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Showing posts with label INORGANIC CHEMISTRY. Show all posts
Showing posts with label INORGANIC CHEMISTRY. Show all posts

Jan 8, 2019

Soft and Hard Acids and Bases (SHAB) Principle

SHAB Principle and Stability of the Complex:

According to Lewis Concept an Acid-Base reaction involves an interaction of a vacant orbital of an acid (A) and a filled or unshared orbital of a base (B).
A + :B
A : B 
Lewis Acid
(Acceptor)

Lewis Base
(Donor)

Adduct or Complex
The species A is called Lewis Acid or a generalized acid and B is called Lewis Base or a generalized base. A strong acid A and a strong base B, will form the stable complex A : B.
A concept known as Principle of Soft and Hard Acids and Bases(SHAB) Principle is very helpful in making a stability of the complex A : B.
According to this principle the complex A : B is most stable when A and B are either both soft or both hard.
The complex is least stable when one of the reactants (namely A and B) is very hard and the other one is very soft. 
In order to arrive at a comparative estimate of the donor properties of different bases, the preferences of a particular base to bind a proton H+ and methyl mercury (II) ion, [CH3HgB]+ was determined. Both the proton and methyl mercury cation can accommodate only one coordinate bond but the two cation vary widely in their preferences to bases. This preferences was estimated from the experimental determination of equilibrium constants for the exchange reactions:
BH+ + [CH3Hg(H2O)]+
[CH3HgB]+ 
+ H3O+
The results indicate that bases in which the donor atom is N, O or F prefer to coordinate to the proton. Bases in which the donor atoms is P, S, I, Br, Cl or C prefer to coordinate to mercury.

Hard and Soft Bases:

The donor atoms in the first group have high Electronegativity, low Polaris ability and hard to oxidize. Such donors have been named ‘hard bases by Pearson, since they hold on to their electrons strongly. 
The donor atoms of second category are of low electronegativity, high polarisability, and are easy to oxidize. Such donors have been called ‘soft’ bases since they are hold on to their valence electrons rather loosely.
In simple terms hardness is associated with a tightly held electron shell with little tendency to polarise. On the other hand softness is associated with a loosely bound polarisable electron shell.
It will be seen that within a group of the periodic table softness of the Lewis bases increases with the increase in size of the donor atoms. 
Thus, among the halide ions softness increases in the order:
-Cl -Br -I -
Thus F - is the hardest and I - is the Softest base.
Classification of Bases
(R = alkyl or aryl group)
Hard Borderline
Soft
H₂O, OH⁻, F⁻, CH₃COO⁻, PO₄⁻³, SO₄⁻², Cl⁻, CO₃⁻², ClO₄⁻, NO₃⁻, ROH, RO⁻, R₂O, NH₃, RNH₂, N₂H₄
C₆H₅NH₂, C₅H₅N, N₃⁻, Br⁻, NO₂⁻, SO₃⁻²,N₂
R₂S, RSH, RS⁻, I⁻, SCN⁻, S₂O₃⁻², R₃P, R₃As, (RO)₃P, CN⁻, RNC, CO, C₂H₄, C₆H₆, H⁻

Hard and Soft Acids:

After having gone through a classification of bases, a classification of Lewis acids is necessary. The preferences of a given Lewis acid towards ligands of different donor atoms is usually determined from the stability constant values of the respective complexes or from some other useful equilibrium constant measurements. When this is done, metal complexes with different donor atoms can be classified into two sets based on the sequences of their stabilities.
Hard acids have small acceptor atoms, are of high positive charge and do not contain unshared pair of electrons in their valence shell, although all these properties may not appear in one and the same acid. These properties lead to high electronegativity and low polarisability. In keeping with the naming of the bases, such acids are termed as 'Hard'Acids.
Hard Acids:
NPOSFClBrI
Soft acids have large acceptor atoms, are of low positive charge and contain ushered pairs of electrons in their valence shell. These properties lead to high polarisability and low electronegativity. Again in keeping with the naming of the bases, such acids are termed 'Soft' Acids
Soft Acids:
NPOSFClBrI
Classification of Lewis Acids as Hard, Intermediate and Soft Acids.
Hard Borderline
Soft
H⁺, Li⁺, Na⁺, K⁺, Be⁺², Mg⁺², Ca⁺², Sr⁺², Mn⁺², Al⁺³, Ga⁺³, In⁺³, La⁺³, Lu⁺³, Cr⁺³, Co⁺³, Fe⁺³, As⁺³, Si⁺⁴, Ti⁺⁴, Zr⁺⁴, Th⁺⁴, U⁺⁴, Ce⁺³, Sn⁺⁴, VO⁺², UO₂⁺², MoO⁺³, BF₃, AlCl₃, SO₃, Cr⁺⁶
Fe⁺, Co⁺², Ni⁺², Cu⁺², Zn⁺², Pb⁺², Sn⁺², Sb⁺³, Bi⁺³, Rh⁺³, B(CH₃)₃, SO₂, NO⁺, GaH₃
Cu⁺, Ag⁺, Au⁺, Tl⁺, Hg⁺, Pd⁺², Cd⁺², Pt⁺², Hg⁺², CH₃Hg⁺, Tl⁺³, BH₃, GaCl₃, InCl₃, I⁺, Br⁺, I₂, Br₂, Zerovalent Metal atoms

