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

Jan 26, 2019

Slater's Rules

Slater's Rule for Effective Nuclear Charge:

Slater proposed some set of empirical rules to calculate the screening constant (σ) of various electrons present in different orbitals of an atom or an ion.
Once we get the value of screening constant it is easy enough to find the effective nuclear charge (Z*).

Screening Effect:

The Valence electron in a multi - electron atom is attracted by the nucleus and repelled by the electrons of inner-shells.
What is the Slater's rule ?
Shielding Effect
The combined effect of this attractive and repulsive force acting on the valence electron is that the Valence - electron experiences less attraction from the nucleus. This is known as the screening effect.
Slater's Rules to Calculate the Screening Constant (σ):
Rules for an electron in the nS, nP level:
(i) The first to do is to write out the electronic configuration of the atom or the ion in the following order and grouping.
(1S) (2S, 2P) (3S, 3P) (3d) (4S, 4P) (4d) (4f) (5S, 5P) etc.
It may be noted that so far as the screening effect is concerned the S and P electrons belonging to the same principal quantum shell have the same effect as advocated by Slater.
Na atom :
(1S)² (2S, 2P)⁸ (3S)¹
(ii) An electron in a certain (nS, nP) level is screened only by electrons in the same level and by the electrons of lower energy level.
Electrons lying above (nS, nP) level do not screen any (nS, nP) electron to any extent. Higher energy electrons have no screening effect on any lower energy electrons.
Examples:
Estimation of screening constant of the valence electron of the Sodium atom,
(1S)² (2S, 2P)⁸ (3S)¹ 
(The Valence electron will be excluded from our Calculation).
Estimation of screening constant of the 3d electron of the Sodium atom,
(1S)² (2S, 2P)⁸ (3S)¹
 (The one 2P electron and 3S electrons will be excluded from our Calculation).
(iii) Electrons of an (nS, nP) level shield a valence electrons in the same group by 0.35 each. This is also true for the electrons of the nd or nf that is for the electrons in the same group.
(iv) Electrons belonging to one lower quantum shell, that is (n-1) shell shield the valence electrons by 0.85 each.
(v) Electrons belonging to (n-2) or still lower quantum shell shield the valence electron by 1.0 each. This means the screening effect is complete.
 
 
Examples:
(a) For the valence electron of Na atom:
Screening Constant (σ 
= (2 × 1) + (8 × 0.85) + (0 × 0.35) 
8.8
(a) For the 2P electron of Na atom:
Screening Constant (σ ) 
= (2 × 0.85) + (7 × 0.35) 
4.15
Na+ Ion:
  (1s)2(2s, 2p)8
Screening Constant 
σ  = (2×0.85)+(8×0.35)
= 4.5
Effective Nuclear Charge
Z⋆ = (11 - 4.5) = 6.5
 Mg+2 : (1s)2(2s, 2p)8
Screening Constant 
σ  = (2×0.85)+(8×0.35)
= 4.5
Effective Nuclear Charge
Z⋆ = (12 - 4.5) = 7.5

 K+ (1s)2(2s, 2p)8(3s, 3p)8
Screening Constant 
σ  = (2×1.0)+(8×0.85)+(8×0.35)
= 11.6
Effective Nuclear Charge
Z⋆ = (19 - 11.6) = 7.4

Valence Electron of F ion:
  (1s)2(2s, 2p)7
Screening Constant 
σ  = (2×0.85)+(6×0.35)
= 3.8
Effective Nuclear Charge
Z⋆ = (9 - 3.8) = 5.2

F- ion (1s)2(2s, 2p)8
Screening Constant 
σ  = (2×0.85)+(8×0.35)
= 4.5
Effective Nuclear Charge
Z⋆ = (9 - 4.5) = 4.5
Estimate the screening constant for the outermost 4S electron of Vanadium.
Vanadium has atomic number 23 and the electronic configuration according to the Slater's Rule is :
(1S)² (2S, 2P)⁸ (3S, 3P)⁸ (3d)³ (4S)²
We have consider only one electron of the two 4S electrons.
∴Screening Constant (σ
(2 ×1.0) + (8×1.0) + (8×0.85) + (3×0.85) (1×0.35) 
19.7.
Rules for an electron in the nd, nf level:
The above rules are quite well for estimating the screening constant of S and P orbitals. However when a d or f electron being shielded the (iv) and (v) rule replaced by a new rules for estimation of screening constant.
The replaced rules are:
All electrons below the nd or nf level contribute 1.0 each towards the screening constant.
 
