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} $ =



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