Explore topic-wise InterviewSolutions in Current Affairs.

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1.

Calculate the oxidation number of the underlined element in the following molecules. Hul(N)O_(3)

Answer»

SOLUTION :`HNO_(3)` Let the OXIDATION number of N in `HNO_(3)` be x. The oxidation number of His +1 while that of each oxygen atom is-2. It is a neutral molecule. THEREFORE, we have
`+1+(x)+(-2)xx3=0`
or x=+5
Hence, the oxidation number of N in `HNO_(3)` is + 5.
Discussion
2.

Calculate the oxidation number of the underlined element in the following molecules. Kul(Mn)O_(4)

Answer»

Solution :`KMnO_(4)` SUPPOSE the oxidation number of Mn in `KMnO_(4)` is x. The oxidatio number of K is +1 because it is an alkali metal and the oxidation number of each O ATOM is -2. Since `KMnO_(4)` is a neutral molecule, we have
`+1+(x)+(-2)xx4=0`
or x=+7
HENCE, the oxidation number of Mn in `KMnO_(4)` is +7.
Discussion
3.

Calculate the oxidation number of the underlined element in the following molecules. Na_(2)ul(S_(2))O_(3)

Answer»

Solution :`Na_(2)S_(2)O_(3)` : Suppose, the oxidation NUMBER of each S atom in `Na_(2)S_(2)O_(3)` is x. Since `Na_(2)S_(2)O_(3)` is a NEUTRAL MOLECULE, we have
`(+1)xx2+(x)xx2+(-2)xx3=0`
or x=+2
HENCE, the oxidation number of each S atom in `Na_(2)S_(2)O_(3)` is +2.
Discussion
4.

Calculate the oxidation number of the underlined element in the following molecules. H_(2)ul(S)O_(4)

Answer»

Solution :`H_(2)SO_(4)` Suppose the oxidation number of S in `H_(2)SO_(4)` is x. The oxidation number of each H atom is +1 while that of each O atom is-2. Since`H_(2)SO_(4)` is a NEUTRAL molecule, the SUM of oxidation NUMBERS of all atoms must be EQUAL to zero.
Therefore, we have `(+1)xx2(x)+(-2)xx4=0`
or `+2+(x)-8=0orx=+8-2=+6`
HENCE the oxidation number of S in `H_(2)SO_(4)` is +6.
Discussion
5.

Calculate the oxidation number of the underlined element in the following ions. ul(S)_(2)O_(3)^(2-)

Answer»

Solution :`S_(2)O_(3)^(2-)` : Let the oxidation number of each S atom in this ion be X. The SUM of oxidation NUMBERS of all the atoms PRESENT must be equal to the CHARGE present on this ion i.e., equal to -2.
Therefore,
`(x)xx2+(-2)xx3=-2`
or 2x=-2+6
or `x=(-2+6)/(2)=+2`
Hence, the oxidation number of S in `S_(2)O_(3)` ion is +2.
Discussion
6.

Calculate the oxidation number of the underlined element in the following ions. ul(N)H_(4)^(+)

Answer»

Solution :`NH_(4)^(+)` : Let the oxidation number of N in `NH_(4)^(+)` bei X. The oxidation number of each hydrogen is +1. SINCE `NH_(4)^(+)` has a charge equal to +1, the sum of oxidation NUMBERS of all atoms in it must be equal to +1. Therefore,
`(x)+(+1)xx4=+1`
or `x=+1-4=-3`
Hence, the oxidation number of N in `NH_(4)^(+)` ion is -3.
Discussion
7.

Calculate the oxidation number of the underlined element in the following ions. ul(P)O_(4)^(3-)

Answer»

<P>

SOLUTION :`PO_(4)^(3-)` : Let the oxidation NUMBER of P in `PO_(4)^(3-)` be X. The oxidation number of each O atom is -2 .Therefore,
`(x)+(-2)xx4=-3`
or x=-3+8=+5
HENCE, the oxidation number of P in `PO_(4)^(3-)` ion is +5.
Discussion
8.

Calculate the oxidation number of the underlined element in the following ions. ul(Mn)O_(4)^(2-)

Answer»

Solution :`MnO_(4)^(2-)` : Let the oxidation NUMBER of Mn in this ioni be x.
THEREFORE,
`(x)+(-2)xx4=-2`
or x=-2+8=+6.
HENCE, the oxidation number of Mn in `MnO_(4)^(2-)` ion is +6.
Discussion
9.

