Chemistry - Salt Analysis Question with Solution | TestHub
A metal ion (M) solution neither form precipitate with dilute HCl, nor with in acidified aqueous solutions but produces 9 gram black precipitate(X) in alkaline medium with . Black precipitate X is completely dissolved in aqua regia. The aqueous solution obtained after treatment with aqua regia is neutralised with ammonium hydroxide and then acidified with acetic acid and filtered. In the filtrate strong solution of potassium nitrite is added, which gives yellow precipitate. In the last reaction, a paramagnetic gas is released as a byproduct.
Calculate sum of X and Y
if
X = total number of possible isomers for complex of metal M produced
Y = total number of unpaired electrons present in a complex ion of metal M produced
Answer:
Solution:
Identification of Metal Ion: The metal ion is Co²⁺ because it forms a black precipitate (CoS) in an alkaline medium and a characteristic yellow precipitate (K₃[Co(NO₂)₆]) when treated with potassium nitrite.
Complex Formula: The resulting complex ion of the metal is .
Finding Y (Unpaired Electrons): In , cobalt is in the +3 oxidation state (). Since NO₂⁻ is a strong field ligand, all electrons pair up completely (), giving unpaired electrons.
Finding X (Isomers): The total number of possible linkage/structural isomers due to the ambidentate nature of the NO₂⁻ ligand is .
Final Calculation: The sum of and is .
