Redox Reactions - Class 11 Medical Chemistry - Extra Questions

 The order of increasing oxidation number of chlorine for $$Cl_2O_7, Cl_2O, HCl, ClF_3, Cl_2$$ is $$HCl(-1), Cl_2(0), Cl_2O(+1), ClF_3(+3), Cl_2O_7(+7)$$.

The oxidation number of oxygen is superoxide is $$\cfrac{-1}{2}$$.

$$NaB{ H }_{ 4 }\quad \rightarrow \quad N{ a }^{ \\ + }\quad { B{ H }_{ 4 }^{ - } }\\ { B{ H }_{ 4 }^{ - } }\\ x-4\times 1=-1\\ x=-1+4\\ x=3\\ So\quad oxidation\quad number\quad of\quad B\quad in\quad NaB{ H }_{ 4 }\quad is\quad 3.$$

1)The above given equation is a redox reaction.

2)Reaction  between$$BaO_2$$ and$$H_2SO_4$$ is$$BaO_2 + H_2SO_4\Rightarrow BaSO_4 + H_2O_2.$$The most electron negative element in products is oxygen. O.N. of O in $$BaSO_4 , H_2O_2$$ is$$ -2 , -1$$ respectively.

Oxidation number or oxidation state of an element is the total no. of electrons gained or lost by that element in a chemical reaction.

The reaction of iodine with aqueous ammonia to yield explosive crystals, said to be nitrogen triiodide. So oxidation state of Nitrogen is +3.

When copper powder is heated in a china dish, the surface of copper powder becomes coated with a layer of $$\mathrm{CuO}$$, which is black in colour because on heating oxidation of $$\mathrm{Cu}$$ takes place.

N$$_2O_5$$

(i) $$4Na+O_2\rightarrow 2Na_2O$$
(ii) $$H_2SO_4+2NaOH\rightarrow Na_2SO_4+yH_2O$$
(iii) $$2H_2O_2\rightarrow 2H_2O+O_2$$
(iv) $$2Al+2H_3PO_4\rightarrow 2AlPO_4+3H_2$$
(v) $$Ca(OH)_2+2HCl\rightarrow CaCl_2+yH_2O$$

The balanced chemical equation for the reduction of dichromate with sulphur dioxide is shown.
$$\displaystyle K_2Cr_2O_7 + 3SO_2 + H_2SO_4 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + H_2O$$
In $$\displaystyle K_2Cr_2O_7 $$, the oxidation number of Cr is +6.
In $$\displaystyle Cr_2(SO_4)_3$$, the oxidation number of Cr is +3.
The change in the oxidation number of each Cr atom is 3.

In this redox reaction, sulphur dioxide is oxidized to sulphate ions. The oxidation number of S changes from +4 to +6.
The dichromate ion (orange in colour) is reduced to chromate ion (green in colour).
The oxidation number of Cr changes from +6 to +3.
Hence, the oxidation number of the species that has this green color is 3.

Carbon in diamond is in elemental state, so the oxidation state of C in diamond is zero.

Let the oxidation state of $$Mg$$ in $$Mg_3N_2$$ be $$x$$.
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, $$ 3\times x+2\times(3)=0$$
or, $$x=2$$

$$NO$$ in iron complex has +1 oxidation number.
Let oxidation state of Fe be $$x$$.
The complex is $$[Fe(H_2O)_5NO^+]SO_4$$.
Therefore, $$x+1 =+2$$
or, $$x=+1$$.

$$CH_4+O_2\rightarrow CO_2+2H_2$$
$$-4$$                     $$+4$$
Oxidation number of $$C$$ changes by $$4-(-4) =8$$.

Let $$x$$ be the number of peroxy linkage in $$K_3Cr(O_2)_x$$.
The sum of the oxidation numbers of all atoms in neutral molecule is zero.
Therefore, $$3(+1) +5 +x(-2)=0$$
Hence, $$x=4$$
Hence, the number of peroxy linkage in $${ K }_{ 3 }Cr{ \left( { O }_{ 2 } \right)  }_{ x }$$ having oxidation number of $$Cr$$ as $$+5$$ is $$4$$.

Let $$x$$ be the oxidation number of $$Cr$$ in $${ \left( { NH }_{ 3 } \right)  }_{ 2 }Cr{ \left( { O }_{ 2 } \right)  }_{ 2 }$$.

Let the oxidation state of $$I$$ in $$H_4IO_6^-$$ be $$x$$. 
Since the overall charge on the complex is $$-1$$, the sum of oxidation states of all elements in it should be equal to $$-1$$.
Therefore, $$4+x-12 = -1$$
or, $$x= 7$$

Let $$x$$ be the oxidation number of chlorine in $$NaClO_3$$.
Since the overall charge on the complex is $$0$$, the sum of oxidation states of all elements in it should be equal to $$0$$.
So, $$+1+x+3(-2)=0$$
or, $$x= +5$$
Hence, the oxidation number of chlorine in $$NaClO_3$$ is $$+5$$.

Let $$x$$ be the oxidation state of $$N$$ in $$NaNO_3$$.
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, $$+1+x+3(-2) = 0$$
or, $$x=+5$$

Oxidation state of P in $${H}_{3}P{O}_{4}$$ is 5 and oxidation state of P in $${H}_{3}P{O}_{3}$$ is 3 ( taking oxidation state of oxygen is -2 and hydrogen +1). So sum is $$5+3=8$$.

Let the oxidation number of $$S$$ in $$S^{2-}$$ be $$x$$.
The oxidation number of a monoatomic ion is equal to its charge.
Therefore, $$x = -2$$
Hence, the oxidation number of $$S$$ in $$S^{2-}$$ is $$-2$$.

The total number of compounds having $$+5$$ oxidation state of the underlined elements is $$2$$.
The $$+5$$ oxidation state of the underlined element is in (a) $${ K }_{ 4 }\underline { { P }_{ 2 } } { O }_{ 7 }$$ and (c) $${

Rb }_{ 4 }Na\left[ H\underline { { V }_{ 10 } } { O }_{ 28 } \right]$$.
The compounds and the oxidation states of the underlined elements are shown below:
a.  $${ K }_{ 4 }\underline { { P }_{ 2 } } { O }_{ 7 }$$   $$+5$$
b.  $$Na\underline { Au } { Cl }_{ 4 }$$  $$+3$$
c.  $${ Rb }_{ 4 }Na\left[ H\underline { { V }_{ 10 } } { O }_{ 28 } \right]$$  $$+5$$ 
d.  $$\underline { I } Cl$$  $$+1$$
e.  $${ Ba }_{ 2 }\underline { Xe } { O }_{ 6 }$$  $$+8$$
f.  $$\underline { O } { F }_{ 2 }$$  $$+2$$  
g.  $$Ca{ \left( \underline { Cl } { O }_{ 2 } \right)  }_{ 2 }$$ $$+3$$
h.  $$\underline { N } { O }_{ 2 }^{ \ominus  }$$   $$+3$$

The total number of compounds having $$+3$$ oxidation state of the underlined elements is $$3$$.
The $$+3$$ oxidation state of the underlined element is in (b) $$Na\underline { Au } { Cl }_{ 4 }$$, (g)$$Ca{ \left( \underline { Cl

} { O }_{ 2 } \right)  }_{ 2 }$$, and (h)$$\underline { N } { O }_{ 2 }^{ \ominus  }$$.
The compounds and the oxidation states of the underlined elements are shown below:
a.  $${ K }_{ 4 }\underline { { P }_{ 2 } } { O }_{ 7 }$$   $$+5$$
b.  $$Na\underline { Au } { Cl }_{ 4 }$$  $$+3$$
c.  $${ Rb }_{ 4 }Na\left[ H\underline { { V }_{ 10 } } { O }_{ 28 } \right]$$  $$+5$$ 
d.  $$\underline { I } Cl$$  $$+1$$
e.  $${ Ba }_{ 2 }\underline { Xe } { O }_{ 6 }$$  $$+8$$
f.  $$\underline { O } { F }_{ 2 }$$  $$+2$$  
g.  $$Ca{ \left( \underline { Cl } { O }_{ 2 } \right)  }_{ 2 }$$ $$+3$$
h.  $$\underline { N } { O }_{ 2 }^{ \ominus  }$$   $$+3$$

The total number of compounds having zero oxidation state of the underlined elements is $$+3$$.
The zero oxidation state of the underlined element is in (b) $${ H }_{ 2 }\underline { C } O$$, (c) $$\underline { C } { H }_{ 2 }{ Cl }_{ 2 }$$ and (e) $$\underline { { O }_{ 3 } } $$.
The compounds and the oxidation states of the underlined elements are shown below:
a.  $$\underline { S } { O }_{ 3 }^{ 2- }$$  $$+4$$
b.  $${ H }_{ 2 }\underline { C } O$$  $$0$$
c.  $$\underline { C } { H }_{ 2 }{ Cl }_{ 2 }$$  $$0$$
d.  $${ Na }_{ 2 }\underline { { Cr }_{ 2 } } { O }_{ 7 }$$  $$+6$$
e.  $$\underline { { O }_{ 3 } } $$  $$0$$

Let $$x$$ be the oxidation state of $$Ni$$ in $$K_4[Ni(CN)_4]$$.
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,
$$+4 +x -4 = 0$$
or, $$x = 0$$.
Hence,
the oxidation state of $$Ni$$ in $$K_4[Ni(CN)_4]$$ is $$0$$.

