Classification Of Elements And Periodicity In Properties Question 100
Question: $ IE_1 $ and $ IE_2 $ of Mg are 178 and 348 kcal $ mo{{l}^{,-,1}} $ . The energy required for the reaction $ Mg\xrightarrow{{}} Mg^{2+}+2{{e}^{-}} $ is
Options:
A) +170 kcal
B) +526 kcal
C) -170 kcal
D) -526 kcal
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ (i)M{g_{(g)}}\to M{{g}^{\oplus }}_{(g)}+{{e}^{-}},IE_1=178kcalmo{{l}^{-1}} $
$ (ii)Mg^{\oplus } _{(g)}\to Mg _{(g)}^{2+}+e^{-},IE_2=348kcalmol^{-1} $
Adding (i) and (ii), we have $ M{g_{(g)}}\to M{{g}^{2+}}_{(g)}+2{{e}^{-}} $
$ \therefore E=IE_2+IE_2=178+348=526kcal $
BETA