Calculate the oxidation number of the underlined element in the following ions. ul(l)O_(3)^(-)

Answer»

SOLUTION :`IO_(3)^(-)` : LET the oxidation number of I in this ion be x. Therefore,
`(x)+(-2)xx3=-1`
or x=-1+6=+5
HENCE, the oxidation number of I in `IO_(3)^(-)` ion is +5.
Discussion
10.

Calculate the oxidation number of the underlined atoms in the following species. ul(N)H_(2)OH,[ul(Co)(NH_(3))_(5)Cl]Cl_(2),(ul(N_(2))H_(5))_(2)SO_(4),ul(Mg)_(3)N_(2)

Answer»


Solution :Following calculations are CARRIED out in SHORT. x represents the O.N. of the underlined atom.
`NH_(2)OH:x+(+)xx2+(-2)+(+1)=0`
`:.x=-1`
`[Co(NH_(3))_(5)Cl]Cl_(2):x+(0)xx5+(-1)+(-1)xx2=0`
`:.x=+3`
`(N_(2)H_(5))_(2)SO_(4):{(x)xx2+(+1)xx5}xx2+(-2)=0`
`x=-2`
`Mg_(3)N_(2):3(x)+(-3)xx2=0`
`:.x=+2`
Discussion
11.

Calculate the oxidation number of the underlined element in the following ions. ul(Cl)O^(-)

Answer»

SOLUTION :`ClO^(-)` : Let the oxidation number of CL in this ION be x.
Therefore,
`(x)+(-2)=-1`
or x=-1+2=+1
HENCE, the oxidation number of cl in `ClO^(-)` ion is +1
Discussion
12.

Calculate the oxidation number of the underlined element in the following ions. ul(Cr_(2))O_(7)^(2-)

Answer»

Solution :`Cr_(2)O_(7)^(2-)` : Suppose, the oxidation number of CR in this ion is x. The sum of oxidation NUMBERS of all ATOMS present in this ion must be equal to -2 i.e., the charge present on the ion.
Therefore,
`(x)xx2+(-2)xx7=-2`
or 2x=-2+14=+12
x=+6
Hence, the oxidation number of Cr in `Cr_(2)O_(7)^(2-)` is +6.
Discussion
13.

Calculate the oxidation number of the underlined atom in the following species. ul(Fe)(H_(2)O)_(6)Cl_(3)

Answer»

Solution :`FE(H_(2)O)_(6)Cl_(3)` : Let the oxidation NUMBER of Fe in this molecule be x. The sum of oxidation numbers of atoms present in `H_(2)O` is ZERO because it is a NEUTRAL molecule. The oxidation number of each Cl atom is -1. Therefore,
`(x)+(0)xx6+(-1)xx3=0`
or x=+3
Hence, the oxidation number of Fe in the given molecule is +3.
Discussion
14.

Calculate the oxidation number of the underlined atoms in the following compounds and ions : underlineCH_4,Sb_2O_5,underlineC_6H_12O_6

Answer»

SOLUTION :`CH_4,x+4=0,x=-4`
`Sb_2O_5,2x-10=0,2x=10,x=+5`
`C_6H_12O_6,6x+12-12=0,x=0`
Discussion
15.

Calculate the oxidation number of the underlined atom in the following species. [ul(Co)(NH_(3))_(6)]^(3+)

Answer»

Solution :`[Co(NH_(3))_(6)]^(3+)` : Let the oxidation number of Co in the GIVEN ion be x. `NH_(3)` is a neutral molecule. Therefore the SUM of oxidation numbers of atoms PRESENT in it is zero. The sum of oxidation numbers of all atoms present in the given ion must be equal to the CHARGE present on it. Therefore,
`(x)+(0)xx6=+3`
or x=+3
Hence, the oxidation number of Co in `[Co(NH_(3))_(6)]^(3+)` ionis +3.
Discussion
16.

Calculate the oxidation number of the underlined atom in the following species. K_(4)ul(Fe)(CN)_(6)

Answer»

Solution :`K_(4)Fe(CN)_(6)` : Suppose the oxidation number of Fe in this molecule is x. The oxidation number of each K (alkali metal) atom is +1. `CN^(-)` is an ION and bears a CHARGE equal to -1. Therefore, the sum of oxidation numbers of all atoms present in `CN^(-)` ion is equal to -1. Since `K_(4)Fe(CN)_(6)` is a neutral molecule, the sum of oxidation numbers of all atoms present in it will be equal to zero.
Therefore,
`(+1)xx4+(x)+(-1)xx6=0`
or `x=+6-4=+2`
Hence, the oxidation number of Fe in `K_(4)Fe(CN)_(6)` is +2.
Discussion
17.

Calculate the oxidation number of the underlined atom in the following species. Kul(Ag)(CN)_(2)

Answer»

SOLUTION :`KAG(CN)_(2)` : SIMILARLY, if X is the oxidation number of Ag, we have
`(+1)+(x)+(-1)xx2=-0`
x=+1
Hence, the oxidation number of Ag in `KAg(CN)_(2)` is +1.
Discussion
18.

Calculate the oxidation number of the underlined atom in the following ions. ul(S)O_(4)^(2-),[ul(Cr)(H_(2)O)_(6)]^(3+),[ul(Fe)(CN)_(6)]^(3-),CrO_(4)^(2-),BrO_(3)^(-)

Answer»

SOLUTION :`+6,+3,+3,+6,+5`
Discussion
19.

Calculate the oxidation number of the underlined atom in the following molecules. H_(2)ul(C)_(2)O_(4),ul(C)_(6)H_(12)O_(6),ul(Pb)_(3)O_(4),ul(lF)_(7),Hul(Cl)O,ul(O)F_(2),ul(Ni)(CO)_(4),Hul(Au)Cl_(4),BaO_(2),Mg_(3)ul(N_(2)),ul(O_(3)),

Answer»

Solution :`+3,0,+(8)/(3),+7,+1,+2,+0,+3,+2,+3,0`
Discussion
20.

Calculate the oxidation number of the element underline in each of the following cases. A. Al in Al_(2)O_(3) b. P in P_(2)O_(5)

Answer»

Solution :The rule used here is that the algebraic sum of the oxidation numbers of all the atoms a molecule is zero.
a. `Al_(2)O_(3)2xx("oxidation number of "AL)+3xx("Oxidation number of O")=0`
`2xx("Oxidation number of Al")+3(-2)=0`
`2xx("oxidation number of Al")+6`
`:.` oxidation number of Al `=+3`
b. `P_(2)O_(5):2xx`(oxidation no. P)`+5(-2)=0`
`:.` oxidationn of `P+(+10)/2=+5`
Discussion
21.

Calculate the oxidation number of the underlined atom : (i) K ul(Mn)O_4 (ii) ulP_2O_5 (iii) ul(Fe_2)O_3 (iv) ul(Xe)OF_4 (v) ulS_2O_3^(2-)(vi) ul(Cr)_2O_7^(2-) .

Answer»


ANSWER :(i) +7 (ii) +5 (III) +3 (IV) +6 (v) +2 (VI) +6
Discussion
22.

Calculate the oxidation number of sulphur in the following molecules ions . (a) H_2S (b) H_2SO_3 (c) SO_4^(2-) (d) Na_2S_2O_3 (e) S_2O_7^(2) (f) H_2SO_4 (g) S_2O_4^(2-).

Answer»

Solution :(a) `H_2S`. The oxidation number of hydrogen is +1. Let the oxidation number of sulphur be x.
`{:(+1,x,),(H_2,S,),((1xx2),"(x)",),(1xx2+x=0,:.,x=2):}`
Oxidation number of S in `H_2S `is -2.
(B) `H_2SO_3`. Oxidation number of hydrogen is +1 and that of OXYGEN is -2. Let the oxidation number of S be x.
`{:(+1,x,-2),(H_2,S,O_3),((+1xx2),(x),(-2xx3)""):}`
`(+1xx2)+x+(-2xx3)=0`
or `+2+x-6=0`
or `x=+4`
Oxidation number of S in `H_2SO_3` is +4.
(c) `SO_4^(2-)` . The oxidation number of oxygen is -2 and let oxidation number of S be x.
`[{:(x,-2),(S,O_4):}]^(2-)`
`{:(""x,,(-2xx4)),(x+(-2xx4)=-2,"or",x=-2+8),(,"or",x=+6):}`
Oxidation number of S in `SO_4^(2-)` is +6.
(d) `Na_2S_2O_3` . The oxidation number of NA is +1 and that of oxygen is -2. Let the oxidation number of S of x.
`{:(+1,x,-2),(Na_2,S_2,O_3),((1+xx2),("x"xx2),(-2xx3)""):}`
`{:((+1xx2)+("x"xx2)+(-2xx3)=0,,),(""+2+2x-6=0,"or",2x=4),(,,x=+2):}`
Oxidation number of S in `Na_2S_2O_3` is +2.
(E) `S_2O_7^(2-)` . The oxidation number of oxygen is -2and let the oxidation number of sulphur be x.
`[{:(x,-2),(S_2,O_7)]^(2-)`
`("x"xx2)(-2xx7)`
`"x " xx 2 +(-2xx7) =-2`
`2x = -2 +14`
`:.""x = 6`
Oxidation number of S in `S_2O_7^(2-)` is +6.
(f) `H_2SO_4`. The oxidation number of hydrogen is +1 and that of oxygen is -2. Let the oxidation number of S be x.
`{:(+1,x,-2),(H_2,S,O_4),((+1xx2),x,(-2xx4)):}`
`(+1 xx2)+x + (-2xx4_=0`
or `+2+x-8=0`
or `x= +6`
Oxidation number of S in `H_2SO_4` is +6.
(g) `S_2O_4^(2-)` . The oxidation number of oxygen is -2 and let the oxidation number of sulphur be x.
`[{:(x,-2),(S_2,O_4):}]^(2-)`
` ("x" xx2) (-2xx4)`
`"x" xx 2 + (-2xx4) =-2`
`2 x - 8=-2 :. x = +3`
`:.` Oxidation number of S in `S_2O_4^(2-)` is +3.
Discussion
23.