Classify the following as Hard and Soft Acids and Bases. 
(i) H⁻ (ii) Ni⁺⁴ (iii) I⁺ (iv) H⁺

(i) The hydride ion has a negative charge and is far too large in size compared to the hydrogen atom. Its electronegativity is quite low and it will be highly polarisable by virtue of its large size. Hence it is Soft Base.
(ii) Quadrivalent nickel has quite a high positive charge. Compared to bivalent nickel its size will be much smaller. Its electronegativity will be very high and polarisability will be low. Hence it is Hard Acid.
(iii) Mono-positive iodine has a low positive charge and has a large size. It has a low electronegativity and a high polarisability. Hence it is a Soft Acids.
(iv) H⁺ has the smallest size with a high positive charge density. It has no unshared pair of electrons in its valence shell. All these will give a high electronegativity and very low polarisability. Hence H⁺ is a Hard Acid.

Soft and Hard Acids and Bases (SHAB) Principle:

This principle also means that if there is a choice of reaction between an Acid and two Bases and two Acids and a Base,
A Hard Acid will prefer to combine with a Hard Base and a Soft Acid will prefer to combine with Soft Base and thus a more stable product will be obtained.
Hard Acid - Hard Base may interact by strong ionic forces. Hard Acids have small acceptor atoms and positive charge while the Hard Bases have small donor atoms but often with negative charge. Hence a strong ionic interaction will lead to Hard Acid - Base combination.
On other hand a Soft Acid - Soft Base combination mainly a covalent interaction. Soft Acids have large acceptor atoms, are of low positive charge and contain ushered pair of electrons in their valence shell.

Application of Soft and Hard Acids and Bases (SHAB) Principle:

(i) [CoF₆]⁻³ is more stable than [CoI₆]⁻³
It will be seen that Co⁺³ is a hard acid
F⁻ is a hard base and I⁻ is a soft base
Hence [CoF₆]⁻³ (Hard Acid + Hard Base) is more stable 
than [CoI₆]⁻³ (Hard Acid + Soft Base).

AgI₂⁻ is stable, but AgF₂⁻ does not exist. Explain.

We know that Ag⁺ is a soft acid, F⁻ is hard base and I⁻ is soft base
Hence AgI₂⁻ (Soft Acid + Soft Base) is a stable complex and AgF₂⁻ (Soft Acid + Hard Base) does not exist.
(ii) The existence of certain metal ores can also be rationalised by applying SHAB principle. Thus hard acids such as Mg⁺², Ca⁺² and Al⁺³ occur in nature as MgCO₃, CaCO₃ and Al₂O₃ and not as sulphides (MgS, CaS and Al₂S₃), since the anion CO₃⁻² and O⁻² are hard bases and S⁻² is a soft base.
Soft acids such as Cu⁺, Ag⁺ and Hg⁺² on the other hand occurs in nature as sulphides.
The borderline acids such as Ni⁺², Cu⁺² and Pb⁺² occur in nature both as carbonates and sulphides. 
The combination of hard acids and hard bases occurs mainly through ionic bonding as in Mg(OH)₂ and that of soft acids and soft bases occurs mainly by covalent bonding as in HgI₂.

Explains why Hg(OH)₂ dissolved readily in acidic aqueous solution but HgS does not?

