species, smaller positive charge → smaller ionic radius
(3) Amongst isoelectronic species, greater negative charge → larger ionic radius
(4) Atomic radius increases down Group 1
Step 1 - Concept: For isoelectronic species, the size is determined by the proton-to-electron ratio. More protons (higher positive charge on the ion) means a stronger pull on the same number of electrons, resulting in a smaller radius.
Step 2 - Identify Data: Let's evaluate statement (2): "smaller positive charge → smaller ionic radius". Consider Na+, Mg2+, Al3+ (all 10 e⁻). Na+ has the smallest positive charge (+1) but the largest radius. Al3+ has the highest positive charge (+3) and the smallest radius.
Step 3 - Application: Therefore, a smaller positive charge actually results in a larger ionic radius. Statement (2) is factually incorrect.
Conclusion: Option (2) is the wrong statement.
Ionization Enthalpy (IE)
- Core definition & intuition: The energy required to remove an electron from an isolated gaseous atom in its ground state. Generally, IE increases across a period and decreases down a group.
- Crucial detail/Trap: Watch out for half-filled and fully-filled orbital stability. Be (2s2) has a higher IE than B (2p1) because the 2s electrons are more penetrating and stable. Similarly, N (2p3) has a higher IE than O (2p4) due to the extra stability of the half-filled p-subshell.
PYQ Problem 12:
Arrange the following elements in increasing order of first ionization enthalpy: Li, Be, B, C, N (2024) 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: While IE generally increases across a period, we must account for the anomaly between Group 2 and Group 13 elements due to the penetration effect of s-orbitals vs p-orbitals.
Step 2 - Identify Data: Li (2s1) is the lowest. Be (2s2) is exceptionally stable. B (2p1) is easier to ionize than Be. C and N follow the normal trend, with N (2p3) being half-filled and very stable.
Step 3 - Application: The correct increasing order accounting for the Be > B anomaly is: Li < B < Be < C < N.
Conclusion: Option (4) is correct.
PYQ Problem 13: Step 1 - Concept: We must apply both periodic anomalies for Period 2: Be > B (s-orbital stability) and N > O (half-filled p-orbital stability). PYQ Problem 14: Step 1 - Concept: We need to find the option where the stated trend contradicts established chemical principles, specifically looking at the IE anomaly between Group 15 and 16. PYQ Problem 15: Step 1 - Concept: Ionization energy generally increases across a period, but Oxygen (Group 16) will have a lower IE than Nitrogen (Group 15) due to electron-electron repulsion in the paired 2p orbital of O. PYQ Problem 16: Step 1 - Concept: We must apply both the Be > B anomaly (fully filled 2s) and the N > O anomaly (half-filled 2p). PYQ Problem 17: Step 1 - Concept: Electronegativity increases across a period (left to right) and decreases down a group (top to bottom) on the Pauling scale. PYQ Problem 18: Step 1 - Concept: "Increasing electron gain enthalpy with negative sign" means ordering them from least negative (hardest to add an electron) to most negative (easiest to add an electron). Remember the anomaly: F < Cl and O < S. PYQ Problem 19: Step 1 - Concept: We must evaluate each statement based on periodic properties and the specific anomalies of the second-period elements. PYQ Problem 20: Step 1 - Concept: Electron affinity generally becomes more negative (higher affinity) across a period and less negative down a group. However, the second-period elements (F, O) have anomalously low electron affinities due to small size and high electron repulsion. PYQ Problem 21: Step 1 - Concept: To be isoelectronic, the total number of electrons (Atomic Number - Charge) must be identical for both species in the pair. PYQ Problem 22: Step 1 - Concept: For isoelectronic species, ionic radius is inversely proportional to the nuclear charge (Z). More protons = stronger pull = smaller radius. PYQ Problem 23: Step 1 - Concept: We must evaluate the isoelectronic series and the general rule for the radii of an atom versus its ions. An anion is always larger than its parent atom, which is larger than its cation. PYQ Problem 24: Step 1 - Concept: We must classify each oxide given in the options based on its acid-base character. PYQ Problem 25: Step 1 - Concept: To achieve the nearest noble gas configuration (Neon, 10 electrons), these elements will form ions: Na+, O2-, F-, and N3-. Since they are isoelectronic, the size is determined by the nuclear charge (Z).
For the second period elements, the correct increasing order of first ionisation enthalpy is: (2019) NEET 2019
(1) Li
Step 2 - Identify Data: Normal trend would be Li < Be < B < C < N < O < F < Ne. Applying the exceptions: B drops below Be, and O drops below N.
Step 3 - Application: The fully corrected sequence is Li < B < Be < C < O < N < F < Ne.
Conclusion: Option (2) is correct.
In which of the following options, the order of arrangement does not agree with the variation of property indicated against it? (2016) NEET 2016
(1) Li
(3) B
Step 2 - Identify Data: Let's check option (3): B < C < N < O. We know N has a half-filled p3 configuration, making it harder to remove an electron from N than from O. Therefore, the IE of N is greater than O.
Step 3 - Application: The correct order for IE should be B < C < O < N. Because option (3) lists N < O, it incorrectly represents the variation of ionization enthalpy.
Conclusion: Option (3) is the incorrect arrangement.
The correct order of first ionization enthalpy for C, N, O, F is: (2022 Re) NEET 2022
(1) C
Step 2 - Identify Data: Standard values (kJ/mol): C (1086), O (1314), N (1402), F (1681).
Step 3 - Application: Arranging these values in increasing order gives: C < O < N < F.
Conclusion: Option (4) is correct.
Correct order of I.E. among Be, B, C, N, O is: (2001) AIPMT 2001
(1) B
Step 2 - Identify Data: B is lower than Be. O is lower than N. The normal increasing trend for the rest is Be < C < N.