Calculate the oridation numbers of sulphur in H_(2)SO_(5) and in H_(2)S_(2)O_(8)

Answer»

SOLUTION :`H_(2)SO_(5)` has one peroxy BOND. Two oxygen atoms have -1 oxidation state.
Oxidation number of .S. in `H_(2)SO_(5)` is +6.
`H_(2)S_(2)O_(8)` has one peroxy bond. Two oxygen atoms have -1 oxide state.
Oxidation number of each .S. in `H_(2)S_(2)O_(8)` +6.
Discussion
24.

Calculate the oxidation number of nitrogen in NO_(3)^(-). Suggest structure of this compound. Count for the fallacy.

Answer»

Solution :`(x)+[(-2)xx3]=-1` (as `NO_(3)^(-)` bears a charge of -1).
or x=+5
The structure of `NO_(3)` is as follows.

On the BASIS of above structure, we have
`[(-2)xx2]+(x)+(-1)=0`
or x=+5
THUS, this structure GIVES the same O.N. for N in `NO_(3)^(-)`. HENCE, there is no fallacy.
Discussion
25.

Calculate the oxidation numberof sulphur chromium and nitrogen in H_(2)SO_(5),CrO_(5) and suggest structure ofthese three compunds count for the fallacy

Answer»

SOLUTION :(i) O.N of S is `H_(2)SO_(5)`
By conventinal methd the O.n of S in `G_(2)SO_(5) Cr O_(5)` is : 2(+1)+x+5(-2)=0or x =+8
this is impossible because the maximum O.N of S cannot be more than since is has only six eletron in the VELENCE sheel this fallacy s oversome if we calculate the O.N of S by chemical bonding mehtod
The structure of `H_(2)SO_(5)H-underset(O)underset(||)overset(O)overset(||)S-O-O-H`
(ii)O.B cr in `CO_(5)` ACCORDINGTO conventional maximum
this fallacy is REMOVED if we half O.N of Cr by chemical bonding method the structure of ,`CO_(5)` is (ii) O.N `Cr Cro_(5)`
according to conventional maximum O.N of Cr cannot more than six since it chemical bonding configuration this is removed if w calculate O.N Cr by chemical bondng method
Discussion
26.

Calculate the oxidation number of sulphur, chromium and nitrogen in H_(2)SO_(5),Cr_(2)O_(7)^(2-)andNO_(3)^(-). Suggest structure of these compounds. Count for the fallacy.

Answer»

SOLUTION :(i) `H_(2)underlineSO_(5)` :
As per rules : `H_(2)SO_(5)to2(+1)+S+5(-2)=0`
`thereforeS=+8`
Which is impossible as suphur.s oxidation number should not greater than +6.
So, structure should
is as following : `H-O-underset(O)underset(||)overset(O)overset(||)(S)-overset(-1)(O)-overset(-1)(O)-H`
`therefore2(+1)+S+2("Peroxide O")+3(O)=0`
`therefore2(+1)+S+2(-1)+3(-2)=0`
`therefore+2+S-2-6=0`
`thereforeS=+6`
(ii) `Cunderliner_(2)underlineO_(7)^(-2)` :
As per rules : `Cr_(2)O_(7)^(-2)=2Cr+7(-2)=-2`
= `2Cr-14=-2`
Cr = +6
Chemical structure : `overset(-2)(O)=underset(O^(-1))underset(|)overset(O^(-2))overset(||)(Cr)-overset(-2)(O)-underset(O^(-1))underset(|)overset(O^(-2))overset(||)(Cr)=overset(-2)(O)`
(iii) `NunderlineO_(3)^(-)` :
As per rules : `NunderlineO_(3)^(-)=N+3(-2)=-1`
`thereforeN=+5`
Chemical structure :
Discussion
27.