In the case of Hg(OH)₂ and HgS, Hg is a soft acid and OH⁻ and S⁻² is hard base and soft base respectively. Evidently HgS (Soft acid + Soft base) will be more stable than Hg(OH)₂ (Soft Acid + Hard base). More stability of HgS than that of Hg(OH)₂ explain why Hg(OH)₂ dissolved readily in acidic aqueous solution but HgS does not. 
If we arrange the donor atoms of most common Lewis bases in an increasing order of their electronegativity values we get,
Soft and Hard Acids and Bases and Hard Soft Acid Base Principle
Donor Atoms of Most Common Lewis Bases





Jan 3, 2019

Acids and Bases Questions and Answers


What is the acidic order of oxoacids of phosphorus?

When the oxidation state rule applied to the oxoacids of phosphorus (H₃PO₂, H₃PO₃ and H₃PO₄) it is predicted that the acidic Character of these acids should be in the order:
H₃PO₂ㄑH₃PO₃ㄑH₃PO₄, but the experimental observation suggested the reverse order:
H₃PO₂ ≥ H₃PO₃〉H₃PO₄.
The experimental order can be explained when we consider the structures of these acids given as,
Acids and Bases Questions and Answers
Oxoacids-of-Phosphorus
H₃PO₂ is a mono-basic acid. The proton attached to an oxygen has a far greater chance of dissociation than any directly bonded hydrogen. the structure of H₃PO₂ involves one protonated oxygen and another unprotonated oxygen. H₃PO₃ is dibasic and hence has two protonated oxygen's and one unprotonated oxygen. H₃PO₄ is tribasic, has three protonated oxygen's and one unprotonated oxygen. 
In this series therefore the number of unprotonated oxygen's, which are the vehicles for the enhancement of acidity, is the same for all the three acids. But dissociable protons increase from one in H₃PO₂ to three in H₃PO₄. Therefore the overall inductive effect of the unprotonated oxygen decreases from H₃PO₂ to H₃PO₄. Hence the acidity slightly falls off in the order:
H₃PO₂ ≥ H₃PO₃H₃PO₄

pH of a solution is 4.5. Calculate the concentration of H⁺ ion.

We have from the definition,
pH = -log[H⁺] =4.5
or, log[H⁺] = -4.5
∴ [H⁺] = 3.16×10⁻⁵

Calculate the [H⁺], [OH⁻] and pH of a solution prepared by diluting 20 ml of 0.1M HCl to one lit. 

[H⁺] = (20×0.1)/1000 
=0.002 
= 2×10⁻³
[OH⁻] = (1×10⁻¹⁴)/[H⁺]
=(1×10⁻¹⁴)/(2×10⁻³) 
= 0.5×10⁻¹¹
pH = -log[H⁺] 
= -log2×10⁻³ 
= 3-log2 
= 2.7

Calculate the [H⁺], [OH⁻] and pH of a solution obtained by dissolving 28 gm KOH to make 200 ml of a solution.

[OH⁻] = (28×1000)/(56×200) 
= 2.5M 
(Molecular Weight of KOH = 56gm)
[H⁺] = (1×10⁻¹⁴)/[OH⁻] 
= (1×10⁻¹⁴)/(2.5) 
= 4×10⁻¹⁵ 
pH = -log[H⁺] 
= - log4 × 10⁻¹⁵ 
= (15 - log4) 

What will be the effect of adding KNH₂ to liquid ammonia with respect to acidity?

2NH₃ NH₄⁺ + NH₂⁻
KNH₂ K⁺ + NH₂⁻
Due to common ion effect the equilibrium shift to left and decrees the acidity of NH₃.

Arrange in order of Lewis acidity :
(i) BF₃ (ii) BCl₃ (iii) BI₃ (iv) BBr₃

These boron halides have pi-interaction between filled p- orbitals of the halogen and empty p-orbital of boron. The effectiveness of the pi-interaction falls off with increasing size of halogens so that it is the strongest in BI₃. When boron halides receives an electron pair the pi-bond between boron and halogen has to be ruptured in order to make room for a coordinate bond.
H₃NBF₃
Thus with BF₃ it will be hardest to rupture the pi-bond. Therefore the order of the Lewis acidity is:
BF₃BCl₃BBr₃BI₃

Why all alkali are bases but all bases are not alkali?

