The $I E_{1}, I E_{2}, I E_{3}, I E_{4}$ and $I E_{5}$ of an element are 15.1, 24.3, 34.5, 46.8. 162.2, eV respectively. The element is likely to be: (I.E. = Ionisation energy)

N a

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S i

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F

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C a

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Solution

The correct option is B $S i$
Since fifth energy $(I E)$ is very much greater than all of other energy $,$ so element after loosing $4$ electrons $,$ attains nobel gas configuration $,$ so elements will be
$S i \to {1 s}^{2} {2 s}^{2} {2 p}^{6} {3 s}^{2} {3 p}^{2}$
${S i}^{4 +} \to {1 s}^{2} {2 s}^{2} {2 p}^{6}$ $\leftarrow$ Neon configuration $.$

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Similar questions

{I E}_{1} {I E}_{2} {I E}_{3} {I E}_{4} {I E}_{5}

15.1, 24.3, 34.5, 46.8, 162.2, e V

The IP1, IP2, IP3, IP4 and IP5 of an element are 7.1, 14.3, 34.5, 46.8, 162.2 eV respectively. The element is likely to be -

{I P}_{1}, {I P}_{2}, {I P}_{3}, {I P}_{4}

{I P}_{5}

7.1, 14.3, 34.5, 46.8, 162.2 e V

k J / m o l

I E_{1} = 580, I E_{2} = 1815, I E_{3} = 2740, I E_{4} = 11, 600

I E_{5} = 14, 842

I P_{1^{'}} I P_{2^{'}} I P_{3^{'}} I P_{4^{'}}

I P_{5}

7.1, 14.3, 34.5, 46.8, 162.2

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