Periodic classification - II Five Marks

1. Explain any three factors which affect the ionisation energy.

1. Size of atom or ion

Ionization energy α 1 / Size of atom or ion

The ionization energy decreases with the increasing size of atom. The larger the size of atom, lesser is the ionization energy.

This is due to the fact that electrons are tightly held in smaller atoms whereas in large atoms, electrons are held quite loose, i.e., lesser energy is required for removal of electrons from larger atoms than the smaller one.

Hence ionization energy is lower for larger atoms and higher for smaller atoms.

2.  Magnitude of nuclear charge

Ionization energy α Effective nuclear charge

The higher the nuclear charge of protons in the nucleus, the higher is the ionization energy. Because of the higher nuclear charge, the electrons are bound with more force and hence higher energy will be required for their removal.

3.  Effect of number of electrons in the inner shells. (Screening or shielding effect)

Ionization energy α 1 / Shielding effect

The attractive force exerted by the nucleus on the most loosely bound electron is at least partially counterbalanced by the repulsive forces exerted by the electrons present in the inner shells. The electron to be removed is thus shielded from the nucleus by the electrons in the inner shell. Thus, the electron in the valence shell experiences less attraction from the nucleus. Hence the ionization energy will be low. This is another reason why ionization energy decreases in moving down a group.

4.  Effect of shape of orbital. The shape of orbital also influences the ionization potential. As s-electrons remain closer to the nucleus than p-,d-, and f-electrons of the same valence shell, the ionization energy decreases in the order given below:

s > p > d > f

5.  Effect of arrangement of electrons. The more stable the electronic arrangement, the greater is the ionization energy. As the noble gases have the stablest electronic arrangements, they show maximum ionization energy.

2. Explain how electronegativity values help to find out the percentage of ionic character in polar covalent bond.

Pauling estimated the percentage of ionic character in various Aδ-–Bδ+ polar covalent bonds from known (XA–XB) values and has derived the following conclusions:

1. When (XA–XB) = 1.7, the amount of ionic character in Aδ-–Bδ+ bond is 50% and that of covalent character is also 50%. Thus A-B bond is 50% ionic and 50% covalent.

2. When (XA–XB) < 1.7, the amount of ionic character in Aδ-–Bδ+ bond is less than 50% and that of covalent character is more than 50%. Thus Aδ-–Bδ+ bond is predominantly covalent and hence is represented as A–B.

3. When (XA–XB) > 1.7, the amount of ionic character in Aδ-–Bδ+ bond is more than 50% and that of covalent character is less than 50%. Hence Aδ-–Bδ+ bond is predominantly ionic and hence is represented as A– B+.

3. Explain Pauling's method to determine / calculate ionic radii. Or How is ionic radius determined by Pauling's method?

Pauling has calculated the radii of the ions on the basis of the observed inter nuclear distances in four crystals namely NaF, KCl, RbBr and CsI. In each ionic crystal the cations and anions are isoelectronic with inert gas configuration.

NaF crystal: Na+ - 2, 8

                        F– - 2, 8      Ne type configuration

KCl crystal:    K+ - 2, 8, 8

                       Cl– - 2, 8, 8  Ar type configuration

Further the following two assumptions are made to assign the ionic radii.

1. The cations and anions of an ionic crystal are assumed to be in contact with each other and hence the sum of their radii will be equal to the inter nuclear distance between them.

r(C+) + r(A–) = d (C+–A–)                           (1)

Where,

          r(C+)       - radius of cation, C+

         r(A–)        - radius of anion, A–

         d(C+–A–) - internuclear distance between C+ and A– ions in C+A– ionic crystal

2. For a given noble gas configuration, the radius of an ion is inversely proportional to its effective nuclear charge. i.e.

r(C+) a 1 / Z*(C+)                                       (2)

r(A−) a 1 / Z *(A−)                                      (3)

Where,

          Z*(C+) & Z*(A–) are the effective nuclear charges of cation (C+) and anion (A-) respectively.
On combining (2) & (3)

r(C+)  / r(A−) = Z*(A− ) / Z*(C+ )               (4)

Hence the above two equations (1) & (4) can be used to evaluate the values of r(C+) and r(A–) provided that the values of d(C+–A–), Z*(C+) and Z*(A–) are known.

