A) \[1{{s}^{2}},2{{s}^{2}},2{{p}^{5}}\]
B) \[1{{s}^{2}},\text{ }2{{s}^{2}},\text{ }2{{p}^{6}},\text{ }3{{s}^{2}},\text{ }3{{p}^{2}}\]
C) \[1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{1}}\]
D) \[1{{s}^{2}},2{{s}^{2}},2{{p}^{6}}\]
Correct Answer: C
Solution :
| (1) This electron configuration corresponds to fluorine (atomic number 9). Across the period size decreases with increase in nuclear charge. Hence it has higher for ionisation energy but less than next noble gas. |
| (2) This configuration correspond to silicon (3rd period). |
| (3) This correspond to first element of 3rd period i.e. Na is bigger than Si. So it has lower ionisation energy then Si. |
| (4) This electron configuration corresponds to the inert gas i.e. Ne which will have the highest ionisation energy then Si. |
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