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Trends in Ionization Energy

The IP 1, IP ...

Question

The $I P_{1}, I P_{2}, I P_{3}, I P_{4}, I P_{5}$ of an element are 7.1, 14.3, 34.5, 46.8, 162.2 eV respectively. Which element among the following would that be?

A

Na

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B

Si

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C

F

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D

Ca

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Solution

The correct option is B
Si

Among the Ionization Potentials given there is a huge jump from $I P_{4}$ to $I P_{5}$ , which shows that element attains stable inert gas configuration after losing 4 electrons.

It should be an element from $14^{t h}$ group (IVA). Among the given elements, the answer will be Si.

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Similar questions

The $I P_{1}, I P_{2}, I P_{3}, I P_{4}, I P_{5}$ of an element are 7.1, 14.3, 34.5, 46.8, 162.2 eV respectively. Which element among the following would that be?

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

{I P}_{5}

of an element are

7.1, 14.3, 34.5, 46.8, 162.2 e V

The element is likely to be:

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

I P_{5}

of an element are

7.1, 14.3, 34.5, 46.8, 162.2

eV respectively. The element is likely to be:

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 -

Q. Sucessive ionisation energies of an element 'X' are given below (in K. Cal)

$I P_{1}$	$I P_{2}$	$I P_{3}$	$I P_{4}$
165	195	556	595

Electronic configuration of the element 'X' is:

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