JEE Main 2023 · 13 April · morning · Single correct · +4 / −1
Be(OH)₂ reacts with Sr(OH)₂ to yield an ionic salt.
Choose the incorrect option related to this reaction from the following :
- (a) Be is tetrahedrally coordinated in the ionic salt.
- (b) The element Be is present in the cationic part of the ionic salt.
- (c) Both Sr and Be elements are present in the ionic salt.
- (d) The reaction is an example of acid - base neutralization reaction.
Answer and solution
Answer (b) The element Be is present in the cationic part of the ionic salt.
Solution
Chemistry ma’am · 3 steps
Amphoteric Be(OH)₂ plays the acid here, so beryllium ends up inside the anion, not the cation.
- Who is the acid. Be(OH)₂ is amphoteric because the small, highly charged Be²⁺ holds its electrons tightly. Sr(OH)₂ is a strong base, so against it Be(OH)₂ behaves as the acid.
Be(OH)₂ + Sr(OH)₂ → Sr[Be(OH)₄]
- Read the salt. The salt is strontium beryllate. Sr²⁺ is the cation and [Be(OH)₄]²⁻ is the anion. In that anion Be is bonded to four OH⁻ groups, so it is tetrahedral (sp³).
Option Statement Verdict (a) Be is tetrahedral correct: four OH⁻ around Be in [Be(OH)₄]²⁻ (b) Be is in the cation incorrect: Be is in the anion (c) Both Sr and Be are in the salt correct: Sr[Be(OH)₄] (d) Acid-base neutralization correct: amphoteric Be(OH)₂ acts as the acid, Sr(OH)₂ as the base - So. The incorrect statement is (b). Beryllium sits in the anion [Be(OH)₄]²⁻, and strontium is the cation.
The idea to keep. When an amphoteric hydroxide meets a strong base, the amphoteric metal goes into an oxo or hydroxo anion, like [Be(OH)₄]²⁻ or [Al(OH)₄]⁻. Be always caps at four groups, so it is tetrahedral.
Answer from the NTA final answer key. Solution by PrepWiser.