Classification Of Elements And Periodicity In Properties Question 6
Question 6 - 31 January - Shift 1
$Nd^{2+}=$
(1) $4 f^{2} 6 s^{2}$
(2) $4 f^{4}$
(3) $4 f^{3}$
(4) $4 f^{4} 6 s^{2}$
Show Answer
Answer: (2)
Solution:
Formula: Electronic configuration
$Nd(60)=[Xe] 4 f^{4} 5 d^{0} 6 s^{2}$
$Nd^{2+}=[Xe] 4 f^{4} 5 d^{0} 5 s^{0}$
