Anomalous Properties — Li and Be
s-block Elements
5
JEE Qs
8%
Hard
60
min
Focus on understanding the fundamental reasons (size, charge/radius ratio, d-orbitals) behind the anomalous properties and diagonal relationships, as these principles are tested more than rote memorization of individual facts.
✅ Key Points for JEE
- 1Li and Be exhibit anomalous behavior due to their exceptionally small atomic/ionic size, very high ionization enthalpy, high electronegativity, and the highest charge/radius ratio (polarizing power) within their respective groups.
- 2The high polarizing power of Li+ and Be2+ leads to significantly greater covalent character in their compounds compared to the predominantly ionic nature of other group members.
- 3Key anomalous properties include: Li forming a nitride with N2, non-formation of ethynides by Li, the amphoteric nature of BeO/Be(OH)2, unique reactions with oxygen/water, and formation of stable complex ions.
- 4Due to similar charge/radius ratio and polarizing power, Li exhibits a diagonal relationship with Mg, and Be with Al, resulting in many similar properties and reactions.
- 5The absence of d-orbitals in their valence shell restricts their maximum coordination number (e.g., Be can only form 4 bonds) and prevents expansion of octet.
⚠️ Common Mistakes
- ✕Confusing the underlying reasons for anomalous behavior (e.g., small size, high polarizing power) with the specific anomalous properties observed.
- ✕Incorrectly identifying which specific properties are anomalous for Li or Be, or misattributing properties of one to the other (e.g., assuming Li forms amphoteric oxide).
- ✕Failing to recall specific reactions associated with anomalous behavior, such as Li reacting directly with nitrogen or the amphoteric nature of beryllium compounds.
📝 Practice Questions
See allQ49. List - I List - II Reaction Type of redox reaction (A) Li (I) 589.2 Match List - I with List - II. Choose the correct answer from the options (B) Na (II) 455.5 (C) Rb (III) 670.8 (D) Cs (IV) 780.0 given below : (1) (A)-(II), (B)-(IV), (C)-(III), (D)-(I) (2) (A)-(I), (B)-(IV), (C)-(III), (D)-(II) (3) (A)-(IV), (B)-(II), (C)-(I), (D)-(III) (4) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
Q40.Ion having highest hydration enthalpy among the given alkaline earth metal ions is: (1) Be2+ (2) Sr2+ (3) Ba2+ (4) Ca2+
Q32.Match List I with List II (A) Slaked lime (I) NaOH (B) Dead burnt plaster (II) Ca ( OH ) 2 (C) Caustic soda (III) Na2CO3 · 10H2O (D) Washing soda (IV) CaSO4 Choose the correct answer form the options given below: (1) (A) - I, (B) - IV, (C) - II, (D) - III (2) (A) - III, (B) - IV, (C) - II, (D) - I (3) (A) - II, (B) - IV, (C) - I, (D) - III (4) (A) - III, (B) - II, (C) -IV, (D) -I
Q33.Choose the correct statement(s): A. Beryllium oxide is purely acidic in nature. B. Beryllium carbonate is kept in the atmosphere of CO2. C. Beryllium sulphate is readily soluble in water. D. Beryllium shows anomalous behavior. Choose the correct answer from the options given below: JEE Main 2023 (01 Feb Shift 1) JEE Main Previous Year Paper (1) A, B and C only (2) B,C and D only (3) A and B only (4) A only
Q40.The correct order of bond enthalpy ( kJ mol–1 ) is: (1) Si −Si > C −C > Sn −Sn > Ge −Ge (2) Si −Si > C −C > Ge −Ge > Sn −Sn (3) C −C > Si −Si > Sn −Sn > Ge −Ge (4) C −C > Si −Si > Ge −Ge > Sn −Sn
Q37. O −O bond length in H2O2 is X than the O −O bond length in F2O2 . The O −H bond length in H2O2 is Y ¯¯ than that of the O −F bond in F2O2 . Choose the correct option for X and Y from the given below. ¯¯ (1) X - shorter, Y− shorter (2) X -shorter, Y -longer (3) X -longer, Y -longer (4) X-longer, Y - shorter
NCERT Chapters
- Class 11 Chemistry Ch 10: The s-Block Elements