Hydrogen Test

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Hydrogen Test - 58

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Weekly Quiz Competition

Question 1
1 / -0

The various types of hydrides and examples of each type are given below.
Hydride type Compound
(A) Electron deficient (i) $$LiH$$
(B) Saline (ii) $$CH_{4}$$
(C) Electrons-precise (iii) $$NH_{3}$$
(D) Interstitial (iv) $$B_{2}H_{6}$$
(E) Electron rich (v) $$CrH$$

Choose the correct matching from the codes given below.

A
$$(A) - (ii), (B) - (iv), (C) - (v), (D) - (iii), (E) - (i)$$

B
$$(A) - (iv), (B) - (i), (C) - (ii), (D) - (v), (E) - (iii)$$

C
$$(A) - (iv), (B) - (iii), (C) - (v), (D) - (ii), (E) - (i)$$

D
$$(A) - (v), (B) - (iii), (C) - (iv), (D) - (ii), (E) - (i)$$

E
$$(A) - (iv), (B) - (v), (C) - (i), (D) - (ii), (E) - (iii)$$

Solution

(A) Electron deficient $$\rightarrow { B }_{ 2 }{ H }_{ 6 }$$ ($$\because$$ It has $$3c-2e$$ structure, which shows that $$B$$-atom has electron deficient centre)
(B) Saline $$\rightarrow LiH$$ is called saline
(C) Electron precise $$\rightarrow C{ H }_{ 4 }$$ ($$\because$$ It has perfect complete octet $$C$$-atom).
(D) Interstitial $$\rightarrow CrH$$ Here, hydrogen atoms are trapped in the interstitial sites of crystal of $$Cr$$.
(E) Electron rich $$\rightarrow \ddot { N } { H }_{ 3 }$$, because nitrogen has lone pair of electrons in its outermost shell.
Hence, correct order is $$(A) - (iv), (B) - (i), (C) - (ii), (D) - (v), (E) - (iii)$$
Question 2
1 / -0

Hydorgen forms binary compounds called hydrides with several elements in the periodic table.Regarding these hydrides the wrong statement:

A
Among lanthanoids, europioum and ytterbium form ionic hydrides.

B
The density of ionic hydrides is less, while the density of interstitial hydrides is more than the metals from which they are formed.

C
The rate of reaction of alkali metals towards hydrogen decreases from lithium to caesium

D
Hydrogen gas can be purified by the phenomenon of occlusion of hydrogen by palladium metal.
Question 3
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Ionic hydride reacts with water to give:

A
hydride ions

B
acidic solution

C
protons

D
basic solution

Solution

Ionic hydrides give basic solution when reacts with water.
e.g.
$$LiH+{ H }_{ 2 }O\longrightarrow \underset{base}{LiOH}+{ H }_{ 2 }\uparrow $$
Question 4
1 / -0

Hydrogen gas is obtained from:

A
petroleum

B
coal

C
coke

D
natural gas

Solution

Fossil fuels are the dominant source of industrial hydrogen. Carbon dioxide can be separated from natural gas with a 60-70% efficiency for hydrogen production and from other hydrocarbons to varying degrees of efficiency.

Specifically, bulk hydrogen is usually produced by the steam reforming of methane or natural gas.
Question 5
1 / -0

Which of the following is laboratory preparation of dihydrogen?

A
$$3Fe + 4H_{2} O (steam) \rightarrow Fe_{3} O_{4} + 4H_{2}$$

B
$$2Na + 2H_{2}O \rightarrow 2NaOH + H_{2}$$

C
$$CaH_{2} + 2H_{2}O\rightarrow Ca(OH)_{2} + 2H_{2}$$

D
$$Zn + H_{2}SO_{4} (dil.) \rightarrow ZnSO_{4} + H_{2}$$

Solution

In the laboratory, hydrogen gas is prepared by the action of dilute $$H_{2}SO_{4}$$ on granulated zinc.
$$Zn + H_{2}SO_{4} (dil.) \rightarrow ZnSO_{4} + H_{2} $$
Question 6
1 / -0

Which of the following is most soluble in water?

A
$${ CsClO }_{ 4 }$$

B
$${ NaClO }_{ 4 }$$

C
$${ KClO }_{ 4 }$$

D
$${ LiClO }_{ 4 }$$

Solution

$$NaClO_4$$ is the inorganic compound. It is a white crystalline, hygroscopic solid that is highly soluble in water and in alcohol. It usually encountered as the monohydrate. The compound is noteworthy as the most soluble of the common perchlorate salts. It decreases from top to bottom
Question 7
1 / -0

$$H_{2}$$ gas is liberated at cathode and anode both by electrolysis of the given aq. solution except:

A
$$NaH$$

B
$$HCOONa$$

C
$$CaH_{2}$$

D
$$LiH$$
Question 8
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Which of the following statements regarding hydrides iis not correct?

A
Ionic hydrides are crystalline, non-volatile and non-conducting in solid state

B
Electron-deficient hydrides act as Lewis acids or electron acceptors

C
Elements of group$$-13$$ form electron-deficient hydride

D
Elements of group $$15-17$$ form electron-precise hydrides

Solution

Elements of group 15-17 form electron-rich hydrides. Group 14 elements form electron-precise hydrides. Elements of group-13 form electron-deficient hydrides. For example, diborane $$B_2H_6$$ is electron deficient, methane $$CH_4$$ is electron precise whereas $$NH_3$$, $$H_2O$$ and $$HF$$ are electron rich.
Question 9
1 / -0

The order of reactivity of halogens towards hydrogen is:

A
$$F_{2} > Cl_{2} > Br_{2} > I_{2}$$

B
$$I_{2} > Br_{2} > Cl_{2} > F_{2}$$

C
$$Cl_{2} > Br_{2} > I_{2} > F_{2}$$

D
$$Br_{2} > Cl_{2} > F_{2} > I_{2}$$

Solution
Question 10
1 / -0

Match the column I with column II and mark the appropriate choice.
column I column II
A NaH i Interstitial hydride
B $$CH_{4}$$ ii Molecular hydride
C $$VH_{0.56}$$ iii Ionic hydride
D $$B_{2}H_{6}$$ iv Electron-deficient hydride

A
$$(A)\rightarrow (iii), (B)\rightarrow (iv), (C)\rightarrow (ii) (D)\rightarrow (i)$$

B
$$(A)\rightarrow (ii), (B)\rightarrow (iv), (C)\rightarrow (iii) (D)\rightarrow (i)$$

C
$$(A)\rightarrow (i), (B)\rightarrow (ii), (C)\rightarrow (iv) (D)\rightarrow (iii)$$

D
$$(A)\rightarrow (iii), (B)\rightarrow (ii), (C)\rightarrow (i) (D)\rightarrow (iv)$$

Solution

NaH is an ionic compound and is made of sodium cations $$\left ( Na^{+} \right )$$ and hydride anions $$\left ( H^{-} \right )$$. It has the octahedral crystal structure with each sodium ion surrounded by six hydride ions.

$$CH_{4}$$ forms molecular or covalent hydrides. These are mainly formed by p - block elements and some s - block elements.

The third compound has an oxidation number of hydrogen which is zero. So it belongs to Interstitial hydride.

In $$B_{2}H_{6}$$ the Boron atom is surrounded by 6 electrons, so it is electron deficient due to its incomplete octet.

The correct option is D.

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	Hydride type		Compound
(A)	Electron deficient	(i)	$$LiH$$
(B)	Saline	(ii)	$$CH_{4}$$
(C)	Electrons-precise	(iii)	$$NH_{3}$$
(D)	Interstitial	(iv)	$$B_{2}H_{6}$$
(E)	Electron rich	(v)	$$CrH$$

	column I		column II
A	NaH	i	Interstitial hydride
B	$$CH_{4}$$	ii	Molecular hydride
C	$$VH_{0.56}$$	iii	Ionic hydride
D	$$B_{2}H_{6}$$	iv	Electron-deficient hydride

Hydrogen Test - 58

Result

Hydrogen Test - 58

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Question 1

$$(A) - (ii), (B) - (iv), (C) - (v), (D) - (iii), (E) - (i)$$

$$(A) - (iv), (B) - (i), (C) - (ii), (D) - (v), (E) - (iii)$$

$$(A) - (iv), (B) - (iii), (C) - (v), (D) - (ii), (E) - (i)$$

$$(A) - (v), (B) - (iii), (C) - (iv), (D) - (ii), (E) - (i)$$

$$(A) - (iv), (B) - (v), (C) - (i), (D) - (ii), (E) - (iii)$$

Solution

Question 2

Among lanthanoids, europioum and ytterbium form ionic hydrides.

The density of ionic hydrides is less, while the density of interstitial hydrides is more than the metals from which they are formed.

The rate of reaction of alkali metals towards hydrogen decreases from lithium to caesium

Hydrogen gas can be purified by the phenomenon of occlusion of hydrogen by palladium metal.

Question 3

hydride ions

acidic solution

protons

basic solution

Solution

Question 4

petroleum

coal

coke

natural gas

Solution

Question 5

$$3Fe + 4H_{2} O (steam) \rightarrow Fe_{3} O_{4} + 4H_{2}$$

$$2Na + 2H_{2}O \rightarrow 2NaOH + H_{2}$$

$$CaH_{2} + 2H_{2}O\rightarrow Ca(OH)_{2} + 2H_{2}$$

$$Zn + H_{2}SO_{4} (dil.) \rightarrow ZnSO_{4} + H_{2}$$

Solution

Question 6

$${ CsClO }_{ 4 }$$

$${ NaClO }_{ 4 }$$

$${ KClO }_{ 4 }$$

$${ LiClO }_{ 4 }$$

Solution

Question 7

$$NaH$$

$$HCOONa$$

$$CaH_{2}$$

$$LiH$$

Question 8

Ionic hydrides are crystalline, non-volatile and non-conducting in solid state

Electron-deficient hydrides act as Lewis acids or electron acceptors

Elements of group$$-13$$ form electron-deficient hydride

Elements of group $$15-17$$ form electron-precise hydrides

Solution

Question 9

$$F_{2} > Cl_{2} > Br_{2} > I_{2}$$

$$I_{2} > Br_{2} > Cl_{2} > F_{2}$$

$$Cl_{2} > Br_{2} > I_{2} > F_{2}$$

$$Br_{2} > Cl_{2} > F_{2} > I_{2}$$

Solution

Question 10

$$(A)\rightarrow (iii), (B)\rightarrow (iv), (C)\rightarrow (ii) (D)\rightarrow (i)$$

$$(A)\rightarrow (ii), (B)\rightarrow (iv), (C)\rightarrow (iii) (D)\rightarrow (i)$$

$$(A)\rightarrow (i), (B)\rightarrow (ii), (C)\rightarrow (iv) (D)\rightarrow (iii)$$

$$(A)\rightarrow (iii), (B)\rightarrow (ii), (C)\rightarrow (i) (D)\rightarrow (iv)$$

Solution

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