Notes on Periodic Classification of Elements (BSc and Integrated Standard)

Notes on Periodic Classification  of Elements

    Disadvantages of Mendeleev’s Periodic Table:

1.          Position of hydrogen: It has been confirmed that hydrogen has many characters which match with halogen elements too. But it is only placed in Group IA. Thus position of hydrogen is not clear.

2.     Position of isotopes: Since in Mendeleev’s periodic table elements have been arranged according to their atomic masses, isotopes (more than one element) are placed at a one place which is confusing.

3.      Position of the isobar: Since in Mendeleev’s periodic table elements have been arranged according to their atomic masses, different elements having different atomic numbers but same mass number (isobars) are placed at one place which is again confusing.

4.          No attempts were made to separate metals from non-metals.

5.        Dissimilar elements placed together in the same group:  Elements showing different chemical properties have been placed together in the same group. For example Li, Na, k and Cu, Ag, Au have been placed in Group IA.

Modern Periodic Law:

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            Moseley’s experiment confirmed that atomic number of an element and square root of the frequency of X-ray of that element (which is obtained when various elements were bombarded with cathode rays) are directly proportional to each other. This triggered the chemist to modify the old periodic law as:

“The physical and chemical properties of elements are the periodic function of their atomic numbers. Clearly when elements are arranged in increasing order of their atomic numbers similar elements are repeated after regular intervals.”

Periodicity of Elements (Cause of Periodicity):

            The recurrence of elements with similar properties after certain regular intervals when these are arranged in the increasing order of their atomic numbers is called the periodicity.

            Looking at the elements of alkali metals in the modern periodic table, it’s not difficult to realize that every element here has a similar outer electronic configuration. This repetition in similar outer electronic configurations after certain regular interval (as the atomic number increases while arranging the elements in the periodic table) is the cause of periodicity in properties of elements. This is because the chemical behavior of elements is due to the electrons in the outer most shells. Since all the elements in a particular group have similar outer electronic configuration their chemical behavior are almost same.

 ₃Li = 1s² 2s¹                                                               or [He]2s¹

₁₁Na = 1s² 2s² 2p⁶ 3s¹                                                       or [Ne]3s¹

₁₉K = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹                                   or [Ar] 4s¹

₃₇Rb = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 5s¹              or [Kr] 5s¹

Note: The numbers 2, 8, 8, 18, 18, 32 are called Magic Numbers because the properties of elements get repeated in such intervals. For example atomic number of hydrogen is 1 (₁H) , the next element in alkali metals in the periodic table is Li whose atomic number is 3 (1+2=3), the next one is sodium of atomic number 11 (3+8=11), next is potassium of atomic number 19 (11+8=19), next ₃₇Rb (19+18=37) and so on.

       For elements in any group of p-block (except inert gas elements for which it is 8, 8, 18, 18, 32, 32) this is 8, 18, 18, 32 and 32. For elements in the first group of d-block (Sc, Y, La, Ac) this is 18, 18, and 32 and for all other groups in this block this is 18, 32, and 32. For f-block this is 32.

Long Form of the Periodic table (Present Form or Bohr’s table):

Features of Long Form of the Periodic Table:

1.      All elements have been arranged in the increasing order of their atomic numbers.

2.      Elements of equivalent properties due to their similar electronic configuration have been placed together at one place.

3.      For a better classification of elements, the periodic table has been classified into the following sections and sub-sections:

i.        Four Blocks (s, p, d, f)

ii.   Seven Horizontal Rows called periods, each represented by a unique principal quantum number, n (i.e., n=1 for first period, n=2 for second period, and so on….).

iii.      Eighteen vertical columns called groups, numbered as 1 to 1

Classification of elements into blocks and general electronic configuration:

            On the basis of electronic configuration of atoms of various elements (more specifically based upon the name of the orbital which receives the last electron), elements have been divided mainly into three types:

i.         Representative or s and p-block elements

ii.        Transition or d-block elements

iii.       Inner transition or f-block elements.

s-Block or the first (left) block of representative elements:

            The elements (except Helium) in which the last electron enters into the s-subshell of the outer most main energy level are called s-block elements.

