Redox Reactions Test

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Redox Reactions Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

Which is the anhydride of $${HNO}_{3}$$ out of $${N}_{2}O,NO,{N}_{2}{O}_{5}$$?

A
$${N}_{2}{O}_{5}$$

B
$$NO$$

C
$${N}_{2}O$$

D
$${HNO}_{3}$$

Solution

The anhydride of $$HNO_3$$ is $${N}_{2}{O}_{5}$$ (oxidation number of the element $$(N)$$ in oxyacid and its anhydride remains same).
Question 2
1 / -0

Choose the correct option regarding change in oxidation number of the underlined atom in the following conversion.

$$2NaOH+\underline { Br_{ 2 } } \rightarrow NaBr+NaBrO+H_{ 2 }O$$

A
Decreases to $$-1$$ w.r.t $$NaBr$$ and increases to $$-3$$ w.r.t $$NaBrO$$

B
Decreases to $$-1$$ w.r.t $$NaBr$$ and increases to $$+2$$ w.r.t $$NaBrO$$

C
Decreases to $$-1$$ w.r.t $$NaBr$$ and increases to $$+1$$ w.r.t $$NaBrO$$

D
Decreases to $$-2$$ w.r.t $$NaBr$$ and increases to $$+1$$ w.r.t $$NaBrO$$

Solution

The oxidation number of bromine in $$Br_2, NaBr$$ and $$NaBrO$$ is $$0,-1$$ and $$+1$$ respectively.
Hence, during the reaction, the oxidation number of bromine in $$Br_2$$ decreases to $$-1$$ w.r.t $$NaBr$$ and increases to $$+1$$ w.r.t $$NaBrO$$.
Thus, the correct option will be C.
Question 3
1 / -0

$${ VO }_{ 2 }^{ 2+ }$$ and $${ VO }^{ 2+ }$$ are known as vanadyl ion. The oxidation number of vanadium in each ions are, respectively :

A
$$+6,\:-4$$

B
$$-6,\:-4$$

C
$$-6,\:+4$$

D
$$+6,\:+4$$

Solution

The oxidation number of $$V$$ is $$+6$$ in $${ VO }_{ 2 }^{ 2+ }$$ and $$+4$$ in $${ VO }^{ 2+ }$$.
Question 4
1 / -0

Which of the following is a spectator ion in the reaction?
$$2Na+2HCl \rightarrow 2NaCl + H_2$$

A
$$Na^+$$

B
$$Cl^-$$

C
$$H^+$$

D
None of these

Solution

Spectator ions are the ions which donot take part in the reaction.
$$2Na+2HCl \rightarrow 2NaCl + H_2$$
0 +1 -1 +1 -1 0
So, ionic reaction is,
$$2Na +2H^+ \rightarrow 2Na^+ + H_2$$
Only for chloride ions, oxidation state doesnt change, so they are spectator here.
Question 5
1 / -0

Oxidation number of carbon and total number of atoms in carbon suboxide are :

A
$$+5, \:4$$

B
$$+\frac{4}{3},\:3$$

C
$$+\frac{4}{3},\: 5$$

D
$$+4,\: 5$$

Solution

The formula of carbon suboxide is $$C_3O_2\:[O=C=C=C=O]$$.
So, it has a total of $$5$$ atoms.
Let the oxidation state of $$C$$ in $$C_3O_2$$ be $$x$$.
The oxidation state of $$C$$ is $$3x-4=0$$ or, $$x = \frac{4}{3}$$ (since the compound is neutral, sum of oxidation states is equated to zero).
Question 6
1 / -0

Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen?

A
$$HNO_3, NO, NH_4Cl, N_2$$

B
$$HNO_3, NO, N_2, NH_4Cl$$

C
$$HNO_3, NH_4Cl, NO, N_2$$

D
$$NO, HNO_3, NH_4Cl, N_2$$

Solution

The oxidation state of nitrogen in different compounds are as follows:
Compound Oxidation state
$$HNO_3$$    +5
$$NO$$    +2
$$N_2$$        0
$$NH_4Cl$$      -3
Hence, the correct ordering according to the decreasing order of the oxidation state of nitrogen is $$HNO_3,\ NO,\ N_2,\ NH_4Cl$$.
Question 7
1 / -0

Which one of the following is not correct for the reaction?
$$(CN)_2 + H_2O \rightarrow HCN+ HOCN$$

A
It is an auto redox change.

B
Oxidation number of C in $$(CN)_2, HCN$$ and $$HOCN$$ are +3, +2 and +4 respectively.

C
Oxidation number of n in $$(CN)_2, HCN$$ and $$HOCN$$ are +3, +2 and +4 respectively.

D
The resultant solution is acidic.

