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प्रश्न
The CFSE of [CoCl6]3– is 18000 cm–1 the CFSE for [CoCl4]– will be ______.
पर्याय
18000 cm–1
8000 cm–1
2000 cm–1
16000 cm–1
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उत्तर
The CFSE of [CoCl6]3– is 18000 cm–1 the CFSE for [CoCl4]– will be `underline(bb(8000 cm^-1)`.
Explanation:
∆t = `(4/9 ) xx 18000 cm^-1`
= 8000 cm–1
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संबंधित प्रश्न
Draw figure to show the splitting of d orbitals in an octahedral crystal field.
An aqueous pink solution of cobalt (II) chloride changes to deep blue on addition of excess of HCl. This is because:
(i) \[\ce{[Co(H2O)6]^{2+}}\] is transformed into \[\ce{[CoCl6]}^{4-}\]
(ii) \[\ce{[Co(H2O)6]^{2+}}\] is transformed into \[\ce{[CoCl4]}^{2-}\]
(iii) tetrahedral complexes have smaller crystal field splitting than octahedral complexes.
(iv) tetrahedral complexes have larger crystal field splitting than octahedral complex.
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.
Why are low spin tetrahedral complexes not formed?
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:
\[\ce{[CoF6]^{3-}, [Co(H2O)6]^{2+}, [Co(Cn)6]^{3-}}\]
The correct order of increasing crystal field strength in following series:
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 weak ligand for which Δ0 < P.
