Successive (IE) values of three elements of third period are given belowAtomic numbers of X, Y and Z respectively are
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
11, 12, 13
b
11, 12, 14
c
11, 12, 15
d
11, 12, 18
answer is D.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
For X, (E), is very high, thus X+ attains inert gas configuration. It is Na.Na(11) ⟶Na++e−[Ne]3s1 [Ne] Stable, (E)2 is very high For Y,(IE)3 is very high thus, Y2+ attains inert gas configuration . It is MgMg(12) ⟶Mg2++2e−[Ne]3s2 [Ne] Stable, (IE)3 is very high. For Z,(IE)1 is very high. It is [Ar].