Calculate the oxidation number of Sulphur in H_(2)SO_(5). Suggest structure of this compounds. Count for the fallacy.

Answer»

SOLUTION :`+(1)xx2+(x)+[(-2)xx5]=0`
or `x=10-2=8`.
The O.N. of +8 for S is not POSSIBLE because it has only 6 electrons in its valence shell and can EXHIBIT a maximum O.S. = + 6. Hence, in `H_(2)SO_(5)`, two oxygen atoms must be linked TOGETHER. This consideration can remove the fallacy.
Thus, the structure of `H_(2)SO_(5)` should be as given below.
`overset(+1)H-overset(-1)O-underset(-2)underset(O)underset(||)overset(-2)overset(O)overset(||)S-overset(-1)O-overset(-1)O-overset(+1)H`
On the basis of the above structure, we have
`(+1)+(-2) + XX + [(-2) xx 2] + [(-1) 2] +(+1)=0`
or -1 + X-4-2 + 1 = 0
or x=+6.
Discussion
28.

Calculate the oxidation number of Chromium in CrO_(5) . Suggest structure of this compound. Count for the fallacy.

Answer»

Solution :`(x)+[(-2)xx5]=0`
or x=+10
The O.N. of +10 for Cr is not POSSIBLE because the maximum O.S. of Cr is +6 (outer electronic config: `3d^(5)4s^(1)` ). The fallacy is removed if we consider the following structure for `CrO_(5)`.

On the BASIS of above structure, we have
`[(-1)xx4]+(x)+(-2)=0`
or `-4+x-2=0`
or x=+6
Discussion
29.

Calculate the oxidation number of phosphorus in the following species (a) HPO_(3)^(2-) and (b) PO_(4)^(3-)

Answer»

SOLUTION :(a) Let the O.N of P in `HPO_(3)`be x`THEREFORE` 1+x+3(-2)=-2 or x=+3
(b) Let the O.N of P in `PO_(4)^(3-)` be x `therefore`x+4(-2)=-3 or x=+5
Discussion
30.

Calculate the oxidation number of phosphorus in the following species. (a) HPO_(3)^(-2) (b) PO_(4)^(-3)

Answer»

SOLUTION :
Discussion
31.

Calculate the oxidation number of oxygen in the following : OF_2, O_2, Na_2O_2 and CH_3COOH

Answer»


ANSWER :`+2,0,-1,-1`
Discussion
32.

Calculate the oxidation number of nitrogen in the following oxides. N_(2)O,NO,N_(2)O_(3),NO_(2),N_(2)O_(5)

Answer»

SOLUTION :`+1,+2,+3,+4,+5`
Discussion
33.

Calculate the oxidation number of N in NO_3^(-) .

Answer»


ANSWER :`+5`
Discussion
34.

Calculate the oxidation number of lead in Pb_3O_4 ?

Answer»


ANSWER :`+8//3`
Discussion
35.

Calculate the oxidation number of (i) S in H_(2)S, (ii) C in CO_(2) (iii) C in CO_(2) (iii)C in CH_(2)CI_(2), (iv) N in (NH_(4))SO_(4) and (v)P I Na_(3)PO_(4)

Answer»

