Cobalt (III) chloride forms several octahedral complexes with ammonia. Which of the following will not give test for chloride ions with silver nitrate at $25^{\circ} C$ ?

C o C l_{3} .6 N H_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C o C l_{3} .3 N H_{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C o C l_{3} .4 N H_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C o C l_{3} .5 N H_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $C o C l_{3} .3 N H_{3}$
$C o C l_{3} .3 N H_{3}$ doesn‘t ionize to give free $C l^{-}$ , so it doesn‘t answer the test for chloride ions.

Suggest Corrections

Similar questions

Q. Cobalt (III) chloride forms several octahedral complexes with ammonia. Which of the following will not give test for chloride ions with silver nitrate at

25^{\circ} C

Q. When 0.01 mole of a cobalt complex is treated with excess silver nitrate solution, 4.305 g silver chloride is precipitated. The formula of the complex is:

Q. 2.66 g chloride of a metal when treated with silver nitrate solution give 2.87 g of silver chloride. 3.37 g of another chloride of the same metal give 5.74 g of silver chloride when treated with silver nitrate solution. Show that the results are in agreement with a law of chemical combination.

Q. The number of chloride ions which would be precipitated, when

C r C l_{3} .4 N H_{3}

is treated with silver nitrate solution:

Raghav performed an experiment in which he combined 0.5 g of ammonia with 1.1 g of HCl to produce ammonium chloride. Again, he reacted the ammonium chloride produced with 5.2 g of silver nitrate and formed X g of ammonium nitrate and 4 g of silver chloride precipitates out. Find the value of X.

Join BYJU'S Learning Program

Cobalt (III) chloride forms several octahedral complexes with ammonia. Which of the following will not give test for chloride ions with silver nitrate at 25∘ C ?

The correct option is B CoCl3.3NH3CoCl3.3NH3 doesn‘t ionize to give free Cl−, so it doesn‘t answer the test for chloride ions.

Cobalt (III) chloride forms several octahedral complexes with ammonia. Which of the following will not give test for chloride ions with silver nitrate at $25^{\circ} C$ ?

The correct option is B $C o C l_{3} .3 N H_{3}$
$C o C l_{3} .3 N H_{3}$ doesn‘t ionize to give free $C l^{-}$ , so it doesn‘t answer the test for chloride ions.