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

Dec 31, 2018

Application of Dipole Moment

Application of Dipole Moment in Chemistry: 

Different applications are taken one after another.

1.Determination of Partial Ionic Character and Residual Charge on the Constituent Atoms in a Molecule:


Let us consider a molecule A - B having the dipole moment μobs and the bond length l cm. if the shared electron pair lies at the midpoint of the atoms, the bond will be purely covalent and the present of ionic character is zero. 
But if the bond is 100% ionic, and B is more electronegative then A, B will carry unit negative charge and A uni-positive charged respectively. In that case the dipole moment of the molecule would be,
μionic = e × 
= (4.8 × 10⁻¹⁰ esu cm
But the dipole moment of AB is neither zero or nor μionic .
How to Calculate Percent Ionic Character with Dipole Moment and Bond Length
Determination of Partial Ionic Character and Residual Charge

The displacement of HCl is 1.03 Debye and its bond length is 1.27 A. 
calculate the (a) charge on the constituent atom and 
(b) the % of the ionic Character of HCl.

Given, μobs = 1.03 Debye 
= 1.03 × 10⁻¹⁸ esu cm and length ℓ 
= 1.27 × 10 ⁻⁸ cm.
(a). Hence the charge on the constituent atom(q),
μobs/ = (1.03 × 10⁻¹⁸ )/( 1.27 × 10⁻⁸)
= 0.8 × 10⁻¹⁰ esu 
(b) The % of the ionic Character of the molecule,
(μobs/μionic ) × 100 =(1.03×10⁻¹⁸)/(4.8×10⁻¹⁰×1.27×10⁻⁸)×100 
= 16.89%

  • Determination of molecular radius from induced polarization (Pi) :


The induced polarization (Pi)
= (4/3) π N0 αi 
But when the substance is in the gaseous state, 
Pi = {(D0-1)/(D0+2)} M/ρ 
(for the covalent substance) 
The value of D൦ is close to unity under this condition, 
Hence, {(D0 - 1)/3} × 22400 
= ( 4/3 ) π N0 r³
At NTP, M/ρ = molar volume = 22400cc/mole and αi= r³ taking the spherical shape of the molecule. 
or, r³ = (22400/4πN0 ) (D0 - 1) 
2.94 ×10⁻²¹ (D൦ - 1) 
Hence, radius of the molecule r can be determined by measuring D൦ of the substance at NTP.

  • Determination of molecular structure :


(i) Mono-atomic Molecules (A) : 

The mono-atomic inert gases are non-polar and it indicates the symmetrical charge distribution in molecule. 

(ii) Di-atomic Molecules (AX) :

The homonucler diatomic molecules are largely non-polar such as N₂, O₂, Cl₂ indicating that there are symmetrical charge distributions. The bonding electron pair is equally shared by the two bonding atoms. However, Br₂, I₂ have non zero vale of dipole moment and this indicates the unsymmetrical charge distribution,
I+ I-
Compounds Dipole Moment Structure
H2 0 H H
Cl2 0 Cl Cl
Heteronucler diatomic molecules are always polar due to difference of electronegativity of the constituent atoms. The example are , HCl, HBr, HF etc. this indicates that electron pair is not equally shared and shifted to the more electronegative atom.
δ+ δ-
HCl 1.03 D H Cl
δ+ δ-
HBr 0.79 D H Br
δ+ δ-
HI 0.38 D H I
δ+ δ-
HF 2.00 D H F

The Difference between electronegativity of Carbon and Oxygen Is large but the dipole moments of carbon monoxide is very low - Why?