 
Example
Screening Constant for a 3d electron of Vanadium:
Electronic Configuration according to Slater's Rule is:
(1S)² (2S, 2P)⁸ (3S, 3P)⁸ (3d)³ (4S)²
We have consider only two electron of the three 3d electrons.
∴Screening Constant (σ
(2 ×1.0) + (8×1.0) + (8×1.0) + (3×0.35)
 = 18.70
Ion 
Electronic Configuration
 Screening 
Constant
(σ)
Effective 
Nuclear Charge
(Z) 
V(II) 
Ion
  (1s)2
(2s, 2p)8
(3s,3p)8
(3d)3
(2×1.0)+(8×1.0)+(8×1.0)+(3×0.35)
= 19.05
Z=
(23-19.05)
= 3.95
Cu(II) - Ion
(1s)2
(2s, 2p)8
(3s,3p)8
(3d)9
(2×1.0)+(8×1.0)+(8×1.0)+(9×0.35)
= 21.15
Z=
(29-21.15)
= 7.85
Find out the Screening Constant of the 4S and 3d electron of Chromium Atom.
Chromium has atomic number 24 and the electronic configuration according to the Slater's Rule is :
(1S)² (2S, 2P)⁸ (3S, 3P)⁸ (3d)⁵ (4S)¹
∴ Screening Constant (σ) for the 4S electron is :
σ =(2×1.0) + (8×1.0) + (8×1.0) 
(3×0.85) (0×0.35) 
21.05.
And the Screening Constant (σ) for the 3d electron is :
σ =  (2×1.0) + (8×1.0) + 
(8×1.0) + (4×0.35) 
= 19.4.
Calculate the effective nuclear charge of the hydrogen atom.
Hydrogen atom has a single 1S valence electron. There is no other electron to screen it from the nuclear charge of single proton.
Thus, σ = 0 and Z* = 1.0 -0 = 1.0
Thus hydrogen electron sees the full nuclear charge of the nucleus, that is the electron is totally exposed to the proton.






Electron Affinity

Electron Affinity and its Variation:

Electron Affinity of an element is defined as the amount of energy released when an electron is added to a neutral gaseous atom in its lowest energy state that is, ground state to convert it into uni-negative gaseous ion. 
In simple words, the electron affinity of an atom is defined as the energy liberated when a gaseous atom captures an electron.
A
(Neutral Gaseous Atom)
+ e(Electron)
A-
(Gaseous Anion)
+ EA
Electron Affinity
 
Evidently electron affinity is equal in magnitude to the ionisation energy of the species formed. The definition of electron affinity as the exothermicity of the electron capture reaction would lead to the negative sign according to our usual thermodynamic convention. But electron affinity is normally describes with positive sign. 
Negative electron affinities are known. These negative values indicate that the species concerned does not like to have any more electron.
Measurement of Electron Affinity:
Electron affinities are difficult to obtain. These are often obtained from indirect measurements, by analysis of Born - Haber energy cycles in which one step is electron capture or by the direct study of electron capture from heated filaments.
Calculate the electron affinity of chlorine from the Born -  Haber cycle, 
given the following date : 
Lattice Energy = - 774 kJ mol-1
Ionization Potential of Na = 495 kJ mol-1
Heat of Sublimation of Na = 108 kJ mol-1
Energy for Bond Dissociation of chlorine (Cl2) = 240 kJ mol-1 and 
Heat of Formation of NaCl = 410 kJ mol-1.
Born - Haber Cycle for formation of NaCl (S) is:
Na+(g) + Cl-(g)
-UNaCl -INa +ECl
NaCl(g) Na(g) + Cl(g)
-ΔHf -SNa + -½DCl
Na+(g) + ½Cl2(g)
From the above Born - Haber cycle we can written as:
-UNaCl - INa + ECl - SNa - 1/2DCl - ΔHf = 0 
or, ECl = UNaCl + INa + SNa + 1/2DCl + ΔHf 
ECl = -774 + 495 + 108 + 120 + 410 
= 359 kJ mol-1
Factors Influencing the magnitude of Electron Affinity:
The magnitude of Electron Affinity (EA) is influenced by following factors such as,
1. Atomic Size.
2. Effective Nuclear Charge.
3. Electronic Configuration.
Atomic Size:
Larger the atomic radius lesser the tendency of the atom to attract the additional electron towards itself and lesser is the force of attraction exerted by the nucleus on the extra electron being added to the valence shell of the atom. 
Thus EA values decreases with increases atomic atomic radius.
Effective Nuclear Charge:
Higher the magnitude of effective nuclear charge (Z) towards the periphery of an atom greater is the tendency of the atom to attract the additional electron electron towards itself. Again greater the force of attraction exerted by the nucleus on the extra electron being added to the valence shell of the atom. As a result higher the energy released when extra electron is added to form an anion.
Thus the magnitude of Electron Affinity (EA) of  atom increases with increasing Z value.
 
