Redox Reactions Test

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Redox Reactions Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

The pair of compounds in which the metals are in their highest oxidation state is:

A
$$MnO_2, FeCl_3$$

B
$$Mn{ O }_{ 4 }^{ - }, CrO_2Cl_2$$

C
$$MnCl_2 , CrCl_3$$

D
$$[NiCl_4]^{2-}, [ CoCl_4]^-$$

Solution

The charges on oxygen and chlorine atoms are $$-2$$ and $$-1$$ respectively. Therefore, the oxidation states of the central atoms in the given compounds are calculated as follows:

$$MnO_2 , FeCl_3 \rightarrow +4 , +3 $$

$$MnO_{4}^{-} , CrO_2Cl_2 \rightarrow +7 , +6$$

$$MnCl_2 , CrCl_3 \rightarrow +2 , +3$$

$$[NiCl_4]^{-2} , [CoCl_4]^- \rightarrow +2 , +3$$
Question 2
1 / -0

The element that does not show positive oxidation state is:

A
$$O$$

B
$$N$$

C
$$Cl$$

D
$$F$$

Solution

Fluorine is the most electronegative element in the periodic table. For showing positive oxidation state, it should be bonded to the more electronegative element than fluorine. But there is no such element in the periodic table. Hence, fluorine always shows a negative oxidation state and can not show positive oxidation state.
Question 3
1 / -0

Carbon is in the lowest oxidation state in:

A
$$CH_4$$

B
$$CCl_4$$

C
$$CF_4$$

D
$$CO_2$$

Solution

Let the oxidation number of carbon be $$x$$.
Oxidation state of $$H=+1$$, $$Cl=F=-1$$ and of $$O=-2$$.
(A) $$CH_4$$: $$x+4 \times 1=0$$
$$x+4=0$$ or $$x=-4$$
(B) $$CCl_4$$: $$x+4 \times (-1)=0$$
$$x-4=0$$ or $$x=+4$$
(C) $$CF_4$$: $$x+4 \times (-1)=0$$
$$x-4=0$$ or $$x=+4$$
(D) $$CO_2$$: $$x+2 \times (-2)=0$$
$$x-4=0$$ or $$x=+4$$
Lowest oxidation of carbon is $$-4$$ in $$CH_4$$.
Question 4
1 / -0

Which compound amongst the following has the highest oxidation number of $$Mn$$?

A
$$KMnO_4$$

B
$$K_2MnO_4$$

C
$$MnO_2$$

D
$$Mn_2O_3$$

Solution

Let the oxidation number of $$Mn$$ be $$x$$.
Oxidation state of $$K=+1$$, $$O=-2$$.
(A) $$KMnO_4$$: $$1+x+4 \times (−2)=0$$
$$x=+7$$
(B) $$K_2MnO_4$$: $$2 \times 1+x+4 \times (−2)=0$$
$$x=+6$$
(C)$$MnO_2$$: $$x+2 \times (−2)=0$$
$$x=+4$$
(D) $$Mn_2O_3$$: $$2x+3 \times (−2)=0$$
$$x=+3$$
therefore highest oxidation number of $$Mn$$ is +7 in $$KMnO_4$$
Question 5
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Match the column I with column II and mark the appropriate choice.
Column I
( Compound ) Coloumn II
(Oxidation state of $$Fe$$)
( A ) $$K_3[(Fe(OH)
_6]$$ (i) +8/3
( B ) $$ K_2[FeO_4]$$ (ii) +2
( C ) $$FeSO_4, (NH_4)_2SO_46H_2O$$ (iii) +3
( D) $$Fe_3O4$$ (iv) +6