Solution :(i) `S in H_(2)S` Let the oxidation number of S in `H_(2)S` be x writing the oxidation number of each atom above its symbol
`overset(+1)H_(2)overset(x)S`
sum of oxidation number of various ATOMS in `H_(2)S=2(+1)+x=2+x`
But the sum fo the oxidation numbers of various atoms in `H_(2)S` (neutral) is zero (Rule 7)
`therefore` 2+x=0 or x=-2
Thus the oxidation numbr of S in `H_(2)S is -2`
(II) `C in CO_(2)` Let the oxidaion number of C in `CO_(2)` be x writing the oxidaion number o feach atom above its symbol
`therefore`x-4=0or x=+4
Thus the oxidation number of C in `CO_(2)` is +4
(iii)`C in CH_(2)CI_(2)` Let the oxidaition number of C in `CH_(2)CI_(2)` be x writing the oxidatin number of each atom above its sysmbol `overset(x)C overset(+1)H_(2) overset(-1)CI_(2)`
`therefore` sum of hte oxidation numbers of various atoms in `CH_(2)CI_(2)` (neutral ) is zero (Rule 7)
`therefore`x=0
Thusthe oxidation number of C in `CH_(2)CI_(2)` is zero
(iv) N in `(NH_(4))SO_(4)` Let the oxdation number of nitrogen in `(NH_(4))_(2)` be x writing the oxdation number o fhydrogen above its sysmbol and that of `SO_(4)^(2-)` ion above its FROMULA
`therefore` Sum of oxidation numbers of all the atoms in `(NH_(4))SO_(4)=2x+2(+1xx4)+(-2)=2x+6`
But the sum of oxidation numbers of all the atoms in `(NH_(4))_(2)SO_(4)` (neutral ) is zero (Rule 7)
`therefore 2x+6=0 or x =-3`
Thus the oxdation number of nitrogen in `(NH_(4)^(2))` is -3
(v)`P in Na_(3)PO_(4)` Let the oxidation number of P in `Na_(3)PO_(4)` be x writing hte oxidation number of each atom above its symbol `overset(+1)Na_(3) overset(x)P overset(-2)O_(4)`
Sum of the oxidation numbers of various of various atoms in `Na_(3)PO_(4)=3(+1)+x4(-2)=x-5`
Discussion
36.

Calculate the oxidation number of (i) Nin NO_(3)^(-), (ii) P in H_(3)P_(2)O_(7)^(-), (iii) C in CO_(3)^(2-), (iv)CI in CIO_(4)^(-) (v)Cr in Cr_(2)^(2-) (vi) Mn in MnO_(4)^(-) and (vii) Fe in [fe(CN)_(6)]^(4)

Answer»

Solution :`N in NO_(3)^(-3)` Let the oxidaton number of N in `NO_(3)^(-)` be x WRITING the oxdation number of each atom above its symbol`overset(x)N O_(3)^(-3)`
`therefore` sum of the oxdatin number of all th eaoms in `NO_(3)^(-)=x+3(-2)=x-6`
But the sum of xodatin numbers ofall the atoms in `NO_(3)^(-)` ion is equal to the cahrge present on it i.e (Rule 8) `therefore x-6 =-1 or x=+5`
Thus the oxdation number of N in `NO_(3)^(-) is +5`
(ii) `P in H_(3)P+_(2)O_(7)^(-)` Let the oxdiationn number of P in `H_(3)P_(2)O_(7)^(-)` be x writing the oxidation number of each atom above its symbol
`overset(+1)H_(3)overset(x)P_(2)overset(-2)O_(7)`
Sum of hteoxdaion numbers of all the atoms in `H_(3)P_(2)O_(7)^(-)=3(+1)+2(x)+7(-2)` or 2x-11
But the sum of oxdiatin numbers of all the atoms in `H_(3)P_(2)O_(7)^(-)` is equal to the charge present on it i.e (Rule 8)`therefore` 2x-11=-1 or x=+5
Thus the oxdation number of P in `H_(3)P_(2)O_(7)^(-)` is +5
(iii) C in `CO_(3)^(2-)` Let the oxidatin number of C in `CO_(3)^(2-)` be x writing the oxidation number of each atom above its symbol `overset(x)C overset(-2)O_(3) therefore`sum of hte oxidation number ofll the aotms in `CO_(3)^(2-)` ion = x+3(-2)=x-6But the sum of oxidation numbers of all the atoms in `CO_(3)^(2-)` ions is -2 (Rule 8)`therefore` x-6 =-2 or x = +4
thus the oxidaon state of cin `CO_(3)^(2-)` is +4
(iv) `CI` in `CIO_(4)^(-)` let the oxidation numbner of CI in `CI_(4)^(-)` ions is -2 (Rule 8) `therefore` x-6 =-2 or x=+4 above its symbol `overset(x)CI O_(4)^(-2) therefore` sum of oxidation numbers of al the atoms in `CIO_(4)^(-)` ion = x+4(-2)=x-87
But the sum of oxidaiotin numbers of all the atoms in `CIO_(4)^(-)` ion is equal to the charge present on it i.e -1 (Rule 8)`therefore x-8 =-1 or x -+7`
Thus the oxidation number of CI in `CIO_(4)^(-)` is +7
(V) Cr in `Cr_(2)O_(7)^(2-)` Let the oxidation number of Cr in `Cr_(2)O_(7)^(2-)` be x writing the oxidaitn number of each atom above its symbol`overset(x)Cr_(2) overset(-2)O_(7)`
`therefore` Sum of he oxidation numbers of all the atoms in `Cr_(2)O_(7)^(2-)` ions =2(x)+7 (-2)=2x-14
But the sum of oxidation numbers of all the atoms in `Cr_(21)O_(7)^(2-)` is equal to the charge on it i.e -2 (Rule 8)
`therefore` 2x-14=2 or x=6
Thus the oxidation number of Cr in `Cr_(2)O_(7)^(2-)` ion is +6
(vi) Mn in `MnO_(4)^(-)` Let the oxidation number of Mn in `MnO_(4)^(-)` be x writing oxidation number of each atom above its sysmbol we get
`therefore` sum of he oxidation numbers of all the aotms in `MnO_(4)^(-) =x(-2)=x-8`
But the sum of oxidation numbers ofall the atoms in `MnO_(4)^(-)` is -1 (Rule 8) therefore x-8=-1 or x=+7Thus the oxidation number of Mn in `MnO_(4)^(-)` is +7
(vii) Fe in `[Fe(CN)_(6)]^(4-)` Let the oxidation number of Fe in `[Fe(CN)_(6)]^(4-)` be x writing the oxidation number of each atom above its symbol and that of cyanide ion above its formula we get
`overset(x)Fe overset(-1)(CN)_(6)`
`therefore` Sum of oxidation numbers of all the atoms in `[Fe(CN)_(6)]^(4-)`=x+6(-1)=x-6
But the sum of oxidation numbers of all the atoms in `Fe(CN)_(6)]^(4-)` is equal to -4 (Rule 8)
`therefore` x-6 =-4 or x=+2
Thus the oxidation number of Fe in `[Fe(CN)_(6)]^(4-)` is +2
Discussion
37.

