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

Acid-base neutralization reaction

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 hydrogen and hydroxyl ions to form water.

HCl → H⁺ + Cl⁻

NaOH → Na⁺ + OH⁻

H⁺ + OH⁻ ⇆ H₂O

What is an example of an acid-base 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 acid and base

  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 of acids and bases

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.

Autoionization of solvents

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

Protic solvent examples

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

NH₃ + NH₃ ⇆ NH₄⁺ + NH₂⁻

CH₃COOH + CH₃COOH ⇆ CH₃COOH₂⁺ + CH₃COO⁻

Aprotic solvent examples

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,

Neutralize process for liquid ammonia

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

Autoionization of 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
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Acid base neutralization reaction example
Acid-base neutralization

Uses of solvent system theory

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.

Limitations of solvent system theory

  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.

de Broglie relation in chemistry

Classical mechanics of particles, wave motion, and laws of energy had been the main tools for study and explanation of the different physical phenomena up to almost the advent of the twentieth century.

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, protons, atoms could not be regarded simply corpuscles. The periodicity of wave motion must also be assigned to them.

de Broglie was led to this hypothesis from the consideration of quantum theory and the special theory of relativity. He also proposed a relation between momentum and wavelength of a particle in motion.

Particle and wave nature of radiation

de Broglie relation from classical and wave mechanics
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 of this light is E.

∴ E = hν = h (c/λ)
where λ = wavelength of light
c = velocity of light
h = Plank constant = 6.627× 10⁻²⁷.

Theory of relativity from Einstien
E = m c².

Combining these two relations,
mc = h/λ
or, p = h/λ
where p = mc = mv = momentum.

∴ λ = h/p.

de Broglie extended this relationship to the dynamics of a particle and proposed that a wavelength associated with a moving particle and momentum of the particle.

λ = h/p = h/mv
where m = total mass of the particle
v = velocity.

Bohr's model and de Broglie relationship

Angular momentum of moving electrons in Bohr's model

mvr = n (h/2π)
or, mv = n (h/2πm).

where m = mass of an electron,
n = principle quantum number = 1, 2, 3, 4, ...
r = radius of the orbital of an atom.

According to de Broglie

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.
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  1. Chemical equilibrium
  2. Law of mass action
  3. Chemical kinetics
  4. Properties of gases
  5. Heat capacity of gases

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.

How to find 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.
The kinetic energy of an electron

= Ve = ½ mv².
where e is the charge of an electron.

From the de Broglie relation
λ = h/mv.

∴ λ = h/√(2mVe).

An oxidizing and reducing agent and oxidation number

The study oxidation number of an element used to specify some oxidation and reduction reactions and definitions of oxidation and reduction based on loss or gain of electrons limited to the scope. This definition holds goods for ionic compounds.

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

Classically we could still say that hydrogen oxidized to water. In the same seance burning of magnesium in oxygen considered oxidation. Similarly, hydrogen and chlorine react to form a covalent molecule hydrogen chloride.

2H₂ + Cl₂ → 2HCl

Formation of hydrochloric acid, hydrogen oxidized or chlorine reduced but the resulting compound covalent one, the reaction cannot be covered by the electronic concept. To cover such reactions also under oxidation and reduction, the concept of oxidation number developed.

What is the oxidation number of an element?

How to calculate oxidation number of an element in a compound
Oxidation number of an element
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 arbitrary.

The electronegativity concept utilized in adducing a formal charge to an atom. The less electronegative of the partners of a binary compound 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 calculate the oxidation number of an element?

The following general rules are to be observed for the assignment of oxidation numbers.
  1. Atoms of diatomic molecules like hydrogen, chlorine, oxygen, etc or of metallic elements like zinc, copper, sodium, etc are assigned zero oxidation numbers since the same elements of similar electronegativity are involved in the bonding.
  2. Except for metal hydrides, the oxidation number of hydrogen +1. In alkali metal hydrides, lithium hydride, sodium hydride, cesium hydride, etc, the oxidation number of hydrogen -1.
  3. The oxidation number of metal positive.
  4. Oxygen has normally an oxidation number -2. In peroxide and superoxides the oxidation number of oxygen -1 and -1/2 respectively.
  5. The oxidation number of an ion equal to its charge.
  6. 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.

Hydrogen, alkali metal hydride, hydrochloric acid

H ➖H

In hydrogen molecules, two hydrogen atom of the same electronegativity involved for bonding. Thus the oxidation number of the hydrogen molecule zero.

NaH → Na⁺ + H⁻

Alkali metal always possesses a positive oxidation number. In sodium hydride, the oxidation number of sodium and hydrogen +1 and -1 respectively.

HCl → H⁺ + Cl⁻

In hydrochloric acid oxidation number of hydrogen and chlorine +1 and -1 respectively.

Metal oxide, peroxide, monoxide, superoxide of oxygen

CuO → Cu⁺² + O⁻²
In cupric oxide oxidation number of copper and oxygen +2 and -2 respectively.

Hydrogen peroxide (H₂O₂) sodium peroxide (Na₂O₂), the oxidation number of oxygen -1 since hydrogen has to be assigned +1.

The oxidation number of oxygen in water -2, but in hydrogen peroxide, hydrogen +1 state, and oxygen -1 state.

Sodium +1 state in sodium peroxide and the oxidation number of oxygen -1.

Barium +2 state in barium peroxide, the oxidation number of oxygen -1.

Fluorine is more electronegative than oxygen and the oxidation number of fluorine and oxygen in fluorine monoxide -1 and +1 respectively.

Potassium superoxides (KO₂), the oxidation number of oxygen -1/2.

Sodium and magnesium halide

NaCl → Na⁺ + Cl⁻

Sodium and chloride ion charge and oxidation number +1 and -1 respectively.

MgBr₂ → Mg⁺² + 2Br⁻

Here the charge and oxidation number of magnesium and bromide ion are +2 and -1 respectively.
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The oxidation number of compound and ion

In hydrochloric acid oxidation number of hydrogen +1 and the oxidation number of chlorine -1. And the sum of these = (+1) + (-1) = 0.

In the MnO₄⁻ ion sum of the oxidation number of manganese and oxygen equal to -1.

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

Answer
The oxidation number of barium +2, the oxidation number of hydrogen +1 and the oxidation number of oxygen -2.
Let the oxidation number of P x.
∴ (+ 2) + 2{2(+1) + x +2(-2)} = 0
or, 2x - 2 = 0
or, x = +1

Manganese in potassium permanganate

Let the oxidation number of manganese in permanganate x. Thus according to the above rule,
(+1) + x + 4(-2) = 0
or, x = +7
The oxidation number of manganese in permanganate +7

Manganese in manganate ion

Let the oxidation number of manganese in manganate ion x and the oxidation number of oxygen -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
The oxidation number of manganese in manganate ion +6

Chromium in dichromate ion

Let the oxidation number of chromium in Cr₂O₇⁻² x
∴ 2x + 7(-2) = -2
or, x = +6
The oxidation number of chromium in dichromate +6

Sulfur in sulphuric acid

Let the oxidation number of sulfur in sulfuric acid x. According to the rule oxidation number of hydrogen +1 and oxygen -2.

∴ 2(+1) + x + 4(-2) = 0
or, x = +6
The oxidation number of sulfur in sulfuric acid +6

The oxidation number of carbon

Let the oxidation number of carbon in CH₃COCH₃ x. The oxidation number of hydrogen and oxygen +1 and -2 respectively.

∴ 3x + 6(+1) + (-2) = 0
or, x = -(4/3)
The oxidation number of carbon in CH₃COCH₃ 4/3

Phosphorus in H₄P₂O₇

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

The oxidation state of iron in Fe(CO)₅

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

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

Answer
Water is neutral thus the oxidation number zero, the oxidation number of (NO)⁺ +1 and the oxidation number of sulfate ion -2.

Let the oxidation number of iron in [Fe(H₂O)₅(NO)⁺]SO₄ x.
or, x - 1 = 0
or, x = +1
The oxidation number of iron in [Fe(H₂O)₅(NO)⁺]SO₄ +1.

Chromium in [Cr(NH₃)₆]Cl₃ complex

Let the oxidation number of Cr in [Cr(NH₃)₆]Cl₃ x. Ammonia neutral thus the oxidation number zero and the oxidation number of chlorine -1.

∴ x + 0 +3(-1) = 0
or, x = +3
The oxidation number of chromium in [Cr(NH₃)₆]Cl₃ +3

Question
What is the oxidation number of chromium in CrO₅?

Answer
structure and oxidation number of CrO5
Structure of CrO5
Due to the peroxy linkage oxidation number of chromium in CrO₅ +6.

Organic compounds oxidation number zero

Some organic compounds where the oxidation number of carbon on this compound zero.
Let, the oxidation number of carbon in glucose (C₁₂H₂₂O₁₁) x.
∴ 6x + 12(+1) + 6(-2) = 0
or, x = 0

Compound Formula Oxidation Number
Sugar C₁₂H₂₂O₁₁ 0
Glucose C₆H₁₂O₆ 0
Formaldehyde HCHO 0

The energy released for electron capture reaction

Electron affinity of an element defined as the amount of energy released when an electron is added to a neutral gaseous atom in its lowest energy state or ground state to convert it into uni-negative gaseous ion.

Simply, the electron affinity of an atom defined as the energy liberated when a gaseous atom captures an electron.
A(g) + electron → A- (g) + EA

Energy release for electron capture reaction 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 measure. EA 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.

Question
Calculate the electron affinity of chlorine from the Born - Haber cycle, given the following date : lattice energy = - 774 kJ mol⁻¹ , ionization energy of sodium = 495 kJ mol⁻¹, heat of sublimation of sodium = 108 kJ mol⁻¹, energy for bond dissociation of chlorine  = 240 kJ mol⁻¹ and heat of formation of sodium chloride = 410 kJ mol⁻¹.

Answer
Born - Haber Cycle for the formation of sodium chloride
Born Haber Cycle for sodium chloride
Born Haber Cycle
∴ - UNaCl - INa + ECl - SNa - ½DCl - ΔHf = 0

or, ECl = UNaCl+INa+ SNa +½DCl + ΔHf

= -774 + 495 + 108 + 120 + 410

= 359 kJ mol⁻¹

Factors influencing the magnitude of EA

The magnitude of EA influenced by the atomic size, shielding effect and electronic configuration of an atom or molecules.

The atomic size of a molecule

Larger the atomic size lesser the tendency of the atom to attract the additional electron towards itself. Which decreases the force of attraction exerted by the nucleus of an atom on the extra electron being added to the valence shell of the atom.

Thus EA values decrease with increases atomic size.

Shielding effect or effective nuclear charge

Higher the magnitude of effective nuclear charge (Zeff) towards the periphery of an atom greater 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 added to form an anion. Thus the magnitude of the electron affinity of an atom increases with increasing Zeff value.

Electronic configuration of an atom

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

Question
Account for the large decrees in electron affinity from lithium to beryllium despite the increase in nuclear charge.

Answer
The atomic number of lithium and beryllium 3 and 4. The electronic distribution of lithium and beryllium
1S² 2S¹
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 - 0.10 eV while that of phosphorus + 0.70 eV?

Answer
Electronic configuration of nitrogen and phosphorus

1S²2S²2P³
1S²2S²2P⁶3S²3P³
.

Due to the smaller size of nitrogen atom when an extra electron added to the stable half-filled 2P subshell some amount of energy required and hence the electron affinity of nitrogen is negative.

On the other hand, due to the bigger size of a phosphorus atom in compare to nitrogen less amount of energy released when the extra electron added to the stable half-filled 3P subshell and thus electron affinity of phosphorus expressed with a positive sign.

Electron affinity across a group in the periodic table

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 across the group in the periodic table.

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 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 the fluorine atom.
F Cl Br I
- 3.6 eV - 3.8 eV - 3.5 eV - 3.2 eV

Electron affinity across a period in the periodic table

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.
Electron affinity and the periodic table of atoms or molecules
Electron affinity and the periodic table
Question
Why electron gain enthalpy of beryllium and magnesium are almost zero?

Answer
Berrilium and magnesium have their electron affinity values equal to almost zero. Since beryllium and magnesium have completely filled S sublevel.
1S² 2S²

1S² 2S² 2P⁶ 3S²

The additional electron will be entering the 2P - subshell in the case of beryllium atom and 3P - subshell in the case of magnesium atom which is of considerably higher energy than the 2S - subshell.

