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CBSE Class 12 Chemistry 2018 Solved Paper

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Question : 22 of 26
Marks: +1, -0
(a) Write the formula of the following co-ordination compound :
Iron(III) hexacyanoferrate(II)
(b) What type of isomerism is exhibited by the complex [Co(NH3)5Cl]SO4[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Cl}]\mathrm{SO}_4 ?
(c) Write the hybridisation and number of unpaired electrons in the complex [CoF6]3−[\mathrm{CoF}_6]^{3-}. (( Atomic No. of Co=27\mathrm{Co}=27 )
Solution:  
(a) Fe4[Fe(CN)6]3\mathrm{Fe}_4[\mathrm{Fe}(\mathrm{CN})_6]_3
(b) Ionization isomerism.
[Co(NH3)5SO4]Cl   and   [Co(NH3)5Cl]SO4[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{SO}_4]\mathrm{Cl}\;\text{ and }\;[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Cl}]\mathrm{SO}_4
(c) [COF6]3−[\mathrm{COF}_6]^{3-}
In this complex, Co possesses +3 oxidation state.
Co3+=3d6\mathrm{Co}^{3+}=3d^6
Since F−\mathrm{F}^{-}is weak ligand so it cannot push the electrons to pair up and it forms outer orbital high spin complex.
Number of unpaired electrons =4=4
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