4. Explain the various factors that affect / influence electron affinity.

1. Atomic size

Electron affinity α 1 / Size of atom

Smaller the size of an atom, greater is its electron affinity. As the size of atom increases, the effective nuclear charge decreases or the nuclear attraction for adding electron decreases. Consequently, atom will have less tendency to attract additional electron towards itself.
Therefore, Electron affinity α Effective nuclear charge.

In general, electron affinity decreases in going down the group and increases in going from left to right across the period. On moving down the group atomic size increases and on going from left to right in a period atomic size decreases.

2.  Shielding or Screening Effect

Electron affinity α 1 / Shielding effect

Electronic energy state, lying between nucleus and outermost state hinder the nuclear attraction for incoming electron.

Therefore, greater the number of inner lying state less will be the electron affinity.

3.  Electronic Configuration - The electronic configurations of elements influence their electron affinities to a considerable extent.

2np6 configuration in their valence shell and there is no possibility for addition of an extra electron.

Completely / half filled or stable electronic configuration leads to zero or low electron affinity.

Electron affinities of inert gases are zero. This is because their atoms have stable ns

5. How do electronegativity values help to find out the nature of bonding between atoms?

The concept of electronegativity can be used to predict whether the bond between similar or dissimilar atoms is non-polar covalent bond, polar covalent bond (or) ionic bond.

1.  When XA = XB, i.e. XA- XB = 0, then A-B bond is non polar covalent bond or simply covalent bond and is represented as A-B.

Example: H-H bond in H2 molecule is a covalent bond and is represented as H-H bond.

2.  When XA is slightly greater than XB, i.e. XA - XB is small, the A-B bond is polar covalent bond and is represented as Aδ--Bδ+.

Example: The O-H bonds in H2O molecule are polar covalent bonds and are represented as Oδ--Hδ+, since XO > XH and XO - XH is small.

3.  When XA >> XB, i.e., XA - XB is very large, A-B bond is more ionic or polar bond and is represented as A--B+, Since XA >> XB.

Example: Na-Cl bond in NaCl molecule is an ionic bond and is represented as Na+Cl- (Here Cl = A and Na = B).

6. Write notes on Pauling’s and Mulliken’s Scale of Electronegativity. Or Explain the Pauling scale for the determination of electronegativitv. Give the disadvantage of Pauling scale.

1. Pauling’s scale is based on an empirical relation between the energy of a bond and the electronegativities of bonded atoms.

Consider a bond A-B between two dissimilar atoms A and B of a molecule AB. Let the bond energies of A-A, B-B and A-B bonds be represented as EA-A, EB-B and EA-B respectively. It may be seen that the bond dissociation energy of A-B is almost higher than the geometric mean of the bond dissociation energies of A-A and B-B bonds i.e.,

EA-B > √EA-A x  EB-B

Their difference (Δ) is related to the difference in the electronegativities of A and B according to the following equation

             Δ = EA-B - √EA-A x  EB-B

             Δ = (XA– XB)2 (or)

0.208 √Δ = XA – XB

Here, XA and XB are the electronegativities of A and B respectively.

The factor 0.208 arises from the conversion of kcals to electron volt.

Considering arbitarily the electronegativity of hydrogen to be 2.1, Pauling calculated electronegativities of other elements with the help of this equation.

Disadvantage of Pauling’s scale:

Bond energies are not known with any degree of accuracy for many solid elements.

2. Mulliken suggested an alternative approach to electronegativity based on ionization energy and electron affinity of an atom. According to this method electronegativity could be regarded as the average of the ionization energy and electron affinity of an atom

Electronegativity = I.E + E.A / 2

 Mulliken used ionisation energy and electron affinity values measured in electron volts and values were found to be 2.8 times higher than Pauling values. The values of ionisation energy and electron affinity are measured in kJ mol-1 and 1eV =96.48 kJ mol-1.

Therefore the commonly accepted Pauling values are more nearly obtained by

Electronegativity = I.E + E.A / 2 x 2.8 x 96.48 = I.E + E.A / 540

This method has an ordinary theoretical basis and also has advantage that different values can be obtained for different oxidation states of the same element.

Disadvantage of Mulliken’s scale:

Electron affinities with the exception of a few elements are not reliably known

7. Explain the variation of ionisation energy along the group and period.

In a period, the value of ionisation energy increases from left to right with breaks where the atoms have somewhat stable configurations.

This is due to the reason that the nuclear charge increases whereas atomic radius decreases.

In a group, the ionisation energy decreases from top to bottom.

This is due to the effect of the increased atomic radius 

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