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Ionisation Energy

Which of the ...

Question

Which of the following electronic configuration corresponds to an element having the lowest $1 s t$ ionization energy?

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^{5}

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

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Solution

The correct option is B $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}$
The $s$ -block elements have low ionization potential as compared to $p$ -block elements because $s$ -block elements easily lose $1$ electron to get the stable configuration of noble elements while $p$ -block elements gain electrons to get their stable octet configuration.
Hence, option B is correct.

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Similar questions

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

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

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

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

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

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

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

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

A, B, C

D

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

1 s^{2} 2 s^{2} 2 p^{4}

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

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

1^{s t}

2^{n d}

A = 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}

B = 1 s^{2} 2 s^{2} 2 p^{5}

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