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?

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Solution

In the question it is given that
IE1=165
IE2=195
IE3=556
IE=595

From this given detail we cannot find which element it is but we can conclude to which group it belongs to.

There is a large difference in eneergy between IE2 and IE3. From this we can conclude that it is a 2nd group element(alkaline earth metal).

outer electronic configuration=ns2

There is only 2 electron in the outermost shell.when these two electrons are removed and again we are trying to remove an electron,large energy is to be supplayed because the element is now in a noble gas configuration.(in the case of 2nd group elements)

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Similar questions

I E_{1}

250 k J m o l^{- 1}

I E_{2}

820 k J m o l^{- 1}

I E_{3}

1100 k J m o l^{- 1}

I E_{4}

1400 k J m o l^{- 1}

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 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

X

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

X

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