Advertisements
Advertisements
प्रश्न
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) \[\ce{[Fe(CN)6]^{3-}}\]
Advertisements
उत्तर
(i) \[\ce{[Co(NH3)6]^{3+}}\]
(iii) \[\ce{[Fe(CN)6]^{4-}}\]
Explanation:
(i) Molecular orbital electronic configuration of \[\ce{Co^{3+}}\] in \[\ce{[Co(NH3)6]^{3+}}\] is

Number of unpaired electron = 0
Magnetic property = Diamagnetic
(ii) Molecular orbital electronic configuration of \[\ce{Mn^{3+}}\] in \[\ce{[Mn(CN)6]^{3-}}\]

Number of unpaired electron = 2
Magnetic property = Paramagnetic
(iii) Molecular orbital electronic configuration of \[\ce{Fe^{3+}}\] in \[\ce{[Fe(CN)6]^{4-}}\] is

Number of unpaired electron = 0
Magnetic property = Diamagnetic
(iv) Molecular orbital electronic configuration of \[\ce{Fe^{3+}}\] in \[\ce{[Fe(CN)6]^{3-}}\]

Number of unpaired electron = 1
Magnetic property = Paramagnetic
APPEARS IN
संबंधित प्रश्न
On the basis of crystal field theory, write the electronic configuration for d4 ion if Δ0 > P.
On the basis of crystal field theory, write the electronic configuration for d4 ion if ∆0 < P.
Draw figure to show the splitting of d orbitals in an octahedral crystal field.
How does the magnitude of Δ0 decide the actual configuration of d orbitals in a coordination entity?
State the superiority of crystal field theory over valence bond theory.
Write the electronic configuration of Fe(III) on the basis of crystal field theory when it forms an octahedral complex in the presence of (i) strong field, and (ii) weak field ligand. (Atomic no.of Fe=26)
Complete and balance the following reactions:
\[\ce{P4 + H2SO4 ->}\] ______ + ______ + ______
The colour of the coordination compounds depends on the crystal field splitting. What will be the correct order of absorption of wavelength of light in the visible region, for the complexes, \[\ce{[Co(NH3)6]^{3+}}\], \[\ce{[Co(CN)6]^{3-}}\], \[\ce{[Co(H2O)6]^{3+}}\]
Why are low spin tetrahedral complexes not formed?
Arrange following complex ions in increasing order of crystal field splitting energy (∆O):
\[\ce{[Cr(Cl)6]^{3-}, [Cr(CN)6]^{3-}, [Cr(NH3)6]^{3+}}\].
Match the complex ions given in Column I with the hybridisation and number of unpaired electrons given in Column II and assign the correct code:
| Column I (Complex ion) | Column II (Hybridisation, number of unpaired electrons) |
| A. \[\ce{[Cr(H2O)6]^{3+}}\] | 1. dsp2, 1 |
| B. \[\ce{[Co(CN)4]^{2-}}\] | 2. sp3d2, 5 |
| C. \[\ce{[Ni(NH3)6]^{2+}}\] | 3. d2sp3, 3 |
| D. \[\ce{[MnF6]^{4-}}\] | 4. sp3, 4 |
| 5. sp3d2, 2 |
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?
The correct order of increasing crystal field strength in following series:
Crystal field stabilising energy for high spind4 octahedral complex is:-
The magnitude of CFSE depends upon ______
The CFSE of [CoCl6]3– is 18000 cm–1 the CFSE for [CoCl4]– will be ______.
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 weak ligand for which Δ0 < P.
