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Successive io...

Question

Successive ionization potentials of an element 'X' are $I . P_{1} = 520 k J m o l^{- 1}, I . P_{2} = 7, 298 k J m o l^{- 1}$ and $I . P_{3} = 10, 815 k J m o l^{- 1}$ . Predict the group to which X belongs.

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IIA

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IIIA

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IVA

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Solution

The correct option is A IA
${I P}_{2}$ is very high compared to ${I P}_{1}$ , thus the element achieves inert gas configuration on losing $1$ electron. Thus the elements has only $1$ electron in its valence shell. This implies it belongs to group IA.

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1^{s t}

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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}

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