The pair having the same magnetic moment is:
[At. No. : Cr=24, Mn=25, Fe=26, Co=27]
[Cr(H2O)6 ]2+ and [Fe(H2O)6 ]2+
[Mn(H2O)6 ]2+ and [Cr(H2O)6 ]2+
[CoCl4 ]2− and [Fe(H2O)6 ]2+
[CoCl4 ]2− and [Fe(H2O)6 ]2+
A.
[Cr(H2O)6 ]2+ and [Fe(H2O)6 ]2+
| Complex ion | Electronic configuration metal ion | Number of unpaired electrons (n) |
| [Cr(H2O)6 ]2+ | Cr2+ ; [ar] 3d4 |  |
| [Fe(H2O)6 ]2+ | Fe2+ ; [Ar]3d6 |  |
| [Mn(H2O)6 ]2+ | Mn2+ ; [Ar] 3d5 |  |
| [CoCl4 ]2− | Co2+ ; [Ar]3d7 |  |
Which one of the following complexes shows optical isomerism?
cis[Co(en)2Cl2]Cl
trans[Co(en)2Cl2]Cl
[Co(NH3)4Cl2 ]Cl
[Co(NH3)4Cl2 ]Cl
A.
cis[Co(en)2Cl2]Cl
The optically active compound capable of rotating the plane-polarized light to the right or left.
cis[Co(en)2Cl2]Cl is optically active compound.
The number of geometric isomers that can exist for square planar [Pt (Cl) (py) (NH3) (NH2OH)]+ is (py = pyridine)
2
3
4
4
B.
3
[Pt (Cl) (py) (NH3) (NH2OH)]+ is square planar complex.
The structures are formed by fixing a group and then arranging all the groups. Hence, this complex shows three geometrical isomers.
The number of unpaired electrons in[NiCl4]2-, Ni(CO)4 and (Cu(NH3)4]2+ respectively are
0, 2, 1
2, 0, 1
0, 2, 1
2, 2, 0
B.
2, 0, 1
Configuration for metal (M) in various complex.
Ni2+ has 2 unpaired electrons.
Ni0 has zero unpaired electrons with sp3-hybridisation.
Cu2+ has one unpaired electron with dsp2-hybridisation.
Therefore, number of unpaired electrons is (2, 0, 1).
The Colour of KMnO4 is due to
M → L charge transfer transition
d-d transition
L →M charge transfer transition
L →M charge transfer transition
C.
L →M charge transfer transition
KMnO4 → K+ + MnO-4
Therefore,
In MnO4-, Mn has +7 oxidation state having no electron in d- orbitals.
It is considered that higher the oxidation state of metal, greater is the tendency to occur L →M charge transfer because ligand is able to donate the electrons into the vacant d- orbital of metal.
Since, charge transfer is Laporte as well as spin allowed, therefore, it shows colour.
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