However, in CO, there are large difference of electronegativity between C and O but the molecule is very low value of dipole moment. This suggested that the charge density in O-atom is somehow back-donated to C-atom. This explain by forming a coordinate covalent bond directing towards C-atom.
How to Calculate Percent Ionic Character with Dipole Moment and Bond Length
Formation of Coordinate Covalent Bond in Carbon Monoxide

(iii) Tri-atomic Molecules (AX₂):

The molecules, like CO₂, BeCl₂, BeF₂, SnCl₂ etc have zero dipole moment indicating that the molecules have symmetrical linear structure. 
For example, CO₂ has structure, The electric moment of one C - O bond ( known as bond moment ) cancel the electric moment of the other C - O bond.
δ- δ+ δ-
O = C = O
δ- δ+ δ-
Cl = Be = Cl
The electric moment associated with the bond arising from difference of electronegativity is called the bond moment(m). In molecules, the vectorial addition of the bond moments (m) gives the resultant dipole moment(μ) of the molecule.
μ² = m12 + m22 + 2m1m2Cosθ
Where, m1 and m2 are the bond moments.
These helps to calculate the bond angle (θ) in the molecule. Thus in the carbon dioxide molecule, μ = 0 and m1 = m2 
hence, 0 = 2m²(1 + cosθ) 
or, θ = 180° 
that is the molecule is linear.
For Non-Linear Structure
Another type of molecules such as, H₂O, H₂S, SO₂᠌᠌ etc. have μ ≠ 0 indicating that they have non linear structure. The bond angle can be calculated from the bond moments of the molecules.

H₂O molecule has a dipole moment-Explain. Does it invalidate a linear structure ?

Bond Moment and Dipole Moment
Dipole Moment of H₂O and H₂S
For Water (H₂O) , μ = 1.84 D 
and mOH = 1.60 D 
Thus, μ² = 2 m² (1+ cosθ ) 
or, (1.84)² = 2 (1.60)² (1+ cosθ ) 
or, θ = 105° 
The contribution of non-bonding electrons towards the total dipole moment is included within the bond moment.

The bond angle in H₂S is 97° and dipole moment = 0.95 D . Find the bond moment of the S - H link. ( Given, Cos97° = -0.122)

We have, μ = 0.95 D and θ = 97° . 
From the equation, 
μ² = 2 m² (1+ cosθ ) 
Putting the value we have, 
(0.95)² = 2 m² (1 + cos97° ) 
here m = mS-H 
or, 0.9025 = 2 m² (1-0.122) 
or, m² = 0.9025/ (2 × 0.878) 
or, m² = 0.5139 
or, m = 0.72 
Thus the bond moment of the S - H link is 0.72 D

(iv) Tetra Atomic Molecules (AX₃) : 

The molecules like BCl₃, BF₃ etc have dipole moment zero indicating that they have regular planar structure. There halogen atoms are on a plane at the corner of the equilateral triangle and B atom is at the intersection of the molecules.
Bond Moment and Dipole Moment
Regular Planar Structure Of BF 

Show that the bond moment vectors of BF₃ molecule adds up to zero.

Bond Moment and Dipole Moment
Resultant Dipole Moment of BF
While other type of the molecule such as NH₃, PH₃, AsH₃ are polar (μ≠0) indicating that that the molecule have a pyramidal structure in which the three H-atoms are on the plane and N-atom at the apex of the pyramid in NH₃. 
But NF₃ has very small dipole moment though there is great difference of electronegativity between N and F atoms and similar structure of NH₃.
Bond Moment and Dipole Moment
Resultant Dipole Moment of NH and BF
Low value of μ of NF₃ is explain by the fact that resultant bond moment of the three N - F bonds is acting in opposite direction to that of the lone pair placed at the N-atom. But in NH₃, the resultant bond moment is acting in the same direction as that of the lone pair electrons.



Dec 23, 2018

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) + Electron Affinity (EA)

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:
How to determine Electron Affinity
Born - Haber Cycle Applied to NaCl
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 he 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, nS² nP³, nS² nP⁶ valence shell configuration posses very low electron affinity value due to having their stable valence shell configuration. 
Examples:
(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. 

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  : 1S² 2S¹
Be   : 4  : 1S² 2S²
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 ?

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.

Exceptions: 

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 :
Cl〉F〉Br〉I

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.

Exceptions:

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.
A plot of electron affinities of elements up-to Chlorine against atomic number shown as:
What is Electron Affinity
Electron affinities as Function of Atomic Numbers






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.