Maximum number of electrons that can be accommodated in s-subshell = 2 (by the formula 2(2l+1)). Therefore there are only two groups in this block. The first one being consists of the “hydrogen and alkali metals’’ and the second one is of the ‘alkaline earth metals’.

            Consider an element sodium, which is present in the first group (called The Alkali metals) of this block, ₁₁Na = 1s² 2s² 2p⁶ 3s¹      or [Ne]3s¹. It has been previously mentioned that all the elements in a group have equivalent outer electronic configuration. The only difference is that, in hydrogen the last electron enters into 1s¹ (because hydrogen is present in first period), in lithium it is 2s¹ (because hydrogen is present in second period) and in sodium it is 3s¹ (because hydrogen is present in third period) and so on. Thus the general electronic configuration of the alkali metals is ns¹ (‘n’ representing the number of period in which the element is present).

            Again consider an element from the second group of this block (The alkaline earth metals) calcium, ₂₀Ca = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² or [Ar] 4s^2. It can easily be observed that all the elements in the group of alkaline earth metals have the following outer electronic configurations, beryllium (2s², because in 2^nd period), magnesium (3s², because in 3^rd period), calcium (4s², because in 4^th period) and so on. Thus the general electronic configuration of the alkaline earth metals is ns².

              Combining the general electronic configuration of alkali metals and alkaline earth metals, we can write the general electronic configuration of s-block as ns^1-2.

General Characteristics of s-block elements:

1.     Soft metals and low melting points

2.     Low ionization energies

3.   Highly reactive and form univalent (alkali metals) and divalent (alkaline earth  metals) ions readily.

4.      4. Compounds of these elements (except beryllium) are predominantly ionic.

5.      5. Most of the metals (except beryllium and magnesium) impart characteristic color to the flame which can be used to identify them.

6.      6. Strong reducing agent.

7.      Good conductor of heat and electricity.

8.      Metal hydroxides are strong bases.

9.      The last elements in both group-1 and group-2 in this block namely francium and         radium are radioactive.

1        Very low electron affinity. 

p-block elements or the second (right) block of representative elements:

            The elements (except helium) in which the last electron enters into the p-subshell of the outermost main energy level, are called the p-block elements. e.g ₁₇Cl = 1s² 2s² 2p⁶ 3s² 3p⁵ or [Ne] 3s² 3p⁵

            Maximum number of electrons that can be accommodated in p-subshell = 6 (l=1 for p-subshell). Therefore there are six groups in this block. They are boron family, carbon family, nitrogen family, oxygen family, halogen family, and 18^th group elements.

Consider an element from boron family say ₁₃Al = 1s² 2s² 2p⁶ 3s² 3p¹ or [Ne] 3s² 3p¹. Now we can easily predict the outer electronic configurations of all the elements present in boron family; ₅B = 2s² 2p¹ (in 2^nd period), ₁₃Al = 3s² 3p¹ (in 3^rd period), ₃₁Ga = 4s² 4p¹ (in 4^th period) and so on. Thus the general electronic configuration of boron family is ns² np¹ (‘n’ representing the number of period in which the element is present).  Similarly we can write for the carbon, nitrogen, oxygen and halogen family and also for inert gas elements as, ns² np², ns² np³, ns² np⁴, ns² np⁵ and ns² np⁶ respectively. Thus the general electronic configuration for p-block is ns² np^1-6.

General characteristics of p-block elements:

1.      Contains both metals and nonmetals (nonmetallic character increases towards right)

2.      High ionization energies compared to s-block elements

3.      Form mostly covalent compounds

4.      Some of the elements show variable valency

5.      Oxidizing character increases towards right in a period and reducing character increases top to bottom in a group.

