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Consider the following reaction.
Cl2 + H2 → A
H2 + Na → B
H2 + CuO → C + H2O
A, B and C in the above mentioned reaction are respectively-
1. A= HCl; B= NaH2; C= Cu2O
2. A= HCl; B= NaH; C= Cu2O
3. A= No reaction; B= NaH; C= Cu
4. A= HCl; B= NaH; C= Cu
Subtopic: Preparation & Properties |
60%
Level 2: 60%+
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The hydrides of N, O and F has the higher boiling points than the hydrides of their subsequent group members because-
1. Lower molecular masses of NH3, H2O and HF
2. Strong hydrogen bonding in NH3, H2O and HF
3. Strong vanderwall forces in NH3, H2O and HF
4. None of the above
Subtopic: Type of Hydride |
89%
Level 1: 80%+
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Given below are two statements:
| Assertion (A): | Phosphorus with outer electronic configuration 3s23p3 form PH5 . |
| Reason (R): | Hydrogen atom unable to contracts 3d orbital of phosphorus. |
| 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. |
Subtopic: Hydrogen- Types & Isotopes |
Level 3: 35%-60%
Hints
The number of hydrogen-bonded water molecule(s) associated with CuSO4.5H2O are-
1. 2
2. 3
3. 1
4. 5
Subtopic: Water |
77%
Level 2: 60%+
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Subtopic: H2O2 (Hydrogen Peroxide) |
74%
Level 2: 60%+
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Select Chapter Topics:
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Q. No.
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