Calculate the oxidation number of (i) Fe in Fe_3O_4 (ii) S in Na_2S_4O_6 (iii) Pb in Pb_3O_4 (iv) Nin (NH_4)_2SO_4.

Answer»

Solution :(i) FE in `Fe_3O_4`. LET the oxidation number of Fe be X.
`{:(x,-2,3x-8=0),(Fe_3,O_4,3x=8),(("x"xx3),(-2xx4),x=8//3=2.67):}`
(ii) S in `Na_2S_4O_6` . Let the oxidation number of S be x.
`{:(+1,x,-2,2+4x-12=0),(Na_2,S_4,O_6,""4x=10),(("x" xx3),("x" xx4),(-2xx6),""x=10/4=2.5):}`
(iii) Pb in `Pb_3O_4` . Let the oxidation no. of Pb be x.
`{:(x,-2,,3x-8=0),(Pb_3,O_4,,3x=8),(("x" xx3),(-2xx4),"or",x=8/3=2.67):}`
(iv)N in `(NH_4)_2SO_4` . Let the oxidation no. of N be x.
`(overset(x)Noverset(+1)H_4) SO_4^(2-)""2x+2(+1xx4)+(-2)=0`
`:. ""2x+6=0`
`""2x=-6" or"x=-6/2=-3`
Discussion
38.

Calculate the oxidation number of (i) Cr in CrO_(4)^(2-) (ii) C in HCO_(3)^(-) (iii) CI in CIO_(4)^(-) (iv)Cr in Cr_(2)O_(7)^(2-)

Answer»


Answer :`(i)+6 (II)+5(III)+7(IV)+6`
Discussion
39.

calculate the oxidation number of each suphur atom in the following compounds

Answer»

`Na_(2)S_(2)O_(3)`
`Na_(2)S_(4)O_(6)`
`Na_(2)SO_(3)`
`Na_(2)SO_(4)`

Solution :(a) -2 an d+6 r
(b) +5 ,0,0 +5
(c ) LET the O.N of S in `Na_(2)SO_(3)` be X `THEREFORE`2(+1) +x+3 (-2)=0 or x=+4
(d) Let the O.N of S in `Na_(2)SO_(4)` be x `therefore` 2(+1)+x+4(-2)=0 or x=+6
Discussion
40.