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

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

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

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

Solution

Let the oxidation number of $$Fe$$ be $$x$$.
Oxidation number of the ligands are: $$H=K=+1$$, $$O=-2,OH=-2+1=-1,NH_4=+1,SO_4=-2,H_2O=0$$
$$(A)$$ $$K_3[Fe(OH)_6]$$: $$3 \times 1 +x+6 \times (-1)=0$$
$$x+3-6=0$$ or $$x=+3$$ =(iii)
$$(B)$$ $$K_2[FeO_4]$$: $$2 \times 1 +x+4 \times (-2)=0$$
$$x+2-8=0$$ or $$x=+6$$ =(iv)
$$(C)$$ $$FeSO_4.(NH_4)_2SO_4.6H_2O$$: $$x+(-2)+2 \times 1 +(-2)+6 \times 0=0$$ =(ii)
$$x-2+2-2+0=0$$ or $$x=+2$$
$$(D)$$ $$Fe_3O_4$$: $$3 \times x +4 \times (-2)=0$$
$$3x-8=0$$ or $$x=+8/3$$ $$=(i)$$
$$(A)-(iii), (B)-(iv), (C)-(ii), (D)- (i )$$
Question 6
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Various oxidation states of few elements are mentioned, Which of the options is not correctly matched?

A
Phosphorus : +3 to +5

B
Nitrogen : +1 to +5

C
Iodine : -1 to +7

D
Chromium : -3 to +6

Solution

Oxidation number of a element of a group is either equal to the group number (for metals) or 18- group number (for non-metals). Also, it is equal to the number of valence electrons
A. Phosphorous is group 15 and period 3.
Possible oxidation states are: +3 to +5
B. Nitrogen is group 15 and period 2.
Possible oxidation states are: +1 to +5
C. Iodine is group 17 and period 4.
Possible oxidation states are: -1 to +7.
Group 17 are electronegative elements and have -1 oxidation state.
D. Chromium is group 6 and period 4.
Possible oxidation states are: +1 to +6. It is a metal and thus cannot show negative oxidation state.
Question 7
1 / -0

Which of the following is a decreasing order of oxidation states of the central atoms?

A
$$PCl_5 , HIO_4, Cl_2{ O }_{ 7 }^{ 2- },Cl_2O$$

B
$$Cr_2{ O }_{ 7 }^{ 2- }, Cl_2O, HIO_4, PCl_5$$

C
$$HIO_4, Cr_2{ O }_{ 7 }^{ 2- }, PCI_5,Cl_2O$$

D
$$Cr_2{ O }_{ 7 }^{ 2- }, HIO_4,Cl_2O,PCl_5$$

Solution

Let the oxidation number of the central atom be $$x$$.
Oxidation state of side atoms be: $$H=+1$$, $$Cl=I=-1$$ and of $$O=-2$$.
$$PCl_5$$: $$x+5 \times (-1)=0$$
$$x=+5$$
$$HIO_4$$: $$1+x+4 \times (-2)=0$$
$$x=+7$$
$$Cl_2O_7^{-2}$$: $$2x+7 \times (-2)=-2$$
$$x=+6$$
$$Cl_2O$$: $$2x+(-2)=0$$
$$x=+1$$
$$Cr_2O_7^{-2}$$: $$2x+7 \times (-2)=-2$$
$$x=+6$$
Thus the decreasing order of the oxidation state of central atom is as:
$$HIO_4>Cr_2O_7^{-2}=Cl_2O_7^{-2}>PCl_5>Cl_2O$$
Question 8
1 / -0

In which of the following compounds oxidation state of chlorine has two different values?

A
$$CaCl_2$$

B
$$NaCl$$

C
$$CaOCl_2$$

D
$$CCl_4$$

Solution

Ionic compounds can exist in equillibrium with the constituting ions as:
A. $$CaCl_2 \rightarrow Ca^{2+}+2Cl^-$$
Oxidation state of $$Cl$$ is -1
B. $$NaCl \rightarrow Na^{+}+Cl^-$$
Oxidation state of $$Cl$$ is -1
C. $$CaOCl_2 \rightarrow Ca^{2+}+OCl^-+Cl^-$$
Oxidation state of $$Cl$$ is -1 in $$Cl^-$$ and in $$OCl^-$$ it is: $$-2+x=-1\ or\ x=+1$$
D. $$CCl_4$$ is covalent and thus oxidation of all $$Cl$$ is same.
Question 9
1 / -0

In which of the following compounds carbon is in highest oxidation state?

