Hydrogen Test

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Hydrogen Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following hydrides is electron deficient?

A
$$NaH$$

B
$$CaH_{2}$$

C
$$CH_{4}$$

D
$$B_{2}H_{6}$$

Solution

$$ \left (BH_{3} \right )$$ is the chemical formula for the Boron atom. It is surrounded by 6 electrons, so it is electron deficient due to its incomplete octet. B-atom in diborane is $$sp^{3}$$ hybridized. $$B_{2}H_{6}$$ molecule has a total of 8 covalent bonds but only 12 electrons i.e. 6 pairs. So, it has two electron pairs less than the maximum required number for bonding, which makes it electron deficient.

The answer is option D.
Question 2
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What is the trend of boiling points of hydrides of N, O, and F?

A
Due to lower molecular masses $$NH_{3}, H_{2}O$$ and HF have lower boiling points than those of the subsequent group member hydrides

B
Due to higher electronegativity of N, O and F; $$NH_{3}, H_{2}O$$ and HF show hydrogen bonding and hence higher boiling points than the hydrides of their subsequent group members.

C
There is no regular trend in the boiling points of hydrides

D
Due to higher oxidation states of N, O and F, the boiling points of $$NH_{3}, H_{2}O$$ and HF are higher than the hydrides of their subsequent group members.

Solution

The boiling point increases with increase in molecular masses. But in the case of $$N, O$$ and $$F$$ they show the higher boiling point in molecules.
This is due to the higher electronegativity of $$N, O$$ and $$F$$ which leads to hydrogen bonding in $$NH_3, H_2O$$ and $$HF$$ due to which the boiling point of these increases & is higher than the hydrides of their subsequent group members.
Question 3
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In complex hydrides, hydride ions act as ligand and are coordinated to metal ions. These hydrides are good reducing agents. Which of the following hydrides is not a complex hydride?

A
$$LiAIH_{4}$$

B
$$NaBH_{4}$$

C
$$(AlH_{3})_{n}$$

D
$$LiBH_{4}$$

Solution

$$(AlH_{3})_{n}$$ is a polymeric hydride like $$(BeH_{2})_{n}, (MgH_{2})_{n}$$ etc.
Whereas, $$LiAlH_4$$, $$NaBH_4$$ and $$LiBH_4$$ are complex co-ordinate hydrides.
Hence option C is correct.
Question 4
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Peroxodisulphate on hydrolysis yields:

A
water

B
dihydrogen

C
hydrogen peroxide

D
deuterium

Solution

Peroxodisulphate on hydrolysis yield hydrogen peroxide.
$$HO_3SO-OSO_3H(aq) +2H_2O (l)\xrightarrow{hydrolysis}2HSO^-_4(aq)+2H^+(aq)+H_2O_2(aq)$$
Question 5
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Non-stoichiometric hydrides are produced by:

A
palladium, vanadium

B
manganese, lithium

C
nitrogen, fluorine

D
carbon, nickel

Solution

The hydrogen deficient compounds formed by the reaction of $$d-block$$ and $$f-block$$ elements with dihydrogen are called Non-stoichiometric compounds.
The d-block and f-block element form non-stoichiometric hydride because of the vacant d- and f-orbitals along with the small size.
Their elemental composition proportions cannot be represented in integers. They disobey the law of constant composition. Among the elements given, only vanadium and palladium form non-stoichiometric hydrides.

Answer: (A) palladium, vanadium
Question 6
1 / -0

Which of the statements given below are true for the structure of water molecule?
(i) Oxygen undergoes $$sp^{3}$$ hybridization.
(ii) Due to the presence of two lone pairs of electrons on oxygen, the $$H-O-H$$ bond angle is $$118.4^{0}$$
(iii) Due to angular geometry the net dipole moment of water is not zero, u = 1.84 D.

