The correct order of increasing electrical conductance is -
1. BeH2 > CaH2 > TiH2
2. BeH2 < CaH2 > TiH2
3. TiH2 < BeH2 < CaH2
4. BeH2 < CaH2 < TiH2

Subtopic:  Type of Hydride |
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The correct order of increasing reducing nature is -

1. NaH > H2O < MgH2

2. H2O > MgH2> NaH

3. H2O< MgH2 < NaH

4. H2O  MgH2 < NaH

Subtopic:  Type of Hydride |
 54%
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The isotopes of hydrogen and their mass ratio are -

1. Protium, Deuterium, Tritium ; 3:2:3

2. Protium, Deuterium, Tritium ; 1:2:3

3. Protium, Deuterium, Tritium ; 2:2:3

4. Protium, Deuterium, Tritium ; 3:2:1

Subtopic:  Hydrogen- Types & Isotopes |
 93%
From NCERT
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Hydrogen exists in diatomic form rather than monoatomic form under normal conditions -

1. Due to high ionization enthalpy

2. Due to low ionization enthalpy

3. Due to high electron gain enthalpy

4. Due to low electron gain enthalpy

Subtopic:  Hydrogen- Types & Isotopes |
 70%
From NCERT
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The role of an electrolyte in the  preparation of H2 by electrolytic method is:

1. It lowers the availability of ions available in the process for the conduction of electricity.
2. It makes the ions available in the process for conduction of electricity.
3. It cools down the temperature.
4. None of the above

Subtopic:  Preparation & Properties |
 81%
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The products obtained on the completion of the following reactions (i) and (ii) respectively are :

i. C3H8(g) + 3H2O(g) \(\xrightarrow[Catalyst]{\Delta}\) A
ii. \(\mathrm{Zn}_{(s)}+\mathrm{NaOH}_{(\mathrm{aq})} \xrightarrow[]{Heat}\)B


1. A = CO, and H2 ; B = Na2ZnO2
2. A  = CO2 ; B = Na4ZnO4
3. A = CH3CH2CH2OH ; B=  ZnO
4. None of the above


 

 

Subtopic:  Preparation & Properties |
 78%
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The consequences of high enthalpy of H–H bond in dihydrogen is /are :

1. Hydrogen has a low tendency to form H+ions.
2. It forms diatomic molecules (H2), and a large number of covalent hydrides.
3. Hydrogen does not possess metallic characteristics.
4. All of the above.

Subtopic:  Hydrogen- Types & Isotopes |
 80%
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The correct statement about electron-deficient hydride is that it:

1. Does not have sufficient electrons to form a regular bond.
2. Cannot be represented by conventional Lewis structures.
3. Accept electrons.
4. All of the above.


 

Subtopic:  Type of Hydride |
 84%
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The hydrides of carbon (CnH2n + 2) can act as -

1. Lewis acid.

2. Lewis base.

3. Both 1 and 2

4. None of the above

Subtopic:  Type of Hydride |
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“Non-stoichiometric hydrides” are :

1. Hydrogen-deficient compounds formed by the reaction of a hydrogen atom with d-block and f-block elements.
2. Hydrogen-deficient compounds formed by the reaction of dihydrogen with d-block and f-block elements.
3. Electron-deficient compounds formed by the reaction of dihydrogen with d-block and f-block elements.
4. Hydrogen-deficient compounds formed by the reaction of dihydrogen with s-block and p-block elements.

Subtopic:  Type of Hydride |
 51%
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