Classification of Elements and Periodicity — NEET Chemistry MCQs with answers

Classification of Elements and Periodicity is a Class 11 Chemistry chapter in the NEET (UG) syllabus. NEET720 has 455 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Periodic trends: radius, IE, EA, EN · Modern periodic table and nomenclature · Valency and oxidation state trends · Anomalous behaviour and diagonal relationship · Modern Periodic Law · Atomic Radius · Ionization Enthalpy · Electron Gain Enthalpy · Electronegativity · Diagonal Relationship · Lattice Energy · Ionic Bonding · Fajans Rules · Lewis Structures · VSEPR Theory · Hybridization · Molecular Orbital Theory · Dipole Moment · Resonance · Hydrogen Bonding · Valence Bond Theory · Periodic Trends · Bond Length and Bond Order · Nomenclature of Elements · Covalent Character · Bond Parameters · Electronic Configuration Exceptions · Covalent Bonding · Ionic Radius · Bond Angle · Periodic Classification · Bond Length · Periodic Table & Periodicity · Periodic Table · Classification of Elements · Periodic Trends · Periodic Trends · Ionization Enthalpy · Electron Gain Enthalpy · Electronegativity

8 free Classification of Elements and Periodicity practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · medium · Anomalous behaviour and diagonal relationship

   The first element of a group in the p-block or s-block often shows anomalous behaviour compared to other members of its group mainly because it has

   • A.a fully filled f-subshell that shields the nucleus poorly
   • B.the lowest ionization enthalpy in its group
   • C.the highest atomic mass among group members
   • D.no d-orbitals available in the valence shell and a comparatively small size with high charge density
   Show answer and explanation

   Answer: D. no d-orbitals available in the valence shell and a comparatively small size with high charge density

   The first element of a group (period 2 mainly) lacks d-orbitals in its valence shell, has an unusually small size and high nuclear charge density, giving it different bonding capability, maximum covalency, and diagonal similarity to a period-3 element one group to the right.

   Period 2 elements (Li to F) show anomalous behaviour relative to their own group because: (i) they have only 2s and 2p orbitals available (no 2d), restricting their maximum covalency to 4 (e.g., cannot form NF5 or BF6^3- easily as heavier congeners can with d-orbitals); (ii) their atomic/ionic size is exceptionally small with high charge/radius ratio, giving unusually high electronegativity, ionization enthalpy and polarizing power compared to the rest of the group; (iii) absence of d-orbitals also affects pπ-pπ multiple bonding ability (e.g., N2 has a strong pπ-pπ triple bond, unlike P which prefers single bonds). This combination — small size, no valence d-orbitals, high charge density — is the root cause of the first-member anomaly, and it is this same high charge density that drives diagonal relationships (Li-Mg, Be-Al, B-Si).

   Common mistake: Attributing the anomaly to f-orbital shielding or misremembering the IE/mass trend direction

2. Question 2 · easy · Anomalous behaviour and diagonal relationship

   Which pair of elements shows a well-known diagonal relationship due to comparable charge/radius ratios?

   • A.Li and Na
   • B.Be and Ca
   • C.B and Al
   • D.Li and Mg
   Show answer and explanation

   Answer: D. Li and Mg

   Li (group 1, period 2) and Mg (group 2, period 3) have similar charge/radius ratios, giving them comparable chemical behaviour — this is the classic Li-Mg diagonal relationship.

   A diagonal relationship arises when an element in period 2 resembles the element diagonally below-right of it in period 3, because both have similar polarizing power (charge/size ratio), even though they belong to different groups. The three classic pairs are Li-Mg, Be-Al, and B-Si. Options B, C, D list same-group (vertical) pairs, not diagonal pairs, so they are incorrect despite genuine chemical similarities existing within those groups too.

   Common mistake: Confusing same-group similarity with diagonal relationship

3. Question 3 · medium · Anomalous behaviour and diagonal relationship

   Consider the following statements about lithium's anomalous behaviour compared to other alkali metals: (I) LiCl is appreciably soluble in ethanol/pyridine, showing greater covalent character than NaCl. (II) Li reacts directly with N2 on heating to form the nitride Li3N, while Na does not form a nitride under similar conditions. (III) Li2CO3 is thermally less stable than Na2CO3 and decomposes at a comparatively lower temperature on heating, resembling the behaviour of MgCO3. (IV) LiF and Li2O are much less soluble in water than the corresponding NaF and Na2O. How many of the above statements are correct?

