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Q. No.A metal nitrate reacts with \(KI\) to give a black precipitate which upon addition of excess
of \(KI\) converts into orange colour solution. The cation of the metal nitrate is:
1. \(Hg^{2+}\)
2. \(Bi^{3+}\)
3. \(Pb^{2+}\)
4. \(Cu^+\)
Subtopic: Detection of Cations |
Level 3: 35%-60%
Hints
The reagents, \(NH_4Cl\) and aqueous \(NH_3\), will precipitate:
1. \(Ca^{2+}\)
2. \(Al^{3+}\)
3. \(B i^{3+}\)
4. \(Mg^{2+} \)
1. \(Ca^{2+}\)
2. \(Al^{3+}\)
3. \(B i^{3+}\)
4. \(Mg^{2+} \)
Subtopic: Detection of Cations |
76%
Level 2: 60%+
Hints
Which of the following reagents can be used to distinguish between sulphite and sulphate ions present in a solution?
1. \(FeSO_4\)
2. \(Na_2[Fe(CN)_5NO] \)
3. \(BaCl_2 \)
4. \(Na_3[Co(NO_2)_6] \)
1. \(FeSO_4\)
2. \(Na_2[Fe(CN)_5NO] \)
3. \(BaCl_2 \)
4. \(Na_3[Co(NO_2)_6] \)
Subtopic: Detection of Anions |
Level 3: 35%-60%
Hints
\(Na_2CO_3 \) 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
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 |
Level 3: 35%-60%
Hints
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:
1. \(FeSO_4, SO_2, SO_3, Fe_2O_3,FeCl_3\)
2. \(Al_2(SO_4)_3,SO_2,SO_3,Al_2O_3,FeCl_3\)
3. \(FeS,SO_2,SO_3,FeSO_4,FeCl_3\)
4. \(FeS,SO_2,SO_3,Fe_2(PO_4)_3,FeCl_2\)
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:
1. \(FeSO_4, SO_2, SO_3, Fe_2O_3,FeCl_3\)
2. \(Al_2(SO_4)_3,SO_2,SO_3,Al_2O_3,FeCl_3\)
3. \(FeS,SO_2,SO_3,FeSO_4,FeCl_3\)
4. \(FeS,SO_2,SO_3,Fe_2(PO_4)_3,FeCl_2\)
Subtopic: Detection of Cations |
Level 3: 35%-60%
Hints
In this sequence, X, Y, Z are, respectively:
| 1. | Acidified \(H_2O_2;\) Alkaline \(H_2O_2;\) Acidified \(H_2O_2\) |
| 2. | Alkaline \(H_2O_2;\) Acidified \(H_2O_2;~Zn/HCl.\) |
| 3. | Acidified \(H_2O_2;\) Heat; Alkaline \(H_2O_2\) |
| 4. | Alkaline \(H_2O_2;\) Acidified \(H_2O_2;\) on standing |
Subtopic: Detection of Cations |
Level 3: 35%-60%
Hints
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 of the above
The ion present in the salt is:
1. \(Cl^-\)
2. \(NO_3^-\)
3. \(I^-\)
4. All of the above
Subtopic: Detection of Anions |
79%
Level 2: 60%+
Hints
Ammonium dichromate is used in some fireworks. The green-coloured powder blown in the air is due to:
1. \(CrO_3 \)
2. \(Cr_2O_3 \)
3. \(Cr\)
4. \(CrO(O_2) \)
1. \(CrO_3 \)
2. \(Cr_2O_3 \)
3. \(Cr\)
4. \(CrO(O_2) \)
Subtopic: Detection of Cations |
76%
Level 2: 60%+
Hints
A white precipitate obtained in the analysis of a mixture becomes black on treatment with \(\mathrm{NH_4OH}.\) It may be:
| 1. | \(\mathrm{PbCl_2}\) | 2. | \(\mathrm{AgCl}\) |
| 3. | \(\mathrm{HgCl_2}\) | 4. | \(\mathrm{Hg_2Cl_2}\) |
Subtopic: Detection of Cations |
Level 3: 35%-60%
Hints
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
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 |
55%
Level 3: 35%-60%
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