Advertisements
Advertisements
प्रश्न
The hexaquo manganese (II) ion contains five unpaired electrons, while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory.
Advertisements
उत्तर
The configuration of Mn in oxidation state +2 is 3d5. In the presence of H2O (weak field ligand) as ligand, the distribution of these five electrons is \[\ce{t^3_{2{g}} e^2_{{g}}}\] i.e. all the electrons remain unpaired. In the presence of CN– (strong field ligand) as ligand, the distribution is \[\ce{t^5_{2{g}} e^0_{{g}}}\] i.e. two t2g orbitals have paired electrons while the third t2g orbital has one unpaired electron.
APPEARS IN
संबंधित प्रश्न
Draw figure to show the splitting of d orbitals in an octahedral crystal field.
Complete and balance the following reactions:
\[\ce{P4 + H2SO4 ->}\] ______ + ______ + ______
Atomic number of \[\ce{Mn}\], \[\ce{Fe}\] and \[\ce{Co}\] are 25, 26 and 27 respectively. Which of the following inner orbital octahedral complex ions are diamagnetic?
(i) \[\ce{[Co(NH3)6]^{3+}}\]
(ii) \[\ce{[Mn(CN)6]^{3-}}\]
(iii) \[\ce{[Fe(CN)6]^{4-}}\]
(iv) \[\ce{[Fe(CN)6]^{3-}}\]
On the basis of crystal field theory explain why Co(III) forms paramagnetic octahedral complex with weak field ligands whereas it forms diamagnetic octahedral complex with strong field ligands.
Using crystal field theory, draw energy level diagram, write electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following:
\[\ce{[FeF6]^{3-}, [Fe(H2O)6]^{2+}, [Fe(CN)6]^{4-}}\]
Why are different colours observed in octahedral and tetrahedral complexes for the same metal and same ligands?
The CFSE for octahedral [CoCl6]−4 is 18,000 cm−1. What will be the CFSE for tetrahedral [CoCl3]−2?
[Ni(H2O)6]2+ (aq) is green in colour whereas [Ni(H2O)4 (en)]2+ (aq)is blue in colour, give reason in support of your answer.
Considering crystal field theory, strong-field ligands such as CN–:
The correct order of increasing crystal field strength in following series:
Crystal field stabilising energy for high spind4 octahedral complex is:-
The CFSE of [CoCl6]3– is 18000 cm–1 the CFSE for [CoCl4]– will be ______.
What is the spectrochemical series?
On the basis of crystal field theory, write the electronic configuration for d4 with a strong field ligand for which Δ0 > P.
Read the passage carefully and answer the questions that follow.
|
Crystal field splitting by various ligands Metal complexes show different colours due to d-d transitions. The complex absorbs light of specific wavelength to promote the electron from t2g to eg level. The colour of the complex is due to the transmitted light, which is complementary of the colour absorbed. The wave number of light absorbed by different complexes of Cr ion are given below:
|
Answer the following questions:
(a) Out of ligands "A", "B", "C" and "D", which ligand causes maximum crystal field splitting? Why?
OR
Which of the two, “A” or “D” will be a weak field ligand? Why?
(b) Which of the complexes will be violet in colour? [CrA6]3- or [CrB6]3+ and why?
(Given: If 560 - 570 nm of light is absorbed, the colour of the complex observed is violet.)
(c) If the ligands attached to Cr3+ ion in the complexes given in the table above are water, cyanide ion, chloride ion, and ammonia (not in this order).
Identify the ligand, write the formula and IUPAC name of the following:
- [CrA6]3-
- [CrC6]3+
