\(Na_2CO_3 \) cannot be used in place of \((NH_4)_2CO_3 \) for the precipitation of fifth group radicals.
This is because:
1. Magnesium will precipitate
2. The concentration of carbonate ions is very low
3. Sodium ions will react with acidic radicals
4. \(Na^{+} \) ions interfere with the detection of the fifth group
Subtopic: Detection of Cations |
Other Reason
Add Note
Highlight in NCERT
Please attempt this question first.
Hints
Upgrade Your Plan
Upgrade now and unlock your full question bank experience with daily practice.
A metal sulphate (A) on heating evolves two gases (B) and (C) and an oxide (D).
Gas (B) turns \(K_2Cr_2O_7\) paper green while gas (C) forms a trimer in which there
is no \(S-S\) bond. Oxide (D) with conc. \(HCl\) forms a Lewis acid (E) which exists
in a dimer. Compounds (A), (B), (C), (D) and (E), respectively, are:
Aqueous solution of a salt when treated with \(AgNO_3\) solution gives a white precipitate, which dissolves in \(NH_4OH.\)
The ion present in the salt is:
1. \(Cl^-\)
2. \(NO_3^-\)
3. \(I^-\)
4. All the av
Subtopic: Detection of Anions |
72%
Other Reason
Add Note
Highlight in NCERT
Please attempt this question first.
Hints
Upgrade Your Plan
Upgrade now and unlock your full question bank experience with daily practice.
On adding \(KI\) solution in excess to a solution of \(CuSO_4\) we get a precipitate 'P and another liquor M. Select the correct pairs:
1. P is \(Cu_2I_2 \) and M is \(I_2\) solution
2. P is \(CuI_2 \) and M is \(I_2\) solution
3. P is \(Cu_2I_2 \) and M is \(KI_3\) solution
4. P is \(CuI_2 \) and M is \(KI_3\) solution
Subtopic: Detection of Anions |
Other Reason
Add Note
Highlight in NCERT
Please attempt this question first.
Hints
Upgrade Your Plan
Upgrade now and unlock your full question bank experience with daily practice.
Consider the following sequence of tests: \(M^{n+} ~+ HCl \rightarrow \) White precipitate \(\xrightarrow \Delta \) water soluble
The metal ion \((M^{n+})\)) can be: