Hydrogen MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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11.
Statues and paintings coated with white lead turn black on long exposure to atmosphere. The original colour can be restored by treating them with $${H_2}{O_2}.$$ The reason behind this is
A
blackened statues get coated with $$PbS$$ which on reaction with $${H_2}{O_2}$$ is oxidised to white $$PbS{O_4}.$$
B
$${H_2}{O_2}$$ dissolves the coating of white lead and exposes the inner surface.
C
white lead reacts with $${H_2}{O_2}$$ to form white $$PbS{O_4}.$$
D
blackened statues get coated with lead sulphate which reacts with $${H_2}{O_2}$$ to give $$PbS.$$
Answer :
blackened statues get coated with $$PbS$$ which on reaction with $${H_2}{O_2}$$ is oxidised to white $$PbS{O_4}.$$
The statues and paintings coated with white lead get blackened due to action of $${H_2}S$$ present in traces in atmosphere. When these are treated with $${H_2}{O_2},PbS$$ is oxidised to $$PbS{O_4}$$ which is colourless or white.
$$\mathop {PbS}\limits_{{\text{black}}} + 4{H_2}{O_2} \to \mathop {PbS{O_4}}\limits_{{\text{white}}} + 4{H_2}O$$
12.
Hydrogen molecules differs from chlorine molecule in the following respect
A
Hydrogen molecule is non-polar but chlorine molecule is polar
B
Hydrogen molecule is polar while chlorine molecule is non-polar
C
Hydrogen molecule can form intermolecular hydrogen bonds but chlorine molecule does not
D
Hydrogen molecule cannot participate in coordination bond formation but chlorine
molecule can
Answer :
Hydrogen molecule cannot participate in coordination bond formation but chlorine
molecule can
Chlorine has lone pair which it can donate to form coordinate bond, while hydrogen cannot.
13.
Among $$Ca{H_2},N{H_3},NaH$$ and $${B_2}{H_6}$$ which are covalent hydrides ?
A
$$N{H_3}\,{\text{and}}\,{B_2}{H_6}$$
B
$$NaH\,{\text{and}}\,Ca{H_2}$$
C
$$NaH\,{\text{and}}\,N{H_3}$$
D
$$Ca{H_2}\,{\text{and}}\,{B_2}{H_6}$$
Answer :
$$N{H_3}\,{\text{and}}\,{B_2}{H_6}$$
$$N$$ and $$B$$ belong to $$p$$ - block and since the electronegativity difference between these elements and $$H$$ is very less, they form covalent hydrides.
14.
How many hydrogen bonded water molecules are associated with $$CuS{O_4} \cdot 5{H_2}O?$$
A
Five
B
One
C
Four
D
Three
Answer :
One
Only one water molecule which is outside the coordination sphere is hydrogen bonded. Other four molecules of $${H_2}O$$ are coordinated.
$${\left[ {Cu{{\left( {{H_2}O} \right)}_4}} \right]^{2 + }}SO_4^{2 - } \cdot {H_2}O$$
15.
Hydrogen peroxide acts both as an oxidising and as a reducing agent depending upon the nature of the reacting species. In which of the following cases $${H_2}{O_2}$$ acts as a reducing agent in acid medium?
A
$$MnO_4^ - $$
B
$$C{r_2}O_7^{2 - }$$
C
$$SO_3^{2 - }$$
D
$$KI$$
Answer :
$$MnO_4^ - $$
$${H_2}{O_2}$$ acts as a reducing agent only in presence of strong oxidising agents ( i.e., $$MnO_4^ - $$ ) in acidic as well as alkaline medium.
$$2KMn{O_4} + 3{H_2}S{O_4} + 5{H_2}{O_2} \to {K_2}S{O_4} + 2MnS{O_4} + 8{H_2}O + 5{O_2}$$
16.
Hard water is not fit for washing clothes because
17.
$${H_2}{O_2} \to 2{H^ + } + {O_2} + 2{e^ - };\,{E^ \circ } = - 0.68\,V.$$ This equation represents which of the following behaviour of $${H_2}{O_2}?$$
A
Reducing
B
Oxidising
C
Acidic
D
Catalytic
Answer :
Reducing
As $${H_2}{O_2}$$ is loosing electrons, so it is acting as reducing agent.
18.
When a sample of hard water is passed through the layer of sodium zeolite resulting which of the following ions will not be present in the resulting sample of water obtained ?
As $$N{a^ + }\,ions$$ in sodium zeolite are replaced by $$C{a^{2 + }}$$ and $$M{g^{2 + }}\,ions$$ present in hard water, these two ions will not be present.
19.
Chemical $$A$$ is used for water softening to remove temporary hardness. $$A$$ reacts with sodium carbonate to generate caustic soda. When $$C{O_2}$$ is bubbled through a solution of $$A,$$ it turns cloudy. What is the chemical formula of $$A?$$
A
$$CaC{O_3}$$
B
$$CaO$$
C
$$Ca{\left( {OH} \right)_2}$$
D
$$Ca{\left( {HC{O_3}} \right)_2}$$
Answer :
$$Ca{\left( {OH} \right)_2}$$
$$Ca{\left( {OH} \right)_2}$$ is used for the softening of temporary hard water.
\[Ca{{\left( OH \right)}_{2}}\left( aq \right)\xrightarrow{C{{O}_{2}}}\underset{\text{cloudiness}}{\mathop{CaC{{O}_{3}}\left( s \right)}}\,+{{H}_{2}}O\left( l \right)\]
20.
Which of the following reactions of hydrogen with non-metals represents Haber's process?
A
\[2{{H}_{2}}+{{O}_{2}}\xrightarrow{\text{heat}}2{{H}_{2}}O\,;\] $$\Delta H = - 285.9\,kJ\,mo{l^{ - 1}}$$
B
\[3{{H}_{2}}+{{N}_{2}}\xrightarrow[200\,\,atm]{673\,K,\,Fe}2N{{H}_{3}};\] $$\Delta H = - 92.6\,kJ\,mo{l^{ - 1}}$$
C
\[{{H}_{2}}+C{{l}_{2}}\xrightarrow{hv}2HCl\]
D
\[2{{H}_{2}}+C\xrightarrow{{{1100}^{\circ }}C}C{{H}_{4}}\]
Answer :
\[3{{H}_{2}}+{{N}_{2}}\xrightarrow[200\,\,atm]{673\,K,\,Fe}2N{{H}_{3}};\] $$\Delta H = - 92.6\,kJ\,mo{l^{ - 1}}$$
Formation of $$N{H_3}$$ by reaction of hydrogen and nitrogen is known as Haber's process.