| Column - I | Column - II | ||
| (A) | \[He\] | (i) | High electron affinity |
| (B) | \[Cl\] | (ii) | Most electropositive element |
| (C) | \[Ca\] | (iii) | Strongest reducing agent |
| (D) | \[Li\] | (iv) | Hishest lonisation energy |
A) A-iii, B-i, C-ii, D-iv
B) A-iv, B-iii, C-ii, D-i
C) A-ii, B-iv, C-i, D-iii
D) A-i, B-ii, C-iii, D-iv
E) A-iv, B-i, C-ii, D-iii
Correct Answer: E
Solution :
(i) For noble gases (eg. He), ionization energy is highest due to their completely filled electronic configuration. (ii) Generally electron affinity increases in a period (from IA to VII A group) and decreases in a group but electron affinity is highest for chlorine\[(Cl)\](due to smaller size and high electron density of fluorine). (iii) The ionization energy is lowest for Li, so it can lose electrons very easily, thus it behaves as a strongest reducing agent. (iv) Electropositive character generally decreases in a period (from left to right) and increases in a group (on moving down), thus Ca is the most electropositive element among the given. Hence, on the basis of above facts, the correct matches are (A)-iv (B)-i (C)-ii (D)-iii
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