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Q.

Which one of the following has the lowest ionisation energy?


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a

<math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><msup><mi mathvariant="normal">s</mi><mn>2</mn></msup><mn>2</mn><msup><mi mathvariant="normal">s</mi><mn>2</mn></msup><mn>2</mn><msup><mi mathvariant="normal">p</mi><mn>6</mn></msup></math>

b

<math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><msup><mi mathvariant="normal">s</mi><mn>2</mn></msup><mn>2</mn><msup><mi mathvariant="normal">s</mi><mn>2</mn></msup><mn>2</mn><msup><mi mathvariant="normal">p</mi><mn>6</mn></msup><mn>3</mn><msup><mi mathvariant="normal">s</mi><mn>1</mn></msup></math>

c

<math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><msup><mi mathvariant="normal">s</mi><mn>2</mn></msup><mn>2</mn><msup><mi mathvariant="normal">s</mi><mn>2</mn></msup><mn>2</mn><msup><mi mathvariant="normal">p</mi><mn>5</mn></msup></math>

d

<math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><msup><mi mathvariant="normal">s</mi><mn>2</mn></msup><mn>2</mn><msup><mi mathvariant="normal">s</mi><mn>2</mn></msup><mn>2</mn><msup><mi mathvariant="normal">p</mi><mn>3</mn></msup></math> 

answer is B.

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Detailed Solution

concept - Ionisation energy is the amount of energy needed by an isolated gaseous atom to lose an electron in its ground state to form an ion.
 For the lowest ionisation energy, the valence electron should be loosely held and after losing that electron, the ion should attain a stable electronic configuration.
Now, the second configuration has the lowest ionisation potential as the valence electron is in the 3s shell which is comparatively far from the nucleus and after losing it, the ion formed has a configuration <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><msup><mi mathvariant="normal">s</mi><mn>2</mn></msup><mn>2</mn><msup><mi mathvariant="normal">s</mi><mn>2</mn></msup><mn>2</mn><msup><mi mathvariant="normal">p</mi><mn>6</mn></msup></math> which represents completely filled orbitals and thus has higher stability.
Therefore, <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>1</mn><msup><mi mathvariant="normal">s</mi><mn>2</mn></msup><mn>2</mn><msup><mi mathvariant="normal">s</mi><mn>2</mn></msup><mn>2</mn><msup><mi mathvariant="normal">p</mi><mn>6</mn></msup><mn>3</mn><msup><mi mathvariant="normal">s</mi><mn>1</mn></msup></math> has the lowest ionisation potential,
 Hence, the correct answer is 2.
 
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