6.      Generally bad conductor of heat and electricity

Note: Diagonal relationship:

            Elements of 2^nd period are called bridging elements. They resemble in certain properties with the elements of the third period diagonally placed,

            

            From the above figure, Lithium has similarities in properties with Magnesium; beryllium has similarities in properties with aluminum, boron with silicon and so on.

d-block elements or transition elements:

            The block in between s and p-block is called d-block. Since the properties of these elements are intermediate between s and p-block elements these are called transition elements. Maximum number of electrons in a d-subshell is equal to ten. Hence there are 10 groups in this block.

          Elements in which the last electron enters to a penultimate (n-1) d-subshell are called d-block elements. Strictly speaking, elements which contain incompletely filled d-orbital are included in this category.

           This block consists of four periods, n=4, n=5, n=6 and n=7, (i.e., 4^th, 5^th, 6th, and 7^th periods of periodic table) each representing one transition series in which the last electron enters into 3d, 4d, 5d and 6d subshell respectively. They are called “First transition series (n=4), Second transition series (n=5), Third transition series (n=6), Fourth transition series (n=7)”

            Consider an element from first transition series, ₂₁Sc = 1s² 2s² 2p⁶ 3s² 3p⁶   3d¹ 4s²   or [Ar] 3d¹ 4s², we see that, after the filling of 18 electrons in the argon core, the 19^th and the 20^th electron normally enter into the 4s (ns) and the rest into the 3d (n-1d) subshell. This trend is followed by all other elements in all the transition series of this block. Thus the general electronic configuration of d-block elements is (n-1) d^1-10 ns^1-2. 

            Due to the stabilities of half filled and full filled orbital, in some elements like Cr and copper the (n-1)d and ns subshell show irregular filling of electrons  compared to other elements. e.g., ₂₄Cr= [Ar] 3 d⁵ 4s¹ and ₂₉Cu = [Ar] 3 d¹⁰ 4s¹

                        There are some exceptions to the above rule of assigning an element as d-block element, e.g., Lutetium, ₇₁Lu = [Xe] ₅₄ 4f¹⁴ 5d¹ 6s², but this element due to its properties resemble more with inner transition elements (f-block elements) is included in f-block elements.

General characteristics of d-block elements:

1.      Hard, malleable and ductile

2.      Good conductor of heat and electricity

3.      Show variable oxidation states

4.      Magnitude ionization enthalpy is in between those of s and p-block elements

5.      Form both ionic and covalent compounds

6.      Compounds are generally colored and paramagnetic

7.      Strong tendency to form complexes which are coloured

8.      Elements like Cr, V, Mn, Fe, Cu, Ni, Pd, Pt are used as catalysts.

9.      Most of the elements are used in the preparation of alloys.

10.   Complexes are colored.

 f-block elements or inner transition elements:

            We can see from the periodic table that after *₅₇La comes *₇₂Hf. More clearly *₅₈Ce to *₇₁Lu (from 6^th period), these fourteen elements have been brought down the main part of the modern periodic table as “lanthanoids” as all these elements follow Lanthanide, *₅₇La. Similarly *₉₀Th to *₁₀₃Lr (from 7^th period) these fourteen elements have been brought down the main part of the modern periodic table as “Actinoids” as all these elements follow Lanthanide, *₈₉Ac. In all these elements, the s-orbital of the last shell (n) is completely filled, the penultimate (n-1) d-orbitals invariably contains zero or one electron and the ante-penultimate (n=2) f-orbitals (being lower in energy than the said d-subshell) gets progressively filled in.

            Thus the general electronic configuration for f-block elements is (n-2) f ^1-14 (n-1) d ^0-1 ns². (n=6 for lanthanoids and n=7 for actinoids).

            ₅₈Ce = [Xe] ₅₄  4f ¹ 5d¹  6s²,       ₉₀Th = [Rn] ₈₆  5f ¹ 6d ¹  7s²

 General characteristics of the f-block elements:

1.      These are Heavy metals

2.      High melting and boiling points

3.      Show variable oxidation states

4.      Form generally colored compounds

5.      High tendency to form complexes

6.      Most of the elements are radioactive

Thus we found that total number of groups in modern periodic table = 2 (s-block) + 10 (d-block)+ 6(p-block.) = 18.