Calculate the oxidation number of each sulphur atom in the following compounds. (a) Na_(2)S_(2)O_(3) (b) Na_(2)S_(4)O_(6) ( c) Na_(2)SO_(3) (d) Na_(2)SO_(4)

Answer»

Solution :(a) `Na_(2)S_(2)O_(3):" "Na^(+)O^(-)-UNDERSET(O)underset(||)overset(S)overset(uarr)(S)-O^(-)Na^(+)`
Coordinate covalent bond is present between TWO S atoms so oxidation number of S which accept `e^(-)` is -2 and for another we have to calculate `2(+1)+3(-2)+x+1(-2)=0`
`thereforex=+6`
(b) `Na_(2)S_(4)O_(6):" "Na^(+)O^(-)-underset(O)underset(||)overset(O)overset(||)(S)-overset(0)(S)-overset(0)(S)-underset(O)underset(||)overset(O)overset(||)(S)-O^(-)-Na^(+)`
In this molecules, oxidation number of two S atoms, which is INDICATED in STRUCTURE is zero. So oxidation number of another two S is,
`2(+1)+6(-2)+2x+2(0)=0`
`therefore2-12+2x=0`
`thereforex=+5`
( c) `Na_(2)SO_(3)` :
`2(+1)+x+3(-2)=0`
`thereforex=+4`
(d) `Na_(2)SO_(4)` :
`2(+1)+x+4(-2)=0`
`therefore2+x-8=0`
`thereforex=+6`
Discussion
41.

Calculate the oxidation number of carbon in the following compounds. CH_(4),CH_(3)Cl,CH_(2)Cl_(2),CHCl_(3),C Cl_(4)

Answer»

SOLUTION :`+4,-2,0,+2,+4`
Discussion
42.

Calculate the oxidation number of C in the following : C_2H_6, C_4H_(10), CO,CO_2 and HCO_3^(-)

Answer»


ANSWER :`-3,-2.5,+2 , +4 and +4`
Discussion
43.

Calculate the oxidation number ofC in CH_2 Cl_2

Answer»

Solution :C in `CH_2 Cl_2`, Let the oxidation number of C in `CH_2 Cl_2` be X. Writing the oxidation number of each ATOM above its symbol.
`overset (x) (C)overset(+1)(H)_2overset(-1)(CL)_2`(`because` number of H is +1 and that of Cl is -1)
`x+2(+1) +2 (-1)=0`. So x=0.
Discussion
44.

Calculate the oxidation number of all the atoms in the following well known oxidants. KMnO_4,K_2Cr_2O_7, KClO_4

Answer»


ANSWER :`K(+1), MN(+7), O(-2),K(+1) , Cr (+6) , O (-2) ,K+(1)`
Discussion
45.

Calculate the oxidation number of all the atoms in the following species. Sb_(2)O_(5)

Answer»

SOLUTION :`SB:+5,O:-2`
Discussion
46.

Calculate the oxidation number of all the atoms in the following species. (NH_(4))_(2)SO_(4)

Answer»

Solution :`N:-3,H:+1,S:+6,O:-2`
Discussion
47.

Calculate the oxidation number of all the atoms in the following species. C_(12)H_(22)O_(11)

Answer»

SOLUTION :`C:O,H:+1,O:-2`
Discussion
48.

Calculate the oxidation number of all the atoms in the following species. BrF_(3)

Answer»

SOLUTION :`BR:+3,F:-1`
Discussion
49.

Calculate the oxidation number of all the atoms in the following compounds and ions : CO_2,SiO_2,PbSO_4,ClO_4^(-)

Answer»

SOLUTION :`CO_2` : OXIDATION NUMBER of each oxygen atom = -2
Oxidation number of carbon (say x)
`{:(x,-2),(C,O_2):}`
`x+(-2xx2) =0 " or " x = +4`
`:.` Oxidation number of C = +4 , O = -2
`SiO_2` : Oxidation number of each oxygen atom = -2
Oxidation number of silicon (say x)
`{:(x,-2),(Si,O_2):}`
`x +(-2xx2) " or "x =+4`
`PbSO_4` : Oxidation number of each oxygen
=-2
Oxidation number of lead (Pb) = +2
Oxidation number of SULPHUR (say x)
`{:(+2,x,-2),(Pb,S,O_4):}`
Oxidation number of Pb = +2, S = +6 , O = - 2
`ClO_4^(-)` : Oxidation number of chlorine (say x)
`[{:(x,-2),(CL,O_4):}]`
`x+(-2xx4)=-1 " or "x=+7`
`:.` Oxidation number of O = 2 , Cl = +7
Discussion
50.

Calculate the [OH^(-)] of a solution whose pH is 9.62.

Answer»

Solution :`pH+pOH=14, pOH=14-pH=14.9.62=4.38`
`pOH=4.38=-log_(10)[OH^(-)],4.38=-log_(10)[OH^(-)]`
`log_(10)[OH^(-)]=-4.38, [OH^(-)]=` antiolog `(-4.38)=` antilog `(5-4.38-5)`
`=` antilog `(0.62-5)=`antilog `(0.62)xx10^(-5)=4.269xx10^(-5)"mol"dm^(-3)`
Discussion