Coordination Compounds

We are given the compound

%5Ctext%7BK%5BCr(H%7D_2%5Ctext%7BO%7D)_6%5D(%5Ctext%7BC%7D_2%5Ctext%7BO%7D_4)_2%5D%20%5Ccdot%203%5Ctext%7BH%7D_2%5Ctext%7BO%7D

where Cr is in the +3 oxidation state. The magnetic moment is determined by the formula:

%5Cmu%20%3D%20%5Csqrt%7Bn(n%2B2)%7D%20%5C%2C%20%5Ctext%7BB.M.%7D

where

n

is the number of unpaired electrons in the Cr

%5E3

ion. For Cr

%5E3%2B

(which is

d%5E3

), there are 3 unpaired electrons, so:

n%20%3D%203

Substitute the value of

n

into the formula:

%5Cmu%20%3D%20%5Csqrt%7B3(3%2B2)%7D%20%3D%20%5Csqrt%7B3%20%5Ctimes%205%7D%20%3D%20%5Csqrt%7B15%7D%20%3D%203.87%20%5C%2C%20%5Ctext%7BB.M.%7D

Thus, the spin-only magnetic moment of Cr

%5E3%2B

ion is 3.87 B.M.

Chapter Questions
1 MCQs