Number of elements in a period:

            First period: This corresponds to n=1, filling of electrons in 1^st shell (K), K shell contains only one subshell,1s, which can accommodate only 2 electrons, therefore there are only two elements in the first period. They are hydrogen and helium.

            Second period: This corresponds to n=2, , filling of electrons in 2^nd shell (L), L shell contains two subshell,2s and 2p, which can accommodate 2+6=8 electrons, therefore there are eight elements in the second period (Li-Ne).

            Third period: This corresponds to n=3, (M shell), the subshells for n=3 are 3s, 3p, 3d but we know, energy of 3d is greater than 4s. thus 3d is excluded, thus total electrons to be accommodated  are due to 3s (2 electrons) + 3p (6 electrons) = 8 electrons. Therefore there are eight elements in the third period (Na-Ar) again.

            Fourth period: n=4, N shell contains four subshells, 4s, 4p, 4d, 4f. Since energies of 4d and 4f are greater than 5s hence they are excluded. 3d, excluded from third period is included in the fourth period. Thus total electrons to be accommodated are due to 3d (10 electrons) + 4s (2 electrons) + 4p (6 electrons) = 18. Hence number of elements in the fourth period is 18.

            In a similar way we can show how fifth, sixth, and seventh period contain 18, 32 and 32 elements.

General trends in periodic properties:

1.  Electronic configuration: So far we have already discussed the general electronic configuration of elements present in different groups, while discussing about various blocks. The general electronic configurations of all the elements in the same group are similar. The chemical properties of elements depend on the valence electronic configuration.

2.    Atomic radii: It is defined as the distance from the centre of the nucleus to the outer most electron(s).

It is measured by electron diffraction method in angstrom (A°) or picometer (pm) unit.

The difficulties in measuring the exact atomic radius of an atom are that, except inert gas elements no other atoms are generally found in isolated or uncombined state. The exact position of the outer most electrons is uncertain according to Heisenberg’s uncertainty principle. Furthermore the electron density of an atom is affected by the presence of neighboring atoms.

Despite the above limitations, we still need some practical approach to estimate the size of atoms. The hope is alive as the atoms pack up at certain definite distances in solids. This gives the idea about the approximate atomic size or radius of the atom. In this concern, depending upon whether the element is a metal or a nonmetal, three different types of atomic radii can be discussed.

i.    Covalent radius: 

a. Covalent Single Bond Radius For Homonuclear Molecules: It is defined as one half of the distance between the nuclei of two covalently bonded atoms (bonded with a single bond) of the same element in a homonuclear molecule such as H2, F2, Cl2 etc.

r  = 1/2 (Internuclear distance between two bonded atoms)

r  = 1/2 (bond length) , e.g. the internuclear distance between two chlorine atoms in Cl₂ molecule is 1.96 A°. Thus the covalent atomic radius of a chlorine atom is 1.96/2 = 0.98 A°.

b. Covalent Single Bond Radius For Heteronuclear Molecules where the electronegativity difference of corresponding bonded atoms is not so high: The covalent single bond radius of an atom X in a heteronuclear molecule is defined as the difference between the single bond length X - Y and the covalent radius of other single bonded atom, Y.

Covalent radius of X in the single bond X - Y = Bond length of X -Y - Covalent radius of Y in X - Y

For example, the bond length of Si - C bond (d Si - C) is 193 pm and covalent radius of carbon atom (r C) is 77 pm. Thus the covalent single bond radius of Si = (d Si - C) - r C = 193 - 77 = 116 pm

c. Covalent Single Bond Radius For Heteronuclear Molecules where the electronegativity difference is high: When there is a large difference in the electronegativity of two bonded atoms X and Y, the bond length will be shorter than expected and will have some ionic character of the bond. In this case the following relationship may be used:

d X-Y = r A + r B - 0.09 (X A - X B), where d X - Y is the bond length of X - Y and r A and r B are the covalent radius of A and B and X A and the X B are the electronegativities of A and B respectively.

