| The successive ionization energies (IE) for an element A are as follows: |
| \[\operatorname{A}\xrightarrow{I{{E}_{1}}}{{A}^{+}}\xrightarrow{I{{E}_{2}}}{{A}^{2+}}\xrightarrow{I{{E}_{3}}}{{A}^{3+}}\to ....\] |
| If the \[{{\operatorname{IE}}_{1}}\], and \[{{\operatorname{IE}}_{2}}\] values are \[27\text{ }kJmo{{l}^{-1}}\]and \[51\text{ }kJ\,mo{{l}^{-1}}\] respectively, then the value of is \[I{{E}_{2}}\]_____\[kJ\,mo{{l}^{-1}}\] |
A) 21
B) 33
C) 59
D) 63
Correct Answer: B
Solution :
| Successive ionization energy always increases for any species because the e/p ratio gradually decreases. Thus, |
| \[\Alpha \xrightarrow{\Iota {{\Epsilon }_{1}}}{{\Alpha }^{+}}\xrightarrow{\Iota {{\Epsilon }_{2}}}{{\Alpha }^{2+}}\xrightarrow{\Iota {{\Epsilon }_{3}}}{{\Alpha }^{3+}}\to ...\] |
| \[I{{E}_{1}}<I{{E}_{2}}<I{{E}_{3}}\] |
| \[I{{E}_{1}}=27kJ/mo{{l}^{-1}}; I{{E}_{3}} = 51 kJ/mo{{l}^{-1}}\] |
| \[I{{E}_{2}}\] can be exist between \[I{{E}_{1}}\And I{{E}_{3}}\] value Therefore option 'b' is correct, having value 33. |
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