Start Free Mock LoginAlready registered? Login here

Chemistry - PERIODIC TABLE Question with Solution | TestHub

ChemistryPERIODIC TABLEIonisation Energy,Medium2 minQB

ChemistryMediumnumerical

Passage / Comprehension

Consider four 2nd period elements A,B, C, D along with corresponding electron affinities and $IE_{1}$ , $IE_{2}$ and $IE_{3}$ , all in $KJ mol^{- 1}$

	EA	$I E_{1}$	$I E_{2}$	$I E_{3}$
A	328	1681	3374	6050
B	141	1314	3388	5300
C	60	520	7298	11815
D	27	801	2427	3660

The number of electrons in B are :

Answer:

8.00

Solution:

Since EA values are positive and hence Be, N, Ne are ruled out.

Now, C is clearly having 1 valence electron.

So it is Li

Amongst A and B, A has higher EA

and

$IE_{1} (A) > IE_{1} (B)$

$IE_{2} (A) < IE_{2} (B)$

$IE_{3} (A) > IE_{3} (B)$

The above can be satisfied for 2 combinations, either A is carbon and B is boron or A is fluorine and B is oxygen.

But 1st combination is ruled out since in that case D has to be oxygen or fluorine but then it's $IE_{1}$ cannot be lower than that of boron and carbon.

So, A is fluorine and B is oxygen

Stream:JEESubject:ChemistryTopic:PERIODIC TABLESubtopic:Ionisation Energy,

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...