 
Electronic Configuration:
The magnitude of electron affinity depends on the electronic configuration of the atom. The elements having nS2, nS2nP6, nS2nP6 valence shell configuration posses very low electron affinity value due to having their stable valence shell configuration.
(a) Electron Affinity of Be and Mg are -0.60 eV and - 0.30 eV respectively.
These elements posses filled S-Orbital in their valence shell which is stable hence Be and Mg do not prefer to except the additional electron in order to form anion. In such cases energy is required to form anion and consequently the electron affinity of Be and Mg posses negative sign with low magnitude.
From the above discussion it is clear that elements of Group - IIA posses low electron affinity vales with negative sign due to stable nS2 valence shell electronic configuration.
Atom Electron Affinity Atom Electron Affinity
H 0.747 S 2.07
F 3.45 Se 1.70
Cl 3.61 Li 0.54
Br 3.36 Na 0.74
I 3.06 K 0.70
O 1.47 Be -0.60
C 1.25 Mg -0.30
Si 1.63 B 0.20
N -0.10 Al 0.60
P 0.70 Zn -0.90
Account for the large decrees in electron affinity from lithium to beryllium despite the increase in nuclear charge.
Atomic number and the electronic distribution of lithium and beryllium are:
Li 3 1S22S1
Be 4 1S22S2
Lithium has an incompletely filled 2S sub-shell while beryllium has the sub shell filled. Thus lithium can receive an electron in its 2S sub-shell but for beryllium a still higher energy 2P level has to be made of. 
A filled shell or sub-shell leads to some extra stability. Hence beryllium resists an extra electron more than does lithium.
(b) The elements of Group - VA (group - 15) having nS2 nP3 valence shell configuration also posses low electron affinity values:
Why electron affinity of Nitrogen is - 0.10 eV while that of Phosphorus is + 0.70 eV ?
Atomic number and the electronic distribution of Nitrogen and Phosphorus are:
N 7 1S22S22P3
P 15 1S22S22P63S23P3
Due to smaller size of Nitrogen atom when extra electron is added to the stable half - filled 2P orbital some amount of energy is required and hence electron affinity of nitrogen is negative. 
On the other hand, due to bigger size of P atom in compare to nitrogen less amount of energy released when the extra electron is added to the stable half - filled 3P orbitals and thus electron affinity of Phosphorus is expressed with positive sign.
(c) Electron Affinity data on noble gases are not available.
Complete lack of evidence for stable noble gas anion is sufficient prof of their low electron affinity. An electron, even if accepted by a noble gas atom, would have to go into an orbital of next quantum shell and presumably the nuclear charge not high enough to hold an electron placed so far out.

Periodic Variations: 

In a Group:
In a group when move from top to bottom the size of atom generally increases with increasing atomic number and consequently the magnitude of electron affinity decreases in the same direction.
There are some exceptions to this general rule as is evident from the following examples:
(a) Although the elements of the second period of the periodic table are relatively smaller in size than those of the third period elements, but the electron affinity values of elements of second period is smaller than the electron affinity values of third period elements.
This unexpected behavior is explained by saying that the much smaller sizes of the second period elements give a very much higher value of charge densities for the respective negative ions. A high vales of electron density is opposes by the inter - electronic repulsion forces.
For example, electron affinity of fluorine is lower than that of Chlorine.
Thus the magnitude of electron affinity of the elements of group 17 decreases in the order:
ClFBrI
 
 
Electron affinity of Cl is greater than that of F - Explain.
The halogen possess large positive electron affinity values indicating their strong tendency to form anions. This is easily understandable because their electronic configurations are only one electron short of the next noble gas element.
The electron affinity of chlorine is greater than that of F. This is probably due to very small size of F atom which results in a very high charge density and consequently strong repulsion between existing valence shell electrons and the entering electron (new coming electron). On the other hand Cl being bigger size , charge density is small and thus such repulsion is not strong enough. Hence the electron affinity Cl is greater than that of F.
In a Period:
In a period, when we move from left to right Z value towards the new coming electron gradually increases with the increasing atomic number and hence the electron affinity gradually increases in the same direction.
Electron Affinity and its Variation along a Period and a Group.
Electron Affinities as Functions of Atomic Number
Exceptional cases may arise in case of the elements having stable nS², nS²nP³ and extraordinarily stable 1S², nS² nP⁶ valence shell configurations.
For example in case of the elements in 2nd period :
Be ,N and Ne the magnitude of electron affinity decreases with increasing atomic number.




Oxidation Number

Oxidation Number and 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)+]SO4 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)+]SO4 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.
Compound Formula Oxidation Number
Sugar C12H22O11 0
Glucose C6H12O6 0
Formaldehyde HCHO 0
Let, the oxidation number of carbon in Glucose (C12H22O11) is x
∴ 6x + 12(+1) + 6(-2) = 0
or, x = 0
Calculate the oxidation number of the element marked with blue 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 Ca -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:

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 Concept of Oxidation and Reduction.
Schematic Representation of Oxidation and Reduction.
| Oxidation
H2S + Cl2 2HCl + S
Reductant Oxidant Oxidant Reductant
| Reduction
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.
0 0 +2  -2
Mg + O2 2MgO
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
Na + Cl2 2NaCl
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.
| Oxidation Number Decreases
Reduction
SO2 + H2S 2H2O + 3S
| Oxidation Number Increases
Oxidation