Q. No.PCl5 reacts with finely divided silver on heating and a white silver salt is obtained, which dissolves on adding excess aqueous NH3 solution gives "A". Compond A is :
1. \([Ag(NH_{3})_{2}]Cl\)
2. \([Ag(NH_{3})_{2}]Cl_{2}\)
3. \([Ag(NH_{3})_{4}]Cl_{3}\)
4. \([Ag(NH_{3})_{}]Cl_{2}\)
2. \([Ag(NH_{3})_{2}]Cl_{2}\)
3. \([Ag(NH_{3})_{4}]Cl_{3}\)
4. \([Ag(NH_{3})_{}]Cl_{2}\)
Phosphorus forms a number of oxoacids. Out of these oxoacids, phosphinic acid has strong reducing property. its structure and reaction is
1. 
2.
2Ag + H3PO3 + 2HNO3
3.
4.
An amorphous solid“A” burns in the air to form a gas “B” which turns lime water milky. The gas is also produced as a by-product during the roasting of sulphide ore. This gas decolourises acidified aqueous KMnO4 solution and reduces Fe3+ to Fe2+. The solid "A" and the gas "B" are:
1. 'A' = S8: ‘B' = SO2
2. 'A' = S8: ‘B' = SO3
3. 'A' = S10: ‘B' = SO2
4. Not predictable
On heating lead (II) nitrate gives a brown gas “A” The gas “A” on cooling changes to colourless solid “B”. Solid “B” on heating with NO changes to a blue solid ‘C’.
Identify ‘A’, ‘B’ and ‘C and also write reactions involved and draw the structures of ‘B’ and ‘C’.
1. , and
2. , and
3. , and
4. None of above
On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with 3 moles of hydrogen (H2) in the presence of a catalyst gives another gas (C) which is basic in nature. Gas C on further oxidation in moist condition gives a compound (D) which is a part of acid rain. Compounds (A) and (D) are :
1. A = N2 ; D = HNO3
2. A = NH4NO2 ; D = HNO3
3. A = NH4NO2 ; D = NH3
4. A = NH4NO2 ; D = HNO2
| Assertion (A): | R3P = O exists but R3N = O does not (R = alkyl group). |
| Reason (R): | Nitrogen doesn't have d orbitals. |
| 1. | (A) is true and (R) is the correct explanation of (A). |
| 2. | (A) is true but (R) is not the correct explanation of (A). |
| 3. | (A) is true but (R) is false. |
| 4. | Both (A) and (R) are false |
| Assertion (A) | Nitrogen exists as a diatomic molecule and phosphorus as P4. |
| Reason (R) | Phosphorus is smaller in size, so it can form P4. |
| 1. | (A) is true and (R) is the correct explanation of (A). |
| 2. | (A) is true but (R) is not the correct explanation of (A). |
| 3. | (A) is true but (R) is false. |
| 4. | Both (A) and (R) are false |
The disproportionation reaction of is as follows:
X and Y in the above reaction are:
1. H3PO2, HPO2
2. H3PO4, PH3
3. H2PO6, PO3
4. None of the above.
| Assertion (A): | Phosphorous (V) has a tendency to only gain electrons as it is in its highest oxidation state |
| Reason (R): | PCl5 can act as an oxidizing as well as a reducing agent. |
| 1. | (A) is true and (R) is the correct explanation of (A). |
| 2. | (A) is true but (R) is not the correct explanation of (A). |
| 3. | (A) is true but (R) is false. |
| 4. | Both (A) and (R) are false |
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