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Electronic Configuration

The IE 1 IE...

Question

The ${I E}_{1} {I E}_{2} {I E}_{3} {I E}_{4} {I E}_{5}$ of an element are $15.1, 24.3, 34.5, 46.8, 162.2, e V$ respectively. This element is likely to be :(I.E.=Ionisation energey )

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 C Si
It is silicon,
As electronic configuration: $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{2}$
because when you are removing the 4 electrons from shell 3, small changes in IP are seen, but as soon as you remove an electron from stable configuration $(1 s^{2} 2 s^{2} 2 p^{6})$ , the change is drastic.
Hence option B is correct answer.

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Similar questions

I E_{1}, I E_{2}, I E_{3}, I E_{4}

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)

Q. Consider the following successive ionization energies in

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

Which element could have this successive ionization energy data?

I E_{1}, I E_{2}

I E_{3}

of an element are 9.32, 18.21 and 153.83 eV. What information these data convey?

Successive ionization enthalpies of an element 'X' are given below

IE1. IE2. IE3. IE4

165. 195. 556. 595

Electronic configuration of the element'X' is?

{I E}_{1}

{I E}_{2}

{I E}_{3}

{I E}_{4}

N i

P t

are given below

	${I E}_{1} + {I E}_{2}$	${I E}_{3} + {I E}_{4}$
$N i$	$2.49$	$8.80$
$P t$	$2.66$	$6.70$

If true enter 1, else enter 0.
It can be predicted that

{N i}^{2 +}

is more stable than

{P t}^{2 +}

.

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