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`IE_(1), IE_(2)` and `IE_(3)` values are `100, 150` and `1500 eV` respectively. The element can beA. `Na`B. `B`C. `Be`D. `F` |
Answer» Correct Answer - C There is a high jump from `IE_(2)` to `IE_(3)`. Therefore, it is difficult to remove the 3rd valence electron. So, the element must be of group 2, e.g. `Be(2s^(2))`. |
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