A
(i) and (ii)

B
(ii) and (iii)

C
(i) and (iii)

D
only (ii)

Solution

$$H-O-H$$ angle in water is slightly less than the typical tetrahedral angle. It is $$104.5^o$$. If all the substituents were the same (such as in $$CH_4$$), the bond angle would be $$109.5^o$$. In water, $$H-O-H$$ bond angle decreases from $$109.5^o$$ to $$104.5^o$$ as lone pair-lone pair repulsion is more than bond pair-bond pair repulsion and lone pair-bond pair repulsion.
Question 7
1 / -0

Liquid water is denser than ice due to:

A
higher surface tension

B
hydrogen bonding

C
van der Waals forces

D
covalent bonding

Solution

Due to inter-molecular hydrogen bonding, ice forms hexagonal three dimensional crystal lattice in which almost half the space is unoccupied. When ice melts, some of the hydrogen bonds are broken and some of the empty space is occupied by water molecules. Hence, liquid water is higher density than ice.
Question 8
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Which is not a property of water ?

A
It is a colorless and tasteless liquid

B
There is no hydrogen bonding in solid state of water

C
It is an excellent solvent for transportation of ions in plants and animals

D
Frozen water is lighter than liquid water.

Solution

There is extensive hydrogen bonding in solid state of water. In solid state water molecules are arranged in tetraheral manner with hexagonal three dimensional structure.
Question 9
1 / -0

Hydrogen resembles halogens in many respects for which several factors are responsible. Of the following factors which one is the most important in this respect?

A
Its tendency to lose an electron to form a cation

B
It tendency to gain a single electron in its valence shell to attain stable electronic configuration

C
Its low negative electron gain enthalpy value

D
Its small size

Solution

Hydrogen resembles halogens as it has similar tendency to gain a single electron to attain stable noble gas configuration. The electronic configuration of hydrogen is $$1s^1$$. It needs one more electron to attain stable electronic configuration of hellium, a noble gas. The electronic configuration of halogen is $$ns^2np^5$$. It needs one more electron to attain stable electronic configuration of a noble gas.
Question 10
1 / -0

When $$CO_2$$ is bubbled through a solution of barium peroxide in water:

A
carbonic acid is formed

B
$$H_2O_2$$ is formed

C
$$H_2O$$ is formed

D
barium hydroxide is formed

Solution

When $$CO_2$$ is bubbled through a cold solution of barium peroxide in water $$H_2O_2$$ is obtained
$$BaO_2+CO_2+H_2O \rightarrow BaCO_3+H_2O_2$$

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Hydrogen Test - 43

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Hydrogen Test - 43

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Question 1

$$NaH$$

$$CaH_{2}$$

$$CH_{4}$$

$$B_{2}H_{6}$$

Solution

Question 2

Due to lower molecular masses $$NH_{3}, H_{2}O$$ and HF have lower boiling points than those of the subsequent group member hydrides

Due to higher electronegativity of N, O and F; $$NH_{3}, H_{2}O$$ and HF show hydrogen bonding and hence higher boiling points than the hydrides of their subsequent group members.

There is no regular trend in the boiling points of hydrides

Due to higher oxidation states of N, O and F, the boiling points of $$NH_{3}, H_{2}O$$ and HF are higher than the hydrides of their subsequent group members.

Solution

Question 3

$$LiAIH_{4}$$

$$NaBH_{4}$$

$$(AlH_{3})_{n}$$

$$LiBH_{4}$$

Solution

Question 4

water

dihydrogen

hydrogen peroxide

deuterium

Solution

Question 5

palladium, vanadium

manganese, lithium

nitrogen, fluorine

carbon, nickel

Solution

Question 6

(i) and (ii)

(ii) and (iii)

(i) and (iii)

only (ii)

Solution

Question 7

higher surface tension

hydrogen bonding

van der Waals forces

covalent bonding

Solution

Question 8

It is a colorless and tasteless liquid

There is no hydrogen bonding in solid state of water

It is an excellent solvent for transportation of ions in plants and animals

Frozen water is lighter than liquid water.

Solution

Question 9

Its tendency to lose an electron to form a cation

It tendency to gain a single electron in its valence shell to attain stable electronic configuration

Its low negative electron gain enthalpy value

Its small size

Solution

Question 10

carbonic acid is formed

$$H_2O_2$$ is formed

$$H_2O$$ is formed

barium hydroxide is formed

Solution

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