All the bases are dissolved in water to produce alkali whereas all the bases are not dissolved in water but all the alkali are dissolved in water. Thus all alkali are bases but all bases are not alkali.
Na₂O is a base because it dissociate in water to produce NaOH. NaOH, KOH and Ca(OH)₂ dissolved in water to produce OH⁻ ion. Thus all these hydroxide are Alkali.
NaOH Na⁺ + OH⁻
KOH K⁺ + OH⁻
Ca(OH)₂ Ca⁺² + 2OH⁻
Al(OH)₃, Fe(OH)₃, Zn(OH)₂ etc does not dissolved in water but reacts with acids to produce Salt and water. Thus These are bases but but not alkali.





Explain - NH₃ behaves as a base but BF₃ as an Acid.

In NH₃, the central N atom have lone pair of electrons This lone pair coordinate to empty orbital, is termed as a base according to Lewis concept. 
The compounds with less than an octet for the central atom are Lewis acids. In BF₃B in BF₃, contain six electrons in the central atom thus it behaves as an acid.
F₃B + :NH₃ F₃BNH₃
Acid Base Adduct


Explain why tri-covalent phosphorus compounds can serve both as Lewis acids and also as bases?

Tri-covalent phosphorus compounds like PCl₃ have a lone pair of electrons in phosphorus. This lone pair may coordinate to a metal ion thus allowing the compound to serve as a Lewis base. 
Ni + 4PCl₃   [Ni(PCl₃)₄]⁰
Again the quantum shell of phosphorus has provision for d-orbitals which can receive back donated electrons from electron reach low oxidation state of a metal ion. In this latter case tri-covalent phosphorus compound serves as a Lewis acid.

Bi-positive tin can function both as a Lewis acid and a Lewis base. Explain?

The Lewis representation of SnCl₂ shows a lone pair on tin through it is as yet short of an octet. Ligands, particularly donor solvents with lone pairs, may coordinate to tin giving complexes. Here SnCl₂ behaves as a Lewis acid.
Again interaction of platinum group metal compounds with SnCl₂ (SnCl₃⁻) as donor leads to the formation of coordinate complexes. As for examples,
[(Ph₃P)₂PtCl(SnCl₃)], [RuCl₂(SnCl₃)₂]⁻². These are examples of Lewis base behavior of SnCl₂.

Can SiCl₄ and SnCl₄ function as Lewis acids?

Both silicon and tin are members of Group IV of the periodic table and their quantum shells admit of d-orbitals. As a result they can expand their valence shell through SP³d² hybridisation and can give rise to six coordinate complexes. In fact complex like [SiCl₄{N(CH₃)₃}₂], [SiCl₂(Py)₄]Cl₂, [SnCl₄(Bpy)] are known. Thus SiCl₄ and SnCl₄ can function as Lewis acids.

Name the conjugate acids and the conjugate bases of HX⁻ and X⁻².

Conjugate acid of a species is the one that is obtained on the addition of a proton and conjugate base of a species is one that is obtained on the release of a proton.
H₂O + HX⁻   OH⁻ + H₂X
Acid₁ Base₂ Base₁ Acid₂
In the above reaction HX⁻ acts as a base and its conjugated acid is H₂X.
HX⁻ + H₂O   X⁻² + H₃O⁺
Acid₁ Base₂ Base₁ Acid₂
In this reaction HX⁻ acts as an acid thus its conjugated base is X⁻².
In the same way conjugate acid of X⁻² is HX⁻ but X⁻² cannot have any conjugated base because there is no proton that can release.

Arrange these oxides in order of their acidic nature : N₂O₅, As₂O₃, Na₂O, MgO.

Acidic oxides react with water to give oxoacids. The higher the oxidation number and the higher the electronegativity the grater the central element will force to reaction with water to give the oxoacid. Of all these oxoacids nitrogen has the highest oxidation number and electronegativity. Thus the order of acidic nature and oxidation number are-
Acidic nature :
Na₂OMgOAs₂O₃N₂O₅
Oxidation number of Oxide forming element: +1  +2  +3  +5. 

Bisulphate ion can be viewed both as an acid and a base. Explain.

Bisulphate ion is HSO₃⁻. It may lose a proton to give the conjugate base SO₃⁻², thus behaving as an acid. Again it may add on a proton to give the conjugate acid, thus showing its base character.

SO₃ Behaves as an acid and H₂O as a base. Explain.

SO₃ like BF₃ has less than an octet,and will be termed as an acid according to Lewis concept.
But in H₂O oxygen atom contain lone pairs to donate the Lewis acid.
Oxygen and sulphur contain six electrons in their valence shell and therefore regarded as Lewis Acids. The oxidation of SO₃⁻² to SO₄⁻² ion by oxygen and S₂O₃⁻² ion by sulphur are the acid-base reactions.
SO₃⁻²  [O]   ⇆   [OSO₃]⁻²
SO₃⁻² +  [S]       [SSO₃]⁻²

Explains why Hg(OH)₂ dissolved readily in acidic aqueous solution but HgS does not?

