Exams › JEE Advanced › Chemistry
In the reaction BeCl2 + LiAlH4 → X + LiCl + AlCl3, compound X is formed. How many 3-centre 4-electron (3c-4e) bonds are present in the monomer form of X?
- 1
- 2
- 3
- 0
Correct answer: 0
Solution
The reaction gives X = BeH2. In the monomer form, BeH2 is a linear molecule (Be is sp hybridized) with two normal 2c-2e Be-H bonds. The 3c-2e bonds appear in the polymer/dimer form (bridging H), and 3c-4e bonds are seen in hypervalent molecules like PCl5 axial bonds. The monomer BeH2 has zero 3c-4e bonds.
Related JEE Advanced Chemistry questions
- When carbonates from the s-block break down, a gas is released that is denser than air and extinguishes a burning splint. Identify the gas.
- What is the classification of Be2C and Al4C3 based on their chemical composition?
- When heated, magnesium nitrate and barium nitrate break down as follows: 2M(NO3)2 → heat → 2MO + 4NO2(g) + O2(g) (where M = Mg or Ba). Which statement is accurate regarding their decomposition behavior?
- Consider the following reactions of Group IA (alkali) metals with liquid ammonia: M + (x+y) NH3 -> [M(NH3)ₓ]+ + [e(NH3)_y]- [M(NH3)ₓ]+ + e-(in NH3) -> M-NH2(aq) + (1/2) H2(g) Which of the following statements about these solutions is INCORRECT?
- How many of the following elements give a characteristic flame test color: Li, Na, K, Be, Mg, Ca, Ba, Cs?
- Arrange the following carbonates in the correct increasing order of their thermal stability: BeCO3, MgCO3, CaCO3, K2CO3.
⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →