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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?

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}

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:

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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