Q.
Consider the process, Mg→M2+g , ∆H=270 kJ ;If the IP1 of M(g) is 100 kJ/mole then the electron gain enthalpy of M2+(g) will be
see full answer
Start JEE / NEET / Foundation preparation at rupees 99/day !!
21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!
An Intiative by Sri Chaitanya
a
+270 kJ
b
-170 kJ
c
+170 kJ
d
-370 kJ
answer is B.
(Unlock A.I Detailed Solution for FREE)
Ready to Test Your Skills?
Check your Performance Today with our Free Mock Test used by Toppers!
Take Free Test
Detailed Solution
Mg→+100 kJM+g→+170 kJM2+g ; Thus M2+g+e-→M+g , ∆egH =-170 kJ ;
Watch 3-min video & get full concept clarity