Solution

$$(CN)_2 + H_2O \rightarrow HCN+ HOCN$$
$$+3$$ $$+2$$ $$+4$$
So, its a autoredox reaction.
The oxidation number of N in $$(CN)_2, HCN$$ and $$HOCN$$ are +3, +2 and +4 respectively.
Question 8
1 / -0

Determine the oxidation state of the underlined element in $${ K }_{ 4 }\underline { { P }_{ 2 } } { O }_{ 7 }$$.

A
$$+5$$

B
$$+3$$

C
$$+4$$

D
$$+6$$

Solution

Let $$x$$ be the oxidation number of $$P$$ in $$K_4P_2O_7$$.
Since the overall charge on the complex is $$0$$, the sum of oxidation states of all elements in it should be equal to $$0$$.
Therefore, $$4(+1) + 2(x) +7(-2)=0$$.
$$4+2x-14=0$$
$$2x =10$$
$$x = +5$$
Hence, the oxidation number of P in $$K_4P_2O_7$$ is $$+5$$.
Question 9
1 / -0

$${ K }_{ 3 }Cr{ \left( { O }_{ 2 } \right) }_{ x }$$ has peroxy linkage with oxidation number of $$Cr$$ as $$+5$$. The value of $$x$$ is :

A
$$4$$

B
$$3$$

C
$$2$$

D
$$6$$

Solution

$${ K }_{ 3 }Cr{ \left( { O }_{ 2 } \right) }_{ x }$$ has $$Cr{ \left( { O }_{ 2 } \right) }_{ x }^{ 3- }$$ ion.
Therefore, $$+5-2x=-3$$
or, $$-2x=-8$$
or, $$x=4$$.
It is $${ K }_{ 3 }Cr{ \left( { O }_{ 2 } \right) }_{ 4 }$$.
Question 10
1 / -0

Which has maximum oxidation number of the underlined atom in the following?

A
$$\underline { Mn } { O }_{ 4 }^{ 2- }$$

B
$$\underline { Cr } { O }_{ 5 }$$

C
$$\underline { Cr } { O }_{ 2 }{ Cl }_{ 2 }$$

D
Equal

Solution

Oxidation state of the underlined atom in each case is given,
(a) $$\underline { Mn } { O }_{ 4 }^{ 2- }$$ ; $$x=6$$

(b) $$\underline { Cr } { O }_{ 5 }$$ ; $$x=6$$

(c) $$\underline { Cr } { O }_{ 2 }{ Cl }_{ 2 }$$ ; $$x=6$$
Hence, all are equal.

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Re-Attempt Weekly Quiz Competition

Redox Reactions Test - 23

Result

Redox Reactions Test - 23

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Question 1

$${N}_{2}{O}_{5}$$

$$NO$$

$${N}_{2}O$$

$${HNO}_{3}$$

Solution

Question 2

Decreases to $$-1$$ w.r.t $$NaBr$$ and increases to $$-3$$ w.r.t $$NaBrO$$

Decreases to $$-1$$ w.r.t $$NaBr$$ and increases to $$+2$$ w.r.t $$NaBrO$$

Decreases to $$-1$$ w.r.t $$NaBr$$ and increases to $$+1$$ w.r.t $$NaBrO$$

Decreases to $$-2$$ w.r.t $$NaBr$$ and increases to $$+1$$ w.r.t $$NaBrO$$

Solution

Question 3

$$+6,\:-4$$

$$-6,\:-4$$

$$-6,\:+4$$

$$+6,\:+4$$

Solution

Question 4

$$Na^+$$

$$Cl^-$$

$$H^+$$

None of these

Solution

Question 5

$$+5, \:4$$

$$+\frac{4}{3},\:3$$

$$+\frac{4}{3},\: 5$$

$$+4,\: 5$$

Solution

Question 6

$$HNO_3, NO, NH_4Cl, N_2$$

$$HNO_3, NO, N_2, NH_4Cl$$

$$HNO_3, NH_4Cl, NO, N_2$$

$$NO, HNO_3, NH_4Cl, N_2$$

Solution

Question 7

It is an auto redox change.

Oxidation number of C in $$(CN)_2, HCN$$ and $$HOCN$$ are +3, +2 and +4 respectively.

Oxidation number of n in $$(CN)_2, HCN$$ and $$HOCN$$ are +3, +2 and +4 respectively.

The resultant solution is acidic.

Solution

Question 8

$$+5$$

$$+3$$

$$+4$$

$$+6$$

Solution

Question 9

$$4$$

$$3$$

$$2$$

$$6$$

Solution

Question 10

$$\underline { Mn } { O }_{ 4 }^{ 2- }$$

$$\underline { Cr } { O }_{ 5 }$$

$$\underline { Cr } { O }_{ 2 }{ Cl }_{ 2 }$$

Equal

Solution

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