Let $$x$$ be the oxidation state of $$S$$ in $$SO_3^{2-}$$.
Since the overall charge on the complex is $$-2$$, the sum of oxidation states of all elements in it should be equal to $$-2$$.
Therefore, $$ x + 3(-2) = -2$$
$$x = +4$$. 

Let $$x$$ be the oxidation state of $$I$$ in $$IF_3$$.
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,
$$x + 3(-1) = 0$$
or, $$x = +3$$.
Hence,
the oxidation state of $$I$$ in $$IF_3$$ is $$+3$$.

From the figure given below, it can be seen that oxidation number of $$S$$ in $$S_8$$ is zero (elemental form) and co-valency is $$2$$.

Let $$x$$ be the oxidation state of $$S$$ in $$S_2O_4^{2-}$$.
Since the overall charge on the complex is $$-2$$, the sum of oxidation states of all elements in it should be equal to $$-2$$.
Therefore, $$2x + 4(-2) = -2$$
or, $$ x = +3$$

Let $$x$$ be the oxidation state of $$S$$ in $$S_2O_6^{2-}$$.
Since the overall charge on the complex is $$-2$$, the sum of oxidation states of all elements in it should be equal to $$-2$$.
$$2x + 6(-2) = -2$$
$$x = +5$$

The oxidation number of $$Cu$$ decreases from $$+2$$ to $$0$$ and that of $$H$$ increases from $$0$$ to $$+1$$. 

According to given conditions,
$${ Sn }^{ 2+ }\longrightarrow { Sn }^{ 4+ }+2{ e }^{ - };\quad increase\quad in\quad O.N=2$$
$${ Ce }^{ 4+ }+n{ e }^{ - }\longrightarrow { Ce }^{ (4-n) };\quad decrease\quad in\quad O.N=n$$
On balancing, we get
$$\quad n{ Sn }^{ 2+ }+2{ Ce }^{ 4+ }\longrightarrow n{ Sn }^{ 4+ }+2{ Ce }^{ (4-n) }$$
Given,
Millimoles of $${ Sn }^{ 2+ }=100\times 0.1=10$$
Millimoles of $${ Ce }^{ 4+ }=50\times 0.4=20$$
At equivalent point, number of equivalents of both reactant and product are same.
Therefore, $$\cfrac { n }{ 2 } =\cfrac { 10 }{ 20 } $$
or, $$n=1$$
Thus, oxidation state of cerium in the reduced product $$=({ Ce }^{ (4-1) }={ Ce }^{ 3+ })=+3$$.

The given reaction is
$$\displaystyle 6\, CO_2(g)\, +\, 6H_2O(l)\, \rightarrow \, C_6H_{12}O_6(aq)\, +\, 6O_2(g) $$
It is more appropriate to write this reaction as
$$\displaystyle  6\, CO_2(g)\, +\, 12H_2O(l)\, \rightarrow \, C_6H_{12}O_6(aq)\, +\, 6O_2(g)+6H_2O(l)$$
This is because water is produced during photosynthesis and water must be shown on product side. To investigate the path, $$\displaystyle H_2O^{18} $$ is used instead of $$\displaystyle H_2O^{16} $$. Thus, instead of using normal O atom in water molecule, we use radioactively labelled O atom in water molecule. By determining the intermediates and products containing labelled oxygen, we can investigate above paths.

Let X be the oxidation number of I in $$\displaystyle KI_3 $$
$$\displaystyle +1+3X = 0 $$
$$\displaystyle X = -\frac {1}{3} $$
But the oxidation number cannot be fractional.

$$2\overset {0}{Fe_{(s)}}+\overset {0}{{O_2}}_{(g)}+2H_2\overset {-2}{O}_{(l)}\longrightarrow {2\overset {+2}{Fe}(OH)_2}_{(s)}$$

The balanced chemical equation is as follow:

1. Oxidation of F is -1, So to be neutral oxidation state of S will be +6
2. Oxidation state of Na is +1, and oxygen is -2, so to be neutral oxidation state of sulphur will be +2

$$Cl_2(g)+H_2S(g)\longrightarrow 2HCl(g)+S(s)$$

$$Mg+Zn{ ({ NO }_{ 3 }) }_{ 2 }\longrightarrow Zn+Mg{ ({ NO }_{ 3 }) }_{ 2 }$$

Removal of electrons or the loss of oxygen is termed as oxidation whereas gain of electrons or gain of oxygen is termed as reduction. So, (a) is reduction and (b) is oxidation.

Let the Ox.no. of Mn in $$KMnO_4$$ be x.
We know that, OX.no. of $$K=+1$$
Ox.no. of $$O=-2$$
So, Ox. no. K$$+$$ Ox. no. Mn $$+$$ $$4(Ox. no. O)=0$$
or $$+1 + x + 4(-2)=0$$
or $$+1 + x - 8=0$$
or $$x=+8-1=+7$$
Hence, Ox.no. of Mn in $$KMnO_4$$ is $$+7$$.
Similarly, for S in $$Na_2S_2O_3$$,
$$2(Ox. no. Na)+2(Ox. no. S)+3(OX. no. O)=0$$
$$2\times (+1)+2x+3(-2)=0$$
$$x=+2$$
Hence, Ox. no. of S in $$Na_2S_2O_3=+2$$.

Let the oxidation number of $$S$$ be x in $$HSO_3{^-}$$ ion.
We know that, Ox. no. of $$H$$$$=+1$$
Ox. no. of O$$=-2$$
So, Ox. no. $$H$$$$+$$ Ox. no. $$S$$ $$+$$ $$3$$(Ox. no. $$O$$)$$=-1$$
$$+1 + x + 3(-2)=-1$$
or $$+1 + x - 6 =-1$$
or $$x - 5 =-1$$
or $$x=+5-1=+4$$
The oxidation number of $$S$$ in $$HSO^-_3$$ ion is $$+4$$.

Let oxidation number of $$Mn$$ be $$x$$.
We know that, Ox. no. of $$O=-2$$.
So, Ox. no. $$Mn$$+ 4(Ox. no. $$O$$)$$=-1$$
$$x + 4(-2)=-1$$
or $$x - 8 =-1$$
or $$x=+8-1=+7$$
The oxidation number of $$Mn$$ in $$[MnO_4]^-$$ ion is $$+7$$.

$$KMnO_4$$	$$+1+x-8=0$$	Ox.no. of $$Mn$$
	$$x=+7$$	$$+7$$
$$K_2MnO_4$$	$$+2+x-8=0$$
	$$x=+6$$	$$+6$$
$$MnO_2$$	$$x-4=0$$
	$$x=+4$$	$$4$$
$$Mn_2O_3$$	$$2x-6=0$$
	$$x=+3$$	$$+3$$

Thus, the highest oxidation number for $$Mn$$ is in $$KMnO_4$$.
Sometimes, oxidation numbers have such values which at first sight appear strange. For example, the oxidation number of carbon in cane sugar $$(C_{12}H_{22}O_{11})$$, glucose $$(C_6H_{12}O_6)$$, dichloromethane, etc., is zero.
Cane sugar $$(C_{12}H_{22}O_{11})$$
$$12\times x+22\times 1+11(-2)=0$$
$$12x+22-22=0$$
so, $$x=0$$
Glucose $$(C_6H_{12}O_6)$$
$$6\times x+12\times 1+6(-2)=0$$
$$6x+12-12=0$$
So, $$x=0$$
Dichloromethane$$(CH_2Cl_2)$$
$$x+2\times 1+2(-1)=0$$
$$x+2-2=0$$
So, $$x=0$$.

