Consider the following ionisation reactions : I.E. (kJ mol-1) A(g) - A+(g) + e-, A1; B+(g) - B2+(g) + e-, B2; C+(g) - C2+(g) + e-, C2. I.E. (kJ mol-1) B(g) - B+(g) + e-, B1; C(g) - C+(g) + e-, C1

Question

Consider the following ionisation reactions : I.E. (kJ mol-1) A(g) -> A+(g) + e-, A1; B+(g) -> B2+(g) + e-, B2; C+(g) -> C2+(g) + e-, C2. I.E. (kJ mol-1) B(g) -> B+(g) + e-, B1; C(g) -> C+(g) + e-, C1; C2+(g) -> C3+(g) + e-, C3. If monovalent positive ion of A, divalent positive ion of B and trivalent positive ion of C have zero electron. Then incorrect order of corresponding I.E. is :

Prashant Kotian · Accepted Answer

Correct answer: D. Step 1: Recall that electropositive (metallic) character increases down a group and decreases across a period. Step 2: Group 1 alkali metals are the most electropositive elements. Among the choices, Rubidium (Rb) is an alkali metal located in the 5th period. Step 3: Thus, Rubidium is the most electropositive element, corresponding to option (b).

💡 Solution & Explanation

Practice 22,000+ questions like this