Q. No.A brown ring formed in the ring test for ion is due to the formation of:
1. [Fe(H2O)5(NO)]2+
2. FeSO4.NO2
3. [Fe(H2O)4(NO)2]2+
4. FeSO4.HNO3
1. [Fe(H2O)5(NO)]2+
2. FeSO4.NO2
3. [Fe(H2O)4(NO)2]2+
4. FeSO4.HNO3
Which of the following statements is correct?
| 1. | Only type of interactions between particles of noble gases are due to weak dispersion forces. |
| 2. | Ionisation enthalpy of molecular oxygen is very high as compared to xenon. |
| 3. | Hydrolysis of XeF6 is a redox reaction. |
| 4. | Xenon fluorides are not reactive. |
Match the formulas of oxides given in Column I with the type of oxide given in Column II and mark the correct option.
| Column I | Column II |
| A. Pb3O4 | 1. Neutral oxide |
| B. N2O | 2. Acidic oxide |
| C. Mn2O7 | 3. Basic oxide |
| D. Bi2O3 | 4. Mixed oxide |
Codes
| A | B | C | D | |
| 1. | 1 | 2 | 3 | 4 |
| 2. | 4 | 1 | 2 | 3 |
| 3. | 3 | 2 | 4 | 1 |
| 4. | 4 | 3 | 1 | 2 |
Which of the following types of π-bonding is present in the N₂ molecule?
1. \(p\pi-p\pi\)
2. \(p\pi-d\pi\)
3. \(d\pi-d\pi\)
4. Only sigma bond
Which of the following is not a condition used in the Haber process for the manufacture of ammonia?
1. Pressure (around 200 ×105 Pa)
2. Temperature (700 K)
3. Catalyst such as iron oxide.
4. Presence of inert gases.
(a) XeF6 + 3H2O →
(b) XeF6 + H2O
The products formed in the above reactions, respectively, are:
1. XeO3 , XeOF4
2. XeO6 , XeOF4
3. XeO3, XeOF6
4. XeO6 , XeOF6
| Assertion (A): | Both rhombic and monoclinic sulphur exist as S8 but oxygen exists as O2. |
| Reason (R): | Oxygen forms pπ - pπ multiple bonds due to its small size and small bond length but pπ - pπ bonding is not possible in sulphur. |
| 1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are 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): | NaCl reacts with concentrated H2SO4 to give colourless fumes with a pungent smell. But on adding MnO2, the fumes become greenish-yellow. |
| Reason (R): | MnO2 oxidizes HCl to chlorine gas which is greenish-yellow. |
| 1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |
| 3. | (A) is True but (R) is False. |
| 4. | (A) is False but (R) is True. |
Which of the following elements does not exhibit allotropy?
1. Nitrogen
2. Phosphorus
3. Antimony
4. Arsenic
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