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Nomenclature of Alcohols

For an elemen...

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For an element of group no. 13 sidden jump on ionization energy will be observed 1. IE1 and IE4 2.IE1 IE2 3.IE2 IE3. 4.IE4 IE5

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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?

Q. The successive ionization energies for element X is given below

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}

Find out the number of valence electrons for the element X.

{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 +}

.

{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 )

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