Hydrogen Test

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Hydrogen Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

During hydrate formation from aqueous solution water can be associated in different forms. Indicate the wrong combination.
Coordinated water -$$[Cr(H_{2}O)_{6}]^{3+}3Cl^{-}$$
Interstitial water - $$BaCl_{2}.2H_{2}O$$
Hydrogen bonded water --$$[Cu(H_{2}O)_{4}]^{2+}SO_{4}^{2-}.H_{2}O$$

A
(i)

B
(ii)

C
(iii)

D
none of these

Solution

Coordinated water is water that is a ligand in a metal complex. The metal atom and the water molecules are held together by coordinate bonds.
Interstitial water is water that is present in the interstitial sites of a crystal lattice.
Hydrogen-bonded water is the water that that is linked to some other ion by hydrogen bonding.
Question 2
1 / -0

Given below are the elements and the type of hydrides formed by them mark the incorrect match.

A
Phosphourus-molecular hydride

B
Potassium-Ionic hydride

C
Vanadium-Interstitial hydride

D
Nitrogen-Electron-deficient covalent hydride

Solution

Nitrogen forms electron-rich covalent or molecular hydrides. They have more number of electrons than normal covalent bonds. Excess electrons appear as unshared pair of electrons.
Question 3
1 / -0

The oxide that gives $$H_2O_2$$ on treatment with dilute $$H_2SO_4 $$ is:

A
$$PbO_2$$

B
$$BaO_2. 8H_2O$$

C
$$MnO_2$$

D
$$TiO_2$$

Solution

The oxide that gives $$H_2O_2$$ on treatment with dilute $$H_2SO_4 $$ is $$BaO_2. 8H_2O$$.
$$BaO_2. 8H_2O({s}) + H_2SO_4({aq})\rightarrow BaSO_4({s}) + H_2O_2({aq}) + 8H_2O{(l)}$$
Question 4
1 / -0

Only one element of ________ forms hydride.

A
group 6

B
group 7

C
group 8

D
group 9

Solution

The only element of group $$6$$ i.e. Chromium $$(Cr)$$ forms hydride. The other elements do not form hydride due to low affinity towards hydrogen.

Answer: (A) group $$6$$
Question 5
1 / -0

Hydrogen peroxide is obtained by the electrolysis of:

A
water

B
sulphuric acid

C
hydrochloric acid

D
fused sodium peroxide

Solution

$$H_2O_2 $$ is obtained by the electrolysis of a cold 50% solution of $$H_2O_2$$
$$2H_2SO_4\rightarrow 2H^+ + 2HSO^-_4 $$
The electrolysis of $$HSO^-_4 $$ gives $$H_2O_2$$.
Question 6
1 / -0

Which of the following hydrides is electron-precise hydride?

A
$$B_2H_6$$

B
$$NH_3$$

C
$$H_2O$$

D
$$CH_4$$

Solution

Methane $$CH_4$$ is electron-precise hydride.
Diborane $$B_2H_6$$ is electron-deficient hydride
Ammonia $$NH_3$$ and water $$H_2O$$ are electron-rich hydrides.

Elements of group 14 form electron-precise (having required number of electrons to write the Lewis structure ) form precise hydrides.
Question 7
1 / -0

When a substance reacts with water and breaks into acids and bases the process is called:

A
dialysis

B
solvolysis

C
hydrolysis

D
dissolution

Solution

Answer:- (C) hydrolysis
Explanation:-
The chemical breakdown of a compound into acid and base due to reaction with water is termed as hydrolysis.
Question 8
1 / -0

Bentonite, a clay of alumina-silica, is dropped from aeroplanes in the form of a slurry with water for the purpose of:

A
cooling the soil

B
fertilizing the soil

C
fumigation of plants

D
spreading water over fires

Solution

Bentonite helps in extinguishing the fire by spreading water over it.
Question 9
1 / -0

Elements of which of the following group (s) of periodic table do not form hydrides?

A
Groups 7, 8, 9

B
Group 13

C
Groups 15, 16, 17

D
Group 14

Solution

$$\bf{Hint-}$$ Hydrogen having negative charge is called as hydride $$H^-$$.

$$\bf{Explanation-}$$
The first elements of group $$7, 8, 9, 13, 14, 15, 16,$$ and $$17$$ are $$Mn, Fe, Co, B, C, N, O,$$ and $$F$$ from the periodic table.
$$\bullet$$ Consider group-13, Boran forms more hydrides with the general formula $$B_nH_{n+4}$$.
For example: diborane $$(B_2H_6 )$$, pentaborane $$(B_5H_9 )$$, etc.
$$\bullet$$ In group-14, carbon forms hydrides with hydrogen.
For example: methane $$(CH_4)$$, etc.
$$\bullet$$ In group-15, ammonia is an important nitrogen hydride.
$$\bullet$$ In group-16, oxygen forms hydrides are water $$(H_2O)$$, hydrogen peroxide $$(H_2O_2)$$, etc.
$$\bullet$$ In group-17, fluorine forms hydrides as $$HF$$.
$$\bullet$$ Group $$7, 8, 9$$ does not form hydrides out of the periodic table. The property does not form hydrides of these $$7, 8, 9$$ groups are known as the hydride gap.

$$\bf{Conclusion-}$$ Hence, option A is correct.
Question 10
1 / -0

The oxidising property of $$H_2O_2$$ is best explained by assuming that two oxygen atoms in its molecule are_______

A
bonded differently.

B
bonded similarly

C
bonded covalently

D
bonded by hydrogen bonds

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Hydrogen Test - 44

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Hydrogen Test - 44

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SHARING IS CARING

Question 1

(i)

(ii)

(iii)

none of these

Solution

Question 2

Phosphourus-molecular hydride

Potassium-Ionic hydride

Vanadium-Interstitial hydride

Nitrogen-Electron-deficient covalent hydride

Solution

Question 3

$$PbO_2$$

$$BaO_2. 8H_2O$$

$$MnO_2$$

$$TiO_2$$

Solution

Question 4

group 6

group 7

group 8

group 9

Solution

Question 5

water

sulphuric acid

hydrochloric acid

fused sodium peroxide

Solution

Question 6

$$B_2H_6$$

$$NH_3$$

$$H_2O$$

$$CH_4$$

Solution

Question 7

dialysis

solvolysis

hydrolysis

dissolution

Solution

Question 8

cooling the soil

fertilizing the soil

fumigation of plants

spreading water over fires

Solution

Question 9

Groups 7, 8, 9

Group 13

Groups 15, 16, 17

Group 14

Solution

Question 10

bonded differently.

bonded similarly

bonded covalently

bonded by hydrogen bonds

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