A
$$CH_3Cl$$

B
$$CCl_4$$

C
$$CHCl_3$$

D
$$CH_2Cl_2$$

Solution

Let the oxidation number of carbon be $$x$$.
Oxidation number of $$H=+1$$, $$Cl=-1$$
(A) $$CH_3Cl$$: $$x+3 \times 1+(−1)=0$$
$$x+3-1=0$$ or $$x=-2$$
(B) $$CCl_4$$: $$x+4 \times (−1)=0$$
$$x-4=0$$ or $$x=+4$$
(C) $$CHCl_3$$: $$x+1+3 \times (−1)=0$$
$$x+1-3=0$$ or $$x=+2$$
(D) $$CH_2Cl_2$$: $$x+2 \times 1+2 \times (−1)=0$$
$$x+2-2=0$$ or $$x=0$$
the highest oxidation state of carbon is +4 in $$CCl_4$$
Question 10
1 / -0

Which is not true about the oxidation state of the following elements?

A
Sulphur +6 to -2

B
Carbon +4 to -4

C
Chlorine +7 to -1

D
Nitrogen +3 to -1

Solution

Oxidation number of a element of a group is either equal to the group number (for metals) or 18- group number (for non-metals). Also, it is equal to the number of valence electrons
A. Sulphur is group 16 and period 3.
Possible oxidation states are: +6 to -2
B. Carbon is group 14 and period 2.
Possible oxidation states are: +4 to -4
C. Chlorine is group 17 and period 3.
Possible oxidation states are: -1 to +7.
D. Nitrogen is group 15 and period 2.
Possible oxidation states are: +5 to -3.

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Re-Attempt Weekly Quiz Competition

	Column I ( Compound )		Coloumn II (Oxidation state of $$Fe$$)
( A )	$$K_3[(Fe(OH) _6]$$	(i)	+8/3
( B )	$$ K_2[FeO_4]$$	(ii)	+2
( C )	$$FeSO_4, (NH_4)_2SO_46H_2O$$	(iii)	+3
( D)	$$Fe_3O4$$	(iv)	+6

Redox Reactions Test - 42

Result

Redox Reactions Test - 42

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Question 1

$$MnO_2, FeCl_3$$

$$Mn{ O }_{ 4 }^{ - }, CrO_2Cl_2$$

$$MnCl_2 , CrCl_3$$

$$[NiCl_4]^{2-}, [ CoCl_4]^-$$

Solution

Question 2

$$O$$

$$N$$

$$Cl$$

$$F$$

Solution

Question 3

$$CH_4$$

$$CCl_4$$

$$CF_4$$

$$CO_2$$

Solution

Question 4

$$KMnO_4$$

$$K_2MnO_4$$

$$MnO_2$$

$$Mn_2O_3$$

Solution

Question 5

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

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

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

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

Solution

Question 6

Phosphorus : +3 to +5

Nitrogen : +1 to +5

Iodine : -1 to +7

Chromium : -3 to +6

Solution

Question 7

$$PCl_5 , HIO_4, Cl_2{ O }_{ 7 }^{ 2- },Cl_2O$$

$$Cr_2{ O }_{ 7 }^{ 2- }, Cl_2O, HIO_4, PCl_5$$

$$HIO_4, Cr_2{ O }_{ 7 }^{ 2- }, PCI_5,Cl_2O$$

$$Cr_2{ O }_{ 7 }^{ 2- }, HIO_4,Cl_2O,PCl_5$$

Solution

Question 8

$$CaCl_2$$

$$NaCl$$

$$CaOCl_2$$

$$CCl_4$$

Solution

Question 9

$$CH_3Cl$$

$$CCl_4$$

$$CHCl_3$$

$$CH_2Cl_2$$

Solution

Question 10

Sulphur +6 to -2

Carbon +4 to -4

Chlorine +7 to -1

Nitrogen +3 to -1

Solution

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