What can be inferred from the magnetic moment of the complex k4 mn(cn)6?
Answers
Answer
I got 3 unpaired electrons.
The spin-only magnetic moment is written as:
μS=2.00023√S(S+1)
where:
g=2.00023 is the gyromagnetic ratio.
S is the total spin of all electrons in the atom. Paired electrons contribute 0 to S because +12+(−12)=0.
Hence, solving for S allows us to determine the number of unpaired electrons.
There would also be the contribution by the orbital magnetic moment, μL, but to find the unpaired electrons in the ion of a relatively light transition metal (i.e. first/second row transition metals), it's sufficiently accurate to use μS.
For example, for Fe3+, a d5 metal, the calculated μS=2.00023√52(52+1)=5.92, whereas the observed (including spin and orbital magnetic moments) μS+L≈5.9, which is pretty excellent agreement.
What we have is μS≈√15 bohr magnetons. Therefore:
√15=2.00023√S(S+1)
15=2.000232(S(S+1))
152.000232=S2+S
0=S2+S−152.000232
Solve the quadratic formula to get:
S≈1.5⇒32
Therefore, the total spin S is 32, the number of unpaired electrons is 3, and the atomic ion could be either a d3 or d7 configuration, as you would see here:
d3:
↑↓−−−−− ↑↓−−−−− ↑↓−−−−− ↑↓−−−−− ↑↓−−−−−
d
Explanation:
Aur bhai kaisa h.......