Let oxidation number of $$Pt$$ be $$x$$.
We know that Ox. no. of $$Cl=-1$$.
So, Ox. no. $$Pt$$+ 6$$ (Ox. no. Cl)=-2$$
$$x + 6(-1)=-2$$
or $$x - 6 =-2$$
or $$x=+6-2=+4$$
The oxidation number of $$Pt$$ in $$[Pt(Cl)_6]^{2-}$$ ion is $$+4$$.

Writing oxidation numbers of all the atoms,
$$\overset{+2}{Pb}\overset{-2}{S}+\overset{+1}{H_2}\overset{-1}{O_2}\rightarrow \overset{+2}{Pb}\overset{+6}{S}\overset{-2}{O_4}+\overset{+1}{H_2}\overset{-2}{O}$$
The oxidation number of $$S$$ has increased and O has decreased.
$$Pb\overset{-2}{S}\rightarrow Pb\overset{+6}{S}O_4$$        ........(i)
$$H_2\overset{-1}{O_2}\rightarrow H_2\overset{-2}{O}$$       ........(ii)
Increase in Ox. no. of $$S =8$$ units per PbS molecule
Decrease in Ox. no. of $$O =1$$ unit per $$\displaystyle\frac{1}{2}H_2O_2$$ molecule
$$=2$$ units per $$H_2O_2$$ molecule
Multiplying eq. (ii) by $$4$$ as to make increase and decrease equal
$$PbS+4H_2O_2\rightarrow PbSO_4+4H_2O$$
This is the balanced equation.

 Let X be the oxidation number of P in $$NaH_2PO_4$$. The oxidation numbers of Na, H and O are $$ \displaystyle +1$$, $$ \displaystyle +1$$  and $$ \displaystyle -2$$ respectively.
$$ \displaystyle +1+2(+1)+X+4(-2)=0$$
$$ \displaystyle 3+X+-8=0$$
$$ \displaystyle X+-5=0$$

Let X be the oxidation number of Fe in $$[Fe(CN)_6]^{4-}$$.
 The oxidation number of cyano group is -1.
$$ \displaystyle =X+6(-1)=-4$$
$$ \displaystyle =X-6=-4$$
$$ \displaystyle =X=6-4$$

Let X be the oxidation number of Ni in $$[Ni(CN)_6]^{4-}$$.
 The oxidation number of cyano group is -1.
$$ \displaystyle =X+6(-1)=-4$$
$$ \displaystyle =X-6=-4$$
$$ \displaystyle =X=6-4$$
$$ \displaystyle X=+2$$
The oxidation number of Ni in $$[Ni(CN)_6]^{4-}$$ is $$+2$$.

Let X be the oxidation number of  P in $$PH^+_4$$.
The H atoms have oxidation number of +1.
$$ \displaystyle     X+4(+1)=+1$$
$$ \displaystyle     X+4=+1$$
$$ \displaystyle X=-3$$
The oxidation number of P in $$PH^+_4$$ is $$-3$$.

Let X be the oxidation number of  Pt in $$[PtCl_6]^{2-}$$.
The Cl atoms have oxidation number of -1.
$$ \displaystyle     X+6(-1)=-2$$
$$ \displaystyle     X-6=-2$$
$$ \displaystyle X=+4$$
The oxidation number of  Pt in $$[PtCl_6]^{2-}$$ is $$+4$$.

Let X be the oxidation number of  C in $$C_{12}H_{22}O_{11}$$.
The H atoms have oxidation number of +1.
The O atoms have oxidation number of -2.
$$ \displaystyle     12X+22(+1)+11(-2)=0$$
$$ \displaystyle     12X+22-22=0$$
$$ \displaystyle X=0$$
The oxidation number of  C in $$C_{12}H_{22}O_{11}$$ is $$0$$.

Let X be the oxidation number of S in $$H_2S_2O_8$$.
One peroxo linkage is present which is $$O-O$$ linkage. The O atoms of this linkage have oxidation number of -1. The remaining O atoms have oxidation number of -2.
The oxidation number of H is +1.
$$ \displaystyle 2(+1)+2X+2(-1)+6(-2)=0$$
$$ \displaystyle 2X-12=0$$
$$ \displaystyle 2X=12$$
$$ \displaystyle X=+6$$
The oxidation number of S in $$H_2S_2O_8$$ is $$+6$$.

Let X be the oxidation number of  P in $$Mg_2P_2O_7$$.
The Mg atoms have oxidation number of +2.
The O atoms have oxidation number of -2.
$$ \displaystyle   2(+2)+2X+7(-2)=0$$
$$ \displaystyle  2X-10=0$$
$$ \displaystyle  2X=10$$
$$ \displaystyle X=+5$$
The oxidation number of  P in $$Mg_2P_2O_7$$ is $$+5$$.

Let X be the oxidation number of Mn in $$MnO^-_4$$..
The O atoms have oxidation number of -2.
$$ \displaystyle    X+4(-2)=-1$$
$$ \displaystyle    X-8=-1$$
$$ \displaystyle X=+7$$
The oxidation number of  Mn in $$MnO^-_4$$. is $$+7$$.

Let X be the oxidation number of N in $$NO^-_3$$.
The O atoms have oxidation number of -2.
$$ \displaystyle  X+3(-2)=-1$$
$$ \displaystyle X-6=-1$$
 $$ \displaystyle X=6-1$$
$$ \displaystyle X=+5$$
The oxidation number of N in $$NO^-_3$$ is $$+5$$.

Let X be the oxidation number of  S in $$S_2Cl_2$$.
The Cl atoms have oxidation number of -1.
$$ \displaystyle  2X+2(-1)=0$$
$$ \displaystyle  2X-2=0$$
$$ \displaystyle  2X=2$$
$$ \displaystyle X=+1$$
The oxidation number of  S in $$S_2Cl_2$$ is $$+1$$.

Let X be the oxidation number of  Fe in $$Na_2[Fe(CN)_5NO]$$.
The Na atoms have oxidation number of +1.
The NO group has oxidation number of 0.
The CN group has oxidation number of -1.
$$ \displaystyle   2(+1)+X+5(-1)+0=0$$
$$ \displaystyle   2+X-5=0$$
$$ \displaystyle   X-3=0$$
$$ \displaystyle X=+3$$
The oxidation number of Fe in $$Na_2[Fe(CN)_5NO]$$ is $$+3$$.

$$Mn_2O_3$$ has the lowest oxidation number of Mn.
The oxidation number of Mn in $$Mn_2O_3$$ is $$ \displaystyle +3$$.
Let X be the oxidation number of Mn in $$Mn_2O_3$$.
$$ \displaystyle 2X+3(-2)=0$$
$$ \displaystyle 2X-6=0=0$$
$$ \displaystyle X=+3$$.
The oxidation number of Mn in $$KMnO_4, K_2MnO_4, MnO_2$$ is $$ \displaystyle +7$$, $$ \displaystyle +6$$ and $$ \displaystyle +4$$ respectively.

$$H_3PO_4$$ has the highest oxidation number of P.
Let X be has the oxidation number of P in $$H_3PO_4$$.
$$ \displaystyle 3(+1)+X+4(-2)=0$$
$$ \displaystyle 3+X-8=0$$
$$ \displaystyle X=+5$$
The oxidation number of P in $$H_3PO_4$$ is $$ \displaystyle +5$$.
The oxidation number of P in $$PH_3, H_3PO_2, PCl_3$$ is $$ \displaystyle -3$$,
$$ \displaystyle +1$$ and $$ \displaystyle +3$$ respectively.

$$ \displaystyle HCl$$ has the lowest oxidation number of chlorine.
The oxidation number of chlorine in $$HCl$$ is $$ \displaystyle -1$$
The oxidation number of chlorine in $$HClO_4, HOCl, ClF_3, HClO_3$$ is $$ \displaystyle +7$$, $$ \displaystyle +1$$, $$ \displaystyle +3$$ and $$ \displaystyle +5$$ respectively.

