If ionization energy of an element in a group is x $k J m o l^{- 1}$ , then the possible ionization energy value in $k J m o l^{- 1}$ of another element situated below that element will be :

6 x / 5

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2 x / 3

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4 x

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7 x / 5

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Solution

The correct option is B $2 x / 3$
As we move down the group, the ionization energy (IE) decreases as the atomic size increases as electrons increasingly loose bound to the nucleus.
Thus the IE of the element mentioned in question should be less than x.
Thus only possible option is B, $2 x / 3$ .

Hence, the correct answer is option B.

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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}

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

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The element with least ionization energy in group 1 is Caesium.

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