JEE Main 2020 · 3 September · evening · Single correct · +4 / −1
Among the statements (I – IV), the correct ones are:
(I) Be has smaller atomic radius compared to Mg.
(II) Be has higher ionization enthalpy than Al.
(III) Charge/radius ratio of Be is greater than that of Al.
(IV) Both Be and Al form mainly covalent compounds.
- (a) (I), (II) and (IV)
- (b) (II), (III) and (IV)
- (c) (I), (III) and (IV)
- (d) (I), (II) and (III)
Answer and solution
Answer (a) (I), (II) and (IV)
Solution
Chemistry ma’am · 2 steps
Three statements are textbook facts; the fourth wrongly says Be and Al differ in charge/radius ratio.
- Diagonal relationship. Be (group 2, period 2) and Al (group 13, period 3) sit diagonally to each other. They behave alike because their ions have a nearly equal charge/radius ratio, which gives them similar polarising power.
Statement Check Verdict (I) Be smaller than Mg Same group, Be has one shell fewer: about 112 pm against 160 pm True (II) IE of Be above Al Small Be with a stable 2s² pair: about 899 against 577 kJ mol⁻¹ True (III) Charge/radius of Be greater than Al Be²⁺ and Al³⁺ have nearly the same ratio. That equality is why they resemble each other. False (IV) Both mainly covalent High polarising power distorts the anion's electron cloud, so BeCl₂ and AlCl₃ are covalent True - So. Statements (I), (II) and (IV) are correct and (III) is not. The answer is (a).
The idea to keep. Diagonal pairs (Li–Mg, Be–Al, B–Si) resemble each other because their charge/radius ratios are about equal. A statement that makes one ratio larger than the other is false.
Answer from the NTA final answer key. Solution by PrepWiser.