Let X be the oxidation number of  Cr in $$(NH_4)_2Cr_2O_7$$.
The O atoms have oxidation number of -2.
The $$NH_4$$ group has oxidation number of +1.
$$ \displaystyle   2(+1)+2X+7(-2)=0$$
$$ \displaystyle   2X-12=0$$
$$ \displaystyle   2X=12$$
$$ \displaystyle X=+6$$
The oxidation number of Cr in $$(NH_4)_2Cr_2O_7$$ is $$+6$$.

Let X be the oxidation number of Ni in $$Ni(CO)_4$$.
The O atoms have oxidation number of -2.
The C atoms have oxidation number of+2.
$$ \displaystyle    X+4(+2-2)=0$$
$$ \displaystyle    X+4(0)=0$$
$$ \displaystyle    X+0=0$$
$$ \displaystyle X=0$$
The oxidation number of Ni in $$Ni(CO)_4$$ is $$0$$.

Let X be the oxidation number of Cl in $$Ca(ClO_2)_2$$.
The O atoms have oxidation number of -2.
The Ca atoms have oxidation number of+2.
$$ \displaystyle    +2+2(X+2(-2))=0$$
$$ \displaystyle    +2+2(X-4)=0$$
$$ \displaystyle    +2+2X-8=0$$
$$ \displaystyle    2X-6=0$$
$$ \displaystyle    2X=6$$
$$ \displaystyle X=+3$$
The oxidation number of Cl in $$Ca(ClO_2)_2$$ is $$+3$$.

In $$CaOCl_2$$, two $$Cl$$ are in different oxidation states because one $$Cl$$ is attached to oxygen atom exists as $$OCl^{-}$$ and other is attached directly to $$Ca$$ exists as $$Cl^{-}$$.One attached directly to $$Ca$$ has -1 oxidation state and one attached to oxygen has +1 oxidation state. So $$Cl$$ exists in two different oxidation states -1 and +1.

$$Fe_3O_4$$ is a mixed oxide of $$FeO$$ and $$Fe_2O_3$$.Oxidation number of $$Fe$$ in $$FeO$$ and $$Fe_2O_3$$ is +2 and +3 as oxidation number of oxygen is -2.

Each nitrogen have different oxidation state because $$NH_4NO_3$$ exists as $$NH_4^{+}$$ and $$NO_3^{-}$$.In $$NH_4^{+}$$, it has -3 oxidation state and in $$NO_3^{-}$$, it has +5 oxidation state.

Each nitrogen have different oxidation state because $$NH_4NO_2$$ exists as $$NH_4^{+}$$ and $$NO_2^{-}$$.In $$NH_4^{+}$$, it has -3 oxidation state and in $$NO_2^{-}$$, it has +3 oxidation state.

Second and third sulphur atoms are joined to two sulphur atoms so their oxidation number is 0. Two sulphur atoms which are joined to 3 oxygen atoms in which one oxygen is attached to $$Na$$ has -1 oxidation state and other two oxygen atoms has -2 oxidation state so sulphur has +5 oxidation state.

Hydrogen has +1 oxidation number.Two oxygen atoms are linked through peroxide linkage so they have -1 oxidation number and other 3 oxygen atoms have -2 oxidation number. Let sulphur has $$x$$ oxidation number so,

Oxidation number of oxygen is -2 and of methyl is 0 as it is neutral. Oxidation number of $$S$$ is $$2(0)+x-2=0,x=+2$$.

Oxidation state of $$K$$ is +1.Let oxidation number of $$O$$ is $$x$$,

By looking the above structure we can observe that 4 oxygen atoms are linked by peroxide linkage.So their oxidation state is -1 as in peroxide.

Oxidation number of $$Se$$ is +6 and of $$O$$ is -2.Total charge on compound is 0.

The value of n in the molecular formula $$Be_nAl_2Si_6O_{18}$$ is:

Action of ozone on hydrogen peroxide produces oxygen and water.
$$ \displaystyle O_3+H_2O_2 \rightarrow H_2O+2O_2$$

• An oxidation-reduction (redox) reaction is a type of chemical reaction that involves a transfer of electrons between two species. An oxidation-reduction reaction is any chemical reaction in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron.
  
• The formation of hydrogen fluoride is an example of a redox reaction. We can break the reaction down to analyze the oxidation and reduction of reactants. The hydrogen is oxidized and loses two electrons, so each hydrogen becomes positive.
  

$$H_2+2F \rightarrow 2HF$$

Aluminium Powder can be used as a reducing agent to obtain manganese from Mangenese Dioxide. The chemical equation for the reaction:-

• $$2KI+Pb{ (N{ O }_{ 3 }) }_{ 2 }\longrightarrow { 2KNO }_{ 3 }+Pb{ I }_{ 2 }\\ \quad (eq.)\quad \quad (eq.)\quad \quad \quad \quad \quad \quad (eq.)\quad \quad \quad (\therefore settled)\quad$$
• $$Zn+2HCl\longrightarrow ZNC{ l }_{ 2 }+{ H }_{ 2 }\uparrow librated$$ 
• $${ C }_{ 24 }{ H }_{ 48 }\longrightarrow 24{ CO }_{ 2 }+{ 24H }_{ 2 }O$$
  

$$N\overset{+2}{a}_2\overset{+4}{S}_2\overset{-6}{O}_3\longrightarrow N \overset{+2}{a}_2\overset{\!\!+6}{SO_4^{-8}}$$

$$OF_2$$, $$X-2 =0$$ or $$X = 6$$

For $$C_2H_6 , C_4H_10$$

Displacement reaction :
A single-displacement reaction, also known as a single-replacement reaction, is a reaction by which one
(or more) element(s) replaces an/other element(s) in a compound. It can be represented generally as:
$$A + B-C \rightarrow A-C + B$$
All simple metal with acid reactions are single displacement reactions. For example, the reaction
between magnesium, $$Mg$$, and hydrochloric acid, $$HCl$$, forms magnesium chloride, $$MgCl_2$$, and hydrogen,
$$H_2$$.
$$Mg(s) + 2 HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)$$
Decomposition reaction :
A decomposition reaction is a type of chemical reaction where one reactant yields two or more
products.
The general form for a decomposition reaction is
$$AB \rightarrow A + B$$
Water can be separated by electrolysis into hydrogen gas and oxygen gas through the decomposition
reaction: $$2 H_2O → 2 H_2 + O_2$$
Another example is the spontaneous decomposition of hydrogen peroxide into water and oxygen: 2
$$H_2O_2 → 2 H_2O + O_2$$
The decomposition of potassium chlorate into potassium chloride and oxygen is yet another example:
$$2 KClO_3 → 2 KCl + 3 O_2$$
Combination reaction :
A combination reaction (also known as a synthesis reaction) is a reaction where two or more elements
or compounds (reactants) combine to form a single compound (product). Such reactions may be
represented by equations of the following form : $$X + Y \rightarrow XY$$.
Formation of calcium oxide by the combination elements calcium and oxygen.
$$2Ca + O_2 → 2CaO$$
Formation of ammonia by the combination of elements nitrogen and hydrogen.
$$N_2 + 3H_2 \rightarrow 2NH_3 $$

$$Br_2$$ is better oxidising agent than $$I_2$$ hence, $$Br_2$$ oxidises sulphur to its maximum oxidation state that is $$+6$$ while $$I_2$$ does not. This is because $$Br_2$$ is more electromagnetic than $$I_2$$. Hence it easily gets converted to Br.

$$Na_2[Fe(CN)_5NO^+]$$                                 $$CN^-$$ has $$-1$$ oxidation state

$${ LiAlH }_{ 4 }\rightarrow { Li }^{ + }+{ \left[ { AlH }_{ 4 } \right]  }^{ - }$$

In the complex $${ K }_{ 3 }\left[ Fe(CN)_{ 6 } \right] $$ K(Potassium) is Alkaline metal with +1 oxidation state,then for the $${ K }_{ 3 }$$ it's +3. CNis a Negative ligand ,for $${ CN }_{ 6 }$$ it's -6

Let, oxidation state of $$Fe$$ will be $$x$$.