In the case of Hg(OH)₂ and HgS, Hg is a soft acid and OH⁻ and S⁻² is hard base and soft base respectively. Evidently HgS (Soft acid + Soft base) will be more stable than Hg(OH)₂ (Soft Acid + Hard base). More stability of HgS than that of Hg(OH)₂ explain why Hg(OH)₂ dissolved readily in acidic aqueous solution but HgS does not. 

Classify the following as Hard and Soft Acids and Bases.
(i) H⁻ (ii) Ni⁺⁴ (iii) I⁺ (iv) H⁺

(i) The hydride ion has a negative charge and is far too large in size compared to the hydrogen atom. Its electronegativity is quite low and it will be highly polarisable by virtue of its large size. Hence it is Soft Base.
(ii) Quadrivalent nickel has quite a high positive charge. Compared to bivalent nickel its size will be much smaller. Its electronegativity will be very high and polarisability will be low. Hence it is Hard Acid.
(iii) Mono-positive iodine has a low positive charge and has a large size. It has a low electronegativity and a high polarisability. Hence it is a Soft Acids.
(iv) H⁺ has the smallest size with a high positive charge density. It has no unshared pair of electrons in its valence shell. All these will give a high electronegativity and very low polarisability. Hence H⁺ is a Hard Acid.

Jan 1, 2019

Oxidation Number

Oxidation Number Concept of Oxidation and Reduction:

The definition of oxidation and reduction based on loss or gain of electrons is limited to the scope. The definition holds goods for ionic compounds. 
For Examples, formation of water from hydrogen and oxygen, can not be covered by electronic concept since water is not an ionic compound.
2H₂ + O₂ 2H₂O
It may recall classically we could still say that hydrogen is oxidised to H₂O. In the same seance burning of magnesium in oxygen is considered oxidation. Similarly hydrogen and chlorine react react to form a covalent molecule hydrogen chloride.
2H₂ + Cl₂  2HCl
The Above reaction hydrogen is oxidised or chlorine is reduced but the resulting compound is covalent one, thus the reaction can not be covered by the electronic concept. In order to cover such reactions also under oxidation-reduction the concept of oxidation number is developed and it is defined as,

Oxidation Number:

The oxidation number of an element in a compound is the formal charge (positive or negative) which would be assigned to the element, if all the bonds in the compounds were ionic bonds.
All the compounds are treated as though they were ionic merely because of the case of counting oxidation numbers. The oxidation number therefore is arbitrary. 
Electronegativity concept is utilized in adducing a formal charge to an atom. The less electronegative of the partners of a binary compound is arbitrary assigned a positive oxidation number and more electronegative one a negative oxidation number. Florine, being the most electronegative, has always a negative oxidation number. The alkali metals are highly electropositive, have low electronegativity, hence have been assigned positive oxidation numbers.

Method of finding out the Oxidation Number:

The following general rules are to be observed for the assignment of oxidation numbers.
1. Atoms of diatomic molecules like H₂, Cl₂, O₂ etc or of metallic elements are assigned zero oxidation numbers since same elements of similar electronegativity are involved in the bonding.
H ➖H
The oxidation number of the above molecules are zero because two hydrogen atom of same electronegativity are involves for bonding.
2. Except of metal hydrides oxidation number of hydrogen in hydrogen compound is +1.
In alkali metal hydrides, LiH, NaH, CsH etc, the oxidation number of hydrogen, for reasons of varying electronegativity, is -1 
Examples:
NaH Na⁺ + H⁻ 
(Here oxidation number of H is  -1)
HCl H⁺ + Cl⁻ 
(Here oxidation number of H is +1)
3. The oxidation number of metal is positive.
For Examples, CuO Cu⁺² + O⁻² 
(Here Oxidation number of Cu is +2)
4. Oxygen has normally an oxidation number -2. In hydrogen peroxide the oxidation number of oxygen is -1 since hydrogen has to be assigned +1.
In peroxides and super oxides the oxidation number of oxygen is -1 and -1/2 respectively. 
In fluorine monoxide(F₂O) oxygen has an oxidation number +1 because fluorine is more electronegative than oxygen. 
For Examples, CuO  Cu⁺² + O⁻² 
(Here Oxidation number of O is -2).
In H₂O, Oxidation Number of Oxygen is (-2), 
but in H₂O₂, Oxidation Number of H is (+1) and Oxidation Number of Oxygen is (-1). 
In BaO₂, Oxidation Number of Ba is (+2) and thus the Oxidation Number of Oxygen is (-1). 
In Na₂O, Oxidation Number of Na is (+1) and Oxidation Number of oxygen is (-2). 