Note: Like covalent single bond radius we can have also covalent multiple bond radius of C in C = C or of O in O = O using which the bond length of C  = O can approximately be calculated.

Covalent radii can further be studied as Tetrahedral covalent radii and Octahedral covalent radii.

d.    Tetrahedral Covalent Radii: Crystals having arrangement of lattice points similar to either diamond or sphalerite (ZnS called Zinc blende having cubic structure) or wurzite (Sulphide of Zn having hexagonal structure) give rise to covalent radii.

In each of the arrangement where one atom is surrounded by four other atoms (similar or dissimilar atoms) tetrahedrally gives the covalent radius.

Elements like C, Si, Ge, Sn have diamond like arrangement. All other elements which show tetrahedral arrangement are similar to either sphalerite (Cubic type structure) or wurzite (hexagonal structure).

If we consider the crystal of FeS2 in which each sulphur atom is surrounded by three Fe and one S atom tetrahedrally, then we can find the S - S bond distance to be 2008 pm thus the tetrahedral covalent radii of S is 208 / 2 = 104 pm

Thus we found that the tetrahedral covalent radii and covalent single bond radii of 1st and 2nd row elements are almost same.

    Let us now calculate the internuclear distances of elements by adding the tetrahedral radii and compare them with corresponding observed value in cubic or hexagonal crystals.

    AlP (Aluminium phosphide): The sum total of tetrahedral radii of Al and P taken from the above table = 126 + 110 = 236 pm and the observed value is also 236 pm.

    In some other elements the sum total of the tetrahedral radii and the observed radii deviate by 1 pm. e.g., HgTe , The sum is = 148 + 132 = 280 pm but the observed internuclear distance between Hg and Te is 279 pm.

    In some other elements there exist a large deviation from the calculated values. The addition of tetrahedral radii in SiC gives 194 pm whereas the observed value is 189 pm.

e.    Octahedral Covalent Radii: Crystals having arrangement of lattice points similar to pyrite (FeS2) structure fall into this category.

    Arrangement of lattice points where one atom is surrounded by 6 other atoms (similar or dissimilar atoms) octahedrally give rise to octahedral radii.

In FeS2 each S atom is surrounded by 4 atoms tetrahedral and each Fe atom is surrounded by six S atoms octahedrally.

Thus we can find the internuclear distance of Fe - S which is 227 pm. subtracting the tetrahedral radius of S from 227 pm we get the octahedral radius of Fe = 227 - 104 = 123 pm.

Here we can find two interesting observation w.r.t. octahedral covalent radii.

Elements which differ by one unit atomic number and are isoelectronic due to varying oxidation number show almost same octahedral radii having slight deviation of 1 pm as the atomic number increases by 1 pm. For Example: Ni (IV) has an octahedral covalent radii of 121 pm then Co (III) has 122 pm and Fe (II) has 123 pm.

It has been observed that for each additional electron in the (n-1) d subshell (where the type of hybridisation of the central atom / ion is d2sp3 type) the increase in the octahedral radii is about 9 pm. For example, Ni (II) has two electrons extra as compared to Fe (III). Thus Ni (II) should have an octahedral radii which is greater than the octahedral radii of Fe (III) by 2  X 9 or 18 pm. Since Fe (III) has an octahedral radii of  120 pm Ni should have octahedral radii = 120 + 2 X 9 = 138 pm. The observed value of Ni(II) is 139 pm which almost equal to the theoretical value.  

ii.      Van der Waals radius:

It is defined as one half the distance between the nuclei of two identical non bonded isolated atoms or distance between two adjacent identical atoms belonging to two neighboring molecules of an element in solid state. 

To make it clear we can take the example of chlorine in solid state. The internuclear distance between adjacent chlorine atoms of the neighboring molecules is 3.6 A°. Thus the Van der Waal's radius (r _{van der} Waal's) is equal to half of 3.6 A°, i.e., 1.8 A°.

