The spin only magnetic moment value (in Bohr magneton units) of Cr(CO)₆ is
(a) 0
(b) 2.84
(c) 4.90
(d) 5.92
Answers
The spin-only magnetic moment is given by:
μ
S
=
2.0023
√
S
(
S
+
1
)
where
g
=
2.0023
is the gyromagnetic ratio and
S
is the total spin of all unpaired electrons in the system. If there are none... then
μ
S
=
0
.
Spin-only means we ignore the total orbital angular momentum
L
=
∣
∣
∣
∣
∑
i
m
l
,
i
∣
∣
∣
∣
for the
i
th electrons.
By conservation of charge,
Cr
(
CO
)
6
has a
Cr
atom in its
0
oxidation state.
For transition metal complexes, the ligand orbitals belong primarily to the ligands, and the metal's orbitals belong primarily to the metal, because the interacting atoms are going to have significantly different electronegativities.
Hence, the unpaired electrons found (if any) are based on the metal oxidation state.
Cr
(
0
)
brought in
6
post-noble-gas-core electrons, i.e.
5
×
3
d
+
1
×
4
s
=
6
.
The
CO
ligands are
σ
donors (which is obvious, as they must make bonds!) AND
π
acceptors (which is not so obvious)...
Inorganic Chemistry, Miessler et al., pg. 385
Inorganic Chemistry, Miessler et al., pg. 385
...so they are strong-field ligands, which promote low-spin octahedral complexes (large
Δ
o
) by lowering the
t
2
g
orbital energies (the
σ
donor factor raises the
e
*
g
orbital energies). So, we have a low-spin
d
6
complex, which has no unpaired electrons.
Δ
o
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎩
¯¯¯¯¯¯¯¯¯
↑
↓
¯¯¯¯¯¯¯¯¯
↑
↓
e
g
↑
↓
−−−−−
↑
↓
−−−−−
↑
↓
−−−−−
t
2
g
Thus, the spin-only magnetic moment is
0
. Don't believe me? Here's some proof of ZERO unpaired electrons:
Inorganic Chemistry, Miessler et al., pg. 484
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➡️Correct Option -: A
➡️
============================
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