Question 1
Find the oxidation numbers
(i) Boron in $NaBH_4$
(ii) Chromium in $Cr_2O_7^{2-}$
(iii)Maganese in $K_2MnO_4$
(iv)Phosphorous in $H_4P_2O_7$
(v) Carbon in $CH_3COOH$
(vi) Carbon in $CH_3CH_2OH$
(vii) Iron in $Fe_3O_4$
(viii) Nitrogen in $HNO_3$ Solution
(i) Let x be the oxidation number of B
Apply the rules for finding oxidation numbers
$+1 + x + 4 =0$
$x=-5
(ii) Let x be the oxidation number of Cr
Apply the rules for finding oxidation numbers
$2x -7 \times 2=-2$
$x=6$
(iii) Let x be the oxidation number of Mn
Apply the rules for finding oxidation numbers
$2 +x -4 \times 2=0$
$x=6$
(iv) Let x be the oxidation number of P
Apply the rules for finding oxidation numbers
$4 +2x -7 \times 2=0$
$x=5$
(v) Let x be the oxidation number of C
Apply the rules for finding oxidation numbers(Conventional method)
$4 +2x -2 \times 2=0$
$x=0$
But According to the structure
Now we assign electron pair shared between atoms of different electronegativity to more electronegative atom and distribute the electron pair shared between atoms of same element equally. Now count
the number of electrons possessed by each atom. Find out the difference in number of electrons possessed by neutral atom and that possessed by atom in the compound. This difference is the oxidation number
For Carbon-2, it is attached to three hydrogen atom, since Carbon is more electronegative, electron are assigned to Carbon atom, Now C-C will share equal number of electrons, So total electron on C-2 will be
7. Now Neutral atom has 4 electrons,So Oxidation Number is -3
For Carbon-1, it is attached to 2 oxygen atom, since oxygen is more electronegative, electron are assigned to Oxygen atom, Now C-C will share equal number of electrons, So total electron on C-1 will be
1. Now Neutral atom has 4 electrons,So Oxidation Number is +3
(vi)
Let x be the oxidation number of C
Apply the rules for finding oxidation numbers(Conventional method)
$6 +2x -1 \times 2=0$
$x=-2$
But According to the structure
Now we assign electron pair shared between atoms of different electronegativity to more electronegative atom and distribute the electron pair shared between atoms of same element equally. Now count
the number of electrons possessed by each atom. Find out the difference in number of electrons possessed by neutral atom and that possessed by atom in the compound. This difference is the oxidation number
For Carbon-2, it is attached to three hydrogen atom, since Carbon is more electronegative, electron are assigned to Carbon atom, Now C-C will share equal number of electrons, So total electron on C-2 will be
7. Now Neutral atom has 4 electrons,So Oxidation Number is -3
For Carbon-1, it is attached to 1 oxygen atom and 2 hydogen atom, since oxygen is more electronegative, electron are assigned to Oxygen atom, Now C-C will share equal number of electrons, So total electron on C-1 will be
1. Now Neutral atom has 4 electrons,So Oxidation Number is -1
(vii)
Let x be the oxidation number of Fe
Apply the rules for finding oxidation numbers(Conventional method)
$3x-8=0$
x=8/3
In fact, $Fe_{3}O_{4}$ exists as a mixture of FeO and $Fe_{2}O_{3}$
So Fe has oxidation number of +2 and +3.
(viii)Let x be the oxidation number of N
Apply the rules for finding oxidation numbers
$1 +x -3 \times 2=0$
$x=5$
Question 2
Calculate the oxidation number of each sulphur atom in the following compounds:
(a) $Na_2S_2O_3$
(b) $Na_2S_4O_6$
(c) $Na_2SO_3$
(d) $Na_2SO_4$ Solution
(i) Let x be the average oxidation number of S
$2 + 2x -6=0$
$x=2$
if you look at the structure, the oxidation number of terminal sulphur atom is 0 and other S atom is +4, So average comes to +2
(ii) Let x be the average oxidation number of S
$2 + 4x -12=0$
$x=2.5$
if you look at the structure of $S_4O_6^{2–}$ ion
To find out oxidation state of each atom we distribute electrons of electron pair shared between two sulphur atoms equally (i.e. one electron is assigned to each sulphur atom). Both electrons of electron pair shared between
sulphur and oxygen atom are assigned to oxygen as oxygen is more electronegative. Thus we find that each of the central sulphur atoms obtains six electrons. This number is same as that in the outer shell of neutral sulphur atom. Hence, oxidation state of each central sulphur atom is zero. Each of the sulphur atoms attached to oxygen atoms finally obtains only one electron as its share. This number is less by five electrons in comparison to the neutral sulphur atom. So, outer sulphur atoms are in +5 oxidation state. Therefore average oxidation state of sulphur atoms is $\frac { 5 + 0+ 0+5}{4}=2.5$
(iii)
Let x be the average oxidation number of S
$2 + x -6=0$
$x=4$
(iv)
Let x be the average oxidation number of S
$2 + x -8=0$
$x=6$
Question 3
Balance the following equations in acidic Medium by ion electron methods
(a) $MnO_4^{-} (aq) + C_2H_2O_4(aq) -> Mn^{2+} (aq) + CO_2(g) + H_2O(l)$
(b) $MnO_4^{-} (aq) + C_2O_4^{2-} + H^{+}-> Mn^{2+} (aq) + CO_2(g) + H_2O(l)$
Solution
(a)$2MnO_4^{-} (aq) + 5C_2H_2O_4(aq) + 6H^{+} -> 2Mn^{2+} (aq) + 10CO_2(g) + 8H_2O(l)$
(b) $2MnO_4^{-} (aq) + 5C_2O_4^{2-} + 16H^{+}-> 2Mn^{2+} (aq) + 10CO_2(g) + 8H_2O(l)$
Question 4
Balance the following equations in Basic Medium
(a) $MnO_4^{-} (aq) + I^{-}(aq) -> MnO_2 (s) + I_2(s)$
(b) $MnO_4^{-} (aq) + Br^{-}(aq) ->MnO_2 (s) + BrO_3^- (aq)$
Question 5
Balance the following equations by the oxidation number method.
(i) $Fe^{2+} + H^{+} + Cr_2O_7^{2-}-> Cr^{3+} + Fe^{3+} + H_2O$
(ii) $I_2 + NO_3^{-} -> NO_2 + IO_3^-$ Solution
(a)$6Fe^{2+} + 14H^{+} + Cr_2O_7^{2-}-> 6Cr^{3+} + 2Fe^{3+} + 7sH_2O$
Question 6
Identify the redox reactions out of the following reactions and identify the oxidising and reducing agents in them.
(i) $3HCl(aq) + HNO_3 (aq) -> Cl)2(g) + NOCl (g) + 2H_2O(l)$
(ii) $HgCl_2 (aq) + 2KI (aq) -> HgI_2(s) + 2KCl (aq)$
(iii) $Fe_2O_3(s) + 3CO (g) -> 2Fe (s) + 3CO_2(g)$
(iv) $PCl_3(l) + 3H_2O (l) -> 3HCl (aq) + H_3PO_3(aq)$
(v) $4NH_3+ 3O_2 (g) -> 2N_2(g) + 6H_2O (g)$
(vi) $NaCl + KNO_3 ->NaNO_3 + KCl$
(vii) $Zn + 2AgCN -> 2Ag + Zn(CN)_2$
(viii)$Mg(OH)_2 + 2NH_4Cl -> MgCl_2 + 2NH_4OH$ Also Read
