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Ionization Enthalpy

The successiv...

Question

The successive ionization energies for element $X$ are given below:
$I E_{1} = 250 {kJ mol}^{- 1}$
$I E_{2} = 820 {kJ mol}^{- 1}$
$I E_{3} = 1100 {kJ mol}^{- 1}$
$I E_{4} = 1400 {kJ mol}^{- 1}$
Find out the number of valence electrons from the element $X$ .

3

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4

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2

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1

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Solution

The correct option is D $1$
According to the given data, there is a large jump from first to second ionization energy. The second ionization energy is more than thrice the first ionization energy. This indicates that the species attains stable electronic configuration after the removal of one electron. Hence, the number of valence electron is $1$ . This indicates that element $X$ belongs to alkali metal group.

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Similar questions

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

I E_{1}

250 k J m o l^{- 1}

I E_{2}

820 k J m o l^{- 1}

I E_{3}

1100 k J m o l^{- 1}

I E_{4}

1400 k J m o l^{- 1}

(X)

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

I_{5}

800, 2427, 3658, 25024, 32824 {kJ mol}^{- 1}

I . P_{1} = 520 k J m o l^{- 1}, I . P_{2} = 7, 298 k J m o l^{- 1}

I . P_{3} = 10, 815 k J m o l^{- 1}

k J m o l^{- 1}

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