Class 12 Exam > Class 12 Questions > Complexes [Co(SO4)(NH3)5]Br and [CoBr(NH3)5]S...
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Complexes [Co(SO4)(NH3)5]Br and [CoBr(NH3)5]SO4 can be distinguished by :
- a)Conductance measurement
- b)Using BaCl2
- c)Using AgNO3
- d)All
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Complexes [Co(SO4)(NH3)5]Br and [CoBr(NH3)5]SO4can be distinguished by...
Br– + Ba2+ → x
SO42– + Ba2+ → BaSO4(s) (white ppt)
Br–+ Ag+ → AgBr(s) Pale yellow ppt
SO42– + Ag+ → x
2 - ions form both the complex but magnitude of chene is different, second complex is more electrically conducting.
SO42– + Ba2+ → BaSO4(s) (white ppt)
Br–+ Ag+ → AgBr(s) Pale yellow ppt
SO42– + Ag+ → x
2 - ions form both the complex but magnitude of chene is different, second complex is more electrically conducting.
Most Upvoted Answer
Complexes [Co(SO4)(NH3)5]Br and [CoBr(NH3)5]SO4can be distinguished by...
Conductance measurement:
When the two complexes are dissolved in water, they will dissociate into their respective ions. Since [Co(SO4)(NH3)5]Br contains sulfate ions and [CoBr(NH3)5]SO4 contains bromide ions, the conductance of the solutions will differ due to the different ions present.
Using BaCl2:
When barium chloride (BaCl2) is added to the solutions, a white precipitate of barium sulfate (BaSO4) will form in the solution containing [Co(SO4)(NH3)5]Br due to the presence of sulfate ions. No precipitate will form in the solution containing [CoBr(NH3)5]SO4 since it does not contain sulfate ions.
Using AgNO3:
When silver nitrate (AgNO3) is added to the solutions, a yellow precipitate of silver bromide (AgBr) will form in the solution containing [CoBr(NH3)5]SO4 due to the presence of bromide ions. No precipitate will form in the solution containing [Co(SO4)(NH3)5]Br since it does not contain bromide ions.
Therefore, all the given methods - conductance measurement, using BaCl2, and using AgNO3 - can be used to distinguish between the complexes [Co(SO4)(NH3)5]Br and [CoBr(NH3)5]SO4.
When the two complexes are dissolved in water, they will dissociate into their respective ions. Since [Co(SO4)(NH3)5]Br contains sulfate ions and [CoBr(NH3)5]SO4 contains bromide ions, the conductance of the solutions will differ due to the different ions present.
Using BaCl2:
When barium chloride (BaCl2) is added to the solutions, a white precipitate of barium sulfate (BaSO4) will form in the solution containing [Co(SO4)(NH3)5]Br due to the presence of sulfate ions. No precipitate will form in the solution containing [CoBr(NH3)5]SO4 since it does not contain sulfate ions.
Using AgNO3:
When silver nitrate (AgNO3) is added to the solutions, a yellow precipitate of silver bromide (AgBr) will form in the solution containing [CoBr(NH3)5]SO4 due to the presence of bromide ions. No precipitate will form in the solution containing [Co(SO4)(NH3)5]Br since it does not contain bromide ions.
Therefore, all the given methods - conductance measurement, using BaCl2, and using AgNO3 - can be used to distinguish between the complexes [Co(SO4)(NH3)5]Br and [CoBr(NH3)5]SO4.
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Complexes [Co(SO4)(NH3)5]Br and [CoBr(NH3)5]SO4can be distinguished by :a)Conductance measurementb)Using BaCl2c)Using AgNO3d)AllCorrect answer is option 'D'. Can you explain this answer?
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