The total oxidation state will be zero.

$$\therefore 3 + x + -6=0$$

$$\therefore x=+3$$

thus the oxidation state of $$Fe$$ is +3.

Oxidation is a losing of electrons and Reduction is a gaining of electrons.

a)$$CaC{O}_{3} + 2HCl \rightarrow Ca{Cl}_{2} + {H}_{2}O + C{O}_{2}$$

Solution:-

$$HClO$$. Let theOxidation number of $$Cl$$ be x

(i)  $${ K }_{ 4 }\left[ Fe{ \left( CN \right)  }_{ 6 } \right] \rightarrow { 4K }^{ + }{ \left[ { Fe\left( CN \right)  }_{ 6 } \right]  }^{ 4- }$$

$${ PbSO }_{ 4 }\rightarrow { Pb }^{ 2+ }{ \left[ { SO }_{ 4 } \right]  }^{ 2- }$$

Let the oxidation number of sulphur be $$x$$.

$${ K }_{ 2 }{ MnO }_{ 4 }\longrightarrow { 2K }^{ + }{ \left[ { MnO }_{ 4 } \right]  }^{ 2- }$$

$$Br_{3}O_{8}$$ is tribromooctaoxide.

$$Br_2$$ is a stronger oxidizing agent than $$I_2$$. Due to this, it oxidizes $$S_2O_3^{2-}$$ from $$+2$$ to $$+6$$ oxidation state, while $$I_2$$ oxidizes $$S_2O_3^{2-}$$ from $$+2$$ to $$+2.5$$ which is comparatively lower oxidation state. Due to this, same reductant behaves differently in different reactions.

$$^{+6}Cr_2^{-2}O_7^2+^1H_2O^{-2}+^4SO_2^{-2}\longrightarrow ^{+3}Cr^{3+}+^{-2}O\overset {1}{H^{-}}+^{+6}S\overset {-2}{O_4^{2-}}$$

The reduced oxidation state of cerium in the reduced product is $$Ce^{3+}$$

$$Fe^{2+}+Cr_2O_7^{2-}+H^+\longrightarrow Fe^{3+}+Cr^{3+}+H_2O$$

1) For $$K_2MnO_4$$; The oxidation state of Mn in $$KMnO_4$$ is $$+7$$.

This can be obtained by taking into account the valencies of potassium and oxygen atoms.

The valency of $$K $$is $$1$$ but it loses electron so to be specific it is$$ +1$$.

Similarly, oxygen in specific has a valency of $$-2$$.

2) For $$NaBH_4$$; As the oxidation state of Na is $$+1$$ and Hydrogen is $$-1$$ because here hydrogen exists as a hydride.

Hence the oxidation state of boron is $$+3$$.

1) In $$XeF_4$$

$$XeF_4=0$$

$$Xe + 4\times(-1)=0$$

$$Xe= +4$$

2) In $$XeO_2$$

$$XeO_2=0$$

$$Xe + (-2)\times 2=0$$

$$Xe = +4$$

3) Thus, Oxidation state of Xe in both the compound is $$+4$$.

1)Oxidation state of Fe in$$ K_4[Fe(CN)_6 = +2$$.

2)The name of this complex salt is Potassium ferrocyanide.( Or) Potassium Hexa cyano Ferrate(II).

       $$4(K)+Fe+6(CN)=0$$

To calculate the OXIDATION-NUMBER of $$Fe: +4+Fe+6(-1)=0$$

   So $$Fe=+6–4=2$$

1) The oxidation number of Fe is $$0$$.

2) The charge on a complex is the sum of the oxidation state of the metal center and the charges on the ligands.

3) $$CO$$ is a neutral ligand. Hence, the charge on the ligand is$$ 0$$. The total charge on the complex is also $$0$$.

4) Calculate the oxidation number of Fe in the complex,  $$Fe(CO)_5$$ is as follows;

              ( Charge on$$ Fe$$) $$+5$$(Charge on $$CO)=0$$

                 ( Charge on Fe) $$+5(0)=0$$

                  ( Charge on Fe)$$ = 0$$

5)Thus, the oxidation number of Fe in$$  Fe(CO)5 $$is $$0$$.

Rule for balancing equation ion electron method :-

In $$[Cu(NH_3)_4]2^+$$ each $$NH_3$$ is neutral.

So the oxidation number of copper is $$ +2$$.

$$1+10x-56=-5$$

$$10x-55= -5$$

$$10x =50x = +5$$

Thus $$V=+5$$

Total oxidation No. of $$S$$ in $$ {(CH_3)}_2SO$$ 
$$(-1)+(-1)+2=0$$

The oxidation number of $$H = +1$$.

There is one $$H$$ atom: $$(1)(+1) = +1 $$.

The oxidation number of $$O$$ is $$ -2$$.

There are four $$O$$ atoms here:$$ (4)(-2) = -8$$

Therefore, the oxidation state of $$Cl$$ is $$+7$$.

$$K_2FeO_4$$   $$2(+1)+x+4(-2)=0\\2+x-8=0\\x-6=0\\x=+6$$

A reaction in which oxidation and reduction happen simultaneously is called redox reaction. 

$$K^+ = H, O^{-2} = -2$$
$$K_2MnO_4$$
$$(+1 \times 2) + x + (-2 \times 4) = 0$$
$$2 + x - 8 = 0$$
$$x = +6$$

$$MnO_2$$
$$\downarrow$$
$$x + (-2 \times 2) = 0$$
$$x = +4$$

$$KMnO_4$$
$$\downarrow$$
$$(+1) + x + (-2 \times 4) = 0$$
$$x = 8 - 1 = +7$$

Higher the electronegativity, greater the pull on oxidizing agent has for electrons. Higher the pull for electrons, stronger the oxidizing agent.
Here, for $$Cr, Mo, W$$ down the group, electronegativity increases. So, $$WO_4^{2-}$$ is strong oxidizing agent.

Hydrogen acts as a good reducing agent means, when hydrogen gas is passed over hot metallic oxides of copper, lead, iron, etc. it removes oxygen from them and thus reduces them to the corresponding metal. Let us consider the following example, in each of which metallic oxide react with hydrogen. Metallic oxide act as oxidising agents and hydrogen acts as a reducing agents.

'

Solution:-

$$2KI+3H_{2}SO_{4}\rightarrow 2KHSO_{4}+SO_{2}+I_{2}+2H_{2}O$$

$$3C_{2}H_{5}OH+2MnO_{4}^{-}+OH^{-}\rightarrow 3C_{2}H_{3}O^{-}+2MnO_{2}+5H_{2}O$$

Balanced Chemical Reaction:

Balanced Chemical Reaction:

Balanced Chemical Reaction:

Balanced Chemical Reaction:

Balanced Chemical Reaction:

$$ Cr_{2}O_{7}^{2-}+3SO_{3}^{2-}+8H^{+}\rightarrow 2Cr^{3+}+3SO_{4}^{2-}+4H_{2}O$$

$$I_{2}+SO_{2}+2H_{2}O\rightarrow SO_{4}^{2-}+2I^{-}+4H^{+}$$

$$ 2MnO_{4}^{-}+5SO_{2}+2H_{2}O\rightarrow MnO_{4}^{2-}+5SO_{4}^{2-}+4H^{+}$$

$$2MnO_{4}^{-}+3SO_{3}^{2-}+H_{2}O\rightarrow 2MnO_{2}+3SO_{4}^{2-}+2OH$$

$$ 2K_{2}Cr_{2}O_{7}+2H_{2}O+3S\rightarrow 3SO_{2}+4KOH+2Cr_{2}O_{3}$$

$$ 2MnO_{4}^{-}+SO_{3}^{2-}+ 2OH^- \rightarrow 2MnO_{4}^{2-}+SO_{4}^{2-}+H_{2}O$$

$$ 2MnO_{4}^{-}+SO_{3}^{2-}+2OH\rightarrow 2MnO_{4}^{2-}+SO_{4}^{2-}+H_{2}O$$

$$ 10Al+6KMnO_{4}+27H_{2}SO_{4}\rightarrow 6KHSO_{4}+5Al_{2}(SO_{4})_{3}+6MnSO_{4}+24H_{2}O$$

$$ Sn+4NO_{3}^{-}+2H^{+}\rightarrow SnO_{2}+4NO_{2}+2H_{2}O$$

$$ ClO^-+Br^-\rightarrow BrO_{3}^- +3Cl^-$$

Balanced chemical reaction is as follows:

