Of the following ground state electron configurations, the one that represents the element of lowest first ionization energy (potential) is :

1 s^{2}, 2 s^{2}, 2 p^{5}

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1 s^{2}, 2 s^{2}, 2 p^{6}

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1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{1}

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1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}

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1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{1}, 3 p^{1}

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Solution

The correct option is B $1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{1}$
Lowest first ionization energy is of the element belonging to alkali metal group characterized by having $3 s^{1}$ outer electronic configuration.

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Similar questions

{1 s}^{2}, {2 s}^{2}, {2 p}^{5}

{1 s}^{2}, {2 s}^{1}

{1 s}^{2}, {2 s}^{2}, {2 p}^{6}

{1 s}^{2}, {2 s}^{2}, {2 p}^{6}, {3 s}^{2}, {3 p}^{5}

{1 s}^{2}, {2 s}^{2}, {2 p}^{6}, {3 s}^{2}, {3 p}^{6}, {4 s}^{1}

{1 s}^{2}, {2 s}^{2}, {2 p}^{6}, {3 s}^{2} {3 p}^{6}, {4 s}^{2}, {3 d}^{10}, {4 p}^{6}

The lowest first ionization energy would be associated with which of the following electronic configurations

1 s t

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