Advertisements
Advertisements
Question
The hexaquo manganese(II) ion contains five unpaired electrons, while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory.
Advertisements
Solution
Mn(II) possesses a 3d5 electronic configuration. Water is a weak field ligand, resulting in a small Δ0. Therefore, the hexaaqua complex will be a high-spin complex with five unpaired electrons. On the other hand, CN− is a strong field ligand, resulting in a substantial Δ0. Therefore, in its cyano complex, the electrons are paired, resulting in a single unpaired electron.
APPEARS IN
RELATED QUESTIONS
How does the magnitude of Δo decide the actual configuration of d orbitals in a coordination entity?
State the superiority of crystal field theory over valence bond theory.
Why are low spin tetrahedral complexes rarely observed?
Draw the structures of the following :
(1) XeF6
(2) IF7
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) \[\ce{[Fe(CN)6]^{3-}}\]
Atomic number of \[\ce{Mn, Fe, Co}\] and Ni are 25, 26, 27 and 28 respectively. Which of the following outer orbital octahedral complexes have same number of unpaired electrons?
(i) \[\ce{[MnCl6]^{3-}}\]
(ii) \[\ce{[FeF6]^{3-}}\]
(iii) \[\ce{[CoF6]^{3-}}\]
(iv) \[\ce{[Ni(NH3)6]^{2+}}\]
Give the electronic configuration of the following complexes on the basis of Crystal Field Splitting theory.
\[\ce{[CoF6]^{3-}, [Fe(CN)6]^{4-} and [Cu(NH3)6]^{2+}}\].
Arrange following complex ions in increasing order of crystal field splitting energy (∆O):
\[\ce{[Cr(Cl)6]^{3-}, [Cr(CN)6]^{3-}, [Cr(NH3)6]^{3+}}\].
The CFSE for octahedral [CoCl6]−4 is 18,000 cm−1. What will be the CFSE for tetrahedral [CoCl3]−2?
Considering crystal field theory, strong-field ligands such as CN–:
The magnitude of CFSE depends upon ______
Using crystal field theory, write the electronic configuration of d5 ion, if Δ0 > P.
What is the spectrochemical series?
What is crystal field splitting energy?
For octahedral Mn(II) and tetrahedral Ni(II) complexes, consider the following statements:
- Both the complexes can be high spin.
- Ni(II) complex can very rarely below spin.
- With strong field Ligands, Mn(II) complexes can be low spin.
- Aqueous solution of Mn (II) ions is yellow in colour.
The correct statements are:
On the basis of Crystal Field theory, write the electronic configuration for the d5 ion with a strong field ligand for which Δ0 > P.
On the basis of crystal field theory, write the electronic configuration for the d5 ion with a weak 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+