Step 3 - Application: Merging the anomalies into the base trend: B < Be < C < O < N.
Conclusion: Option (1) is correct.Electron Gain Enthalpy (EGE) and Electronegativity
Arrange in increasing order of electronegativity: N, O, F, C, Si (2024) NEET 2024
(1) O
Step 2 - Identify Data: Si is lowest (Group 14, Period 3). C, N, O, F are in Period 2. Their Pauling values are approximately: Si (1.8), C (2.5), N (3.0), O (3.5), F (4.0).
Step 3 - Application: Arranging them from lowest to highest gives: Si < C < N < O < F.
Conclusion: Option (3) is correct.
Which of the following represents the correct order of increasing electron gain enthalpy with negative sign for O, S, F, Cl? (2010 Pre) AIPMT 2010
(1) S
Step 2 - Identify Data: The approximate EGE values (kJ/mol) are: O (-141), S (-200), F (-328), Cl (-349).
Step 3 - Application: Ordering these from least negative to most negative yields: O < S < F < Cl.
Conclusion: Option (3) is correct.
Which statement is wrong? (2000) AIPMT 2000
(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 2 - Identify Data: Let's check statement (1). F2 has a lower bond dissociation energy than Cl2 because the extremely small size of the F atom leads to severe lone-pair to lone-pair repulsion in the F-F bond, weakening it.
Step 3 - Application: Since F2 bond energy < Cl2 bond energy, statement (1) claims the exact opposite and is therefore false. Statements (2), (3), and (4) are all factually correct.
Conclusion: Option (1) is the wrong statement.
Which of the following elements has the maximum electron affinity? (1999) AIPMT 1999
(1) I (2) Br (3) Cl (4) F
Step 2 - Identify Data: Comparing the halogens: Cl has the highest (most negative) electron affinity (-349 kJ/mol), followed by F (-328 kJ/mol), Br, and I.
Step 3 - Application: Chlorine has the maximum electron affinity among all elements in the periodic table.
Conclusion: Option (3) is correct.Isoelectronic Species and Ionic Radii
From the following pairs of ions, which one is not an iso-electronic pair? (2021) NEET 2021
(1) Na+, Mg2+ (2) Mn2+, Fe3+ (3) Fe2+, Mn2+ (4) O2-, F-
Step 2 - Identify Data:
- Na+ (11-1=10) and Mg2+ (12-2=10) → Isoelectronic.
- Mn2+ (25-2=23) and Fe3+ (26-3=23) → Isoelectronic.
- Fe2+ (26-2=24) and Mn2+ (25-2=23) → Not isoelectronic.
- O2- (8+2=10) and F- (9+1=10) → Isoelectronic.
Step 3 - Application: Only the pair Fe2+ and Mn2+ has a mismatched electron count (24 vs 23).
Conclusion: Option (3) is correct.
The species Ar, K+ and Ca2+ contain the same number of electrons. In which order do their radii increase? (2015) NEET 2015
(1) Ca2+ < Ar < K+ (2) Ca2+ < K+ < Ar (3) K+ < Ar < Ca2+ (4) Ar < K+ < Ca2+
Step 2 - Identify Data: All three have 18 electrons. Their atomic numbers (Z) are: Ca = 20, K = 19, Ar = 18.
Step 3 - Application: Since Ca2+ has the highest nuclear charge (20), it is the smallest. Ar has the lowest nuclear charge (18), so it is the largest. The increasing order of radii is Ca2+ < K+ < Ar.
Conclusion: Option (2) is correct.
Which of the following orders of ionic radii is correctly represented? (2014) NEET 2014
(1) Na+ > F- > O2- (2) O2- > F- > Na+ (3) Al3+ > Mg2+ > N3- (4) H- > H > H+
Step 2 - Identify Data: Let's check option (4): H- has 2 electrons and 1 proton. H has 1 electron and 1 proton. H+ has 0 electrons and 1 proton. The size strictly decreases as electrons are removed due to reduced shielding and repulsion.
Step 3 - Application: Therefore, H- > H > H+ is a universally correct representation of ionic/atomic radii. (Note: Option 2 is also technically correct for the 10-electron series, but Option 4 is the primary intended answer in this specific AIPMT context).
Conclusion: Option (4) is correct.Chemical Properties and Oxides
The correct sequence containing neutral, acidic, basic and amphoteric oxide each, respectively, is (2023-Manipur) NEET 2023
(1) NO, ZnO, CO2, CaO (2) ZnO, NO, CaO, CO2 (3) NO, CO2, ZnO, CaO (4) NO, CO2, CaO, ZnO
Step 2 - Identify Data:
- NO (Nitric oxide) is a neutral oxide.
- CO2 (Carbon dioxide) is a non-metal oxide, hence acidic.
- CaO (Calcium oxide) is a metal oxide, hence basic.
- ZnO (Zinc oxide) reacts with both acids and bases, hence amphoteric.
Step 3 - Application: Matching these to the required sequence (neutral, acidic, basic, amphoteric) gives: NO, CO2, CaO, ZnO.
Conclusion: Option (4) is correct.
The element expected to form largest ion to achieve nearest noble gas configuration is: (2023) NEET 2023
(1) Na (2) O (3) F (4) N
Step 2 - Identify Data: The atomic numbers (number of protons) are: Na = 11, O = 8, F = 9, N = 7.
Step 3 - Application: The ion with the fewest protons will hold the 10 electrons the least tightly, resulting in the largest radius. Nitrogen has the lowest Z (7), so N3- will be the largest ion.
Conclusion: Option (4) is correct.Select an Option