See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

(A)     2 2 4 Be OH 2OH Be OH       (B) Isoelectric point can be calculated by using relation 1 2 a a pK pK PI 2   and 2 3 a a pK pK PI 2   (C) Fact (D)   2 2 4 BeF 2NaF Na BeF  