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प्रश्न
Explain the geometry of `[Co(NH_3)_6]^(3+)` on the basis of hybridisation. (Z of Co = 27)
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उत्तर
Octahedral complex, `[Co(NH_3)_6]^(3+)`. In this complex ion, the oxidation state of cobalt is +3.
It has the electronic configuration as 3d6. This complex involves the d2sp3 hybridisation.
The six pairs of electrons, one from each NH3 molecule, occupy the six hybrid orbitals. It proves that complex has octahedral geometry. An absence of unpaired electron makes this complex diamagnetic in nature.
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संबंधित प्रश्न
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Discuss the nature of bonding in the following coordination entity on the basis of valence bond theory:
[Co(C2O4)3]3−
Write the hybridisation and number of unpaired electrons in the complex `[CoF_6]^(3-)`. (Atomic No. of Co = 27)
Using valence bond theory, explain the following in relation to the complexes given below:
\[\ce{[Co(NH3)6]^{3+}}\]
(i) Type of hybridisation.
(ii) Inner or outer orbital complex.
(iii) Magnetic behaviour.
(iv) Spin only magnetic moment value.
Using valence bond theory, explain the following in relation to the complexes given below:
\[\ce{[FeCl6]^{4-}}\]
(i) Type of hybridisation.
(ii) Inner or outer orbital complex.
(iii) Magnetic behaviour.
(iv) Spin only magnetic moment value.
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[CoF6]3– [Atomic number of Co = 27]
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[Given: Atomic number of Co = 27]
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Statement I: Both [Co(NH3)6]3+ and [CoF6]3− complexes are octahedral but differ in their magnetic behavior.
Statement II: [Co(NH3)6]3+ is diamagnetic whereas [CoF6]3− is paramagnetic.
In the light of the above statements, Choose the correct answer form the options given below:
