May 2019

Definition of acids and bases

Acids and bases posses in a sense some opposite properties. Any of the classes of substances whose aqueous solutions are characterized by the sour taste, the ability to turns blue litmus, and the ability to react with bases and certain metals to form salts.

In chemistry, a base is a substance that in aqueous solution is slippery to touch, that's bitter, changed the color of the indicator, (litmus to blue) reacts with an acid to form salts.

The concepts of acid and base are developed one after another tend to make the definitions more and more broad-based. We need to familiar with the following five concepts :
  1. Arrhenius concept.
  2. Solvent system concept.
  3. Protonic concept
  4. Lewis concept

Arrhenius concept for acid-base neutralization

Arrhenius was one of the early exponents of electrolytic dissociation theory to define acids and bases. His classification of acids and bases was based on the theory that acids when dissolved in water, dissociate hydrogen ions and anions.
Bases when dissolved in water dissociate into hydroxyl ions and cations.

Thus sodium hydroxide neutralizes hydrochloric acid can be represented by a reaction involving the combination of H⁺ and OH⁻ ions to form H₂O.

HCl → H⁺ + Cl⁻

NaOH → Na⁺ + OH⁻

H⁺ + OH⁻ ⇆ H₂O

What is neutralization reaction?

Arrhenius concept of acids and bases was highly useful in explaining the acid-base neutralization process in an aqueous medium.

The heat liberated during neutralization HCl, HClO₄, HNO₃, HBr, HI, and H₂SO₄) and strong base (NaOH, KOH, RbOH, and Ca(OH)₂) is the same, namely 13.4 kcal/mole (56 KJ/mole)
This indicating that the ion participating in the reaction must be the same (that is H⁺ and OH⁻) and that other ions (Na⁺, K⁺, Rb⁺, Ca⁺², Cl⁻, Br⁻, I⁻, ClO₄⁻ or SO₄⁻²) take no part. This can be explained as follows.

HA(aq) + BOH(aq) ⇆ BA(aq) + H₂O

Where HA and BOH are strong electrolytes and completely dissociate in the solution.

(H⁺+A⁻)+( B⁺+OH⁻)⇆(B⁺+A⁻)+ H₂O

Canceling likes ions, we have,

H⁺ + OH⁻ ⇆ H₂O

Since the neutralization reaction of all strong acids and bases used to the formation of 1-mole water from H⁺ ion and OH⁻ ion. The enthalpy change (ΔH) will also be the same.

Limitations of Arrhenius concept

  1. According to this concept, HCl is an acid only when dissolved in water and not in some other solvent such as benzene or when it exists in the gaseous state.
  2. It cannot account for the acidic and basic character of the materials in non-aqueous solvents, as for example, NH₄NO₃, in liquid ammonia is an acid, though it does not give H⁺ ions. Similarly, many organic materials in NH₃, which does not give OH⁻ ions at all, are actually known to show basic character.
  3. The neutralization process limited to those reactions which can occur in aqueous solutions only, although the reactions involving salt formation do occur in many other solvents and even in the absence of solvents.
  4. It cannot explain the acidic character of certain salts such as AlCl₃ in aqueous solution.

Solvent system concept for acid-base neutralization

Protic solvent molecules may also dissociate into two oppositely charged ions. We consider protic solvent water, its characteristic cation and anion are H⁺ and OH⁻.

H₂O ⇆ H⁺ + OH⁻

Since we know that a bare proton will readily polarize other anions or molecules we write an H⁺ as H₃O⁺ indicating that it is a solvated proton that exists in the solution. So that the overall dissociation of solvent will be:

H₂O + H₂O ⇆ H₃O⁺ + OH⁻

Thus all those compounds which can give H₃O⁺ ions in H₂O will act as acids and all those compounds which can give OH⁻ ions in H₂O will act as bases.

NH₃ + NH₃ ⇆ NH₄⁺ + NH₂⁻

Thus those compounds which give NH₄⁺ ions in liquid NH₃ will act as acids and all those compounds which can give NH₂⁻ ions in liquid NH₃ acts as bases.

Thus the dissociation (or autoionization) of non-aqueous solvents is directly responsible for the nature of the chemical reactions that can be initiated in such solvents.

According to the solvent system concept, An acid is a substance which by dissociation in the solvent forms the same cation as does the solvent itself due to auto-ionization.
A base is one that, gives on dissociation in the solvent the same anion as does the solvent itself on its ionization.