Note: since inert gas elements normally do not form chemical compounds, their atomic radii are usually expressed in terms of Van der Waal's radii which are sometimes referred to as Inert Gas Radii.

 iii.    Metallic radii: We know that metallic atoms or ions arrange themselves in a regular pattern to form a crystal lattice. The metallic radii may be defined as one half of the internuclear distance between the two adjacent metal ions, in the metal lattice.

Note:  the magnitude of various atomic radii is in the order as follows:

                                    Van der Waal's  >  metallic  >  covalent

Factors on which atomic radii depend:

a.         Hybridization: This can be demonstrated by taking suitable examples.

    Radius of carbon atom is 77pm in methane. Here ''C’’ is sp³ hybridised (s character 25%). Radius of carbon atom is 67pm in ethane. Here “C” is sp²  hybridised (s character 33%). Radius of carbon atom is 60pm in ethyne. Here “C” is sp  hybridised (s character 50%).

Greater is the s-character of an atom, smaller the atomic radius of that atom.

b.         Nuclear charge: Greater is the nuclear charge (increase in atomic number means increase in number of protons and hence increase in the nuclear charge) of the nucleus, it attracts the electrons more towards it and the atomic size decreases.

c.       Number of orbits (shielding effect): With increase in the number of orbits, distance between the nucleus and last orbit increases and the electrons present in between nucleus and the last orbit tend to decrease the attractive force of the nucleus by screening the nucleus (shielding or screening effect). Thus the atomic radius increases.

d.    Bond order: Greater is the bond order between the bonded atoms, shorter will be bond length due to greater extent of overlapping. This is because their p orbitals overlap to a greater extent and the bonded atoms come closer.

e. Other factors which affect the atomic radius are the nature of bond (whether ionic or covalent or metallic) and the oxidation states of the bonded neighbouring atoms.

Variation of atomic radii:

Variation along a period: As we move from left to right across a period, there is regular decrease in atomic radii of the representative (s and p-block) elements. This is due to the fact that number of energy shells remains the same in a period (electrons are added in the same shell) but nuclear charge increases gradually as the atomic number increases. Thus the force of attraction of the nucleus on the electrons increases which brings contraction in size.

This can also be explained on the basis of effective nuclear charge which increases gradually in a period. (Click and zoom the figure)

Li(1.23 A0)	Be(0.89 A0)	B(0.80 A0)	C(0.77 A0)	N(0.75 A0)	O(0.73	F(0.72	Ne(1.2A0)
Na(1.54A0)	Mg(1.36A0)	Al(1.20 A0)	Si(1.17 A0)	P(1.10 A0)	S(1.04	Cl(0.99	Ar(1.91A0)
K(2.03 A0)	Ca(1.74 A0)	Ga(1.26A0)	Ge(1.22A0)	As(1.20A0)	Se(1.16	Br(1.14	Kr(2.0 A0)

Variation in a group: atomic radii in a group increases as the atomic number increases. The increase in size is due to extra energy shells (a new shell is introduced in elements each time when we move down the group) which dominates over the increased nuclear charge.

The atoms of 18th group elements do not form chemical bonds. Hence, their van der Waaals’ radii are considered which are always greater than the covalent radii of halogens. The following table illustrates the above three points:

Variation in transition elements: Though, in vertical columns of transition elements, there is an increase in the size from first member to second member as expected, there is an abnormal change when we move from second to third member of a column. Either a very small change or no change is observed. This is due to the Lanthanide contraction. In the lanthanide elements the last electron (differentiating electrons) enters into 4f subshells which do not effectively screen the nucleus. This result in the increase of effective nuclear charge and the size gradually decrease in size. The covalent radii in angstrom units are given below:

Sc(1.44)	Ti(1.32)	V(1.22)	……………………..Cu(1.17)
Y(1.62)	Zr(1.45)	Nb(1.34)	……………………..Ag(1.34)
La(1.69)	Hf(1.45)	Ta(1.34)	……………………..Au(1.34)

3.    Ionic Radii: It is defined as the distance between the nucleus and the outer most electron of that ion in an ionic bond or it is the distance from the nucleus of an ion up to which it has an influence on its electron cloud.