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 can easily understandable because their electronic configurations are only one electron short of the next noble gas element.

The electron affinity of chlorine greater than that of fluorine. Due to the very small size of fluorine atom which results in a very high charge density and consequently strong repulsion between existing valence shell electrons and the entering electron or new coming electron.

On the other hand, chlorine is a bigger size, charge density small and thus such repulsion not strong enough. Hence the electron affinity chlorine greater than that of fluorine.

Electron affinity of noble gases

Inert gases in which the nS and nP orbitals are completely filled or nS² nP⁶ electronic 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 the Slater's rules for shielding electrons?

A study of valence electrons for a multi-electron atom attracted by the nucleus of an atom and repelled by the electrons from inner-shells.
Shielding effect and Slater's rules
Shielding electrons and Slater's rules
This attractive and repulsive force acting on the valence electrons experience less attraction from the nucleus of an atom. Larger the number of inner electrons, lesser will be the attraction between the nucleus and outer electrons.

The inner electrons which shielded the force of attraction between the nucleus and valence electron are known shielding electrons and the effect is known as the shielding effect or screening effect.

To study shielding electron, 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.

Thus the nuclear charge of an atom or ion is less than the actual nuclear charge and can easily be calculated by the following equation.

∴ Effective nuclear charge = Z - σ
where σ = shielding constant.

How to calculate the effective atomic number?

This article is an important article for school and college-level courses in different schools and universities.

Effective nuclear change means the net positive charge which affects the attraction of outer electrons from the nucleus of polyelectronic atom. The term effective is used because the shielding electrons prevent the attraction of outer orbital electrons of an atom.

Slater's rules are applicable for calculating the value shielding constant and effective nuclear charge of an electron in an atom or ion.

nS and nP level electron of an atom or ion

  1. The first we write 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.

    The shielding effect concerned the nS and nP electron belonging to the same principal quantum level have the same effect 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, or ion.

    Electrons lying above nS, nP level do not shield any nS, nP electron to any extent.
  3. Electrons of higher energy levels did not shield the lower-energy level of an atom. For calculating the value screening constant of the valence electron of a sodium atom, the electronic configuration of the sodium atom

    (1S)² (2S, 2P)⁸ (3S)¹.
    The value of the shielding constant for the 3P-electrons, the valence electron or 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 rule also true for the electrons of the nd or nf atomic level of atom or ion.

  5. Electrons belonging to one lower quantum level, that is (n-1) level 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.
Shielding effect in nS, nP subshell of an atom
Shielding effect in nS, nP subshell

Screening constant for sodium atom
= (2 × 1) + (8 × 0.85) + (0 × 0.35)
= 8.8.

∴ Z effective of sodium atom = (11 - 8.8)
= 2.2.
Question
How to calculate the effective nuclear charge of the hydrogen atom?

Answer
The hydrogen atom has a single 1S valence electron. There are no other electrons to shield it from the nuclear charge of a single proton.

∴ Shielding constant = 0
Zeff = 1.0 - 0 = 1.0.

Thus hydrogen electron sees the full nuclear charge of the nucleus and the electron totally exposed to the proton.

Sodium, potassium, and magnesium ion

Electronic configuration of sodium ion or Na⁺
(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 of potassium ion or K⁺
= (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 ion or Mg⁺²
(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
Carbon (1S)² (2S, 2P)⁴
Nitrogen (1S)² (2S, 2P)⁵
Oxygen (1S)² (2S, 2P)⁶
In carbon atom, the 2P electron shield 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

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

Chemistry articles for school-college courses

nd, nf-sublevel of an atom or ion

Previous rules are quite well for estimating the screening constant of S and P orbitals. However when shielding by d subshell or f subshell the four and five rule replaced by new rules for the estimation of screening constant.

The new 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 and 3d electron of the vanadium atom

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

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

The 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.

Why the electron of the 4S energy level lost first?

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

∴ Screening constant 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 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 lower than 4S subshell. From this study, it is clearly indicated that 3d electron more tightly bound within the nucleus than 4S electron.

Thus during the ionization less tightly bound electron losses first. So during the ionization process, 4S electron of energy level loses first than the 3S energy level.

Chemistry 1

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