Coordination Compounds
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New answer posted
8 months agoContributor-Level 10
The colour of the particular complex compound depends on the crystal field splitting energy (CFSE). This CFSE depends on the nature of the ligand attached to the metal atom. In case of [Fe (CN)6]4– and [Fe (H2O)6]2+ the colour differs due to differences in CFSE .
CN- is a strong field ligand so will have high CFSE than H2O with a low value of CFSE. There is absorption of the energy from the visible region for the d-d transition and corresponding complimentary colour is observed. Thus there is the colour difference.
New answer posted
8 months agoContributor-Level 10
In case of [Ni (H2O)6]2+ H2O is a weak field ligand, so it does not cause the pairing of the unpaired electron of Ni2+ ion. Thus there is possibility of the intra d-d transition from the d orbital of lower energy to that of higher energy. Thus the light is absorbed from the visible region and complimentary colour is observed. But in case of [Ni (CN)4]2– CN- is strong field ligand.
Therefore it will cause pairing of the unpaired electrons of Ni2+ ion. There are no unpaired electrons present, so there is no d-d transition and hence it is colourless.
New answer posted
8 months agoContributor-Level 10
Overall charge balance in [Cr (NH3)6]3+ complex:
X + 6 (0) = + 3 X = + 3
Cr is in + 3 oxidation state.
Electronic configuration of Cr in + 2 state: 3d3 . Now ammonia is a weak field ligand so it not causes pairing of the unpaired electron and undergoes hybridisation to form 6 sp3d2 hybrid orbitals filled by the six ammonia ligands. It's geometry is octahedral with unpaired electrons and hence is paramagnetic complex.
In case of [Ni (CN)4]2– ion :
Overall charge balance in [Ni (CN)4]2–complex:
X + 4 (-1) = -2 X = + 2
Ni is in + 2 oxidation state.
Electronic configuration of Ni in + 2 state: 3d8. Now cyanide ion is a strong field ligand so i
New answer posted
8 months agoContributor-Level 10
The difference between the energies of the two set of the d orbitals is called as crystal field splitting energy (CFSE). The degenerate d orbitals split into two levels i.e t2g and eg level due to the presence of the ligands. This splitting of the degenerate orbitals due to the ligand is called as crystal field splitting and the energy difference between the two levels is called as crystal field splitting energy.
After the splitting of the degenerate orbitals has taken place the filling of the electrons takes place. Now first 3 electrons goes into the lower energy three t2g orbitals. The fourth electron can be filled in two ways:
It can
New answer posted
8 months agoContributor-Level 10
The strong ligands have higher splitting power of d orbitals of the central metal ion, whereas weak ligand has relatively lower splitting power of d orbitals of the central metal ion. The energy difference between t2g and eg sets of d orbitals is CFSE. The strength of the ligands depend on the magnitude of Δ . Strong ligands have larger value of CFSE and in case of weak ligands the CFSE values are smaller. The common ligands can be arranged in a series in the order of their decreasing field strength, as follows.
This series depends on the power of splitting the d orbitals and is called spectrochemical series, The order of field s
New answer posted
8 months agoContributor-Level 10
In octahedral complex the splitting of the d orbital will be such a way that the dx2-y2 and dz2 orbitals which face towards the axes along the direction of the ligand will experience more repulsion and will be raised in the energy and the other three orbitals which are directed between the axes are lowered in energy.
New answer posted
8 months agoContributor-Level 10
(i) In the coordination entity iron exists in + 2 oxidation state. Overall charge balance:
X + 6 (-1) = -4 X = + 2.
Its electronic configuration is: 3d6
CN- is strong field ligand so it causes pairing of the unpaired electron and undergoes hybridisation to form 6 d2sp3 hybrid orbitals to be filled by the six cyanide ions. It's geometry is octahedral with no unpaired electrons and hence is diamagnetic complex.
New answer posted
8 months agoContributor-Level 10
CuSO4 + KCN? K2 [Cu (CN)4] + K2SO4
[Cu (H2O)4]2+ + 4CN-? [Cu (CN)4]2- + 4H2O
The coordination entity formed is K2 [Cu (CN)4] .
IUPAC name of the coordination entity is potassium tetracyanocuprate (II). It is a very stable complex. The copper atom present inside the coordination sphere does not separate out to form copper ions and cyanide ions due to strong bond between them.It does not ionize to give Cu2+ ions and hence on adding H2S, since there are no copper ions present so no precipitate of copper sulfide is formed.
New answer posted
8 months agoContributor-Level 10
Copper sulphate exists as [Cu (H2O)4]SO4 . It is blue in colour due to presence of the [Cu (H2O)4]2+ ions.
When KF is added water is replaced by fluoride ion and green colour is due to [Cu (F)4]2- ions.
When KCl is added water is replaced by chloride ion and bright green colour is due to presence of [CuCl4]2- ions.
In both the cases water, weak field ligand is replaced by fluoride and chloride ions.
New answer posted
8 months agoContributor-Level 10
There are four different types of ligands present in the complex. So, by fixing the position of 2 ligands we get 2 geometrical isomers and by changing the position of fixed isomer we get one more geometrical isomer.
Since it is tetrahedral complex so it should be optically active . But however it has not been possible to resolve optically active d and l forms of such a complex due to its complicated nature.
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