If for an element $(X)$ , the values of successive ionization energies $I_{1}, I_{2}, I_{3}, I_{4}$ and $I_{5}$ are $800, 2427, 3658, 25024, 32824 {kJ mol}^{- 1}$ respectively, then the number of valence electrons present are:

3

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5

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2

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4

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Solution

The correct option is A $3$
Difference between successive ionization energies:
$I_{2} - I_{1} - 1627 {kJ mol}^{- 1} I_{3} - I_{2} = 1231 {kJ mol}^{- 1} I_{4} - I_{3} = 21366 {kJ mol}^{- 1} I_{5} - I_{4} = 7800 {kJ mol}^{- 1}$
As the difference between $I_{3}$ and $I_{4}$ is maximum, so the element $(X)$ has three valence electrons as after losing three electrons, it acquires stable inert gas configuration because after attaining a stable configuration, energy to ionize an electron increases abruptly.

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Similar questions

I_{1} < I_{2} <<<< I_{3} < I_{4} < I_{5}

X

I E_{1} = 250 {kJ mol}^{- 1}

I E_{2} = 820 {kJ mol}^{- 1}

I E_{3} = 1100 {kJ mol}^{- 1}

I E_{4} = 1400 {kJ mol}^{- 1}

X

k J m o l^{- 1}

I_{n} = \int_{0}^{π / 4} t a n^{n} x d x

I_{2} + I_{4}, I_{3} + I_{5}, I_{4} + I_{6}, I_{5} + I_{7}

A B C D

O

1, 2, 3

4

I_{1}, I_{2}, I_{3} & I_{4}

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