Chemical Bonding And Molecular Structure Question 76
Question: The magnetic moment of $ K_3[Fe{{(CN)}_6}] $ is found to be 1.7 B.M. How many unpaired electron (s) is/are present per molecule [Orissa JEE 2003]
Options:
A) 1
B) 2
C) 3
D) 4
Show Answer
Answer:
Correct Answer: A
Solution:
$ K_3[Fe{{(CN)}_6}] $
$ Fe_{26}=4s^{2}3d^{6} $
$ F{e^{3+}}=3d^{5}4s^{0} $ =