Solution:

Balanced chemical reaction is as follows:

$$ 4Zn+NO_{3}^{-}+7OH^- \rightarrow NH_3 + {4ZnO_2}^{2-} + 2H_2O$$

Solution:

$$2ClO_{2}+SbO_{2}^{-}+2OH\rightarrow 2ClO_{2}^{-}+Sb(OH)_{6}^{-}+2H_{2}O$$

Balanced chemical reaction is as follows:

Solution:

Solution:

$$ Cl_{2}+SeO_{3}^{2-}+ H_{2}O \rightarrow SeO_{4}^{2-}+2Cl^{-}+2H^{+}$$

$$ 3CuO +2NH_{3}\rightarrow 3Cu+N_{2}+3H_{2}O $$

$$6Cu_{3}P+124H_{+}+11Cr_{2}O_{7}^{2-}\rightarrow 18Cu^{2+}+6H_{3}PO_{4}+22Cr^{3+}+53H_{2}O$$

Balanced Chemical Reaction:

$$K_{4}Fe(CN)_{6}+61Ce(NO_{3})_{4}+258KOH\rightarrow 61Ce(OH)_{3}+Fe(OH)_{3}+36H_{2}O+6K_{2}CO_{3}+259KNO_{3}$$

$$ As_{2}S_{5}+40HNO_{3}\rightarrow 2H_{3}AsO_{4}+5H_{2}SO_{4}+40NO_{2}+12H_{2}O $$

$$ PbS+4O_{3}\rightarrow PbSO_{4}+4O_{2}$$

Balanced chemical reaction is as follows:

Balanced chemical reaction is as follows:

$$K_2Cr_2O_7+H_2SO_4+KI\rightarrow I_2+K_2SO_4+Cr_2(SO_4)_3+H_2O$$

This is an oxidation-reduction (redox) reaction:

$$6I^-\ -6e^-\rightarrow 6I^0$$    (oxidation)

$$2Cr^{VI}+6e^-\rightarrow 2Cr^{III}$$   (reduction)

$$KI$$ is a reducing agent, $$K_2Cr_2O_7$$ is an oxidizing agent

$$ (NH_{4})_{2}Cr_{2}O_{7}= N_{2}+4H_{2}O +Cr_{2}O_{3} $$

$$ 2NaOH+Cl_{2}\to NaCl+NaClO+H_{2}O$$

Balanced chemical reaction is as follows:

$$3MnO_{4}^{2-}+4H^{+} \to 2MnO_{4}^{-} + MnO_{2} +H_{2}O $$

$$ Br_{2}+2OH^{-}\rightarrow BrO^{-}+Br^{-}+H_{2}O$$

$$ Cr_{2}O_{7}^{2-}+3SO_{2}+5H^{+} \to 2Cr^{3+}+3HSO_{4}^{-}+H_{2}O$$

(a) lighter

$$\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \rightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)$$
Redox Reaction

(a) Sodium (Na) and Potassium (K) give hydrogen with cold water. 
(b) Iron 
(c) Oxidation 
(d) Copper oxide (CuO)

Na or Mg are more reactive metals as compared to carbon. So, their oxides are more stable.

$$2H_{2}S(g) +O_{2}(g)  \space    \rightarrow \space2S(s) + 2H_{2}O(l)$$

$$2 H g O(s) \stackrel{H e a t}{\longrightarrow} 2 H g(l)+O_{2}(g)$$

The reactions in which oxidation and reduction are taking place at the same time are called redox reactions.
For examples:
In the example given below, $$Fe_2O_3$$ is reduced to $$Fe$$ by losing oxygen atoms.
Thus, $$Fe$$ undergoes reaction.
$$CO$$ is oxidized to $$CO_2$$ by gaining oxygen atom. Thus, $$C$$ undergoes oxidation.
$$CuO + H_2 \rightarrow Cu + H_2O$$
In the given reaction, $$CuO$$ is reduced to $$Cu$$ by losing oxygen. Thus, $$Cu$$ undergoes reaction.
$$H_2$$ is oxidized to $$H_2O$$ by gaining oxygen atoms. Thus, $$H_2$$ undergoes oxidation.

(a) Catalytic hydrogenation: Catalytic hydrogenation is a process by which hydrogen gas is passed through vegetable oils in the presence of catalyst like Ni, Pt or Pd to convert them into solid vanaspati ghee.

(b) Oxidation: A reaction in which a substance combine with oxygen or in which hydrogen is removed is called oxidation reaction.
Example: $$H_2S + Cl \longrightarrow 2HCl +S$$

(c) Reduction: Those reactions in which hydrogen combines with a substance or oxygen is removed from a substance, are known as reduction reactions.
Example: $$2HgO \overset{heat}{\longrightarrow} 2 Hg + O_2$$

(d) Redox reaction: Redox reactions are those in which reduction and oxidation both takes place simultaneously i.e. one substance is reduced while the other gets oxidised.

Step-1: (LEO= Loss of Electron Oxidation)
(GER+Gain of Electron Reduction) 

Step-2: $$ Fe^{2+} \rightarrow F4^{3+} ...(1)(2 \rightarrow 3=1 $$ unit $$ ON )$$,
$$ Cr_2O_7^{2+} \rightarrow 2Cr^{3+} ...(2)(+6 \rightarrow +3=3 $$ units $$ 2 sets=> 6 $$ units $$ ON)$$

Step-3: Cross multiplication i.e., $$ (1) \times 6  $$ and $$(2) \times 1 $$
(1)$$ \times 6 \Rightarrow 6Fe^{2+} \rightarrow 6Fe^{3+} ...(3) $$
(2) $$ x  1 \Rightarrow  Cr_2O^{-2}_7 \rightarrow 2Cr^{3+} ......(4) $$

Step -4 : add $$ (3) + (4) \Rightarrow 6 Fe^{2+} + Cr_2O^{2-}_7  \rightarrow 6Fe^{3+} + 2Cr^{3+} $$

Step - 5 : balance oxygen and then hydrogen 
$$ 6Fe^{2+} + Cr_2O^{2-}_7 + 14H^+ \rightarrow 6 Fe^{3+} + 2Cr^{3+} + 7H_2O $$

In redox reaction, there is the transfer of electrons, which results in the bond formation. The electropositive atom undergoes oxidation, while the electronegative atom undergoes reduction.

It indicates that the valency of lead in this case is $$2$$ and it forms $$+2$$ ions.

$$ \underline {Si}O^{2-}_3 $$ ( oxidation number of $$ Si $$ ) $$ +3 (-2) = -2 ; $$ ( Oxidation number of Si ) = $$-2+ 6 =  +4 $$

$$x+3+4(-1)=0$$
$$x+3-4=0$$
$$x-1=0$$
$$x=1$$
$$\therefore$$ The oxidation number of $$Li \ in \ LiAlH_{4}$$ is $$+1$$.

In this structure, the oxidation number of $$Z$$ is zero.

Suppose oxidation number of $$Ni$$ in $$[Ni(CN)_4]^{2-}=x$$.
$$x+4(-1)=-2$$
$$x-4=-2$$
$$x=-2+4=+2$$
$$\therefore$$ Oxidation number of $$Ni$$ in $$[Ni(CN)_4]^2 =+2$$

Suppose oxidation number of $$Mo$$ in $$(NH_4)_2 MoO_4=x$$
$$2(+1)+x+4(-2)=0$$
$$2+x-8=0$$
$$x-6=0$$
$$x=+6$$

Lets the oxidation number of oxygen in $$KO_2=x$$
$$1+2x=0$$
$$2x=-1$$
$$x=-1/2$$
So, the oxidation number of oxygen in $$KO_2$$ is $$-1/2$$

$$Cu$$ in the reactant has its oxidation state as 0.

Oxidation state of Cu after the chemical reaction is $$2+$$.

Hydrogen decreases from $$+1$$ to $$0$$ oxidation no.

$$0$$ to $$+2$$

The charge of ions in ionic compound is the oxidation number. Eg. In $$FeCl_2$$ the oxidation number of $$Fe$$ is $$+2$$ and $$FeCl_3$$ is $$+3$$.

