Learn Redox Reaction MCQ questions & answers in Physical Chemistry are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
51.
Oxidation numbers of $$Mn$$ in its compounds $$MnC{l_2},Mn{\left( {OH} \right)_3},Mn{O_2}$$ and $$KMn{O_4}$$ respectively are
A
+2, +4, +7, +3
B
+2, +3, +4, +7
C
+7, +3, +2, +4
D
+7, +4, +3, +2
Answer :
+2, +3, +4, +7
$$\eqalign{
& MnC{l_2} \to x + \left( { - 2} \right) = 0 \Rightarrow x = + 2 \cr
& Mn{\left( {OH} \right)_3} \to x + \left( { - 3} \right) = 0 \Rightarrow x = + 3 \cr
& Mn{O_2} \to x + \left( { - 4} \right) = 0 \Rightarrow x = + 4 \cr
& KMn{O_4} \to + 1 + x + \left( { - 8} \right) = 0 \Rightarrow x = + 7 \cr} $$
53.
Mark the correct statement from the following,
A
Copper metal can be oxidised by $$Z{n^{2 + }}$$ ions.
B
Oxidation number of phosphorus in $${P_4}$$ is 4.
C
An element in the highest oxidation state acts only as a reducing agent.
D
The element which shows highest oxidation number of +8 is $$Os$$ in $$Os{O_4}.$$
Answer :
The element which shows highest oxidation number of +8 is $$Os$$ in $$Os{O_4}.$$
$$Cu$$ cannot be oxidised by $$Z{n^{2 + }}$$ ion, oxidation number of phosphorus is zero. An element in its highest oxidation state acts as an oxidising agent.
54.
In which of the following reactions, the underlined substance has been reduced?
A
$$\underline {CO} + CuO \to C{O_2} + Cu$$
B
$$\underline {CuO} + 2HCl \to CuC{l_2} + {H_2}O$$
C
$${\underline {4H} _2}\underline O + 3Fe \to 4{H_2} + F{e_3}{O_4}$$
D
$$\underline C + 4HN{O_3} \to $$ $$C{O_2} + 2{H_2}O + 4N{O_2}$$
$$\mathop {{H_2}O}\limits^{ + 1} \to \mathop {{H_2}}\limits^0 $$
Oxidation number of $$H$$ decreases from +1 to 0. Hence, $${H_2}O$$ is reduced to $${H_2}.$$
55.
The largest oxidation number exhibited by an element depends on its outer electronic configuration. With which of the following outer electronic configurations the element will exhibit largest oxidation number?
A
$$3{d^1}4{s^2}$$
B
$$3{d^3}4{s^2}$$
C
$$3{d^5}4{s^1}$$
D
$$3{d^5}4{s^2}$$
Answer :
$$3{d^5}4{s^2}$$
$$3{d^1}4{s^2}$$ will exhibit + 3, $$3{d^3}4{s^2} = + 5,3{d^5}4{s^1} = + 6$$ and $$3{d^5}4{s^2} = + 7$$ oxidation numbers.
56.
The number of electrons involved in the reduction of one nitrate ion to hydrazine is
A
8
B
5
C
3
D
7
Answer :
7
$$\mathop N\limits^{ + 5} O_3^ - \to \mathop {N_2^{}}\limits^{ - 2} {H_4}$$ So, for reduction of $$1\,mole$$ of $$NO_3^ - $$ number of electrons required is 7.
57.
Which type of redox reaction is shown by the following reaction?
$$C{l_{2\left( g \right)}} + 2KB{r_{\left( {aq} \right)}} \to 2KC{l_{\left( {aq} \right)}} + B{r_{2\left( l \right)}}$$
A
Decomposition reaction
B
Metal displacement reaction
C
Non-metal displacement reaction
D
Disproportionation reaction
Answer :
Non-metal displacement reaction
Chlorine displaces $$B{r^ - }$$ ions in an aqueous solution.
58.
In an oxidation process for a cell, $${M_1} \to M_1^{n + } + n{e^ - },$$ the other metal $$\left( {{M_2}} \right)$$ being univalent showing reduction takes up ______ electrons to complete redox reaction.
A
$$\left( {n - 1} \right)$$
B
$$1$$
C
$$n$$
D
$$2$$
Answer :
$$n$$
The reaction shows $$M_2^{n + } + n{e^ - } \to {M_2}$$
i.e., electrons released at anode = electrons used at cathode.
59.
A compound contains atoms $$X,Y$$ and $$Z.$$ The oxidation number of $$X$$ is +2, $$Y$$ is +5 and $$Z$$ is -2. The possible formula of the compound is
A
$$XY{Z_2}$$
B
$${Y_2}{\left( {X{Z_3}} \right)_2}$$
C
$${X_3}{\left( {Y{Z_4}} \right)_2}$$
D
$${X_3}{\left( {{Y_4}Z} \right)_2}$$
Answer :
$${X_3}{\left( {Y{Z_4}} \right)_2}$$
Sum of the oxidation numbers of atoms in it, is zero.
60.
Carbon is in the lowest oxidation state in
A
$$C{H_4}$$
B
$$CC{l_4}$$
C
$$C{F_4}$$
D
$$C{O_2}$$
Answer :
$$C{H_4}$$
Oxidation number of carbon in the given compounds :
$$\eqalign{
& {\text{In}}\,\,\mathop {C{H_4}}\limits^{x + 1} = x + 4\left( { + 1} \right) = 0 \Rightarrow x = - 4 \cr
& {\text{In}}\,\,\mathop C\limits^x \mathop {C{l_4}}\limits^{ - 1} = x + 4\left( { - 1} \right) = 0 \Rightarrow x = + 4 \cr
& {\text{In}}\,\,\mathop C\limits^x \mathop {{F_4}}\limits^{ - 1} = x + 4\left( { - 1} \right) = 0 \Rightarrow x = + 4 \cr
& {\text{In}}\,\,\mathop C\limits^x \mathop {{O_2}}\limits^{ - 2} = x + 2\left( { - 2} \right) = 0 \Rightarrow x = + 4 \cr} $$