   • A.All four
   • B.Only three
   • C.Only two
   • D.Only one
   Show answer and explanation

   Answer: A. All four

   All four are genuine, textbook-recognized anomalies of lithium arising from its exceptionally small size and high polarizing power/charge density: covalent character in LiCl, unique nitride formation, lower thermal stability of the carbonate (paralleling Mg, its diagonal partner), and low solubility of the small-anion salts LiF/Li2O.

   Lithium's unusually small ionic size and high charge density make it behave differently from the rest of Group 1, often resembling Mg (its diagonal partner) instead. (I) Li+, with high polarizing power, distorts the Cl- electron cloud strongly, giving LiCl significant covalent character, so it dissolves in organic solvents like ethanol — NaCl does not. (II) Only Li among alkali metals combines directly with N2 on heating to give Li3N; this is linked to its high lattice energy favourably compensating the endothermic nitride formation, an anomaly not seen for Na, K, etc. (III) Small, highly polarizing Li+ destabilizes the large CO3^2- ion (through polarization/covalent character in the C-O bonds), so Li2CO3 decomposes to Li2O + CO2 at a lower temperature than Na2CO3 — exactly the behaviour also seen for MgCO3, reinforcing the diagonal relationship. (IV) LiF and Li2O have high lattice energies (from Li+'s small size pairing with small F-/O2- ions) that are not compensated well by hydration energy, making them poorly soluble compared to the corresponding Na salts. All four statements are therefore correct.

   Common mistake: Assuming smaller cation always gives a more thermally stable carbonate, missing the polarization-driven destabilization

4. Question 4 · medium · Anomalous behaviour and diagonal relationship

   Beryllium shows a diagonal relationship with aluminium. Which observation below is direct evidence of this similarity, rather than of Be's normal Group 2 trend?

   • A.Be has a smaller ionization enthalpy than Mg
   • B.Be reacts with water more readily than Ca
   • C.Be(OH)2 is amphoteric, dissolving in both NaOH and HCl, just like Al(OH)3
   • D.BeO has a higher melting point than MgO
   Show answer and explanation

   Answer: C. Be(OH)2 is amphoteric, dissolving in both NaOH and HCl, just like Al(OH)3

   The amphoteric nature of Be(OH)2, matching Al(OH)3, is a hallmark of the Be-Al diagonal relationship; Mg(OH)2 by contrast is only basic, showing this is not a normal group-2 property.

   Diagonal relationship evidence must show Be behaving like Al rather than like Mg/Ca (its own group). Amphoterism is exactly such evidence: Be(OH)2 dissolves in excess NaOH to give beryllate ion [Be(OH)4]2-, just as Al(OH)3 dissolves to give aluminate [Al(OH)4]-; Mg(OH)2, in contrast, is a normal base and does not dissolve in excess alkali. Options B and D discuss comparisons within Group 2 (Be vs Mg, Be vs Ca) and are not diagonal evidence; D is also factually incorrect since Be, due to a protective oxide layer and high charge density leading to covalent, kinetically inert bonding, does not react with water even on boiling. Option C misstates the IE trend (Be has higher IE than Mg, consistent with small size/high nuclear charge, not lower).

   Common mistake: Citing a Group 2 intra-family trend as diagonal-relationship evidence

5. Question 5 · medium · Anomalous behaviour and diagonal relationship

   A student argues: "Boron and silicon must be chemically similar because they are adjacent in the periodic table." Which statement correctly evaluates this reasoning?

   • A.The reasoning is flawed; B and Si are diagonal neighbours (not adjacent in the same row or column), and their similarity arises from comparable charge/radius ratios, not mere proximity
   • B.The reasoning is correct; any two elements close together in the periodic table show strong chemical similarity
   • C.The reasoning is flawed because B and Si actually belong to the same group
   • D.The reasoning is flawed because B and Si show no significant chemical similarities
   Show answer and explanation

   Answer: A. The reasoning is flawed; B and Si are diagonal neighbours (not adjacent in the same row or column), and their similarity arises from comparable charge/radius ratios, not mere proximity

   B and Si are a diagonal pair, not simply 'adjacent'; the similarity is real but stems from comparable charge/radius ratio and covalent character, not from mere positional closeness, which is a common misconception.

