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The IE1 , IE2 , IE3 and IE4 of an eleme...

The `IE_1 , IE_2 , IE_3` and `IE_4` of an element are 5.7, 11.03, 42.89, 57 eV respectively. The element is likely to be

A

P

B

Sr

C

F

D

Cs

Text Solution

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The correct Answer is:
To determine the element based on the provided ionization energies (IE1, IE2, IE3, and IE4), we can analyze the given values step by step. ### Step 1: Analyze the Ionization Energies The provided ionization energies are: - IE1 = 5.7 eV - IE2 = 11.03 eV - IE3 = 42.89 eV - IE4 = 57 eV ### Step 2: Identify Trends in Ionization Energies 1. **First Ionization Energy (IE1)**: This is relatively low, indicating that the first electron can be removed easily. 2. **Second Ionization Energy (IE2)**: This is higher than IE1 but still not excessively high, suggesting that the removal of the second electron is also relatively easy. 3. **Third Ionization Energy (IE3)**: This shows a significant jump compared to IE2, indicating that after removing two electrons, the atom reaches a more stable electronic configuration, making it much harder to remove the third electron. 4. **Fourth Ionization Energy (IE4)**: The value is high, but it is not as high as IE3, indicating that removing the fourth electron is still difficult but not as much as the third. ### Step 3: Determine the Likely Element The significant increase between IE2 and IE3 suggests that the element likely achieves a stable electronic configuration after losing two electrons. This is characteristic of elements that have a full outer shell after losing two electrons, which is typical for alkaline earth metals. ### Step 4: Identify the Element - **Strontium (Sr)** is an alkaline earth metal with the electronic configuration of [Kr] 5s². - When it loses two electrons, it achieves the stable configuration of [Kr], which is noble gas configuration. - The large jump in ionization energy from IE2 to IE3 indicates that the element has achieved a stable noble gas configuration after losing two electrons. ### Conclusion Based on the analysis of the ionization energies, the element is likely to be **Strontium (Sr)**. ---
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