-3 is the oxidation number of $$N$$ in $$NI_3$$

The list of compounds and the oxidation state of nitrogen in the compound is given below:
$$ {Na}_{}{N}_{3} \implies 1(+1) + 3(x) = 0 \implies x = -\dfrac13 $$
$$ {N}_{2}{H}_{2} \implies 2(x) + 2(+1) = 0 \implies x = -1 $$
$$ NO \implies 1(x) + 1(-2) = 0 \implies x = + 2 $$
$$ {N}_{2}{O}_{5} \implies 2(x) + 5(-2) = 0 \implies x = +5 $$

Let the reduced species be $$Z$$ and its oxidation number be $$x$$.
The reaction for reduction of $$ClO_4^-$$ by $$H_2O_2$$ is as follows:

Compound                           Oxidation state of N
$${ N }_{ 2 }O$$                                         +1
$$NO$$                                           +2
$$H{ NO }_{ 2 }$$                                     +3
$$ {NO }_{ 2 }$$                                         +4
$$H{ NO }_{ 3 }$$                                    +5

(A) $${ H }_{ 2 }\underline { { S }_{ 2 } } { O }_{ 8 }$$ contains peroxy linkage.
The oxidation number of the underlined atom is $$6$$
$${ H }_{ 2 }{ O }_{ 2 }$$ is the hydrolysis product
 It is named as Marshall's acid or peroxodisulphuric acid.
(B) $$\underline { Cr } { O }_{ 5 }$$ Contains peroxy linkage.
The oxidation number of the underlined atom is $$6$$
(C) $${ H }_{ 2 }\underline { S } { O }_{ 5 }$$ contains peroxy linkage.
The oxidation number of the underlined atom is $$6$$
$${ H }_{ 2 }{ O }_{ 2 }$$ is the hydrolysis product
(D) In the compound $${ K }_{ 2 }\underline { Cr } { O }_{ 4 }\ $$, the oxidation number of the underlined atom is $$6$$
(E) $${ \left( { NH }_{ 3 } \right)  }_{ 3 }\underline { \overset { IV }{ Cr }  } { O }_{ 4 }$$ contains peroxy linkage.

a.  $${ Fe }^{ 3+ }$$ is reduced to $${ Fe }^{ 2+ }$$ and $${ H }_{ 2 }S$$ is oxidised to $$S$$.
b.  $${ I }^{ \ominus  }$$ is oxidised to $${ I }_{ 2 }$$ and $${ H }^{ \oplus  }$$ is reduced to $${ H }_{ 2 }O$$.
c.  $$Bi$$ is oxidised to $${ Bi }^{ 3+ }$$ and $${ H }^{ \oplus  }$$ is reduced to $${ H }_{ 2 }O$$.
d.  $${ I }^{ \ominus  }$$ is oxidised to $${ I }_{ 2 }$$, and $${ O }_{ 2 }$$ and $${ H }_{ 2 }O$$ are reduced to $$\overset { \ominus  }{ O } H$$.
e.  $${ Au }^{ \oplus  }$$ is reduced to $$Au$$ and $$Cu$$ is oxidised to $${ Cu }^{ 2+ }$$.
All the above are redox reactions.
Hence, the total number of redox reactions is $$5$$.

6 mol of $$\displaystyle Cl_2 $$ undergo a loss and gain of 10 mol of electrons to form two oxidation states of Cl which are $$\displaystyle Cl^{5+} $$ and $$\displaystyle Cl^- $$
The reduction half reaction is as follows:
$$\displaystyle 5Cl_2  + 10e^- \rightarrow 10Cl^- $$
The oxidation half reaction is as follows:
$$\displaystyle Cl_2 \rightarrow  2Cl^{+5} + 2e^- $$
The overall reaction is given below:
$$\displaystyle 6Cl_2 \rightarrow 10Cl^- + 2Cl^{+5} $$
Hence, the oxidation state of $$Cl$$ having positive charge is $$+5$$.

$$A :   N$$ has lowest oxidation state of $$0$$ in $$N_2$$ element. and highest oxidation state of $$+5$$ in $$HNO_3$$, and other O.S. are $$-1/3$$ in $$HN_3$$, and $$+3$$ is found in $$HNO_2$$.

(A) For the reaction $${ H }^{ \oplus  }+U{ \left( S{ O }_{ 4 } \right)  }_{ 2 }+KMn{ O }_{ 4 }\longrightarrow U{

O }_{ 2 }S{ O }_{ 4 }+MnS{ O }_{ 4 }$$
the molar ratio of oxidants and reductant is $$\displaystyle\frac { 2 }{ 5 }$$.
(B) For the reaction $$Cr{ I }_{ 3 }+{ Cl }_{ 2 }+\overset { \ominus  }{ O } H\longrightarrow Cr{ O }_{ 4 }^{ 2-

}+I{ O }_{ 4 }^{ \ominus  }+{ Cl }^{ \ominus  }$$
the molar ratio of oxidants and reductant is $$\displaystyle\frac { 27 }{ 2 }$$.
(C) For the reaction $$HAs{ O }_{ 3 }^{ 2- }+Br{ O }_{ 3 }^{ \ominus  }+{ H }^{ \oplus  }\longrightarrow { H }_{

3 }As{ O }_{ 4 }+{ Br }^{ \ominus  }$$
the molar ratio of oxidants and reductant is $$\displaystyle\frac { 1 }{ 3 }$$.
(D) For the reaction $${ O }_{ 2 }+N{ H }_{ 3 }\longrightarrow NO+{ H }_{ 2 }O$$
the molar ratio of oxidants and reductant is $$\displaystyle\frac { 5 }{ 4 }$$.
(E) For the reaction $${ Sb }_{ 2 }{ O }_{ 3 }+I{ O }_{ 3 }^{ \ominus  }+{ H }^{ \oplus  }\longrightarrow HSb{

\left( OH \right)  }_{ 6 }+ICl$$
the molar ratio of oxidants and reductant is $$1$$.
(F) For the reaction $$W{ O }_{ 3 }+Sn{ Cl }_{ 2 }+{ H }^{ \oplus  }\longrightarrow { W }_{ 3 }{ O }_{ 8 }+Sn{

Cl }_{ 6 }^{ 2- }$$
the molar ratio of oxidants and reductant is $$3$$.

The total number of elements having the lowest oxidation state of zero is $$4$$.
The maximum oxidation state is the group number. The minimum oxidation state for metals is zero while for non-metals, it is equal to the group number minus $$8$$. The maximum and the minimum oxidation states are given below:
Compound  Maximum  Minimum
a.  $$Ta$$           +5                0
b. $$Te$$           +6                -2
c.  $$Tc$$          +7                  0
d.  $$Ti$$          +4                  0
e.  $$Tl$$          +3                  0

The oxidation number of Fe has been calculated below:
$${ Fe }_{ 4 }^{ 3+ }{ \left[ { Fe }^{ 2+ }{ \left( CN \right)  }_{ 6 } \right]  }_{ 3 }$$
Average oxidation number of $$Fe$$ is $$\displaystyle\frac { 3\times 4+2\times 3 }{ 7 } = \displaystyle\frac { 18 }{ 7 }$$
$${ Fe }_{ 2 }^{ 2+ }\left[ { Fe }^{ 2+ }{ \left( CN \right)  }_{ 6 } \right]$$
Average oxidation number of $$Fe$$ is $$\displaystyle\frac { 2\times 2+2 }{ 3 } = \displaystyle\frac { 6 }{ 3 } = 2$$
$${ Fe }_{ 3 }^{ 2+ }{ \left[ { Fe }^{ 3+ }{ \left( CN \right)  }_{ 6 } \right]  }_{ 2 }$$
Average oxidation number of $$Fe$$ is $$\displaystyle\frac { 2\times 3+3\times 2 }{ 5 } = \displaystyle\frac { 12 }{ 5 }$$
$${ Fe }^{ 3+ }\left[ { Fe }^{ 3+ }{ \left( CN \right)  }_{ 6 } \right]$$
Average oxidation number of $$Fe$$ is $$\displaystyle\frac { 3+3 }{ 2 } = \displaystyle\frac { 6 }{ 2 } = 3$$
$${ Na }_{ 2 }^{ 2+ }{ \left[ { Fe }^{ 2+ }{ \left( { CN }^{ 5- } \right)  }_{ 5 }N{ O }^{ 1+ } \right]  }^{ 2- }$$
Oxidation state of $$Fe$$ is $$+2$$.
$${

\left[ { Fe }^{ 1+ }\left( { NO }^{ 1+ } \right) { \left( \overset { 0

}{ { H }_{ 2 }O }  \right)  }_{ 5 } \right]  }^{ 2+ }S{ O }_{ 4 }^{ 2-

}$$
Oxidation state of $$Fe$$ is $$+1$$.
$${ Fe }_{ 3 }{ O }_{ 4 }$$:
$$3x - 8 = 0$$
$$\Rightarrow x = \displaystyle\frac { 8 }{ 3 }$$
Therefore, oxidation state of $$Fe$$ is $$\displaystyle\frac { 8 }{ 3 }$$.