   The student's reasoning conflates 'adjacent' with 'diagonal' and misattributes the cause. B (Group 13, Period 2) and Si (Group 14, Period 3) are diagonally placed, and their genuine similarities — both form acidic, covalent, network-type oxides (B2O3, SiO2), both give volatile, thermally unstable, spontaneously flammable hydrides (boranes, silanes), both are semiconductors/metalloids — arise because their comparable charge/radius ratios give similar polarizing power and covalent bonding tendency, not because they happen to sit near each other on the chart. Option D wrongly denies real, well-documented similarities. Option C misstates their group numbers.

   Common mistake: Believing any positional closeness in the periodic table implies chemical similarity

6. Question 6 · easy · Anomalous behaviour and diagonal relationship

   Unlike other alkaline earth metal carbonates and hydroxides, which decompose only on strong heating, beryllium and magnesium hydroxides/carbonates decompose relatively easily on heating. This shared behaviour of Be and Mg is best explained by their

   • A.stronger metallic bonding in the solid state
   • B.small cationic size and high polarizing power, which are comparatively closer to each other than to the larger, less polarizing heavier members (Ca, Sr, Ba)
   • C.identical number of valence electrons across all Group 2 members
   • D.greater atomic mass compared to the heavier alkaline earth metals
   Show answer and explanation

   Answer: B. small cationic size and high polarizing power, which are comparatively closer to each other than to the larger, less polarizing heavier members (Ca, Sr, Ba)

   Be2+ and Mg2+ are small, highly polarizing cations relative to Ca2+, Sr2+, Ba2+; this polarizing power destabilizes the large carbonate/hydroxide anions, lowering their decomposition temperature compared to the heavier, larger, less polarizing Group 2 cations.

   Down Group 2, cationic radius increases and polarizing power (charge/radius ratio) decreases from Be2+ to Ba2+. Be2+ and Mg2+, being the smallest members, have significantly higher polarizing power than Ca2+/Sr2+/Ba2+. High polarizing power distorts the large CO3^2-/OH- anion's electron cloud, weakening the ionic lattice and inducing covalent character, which lowers the thermal decomposition temperature of the carbonate/hydroxide. Since Be2+ and Mg2+ are both 'small and polarizing' relative to the rest of the group, they behave similarly to each other and differently from Ca, Sr, Ba — this is the standard explanation for the trend in thermal stability of Group 2 carbonates/hydroxides increasing down the group.

   Common mistake: Attributing the trend to atomic mass or metallic bonding instead of ionic polarizing power

7. Question 7 · easy · Anomalous behaviour and diagonal relationship

   Which of the following statements about beryllium is INCORRECT?

   • A.Be forms a large number of covalent compounds, unlike the predominantly ionic compounds of Mg, Ca, Sr, Ba
   • B.BeO and Be(OH)2 are amphoteric, unlike the basic oxides/hydroxides of the rest of the group
   • C.Be2+ readily attains a coordination number of 6, exactly like Mg2+ and the heavier Group 2 cations, since ionic size has no bearing on coordination number
   • D.Be salts undergo appreciable hydrolysis in water because of the high polarizing power of the small, highly charged Be2+ ion
   Show answer and explanation

   Answer: C. Be2+ readily attains a coordination number of 6, exactly like Mg2+ and the heavier Group 2 cations, since ionic size has no bearing on coordination number

   Be2+ is restricted mainly to coordination number 4 (using only 2s/2p orbitals, sp3 hybridization) because it lacks valence d-orbitals and is too small to accommodate 6 ligands; Mg2+ and heavier Group 2 cations readily show coordination number 6, so option C's claim is false.