Oxidation number of P in Ba(H₂PO₂)₂ is - (a)+3, (b)+2, (c) +1, (d) -1.

(c) +1
Oxidation number of Ba is +2, oxidation number of hydrogen is +1 and oxidation number of oxygen is -2.
Let the oxidation number of P is x.
∴ (+2)+2{2(+1)+x+2(-2)} = 0
or, 2x-2=0; 
or, x=+1 
5. The oxidation number of an ion is equal to its charge.
For examples, NaCl Na⁺ + Cl⁻
The charge and oxidation number of Na⁺ is +1 and the charge and oxidation number of Cl⁻ is -1
Similar way, MgBr₂ Mg⁺² + 2Br⁻
Here the charge of Mg⁺² is +2 and the oxidation number also +2 and the charge of Br⁻ is -1, thus the oxidation number also -1. 
6. The algebraic sum of the oxidation numbers of all the atoms in a compound must be zero.
For Examples:
In HCl oxidation number of hydrogen is +1 and oxidation number of chlorine is -1.
And the sum of these = (+1) + (-1) = 0

Oxidation Number of an Element in a compound:

1. Oxidation Number of Mn in KMnO₄:
Let the oxidation number of Mn in KMnO₄ is x.
Thus according to the above rule, 
(+1) + x + 4(-2) = 0
or, x = +7
Thus, the oxidation number of Mn in KMnO₄ is +7
2. Oxidation Number of Mn in MnO₄⁻² :
Let the oxidation number of Mn in MnO₄⁻² is x and the oxidation number of oxygen is -2(according to the above rule).
Thus the sum of the oxidation number of MnO₄⁻² = Charge of the MnO₄⁻².
x +4(-2) = -2
or, x = +6
Thus, the oxidation number of Mn in MnO₄⁻² is +6
3. Oxidation Number of Cr in Cr₂O₇⁻² : 
Let the oxidation number of Cr in Cr₂O₇⁻² is x.
∴ 2x + 7(-2) = -2
or, x = +6
Thus, the oxidation number of Cr in Cr₂O₇⁻² is +6 .
4. Oxidation Number of S in H₂SO₄ :
Let the oxidation number of S in H₂SO₄ is x. According to the rule oxidation number of hydrogen is +1 and oxygen is -2
∴ 2(+1) + x + 4(-2) = 0
or, x = +6
Thus, the oxidation number of S in H₂SO₄ is +6
5. Oxidation Number of C in CH₃COCH₃:
Let the oxidation number of C in CH₃COCH₃ is x. And the oxidation number of hydrogen and oxygen is +1 and -2 respectively.
∴ 3x + 6(+1) + (-2) = 0
or, x = -(4/3)
Thus, the oxidation number of C in CH₃COCH₃ is -(4/3). 
6. Oxidation Number of S in Na₂S₂O₃:
Let the oxidation number of S in Na₂S₂O₃ is x
∴ 2(+1) + 2x + 3(-2) = 0
or, x = +2
Thus, the oxidation number of S in Na₂S₂O₃ is +2.
7. Oxidation Number of S in Na₂S₄O₆:
Let the oxidation number of S in Na₂S₄O₆ is x
∴ 2(+1) + 4x + 6(-2) = 0
or, x = +2.5
Thus, the oxidation number of S in Na₂S₄O₆ is +2.5.
8. Oxidation Number of Cr in [Cr(NH₃)₆]Cl₃
Let the oxidation number of Cr in [Cr(NH₃)₆]Cl₃ is x. NH₃ is neutral thus the oxidation number is zero and the oxidation number of Chlorine is -1
∴ x + 0 +3(-1) = 0
or, x = +3
Thus, the oxidation number of Cr in [Cr(NH₃)₆]Cl₃ is +3.

Calculate the oxidation number of Iron in [Fe(H₂O)₅(NO)⁺]SO₄.