Auto-ionization of solvents

Auto-ionization of some protonic and non-protonic solvents are given below

Dissociation of a protic solvent

H₂O + H₂O H₃O⁺ + OH⁻
Acid Base Acid Base

NH₃ + NH₃ ⇆ NH₄⁺ + NH₂⁻

CH₃COOH + CH₃COOH ⇆ CH₃COOH₂⁺ + CH₃COO⁻

Dissociation of an aprotic solvent

SO₂ + SO₂ SO⁺² + SO₃⁻²
Acid Base Acid Base

BrF₃ + BrF₃ ⇆ BrF₂⁺ + BrF₄⁻

N₂O₄ + N₂O₄ ⇆ 2NO⁺ + 2NO₃⁻

Just as with the Arrhenius definition, neutralization reaction between an acid and a base to produce salt and solvent. Neutralization of some non-aqueous solvents are,

Neutralization process for liquid NH₃

Solvent: NH₃ + NH₃ ⇆ NH₄⁺ + NH₂⁻

Acid: NH₄Cl ⇆ NH₄⁺ + Cl⁻

Base: KNH₂ ⇆ K⁺ + NH₂⁻

Acid-base neutralization reaction

NH₄Cl + KNH₂ KCl + 2NH₃
Acid Base Salt Solvent

Neutralization process for liquid SO₂

Solvent: SO₂ + SO₂ ⇆ SO⁺² + SO₃⁻²

Acid: SOBr₂ ⇆ SO⁺² + 2Br⁻

Base: K₂SO₃ ⇆ 2K⁺ + SO₃⁻²

Acid-base neutralization reaction

SOBr₂ + K₂SO₃ 2KBr + 2SO₂
Acid Base Salt Solvent

Acids-bases neutralization reactions
Acid-base neutralization

Utility of the solvent system concept

Evidently, this concept of the solvent system can be used to explain the acid-base neutralization reactions occurring in protic and aprotic solvents(protonic or non-protonic both).
Thus this theory can simply be said to be an extension of the Arrhenius theory.

Solvent system concept limitations for acids and bases

  1. This theory does not consider a number of acid-base reactions included in the protonic definition.
  2. It limits acid-base phenomena to the solvent system only. Thus it does not explain the acid-base reactions which may occur in the absence of solvent.
  3. It can not explain acid-base neutralization reactions occurring without the presence of ions.

Particle and wave nature of an electron

In 1924 de Broglie's relation pointed out that just as a light electron also has both particle and wave nature. 
    According to de Broglie, this dual nature - wave and particle - should not be confined to radiations alone but should also be extended to matter.
    He suggested that electrons travel in waves, analogous to light waves. His idea could be fitted to drive the same relation that Bohr arrived at from his particle treatment of electrons.

Momentum and wavelength of a particle

de Broglie relation
de Broglie relation
    de Broglie proposed a relation between momentum and wavelength of a particle in motion. He considered the light of frequency ν, the energy is given by
E = hν = h (c/λ)
    Where λ = wavelength, c = velocity of light, and h = Plank constant = 6.627× 10⁻²⁷.
    Again from the famous mass-energy equivalence relation from Einstien,
E = m c²

The momentum of the photon is
p = mv = mc
    Combining these two relations, we have
mc = h/λ
or, p = h/λ
∴ λ = h/p
    de Broglie extended this relationship to the dynamics of a particle and proposed that a wavelength λ is associated with a moving particle and is related to its momentum as
λ = h/p = h/mv
    Where m is the total mass of the particle and v is its velocity.

Bohr's model and de Broglie relation

    Bohr's model for the angular momentum of n-th orbital of moving electrons
mvr = n (h/2π)
or, mv = n (h/2πm)
    where m = mass of an electron, n = 1, 2, 3, 4, ..., and r = radius of the orbital of an atom.
    Again according to de Broglie relation
λ = h/mv
or, mv = h/λ
    where λ = wavelength of the moving electron.
    Combining these two relations
nh/2πr = h/λ
or, 2πr = n λ
    Thus a standing produces a stationary pattern, its profile being fixed within the space allowed to it. It does not travel beyond the allowed space.

Electron diffraction experiment

    de Broglie's suggestion of matter waves and its confirmation by Davisson and Germer's electron diffraction experiment conclusively proves that electrons are not is not an ordinary particle.
    From the evidence by the experiment of determination of mass and e/m electron has particle nature.
    Electron diffraction experiment by Davison and Gramer's given the evidence of the wave nature of the electron.

The kinetic energy of an electron

    If the particle is an electron and if it is subjected to the potential difference V so as to acquires a velocity v, then,
Kinetic energy of an electron = Ve = ½ mv²
where e is the charge of an electron.

Again from the de Broglie relation
λ = h/mv

From these two relations
λ = h/√(2mVe)

How to find oxidation number?
Oxidation number

    Oxidation-reduction reaction

    Examples, the formation of water from hydrogen and oxygen, can not be covered by the electronic concept since water is not an ionic compound.
2H₂ + O₂ → 2H₂O
    It may recall classically we could still say that hydrogen is oxidized to H₂O. In the same seance burning of magnesium in oxygen is considered oxidation. Similarly, hydrogen and chlorine react to form a covalent molecule hydrogen chloride.
    2H₂ + Cl₂ → 2HCl
    The above reaction hydrogen is oxidized or chlorine is reduced but the resulting compound is covalent one, thus the reaction cannot 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.
    The electronegativity concept is utilized in adducing a formal charge to an atom. The less electronegative of the partners of a binary compound is arbitrarily 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 they have been assigned positive oxidation numbers.

How to find oxidation number?

    The following general rules are to be observed for the assignment of oxidation numbers.
    Atoms of diatomic molecules like H₂, Cl₂, O₂, etc or of metallic elements like Zn, Cu, Na, etc are assigned zero oxidation numbers since the 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 the same electronegativity is involved for bonding.
    Except for metal hydrides, the oxidation number of hydrogen is +1. In alkali metal hydrides, LiH, NaH, CsH, etc, the oxidation number of hydrogen is -1.
NaH → Na⁺ + H⁻
(Here oxidation number of H is -1)
HCl → H⁺ + Cl⁻
(Here oxidation number of H is +1)
    The oxidation number of metal is positive.
CuO → Cu⁺² + O⁻²
(Here oxidation number of Cu is +2)
    Oxygen has normally an oxidation number -2.
CuO → Cu⁺² + O⁻²
here the oxidation number of O is -2
    Peroxide (H₂O₂, Na₂O₂), the oxidation number of oxygen is -1 since hydrogen has to be assigned +1.
    H₂O, the oxidation number of oxygen is -2, but in H₂O₂, the oxidation number of H is +1 and the oxidation number of oxygen is -1.
    Na₂O₂, the oxidation number of Na is +1 and the oxidation number of oxygen is -1.
    BaO₂, the oxidation number of oxygen is -1 because the oxidation number of Ba is +2.
    Fluorine monoxide(F₂O) oxygen has an oxidation number +2 because fluorine is more electronegative than oxygen.
    Superoxides (KO₂), the oxidation number of oxygen is -1/2.
    The oxidation number of an ion is equal to its charge.
NaCl → Na⁺ + Cl⁻
    The charge and oxidation numbers of Na⁺ and Cl⁻ are +1 and -1 respectively.
MgBr₂ → Mg⁺² + 2Br⁻
    Here the charge and oxidation number of Mg⁺² and Br⁻ are +2 and -1 respectively.
    The algebraic sum of the oxidation numbers of all the atoms in a compound must be zero and the oxidation number of many atomic ions equal to its charge.
    In HCl oxidation number of hydrogen is +1 and the oxidation number of chlorine is -1. And the sum of these = (+1) + (-1) = 0.
    In the MnO₄⁻ ion sum of the oxidation number of Mn and oxygen equal to -1.

Oxidation number of elements

Manganese in potassium permanganate

    Let the oxidation number of Mn in KMnO₄ is x. Thus according to the above rule,
(+1) + x + 4(-2) = 0
or, x = +7
Oxidation number of Mn in KMnO₄ is +7

Manganese in manganate ion

    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
Oxidation number of Mn in MnO₄⁻² is +6

Chromium in dichromate ion

Let the oxidation number of Cr in Cr₂O₇⁻² is x
∴ 2x + 7(-2) = -2
or, x = +6
Oxidation number of Cr in Cr₂O₇⁻² is +6

Sulfur in sulphuric acid

    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
Oxidation number of S in H₂SO₄ is +6

Oxidation state of carbon 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)
Oxidation number of C in CH₃COCH₃ is 4/3

Oxidation number of P in H₄P₂O₇

Let the oxidation number of P in H₄P₂O₇ is x.
∴ 4(+1) + 2x + 7(-2) = 0
or, x = +5
Oxidation number of P in H₄P₂O₇ is +5

Oxidation state of iron in Fe(CO)₅

The oxidation number CO is zero.
The oxidation number of Fe also zero.

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
Oxidation number of Cr in [Cr(NH₃)₆]Cl₃ is +3

Oxidation number of an element in a compound is zero

    Some organic compounds where the oxidation number of carbon on this compound is zero.
Let, the oxidation number of carbon in Glucose (C₁₂H₂₂O₁₁) is x.
∴ 6x + 12(+1) + 6(-2) = 0
or, x = 0
Compound Formula Oxidation Number
Sugar C₁₂H₂₂O₁₁ 0
Glucose C₆H₁₂O₆ 0
Formaldehyde HCHO 0

Questions answers

Question
    Oxidation number of P in Ba(H₂PO₂)₂ is - (a)+3, (b)+2, (c) +1, (d) -1.
Answer
    The oxidation number of Ba is +2, the oxidation number of hydrogen is +1 and the 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

Question
    Calculate the oxidation number of Iron in [Fe(H₂O)₅(NO)⁺]SO₄.
Answer
    H₂O is neutral thus the oxidation number is zero, the 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.
or, x - 1 = 0
or, x = +1
Thus, the oxidation number of Fe in [Fe(H₂O)₅(NO)⁺]SO₄ is +1.
Question
    What is the Oxidation state of chromium in CrO₅?
Answer
Oxidation number of CrO5
Structure of CrO5
    Due to the peroxy linkage oxidation state of Cr in CrO₅ is +6.

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(g) + Electron→A- (g) + Electron Affinity
    Evidently, electron affinity is equal in magnitude to the ionization 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 described with a 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.

Factors Influencing Electron Affinity

    The magnitude of Electron Affinity (EA) is influenced by the following factors such as,
  1. Atomic size.
  2. Effective nuclear charge.
  3. Electronic configuration.

Electron affinity vs 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 Electron affinity values decrease with increases atomic radius.

Electron affinity vs effective nuclear charge

    Higher the magnitude of effective nuclear charge (Zeff) towards the periphery of an atom greater is the tendency of the atom to attract the additional 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 energy released when an extra electron is added to form an anion. Thus the magnitude of Electron Affinity of an atom increases with increasing Zeff value.

Electron affinity vs electronic configuration

    The magnitude of electron affinity depends on the electronic configuration of the atom. The elements having,
    nS², nS², nP⁶, nS², nP⁶
    valence shell configuration posses very low electron affinity value due to having their stable valence shell configuration.
Electron Affinity
Electron affinity of different elements

Electron affinity trend

Electron affinity across 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:
    Although the elements of the second period of the periodic table are relatively smaller in size than those of the third-period elements, the electron affinity values of elements of the second period are 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 value of electron density is opposed by the interelectronic repulsion forces.
    Electron Affinity of fluorine is lower than that of Chlorine. The lower values of electron affinity for Florine due to the electron-electron repulsion in relatively compact 2P-Orbital of Florine atom. Thus the electron affinity values of halogens are:
F Cl Br I
- 3.6 eV - 3.8 eV - 3.5 eV - 3.2 eV

Electron affinity across a period

    In a period, when we move from left to right Zeff value towards the new coming electron gradually increases with the increasing atomic number and hence the electron affinity gradually increases in the same direction.
    A plot of electron affinities of elements up-to chlorine against atomic number shown as,
Electron affinities as functions of atomic number
Electron affinities as functions of atomic number

Electron affinity questions answers

Question
    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.
Answer
    Born - Haber Cycle for the formation of NaCl (S) is:
Born Haber Cycle for the calculation of electron affinity
Born Haber Cycle
From the above Born - Haber cycle we can write as,

-UNaCl - INa + ECl - SNa - ½DCl - ΔHf = 0

or, ECl = UNaCl+INa+ SNa +½DCl + ΔHf

∴ ECl = -774 + 495 + 108 + 120 + 410

= 359 kJ mol-1
Question
    Account for the large decrees in electron affinity from lithium to beryllium despite the increase in nuclear charge.
Answer
    Atomic number and the electronic distribution of lithium and beryllium are:
Li 3 1S² 2S¹

Be 4 1S²2S²
    Lithium has an incompletely filled 2S sub-shell while beryllium has the subshell 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.
Question
    Why electron affinity of Nitrogen is - 0.10 eV while that of Phosphorus is + 0.70 eV?
Answer
    Atomic number and the electronic configuration of Nitrogen and Phosphorus are
N 7 1S²2S²2P³

P 15 1S²2S²2P⁶3S²3P³
    Due to the smaller size of Nitrogen atom when an extra electron is added to the stable half-filled 2Porbital some amount of energy is required and hence the electron affinity of nitrogen is negative.
    On the other hand, due to the 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 a positive sign.
Question
    Explain why the electron affinity of chlorine is more than fluorine?
Answer
    The halogen possesses large 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 the 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 a 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.
Question
    Why electron gain enthalpy of Be and Mg are almost zero?
Answer
    Be and Mg have their electron affinity values equal to almost zero. Since Be and Mg have completely filled S orbitals.
Be 4 1S² 2S²

Mg 12 1S² 2S² 2P⁶ 3S²
    The additional electron will be entering the 2P-orbital in the case of Be and 3P-orbitals in the case of Mg which are of considerably higher energy than the 2S- Orbitals respectively.
Question
    Why the electron affinity of noble gases is zero?
Answer
    Inert gases in which the nS and nP orbitals are completely filled (nS² nP⁶ configuration) the incoming electron must go into an electron shall have the larger values for the principal quantum number, n. Thus inert gas has its electron affinity values equal to zero.

What is shielding effect?

Valence electrons for a multi-electron atom are attracted by the nucleus of an atom and repelled by the electrons from inner-shells. Slater's rules used for calculating shielding.
Shielding effect and Slater's rules
Shielding electrons and Slater's rules
The combined effect of this attractive and repulsive force acting on the valence electrons is that the Valence electrons experience less attraction from the nucleus of an atom. This is known as the shielding effect.

Slater's set some empirical rules to calculate the shielding or screening 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 an effective nuclear charge (Zeff).

Shielding electrons in nS, nP level

Slater's rules are applicable for calculating screening constant and Z effective of an electron in atom, ions or molecule and the rules are
  1. The first to do is to write out the electronic configuration of elements of the atom or 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 shielding effect is concerned the S and P electron belonging to the same principal quantum shell have the same effect as advocated by Slater.
    (1S)²(2S, 2P)⁸(3S)¹
  2. Electron in a certain nS, nP level is screened only by electrons at the same level and by the electrons of lower energy levels of an atom.
    Electrons lying above nS, nP level do not screen any nS, nP electron to any extent.
  3. Electrons of higher energy levels of an atom no shielding effect on any lower-energy level of an atom.
    Thus for calculating the screening constant of the valence electron of a sodium atom, the electronic configuration of the sodium atom is
    (1S)² (2S, 2P)⁸ (3S)¹
    Calculation of the screening constant, the valence electron of an atom will be excluded from our calculation.
    Thus for calculating the screening constant of the 2P electron of the sodium atom, one 2P electron, and one 3S electron will be excluded from our calculation.
  4. Electrons lying nS and nP atomic level shield a valence electron in the same group by 0.35 each. This is also true for the electrons of the nd or nf atomic level of atom or ion.
  5. Electrons belonging to one lower quantum shell, that is (n-1) shell shield the valence electron by 0.85 each.
  6. Electrons belonging to (n-2) or still lower quantum shell shields the valence electron by 1.0 each. This means the shielding effect is complete.
Shielding effect in nS, nP subshell of an atom
Shielding effect in nS, nP subshell

Effective nuclear charge of sodium atom

Screening constant (σ) = (2 × 1) + (8 × 0.85) + (0 × 0.35)
= 8.8.

∴ Z effective of sodium atom = (11 - 8.8)
= 2.2.

Question
Calculate the effective nuclear charge of the hydrogen atom.

Answer
The hydrogen atom has a single 1S valence electron. There is no other electron to screen it from the nuclear charge of a single proton.
Thus, σ = 0 and Zeff = 1.0 - 0 = 1.0.

Thus hydrogen electron sees the full nuclear charge of the nucleus of an atom, that is the electron is totally exposed to the proton.

Sodium, potassium, and magnesium ion

Electronic configuration of sodium (Na⁺) ion
(1S)²(2S, 2P)⁸
Screening constant of sodium ion
= (2 × 0.85) + (8 × 0.35)
= 4.5.
Effective nuclear charge of sodium ion
= (11 - 4.5)
= 4.5.

Electronic configuration of potassium ion
(1S)²(2S, 2P)⁸(3S, 3P)⁸
Screening constant (σ) for potassium ion = (2× 1) + (8 × 0.85) + (8 × 0.35)
= 11.6.
Effective nuclear charge of potassium ion
= (19 - 11.6)
= 7.40.

Electronic configuration of magnesium (Mg⁺²) ion
(1S)²(2S, 2P)⁸
Screening constant of magnesium ion
= (2 × 0.85) + (8 × 0.85)
= 4.50.
Effective nuclear charge of Magnesium ion
= (12 - 4.50)
= 7.50.

Valence electron of fluorine and fluoride ion

Electronic configuration of fluorine atom
(1S)²(2S, 2P)⁷

Screening constant (σ) of fluorine atom
= (2 × 0.85) + (6 × 0.35)
= 3.80.

Effective nuclear charge for valence electron of fluorine
= (9 - 3.8)
= 5.20.

Electronic configuration of fluoride ion
(1S)²(2S, 2P)⁸

Screening constant (σ) of  fluoride ion
= (2 × 0.85) + (8 × 0.35)
= 4.50.

Effective nuclear charge of fluoride ion
= (9 - 4.50)
= 4.50.
Question
Comment on the variation in effective nuclear charge for a 2P electron from carbon to oxygen.

Answer
Electronic distribution according to the Slater's rule is:
Carbon (1S)² (2S, 2P)⁴
Nitrogen (1S)² (2S, 2P)⁵
Oxygen (1S)² (2S, 2P)⁶
In carbon, the 2P electron is screened by 1S² 2S² 2P¹ electrons while in nitrogen and oxygen this is done by 1S² 2S² 2P² and 1S² 2S² 2P³ electrons respectively.

Zeff of nitrogen = Zeff of carbon + (1 nuclear charge) - shielding due to one 2P electron
Zeff of nitrogen = Zeff of carbon + 1 - 0.35
= Zeff of carbon + 0.65 and Zeff of oxygen
= Zeff of nitrogen + 0.65

Thus effective nuclear charge will go up by the same amount from carbon to nitrogen and then to oxygen.

Shielding electrons in nd, nf level

The above rules are quite well for estimating the screening constant of S and P orbitals. However when d subshell or f subshell is being shielded the four and five rule replaced by rules for estimation of screening constant.
The replaced rules are, all electrons below the nd subshell or nf subshell contribute 1.0 each towards the screening constant.
Shielding effect in nd, nf subshell of an atom
Shielding effect in nd, nf subshell

4S electron of vanadium and 3d electron of vanadium

Vanadium has atomic number 23 and the electronic configuration according to the Slater's rules is,
(1S)² (2S, 2P)⁸ (3S, 3P)⁸ (3d)3 (4S)²

∴ Screening constant (σ) for 4S electron of vanadium
= (2 ×1.0) + (8×1.0) + (8×0.85) + (3×0.85) + (1×0.35)
= 19.7.

Effective nuclear charge for 4S electron of vanadium
(23 - 19.7)
= 3.3.

Screening constant (σ) for 3d electron of vanadium
= (2 ×1.0) + (8×1.0) + (8×1.0) + (2 ×0.35)
= 18.70.

Effective nuclear charge for 3d electron of vanadium
= (23 - 18.70)
= 4.30

d subshell of vanadium(II) ion

Electronic configuration of vanadium ion
(1S)²(2S, 2P)⁸(3S, 3P)⁸(3d)³

Screening constant (σ)  of vanadium(II) ion
= (2 ×1.0) + (8×1.0) +(8×1.0) + (3×0.35)
= 19.05.

Effective nuclear charge of vanadium(II) ion
= (23 - 19.05)
= 3.95.

4S and 3d energy level of an atom

In the first transition series electron filling up process begins in the 3d level below a filled 4S level. During the ionization process, 4S electron will be lost first. This can be explained with the reference of chromium.
Chromium has atomic number 24 and the electronic configuration
(1S)² (2S, 2P)⁸ (3S, 3P)⁸ (3d)⁵ (4S)¹

∴ Screening constant (σ) for 4S electron of chromium
= (2×1.0) + (8×1.0) + (8×0.85) + (5×0.85) + (0×0.35)
= 21.05.

Effective nuclear charge for 4S electron of chromium atom
= (24 - 21.05)
= 2.95.

∴ Screening constant (σ) for 3d electron of chromium atom
= (2×1.0) + (8×1.0) + (8×1.0) + (3×0.85) + (4×0.35)
= 19.40

Effective nuclear charge for 4S electron of chromium atom
= (24 - 19.40)
= 4.60.
Shielding of 3d subshell electron is lower than 4S subshell. Thus for the first transition series, 3d electron is more tightly held than 4S electron.
Hence during the ionization 4S subshell lost electron in the 3d subshell.

Popular Posts

Learn organic chemistry

Contact us

Name

Email *

Message *

Powered by Blogger.