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Which of the following electronic config...

Which of the following electronic configurations has the lowest value of ionisation energy? Explain.
a. `1s^(2) 2s^(2)2p^(6)`
b. `1s^(2) 2s^(2)2p^(5)`
c. `1s^(2) 2s^(2)2p^(6) 3s^(1)`

A

`II lt IV lt III lt I`

B

`I lt II lt III lt IV`

C

`I lt III lt II lt IV`

D

`IV lt III lt II lt I`

Text Solution

AI Generated Solution

The correct Answer is:
To determine which of the given electronic configurations has the lowest value of ionization energy, we need to analyze each configuration and understand the factors that influence ionization energy. ### Step 1: Understand Ionization Energy Ionization energy is the energy required to remove an electron from an isolated gaseous atom. The lower the ionization energy, the easier it is to remove an electron. ### Step 2: Analyze Each Configuration 1. **Configuration a: `1s² 2s² 2p⁶`** - This configuration represents a noble gas (Neon). Noble gases have fully filled electron shells, which makes them very stable. Removing an electron from a stable, fully filled configuration requires a significant amount of energy. Therefore, this configuration will have a high ionization energy. 2. **Configuration b: `1s² 2s² 2p⁵`** - This configuration corresponds to a halogen (Fluorine). Halogens have one electron less than a noble gas configuration and are quite electronegative. While they have a lower ionization energy than noble gases, they still have relatively high ionization energy compared to alkali metals because they are close to achieving a stable noble gas configuration. 3. **Configuration c: `1s² 2s² 2p⁶ 3s¹`** - This configuration represents an alkali metal (Sodium). Alkali metals have a single electron in their outermost shell, which they can easily lose to achieve a stable noble gas configuration. Therefore, they have the lowest ionization energy among the three configurations because it is energetically favorable for them to lose that one electron. ### Conclusion Based on the analysis: - Configuration **c (`1s² 2s² 2p⁶ 3s¹`)** has the lowest value of ionization energy because it is an alkali metal configuration, which easily loses its outermost electron. ### Final Answer The electronic configuration with the lowest value of ionization energy is **c: `1s² 2s² 2p⁶ 3s¹`**. ---
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