   This is a 'find the incorrect statement' question. (A) is true: Be's small size and high charge density give its bonds significant covalent character, unlike the largely ionic bonding of heavier Group 2 elements. (B) is true: BeO and Be(OH)2 are amphoteric (react with both acids and bases), a hallmark anomaly linked to its diagonal relationship with Al; Mg(OH)2 and beyond are only basic. (C) is FALSE and is the answer: because Be lacks valence d-orbitals (period 2, only 2s/2p available) and has a very small ionic radius, Be2+ is essentially restricted to coordination number 4 (as in [BeF4]2-, [Be(H2O)4]2+), unlike Mg2+ and heavier Group 2 cations, which comfortably reach coordination number 6 using larger orbital sets/size. Ionic size absolutely does affect coordination number, contrary to the claim. (D) is true: high polarizing power of small, doubly-charged Be2+ causes extensive hydrolysis of its salts in water (e.g., BeCl2 solutions are acidic).

   Common mistake: Assuming coordination number is unaffected by ionic size/orbital availability and that all Group 2 cations behave alike

8. Question 8 · hard · Anomalous behaviour and diagonal relationship

   Boron chloride, BCl3, is a monomeric planar molecule and acts as a strong Lewis acid, while aluminium chloride, AlCl3, exists as a dimer (Al2Cl6) in the vapour phase/non-polar solvents. Despite this structural difference, both are classified as showing the diagonal-relationship-consistent property of being strong Lewis acids with electron-deficient bonding character. Why does AlCl3 dimerize while BCl3 does not, even though both are electron-deficient trihalides?

   • A.AlCl3 dimerizes because aluminium has a lower oxidation state than boron in these compounds
   • B.Aluminium, being larger and in period 3, can accommodate a fourth chlorine via a coordinate bond using an available empty orbital, forming bridging Cl atoms; boron's much smaller size and period-2 orbital limitations make such bridging in BCl3 geometrically and electronically unfavourable, so it satisfies its electron deficiency instead through strong Cl-to-B pπ-pπ back donation
   • C.BCl3 does not dimerize because chlorine cannot bridge between two boron atoms under any circumstances
   • D.Boron is more electronegative than aluminium, so BCl3 refuses to dimerize
   Show answer and explanation

   Answer: B. Aluminium, being larger and in period 3, can accommodate a fourth chlorine via a coordinate bond using an available empty orbital, forming bridging Cl atoms; boron's much smaller size and period-2 orbital limitations make such bridging in BCl3 geometrically and electronically unfavourable, so it satisfies its electron deficiency instead through strong Cl-to-B pπ-pπ back donation

   BCl3's boron, being small and period-2, can use empty 2p orbital to accept back-donated lone pair density from filled Cl 3p orbitals (pπ-pπ back bonding), largely satisfying its electron deficiency without dimerizing; Al, being larger with weaker back bonding, instead dimerizes via chlorine bridges to satisfy its electron deficiency.

   Both B and Al in their trichlorides are electron-deficient (6 electrons around the central atom), but they resolve this deficiency differently. In BCl3, boron's empty 2p orbital is well-matched in size/energy with chlorine's filled 3p orbitals, allowing significant pπ(Cl)-pπ(B) back donation that partially satisfies boron's electron deficiency and also shortens/strengthens the B-Cl bond; this makes BCl3 stable as a planar monomer and a comparatively weaker Lewis acid than expected from electron deficiency alone (still a good Lewis acid, but the back bonding partially quenches its acidity, e.g., weaker Lewis acid than BF3's back-bonding logic aside). Aluminium's larger 3p orbital overlaps poorly with Cl's 3p orbitals for effective pπ-pπ back bonding, so instead AlCl3 dimerizes: one lone pair from a Cl atom on one AlCl3 unit donates into an empty orbital on the Al of a neighbouring unit, forming two bridging Cl atoms and completing each Al's coordination/electron requirement via a coordinate covalent bridge, giving Al2Cl6. This is a direct consequence of the size/period-2-vs-period-3 anomaly discussed for diagonal-relationship elements, even though the final Lewis acidic character of both compounds is broadly similar in application (both catalyze Friedel-Crafts type reactions).

   Common mistake: Attributing the BCl3/AlCl3 structural difference to electronegativity or oxidation state rather than orbital size/back-bonding effects

Questions about Classification of Elements and Periodicity for NEET

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