H₂O is neutral thus the oxidation number is zero, oxidation number of (NO)⁺ is +1 and the oxidation number of SO₄ is -2.
Let the oxidation number of Fe in [Fe(H₂O)₅(NO)⁺]⁺² is x.
∴ x + 0 + (+1) + (-2) = 0 
or, x-1 = 0
or, x = +1 
Thus, the oxidation number of Fe in [Fe(H₂O)₅(NO)⁺]⁺² is +1.

Oxidation Number of an element in a compound is zero:

Some organic compound where the oxidation number of carbon on this compound is zero. As for Example, In Sugar (C₁₂H₂₂O₁₁), Glucose (C₆H₁₂O₆), Formaldehyde (HCHO) etc. the oxidation number of carbon=
Let, the oxidation number of carbon in Glucose (C₆H₁₂O₆) is x
∴ 6x + 12(+1) + 6(-2) = 0
or, x = 0

Calculate the oxidation number of the element marked with line in the following compounds, (i) K₂CrO₄, (ii) HOCl, (iii) BaO₂, (iv) ClNO, (v) NaNH₂, (vi) NaN₃, (vii) CH₂Cl₂, (viii) Ca(OCl)Cl, (ix) Ba(MnO₄)₂ (x) CaH₂.

Compound
Element Oxidation Number
 K2CrO4
Cr +6
HOCl
Cl +1
 BaO2
Ba +2
ClNO
N +3
 NaNH2
N - 3
 NaN3
N - (1/3)
 CH2Cl2
C 0
Ca(OCl)Cl
Cl +1
 Ba(MnO4)2
Mn +7
 CaH2
H -1

What is the Oxidation state of chromium in Cr2O5?

Due to peroxy linkage oxidation state of Cr in Cr2O5 is +6.

Oxidation Number and concept of Oxidation and Reduction:

The oxidation number of CH₄, CH₃Cl, CH₂Cl₂, CHCl₃, CCl₄ and CO₂ are -4, -2, 0, +2, +4 and +4 respectively. In H₂, the oxidation number of hydrogen is zero but in H₂O it is +1. Similarly magnesium in elementary state has a zero oxidation number but in MgCl₂, the oxidation number is +2.
From the above we can define Oxidation and Reduction according to the Oxidation number increase or decrees,

Oxidation:

Oxidation may be defined as a process in which an increase in oxidation number by atoms or ions occurs. A reagent which can increase the oxidation number of an element or ion is called an oxidising agent. 

Reduction:

Reduction is a process in which a decrease in oxidation number by atoms or ions occurs. A reagent that lowers the oxidation number of an element or ion is called a reducing agent.

Schematic Representation of Oxidation and Reduction According to Electronic Concept and Oxidation Number Concept:

What is the Oxidation Number? How can we find out the oxidation number of an element in a Compound?
Schematic Representation of Oxidation and Reduction
Oxidation and reduction are always found to go hand to hand during a radox reaction. Whenever an element or a compound is oxidised, another element or another compound must be simultaneously reduced. 
An oxidant is reduced and simultaneously the reductant is oxidised.
Oxidation Number and Oxidation Number of an Elements in a Compound.
Representation of oxidant and reductant

1.Magnesium metal burns in oxygen to produce magnesium oxide:

Before the reaction oxidation number of magnesium and oxygen both zero. But after the reaction the oxidation number of magnesium and oxygen is +2 and -2 respectively. Thus the oxidation number of magnesium is increases and oxidation number of oxygen is decreases. So magnesium is oxidised and oxygen is reduced.
How can we define oxidation and Reduction According to Oxidation Number Concept?
Representation of Oxidation and Reduction

2. Reaction between Sodium and Chlorine:

From same way we can explain this reaction. Here oxidation number of Sodium and Chlorine is zero before the reaction and +1 and -1 after the reaction. Thus sodium oxidised and chlorine reduced.
0
0
+1
-1
2Na
+
Cl₂
2
Na
Cl

Why sulphur dioxide has properties of Oxidation and reduction?

This can be explain by the oxidation state of sulphur. The oxidation numbers of sulphur in the compounds H₂S, SO₂, and SO₃ are -2, +4 and +6 respectively. Thus the highest oxidation state of sulphur is +6 and lowest is -2. The oxidation state of free sulphur element is 0. In SO₂, the oxidation state of sulphur is +4, this oxidation state is middle of 0 and +6.
Thus, the oxidation state can increases from +4 to +6 and decreases from +4 to 0.