(A) $$12$$ mL of $$0.05 M$$ $$SeO_{2}$$ reacted exactly with $$24$$ mL of $$0.1 M$$ $$CrSO_{4}$$. In this reaction, $$Cr^{2+}$$ is oxidised to $$Cr^{3+}$$.
The oxidation state of Se changes from +4 to 0 during the reaction.
$$SeO_{2}=12\times 0.05 = 0.6$$ $$mmol$$ $$(n = ?)$$
$$CrSO_{4}=24\times 0.1 = 2.4$$ $$mmol$$ $$(Cr^{2+}\rightarrow Cr^{3+}+e^{-})$$
$$mEq. SeO_{2}=mEq.$$ of $$CrSO_{4}$$
$$0.6\times n = 2.4\times 1 (n = 1)$$
$$\displaystyle n=\frac {2.4}{0.6}=4$$
So, $$Se^{4+}$$ is reduced to $$Se^{0}$$.
$$\overset{+4}{Se}+4e^{-}\rightarrow Se$$

(B) $$40$$ mL of $$1.0 M$$ $$Ce^{4+}$$ is required to react with $$20 mL$$ of $$1.0 M$$ $$Sn^{2+}$$ to $$Sn^{4+}$$.
The oxidation state of cerium in the reaction product is +3.
$$Ce^{4+}=40\times 1.0=40$$ $$mmol$$ $$(n =?)$$
$$Sn^{2+}=20\times 1.0 = 20$$ $$mmol$$ $$(n = 2)$$
                            $$(Sn^{2+}\rightarrow Sn^{4+})$$
$$mEq$$ of $$Ce^{4+}=mEq$$ of $$Sn^{2+}$$
$$40\times n=20\times 2$$
$$n=1$$
So, $$Ce^{4+}$$ is reduced to $$Ce^{3+}$$
$$Ce^{4+}+e^{-}\rightarrow Ce^{3+}$$
Oxidation state of $$Ce$$ in reaction product $$=+3$$.

(C) $$30$$ mL of $$1.0 MI^{\circleddash}$$ is required to react with $$10 mL$$ of $$1.0 M S_{2}{O_8}^{2-}$$ to react in acidic medium. In this reaction, $$I^{\circleddash}$$ is oxidised to $${I_3}^{\circleddash}$$. The oxidation state of $$S$$ in the reaction product is +4.
$$\displaystyle \underline{{\underset{\displaystyle 3x=-3}{3I^{\circleddash}}}\rightarrow{\underset{\displaystyle 3x=-1}{{I_3}^{\circleddash}}}}+2e^{-}$$
    $$-1-(-3)=2$$
$$mmol$$ of $$I^{\circleddash}=30\times 1.0=30$$
$$mmol$$ of $$S_{2}{O_8}^{2-}=10\times 1=10$$
$$mEq$$ of $$S_{2}{O_8}^{2-}=mEq$$ of $$I^{\circleddash}$$
$$10\times n=30\times 2$$
$$n=6$$
In $$S_{2}{O_8}^{2-}(2x-16=-2,2x=14)$$
The change of $$e^{-}$$ should be $$=6$$.
$$\displaystyle6e^{-}+{\underset{\displaystyle 2x=14}{S_{2}{O_8}^{2-}}}\rightarrow {\underset{\displaystyle (2y)}{2S}}$$
            $$\underline{\uparrow   14-2y=6  \uparrow}$$
                  $$2y=8$$
                  $$y=4$$
Oxidation state of $$S$$ in the product $$= +4$$.
Hence, the reaction is $$6e^{-}+S_{2}{O_8}^{2-}\rightarrow 2SO_{2}$$.

(D) $$25$$ mL of $$2.0 M Zn$$ is oxidised to $$[ZnCl_{4}]^{2-}$$ by $$100 mL$$ of $$1.0 M [CrCl_{6}]^{3-}$$. The oxidation state of $$Cr$$ in the reaction product is +2.
$$Zn\rightarrow [\overset{+2}{Zn}Cl_{4}]^{2-}+2e^{-}(n=2)$$
$$mmol$$ of $$Zn=25\times 2=50$$
$$mmol$$ of $$[CrCl_{6}]^{3-}=100\times 1=100$$
$$mEq$$ of $$[CrCl_{6}]^{3-}=mEq$$ of $$Zn$$
$$100\times n=50\times 2$$
$$n=1$$
$$e^{-}+[Cr^{3+}Cl_{6}]^{3-}\rightarrow Cr^{2+}$$
Oxidation state of $$Cr$$ in reaction product $$=+2$$.

(E) An acid solution of $$0.1$$ mol of $$KReO_{4}$$ was reduced with $$Zn$$ and then titrated with $$0.8 Eq$$ of acidic $$KMnO_{4}$$ solution for the reoxidation of all the $$Re$$ to $$Re{O_4}^{\circleddash}$$.
The oxidation state to which $$Re$$ was reduced by $$Zn$$ is -1.
$$Eq$$ of $$KReO_{4}\equiv Eq$$ of $$Mn{O_4}^{\circleddash}$$
$$0.1\times n\equiv0.8$$ eq.
$$\displaystyle n=\frac {0.8}{0.1}=8$$
Hence, there is a $$8e^{-} $$ reduction.
$$\displaystyle 8e^{-}+\overset{+7}{Re}{O_4}^{\circleddash}\rightarrow Re^{-1}$$
Oxidation state of $$Re=-1$$.

The IUPAC name of $$CrO_5$$ is chromium (VI) peroxide or chromium (VI) oxide peroxide or monochromium (VI) pentaoxide.

Let X be the oxidation state of P in $$\displaystyle NaH_2PO_4 $$.

The order of increasing oxidation number of nitrogen: $$NH_3, N_3H, N_2O, NO, N_2O_5$$ is $$NH_3(-3), N_3H(-1/3), N_2O(+1), NO(+2), N_2O_5(+5)$$.
The numbers in bracket represents the oxidation number of nitrogen. With increase in the number of electronegative  O atoms, the oxidation number of nitrogen increases.
The oxidation numbers of H and O are $$ \displaystyle +1$$ and $$ \displaystyle -2$$ respectively.

Butanone reacts with 2,4 dinitro-phenyl hydrazine to form 2,4 dinitro-phenyl hydrazone derivative of butanone. A molecule of water is lost.

$${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }+{ Fe }^{ 2+ }\rightarrow { Fe }^{ 3+ }+{ Cr }^{ 3+ }$$

(a) $$\overset { 0 }{ P } \left( s \right) +\overset { +5 }{ P } { O }_{ 4 }^{ 3- }\left( aq \right) \rightarrow \overset { +3 }{ HP } { O }_{ 3 }^{ 2- }\left( aq \right) $$

$$AlF_3$$ becomes soluble due to the formation of $$[AlF_6]^{3-}$$ an addiction of KF

The Ionisation energy of $$Al$$ is lesser than $$C$$ as after removal of electron $$Al$$ will gain $$3s^2$$ electronic configuration that is much stable than $$3p^1$$ . 

Balanced chemical reaction is as follows:

Balanced chemical reaction is as follows:

Balanced Chemical Reaction:

This is an example of oxidation reaction since two $$ H- $$ atoms have been removed.

First two Reactions are irreversible reactions. Third Reaction is a reversible Reaction.