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Given Δ H I...

Question

Given $Δ$ H (Ionization enthalpy) for the process is 19800 kJ/mole & $I E_{1}$ for Li is 520, then $I E_{2}$ & $I E_{3}$ of $L i^{+}$ are _______ respectively.
[Note : approx. value]

7505, 11775

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520, 19280

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11775, 19280

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Data insufficient

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Solution

The correct option is C 7505, 11775
Ionization enthalpy for process is $19800 K J / m o l$
If ${I E}_{1}$ for $L i$ is $520$
${I E}_{2}$ for $L i$ should be greater than $520$
Now,
$19800 - 520 = 19280$
$\to$ So add the value given in options to get $19280.$
Option $(A)$ $\Rightarrow$ $7505 + 11775 = 19280$
Option $(B)$ $\Rightarrow$ $520 + 19280 = 19800$
Option $(C)$ $\Rightarrow$ $11775 + 19280 = 31055$

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Similar questions

IE (in kJ

{m o I}^{- 1}

) of

L i

and

B e

are given below:

	Z	${(I E)}_{1}$	${(I E)}_{2}$	${(I E)}_{3}$
$L i$	3	520.1	$X (1)$	11810
$B e$	4	899.3	$Y (2)$	14810

From above table

X (1)

is higher than

Y (2)

The ionization enthalpies of Li and Sodium are 520 KJ/mole and 495 KJ/mole respectively. The energy required to convert all the atoms present in 7mg of Li vapours and 23 mg of Sodium vapours to their respective gaseous cations respectively are

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.

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

, and

I E_{5} = 14, 842

Which element could have this successive ionization energy data?

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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Given Δ H (Ionization enthalpy) for the process is 19800 kJ/mole & IE1 for Li is 520, then IE2 & IE3 of Li+ are _______ respectively. [Note : approx. value]

The correct option is C 7505, 11775Ionization enthalpy for process is 19800KJ/molIf IE1 for Li is 520IE2 for Li should be greater than 520Now,19800−520=19280→So add the value given in options to get 19280.Option (A)⇒ 7505+11775=19280Option (B)⇒ 520+19280=19800Option(C)⇒ 11775+19280=31055

Given $Δ$ H (Ionization enthalpy) for the process is 19800 kJ/mole & $I E_{1}$ for Li is 520, then $I E_{2}$ & $I E_{3}$ of $L i^{+}$ are _______ respectively.
[Note : approx. value]