    The internuclear distance between the two oppositely charged ions touching each other in a crystal can be determined using X-ray technique. But measuring ionic radii is not easy because it is nearly impossible to define a clear boundary between the cation and anion and up to which the electron clouds of the oppositely charged ions are spread.

     Still the hope remains alive as X ray crystallography can measure the electron density (ED in electrons per angstrom cube) in the region surrounding the nucleus which is known as the electron density map (the ED map).

    In the first step the electron density around the nucleus is measured. As we go away from the nucleus the electron density around the nucleus decreases and falls down to a considerably low value of 0.2 electrons per angstrom cube, as is shown by the largest contour ring around any nucleus. In the next step, an arbitrary point is intuitively chosen from this low ED region and the ED is suitably integrated inward (towards the nucleus) from that arbitrary point. In case of NaCl the integration indicated about 10.05 electrons around Na nucleus and 17.70 electrons around Cl nucleus. This is in fact a very close resemblance to the electrons present in Na ion (10 electrons) and in Chloride ion (18 electrons). Thus the arbitrary point chosen can be considered to be the boundary between the Na and Chloride ion and help to determine the ionic radii.

Note: The size of a cation is always smaller that of the corresponding atom due to (i) decrease in the number of electrons (ii) increase in the effective nuclear charge.

Similarly the size of an anion is always greater than the corresponding atom.

Note: in a set of species having same number of electrons (isoelectronic species), the size decreases as the charge on the nucleus increases. Ex: O^2-> Ne > Mg2+ > Al³⁺.

4.    Ionization enthalpy or ionization potential or ionization energy (I.E/I.P/Δ_i H): it is defined as the minimum potential difference maintained in a discharge tube to remove the most loosely bound electrons from an isolated gaseous atom to form gaseous cation or the minimum amount of energy required to remove the most loosely bound electrons from an isolated gaseous atom to form a monovalent positive ion. It is also known as the first ionization enthalpy or energy.

Ionization potential may be expressed either in terms of ev/atom or kcal mol^-1 or kj mol^-1.

1 ev = 1.602 x 10^-19 j per atom = 1.602 x 10^-19 x 6.02 x 10²³ x 10^-3 kj per mole = 96.49 kj mol-¹ = 23.06 kcal per mole

            Like the removal of first electron from neutral gaseous atom, we can remove second, third and successive electrons from positive ions one after another. The amount of energies required to do so are termed as second, third, fourth….ionization energy respectively.

            The increasing order of the successive ionization energies is, I.E₁   ˂   I.E₂   ˂   I.E_3. This is due to the successive increase in the effective nuclear charge.

Factors affecting ionization enthalpy:

a.  Size of the atom: ionization energy decreases as the atomic radius increases. This is because the distance between the nucleus and the outer most electron increases and the attractive force on the outer most electron decreases and that electron can now be removed with a lesser amount of energy.

b.   Nuclear charge, shielding or screening effect and effective nuclear charge: the ionization energy increases with increase in nuclear charge (increased numbers of protons in the nucleus) because the attractive force on the outer most electron increases.

     In multi-electron atoms, the attractive force exerted by the nucleus on the valence shell is partly reduced by the presence of inner shell electrons. Thus the valence electrons do not feel the full charge of the nucleus. The actual charge felt by the valence shell electrons is called the effective nuclear charge and the partial reduction of the nuclear charge by the inner shell electrons is called the shielding or screening effect. They are related as follows:

      The sigma star (σ *) is called the screening effect constant. It increases as the number of inner shell electrons increases and can be calculated by Slater’s rule:

1.      For the ns or np orbital electrons:

i.     Write the electronic configuration in the following order and group as:

(1s)  (2s, 2p)  (3s, 3p)  (3d)  (4s, 4p)  (4d, 4f)  (5s, 5p)  (5d, 5f)  (6s, 6p) ….

ii.   ii.     Electrons present in group which are right to the (ns, np) group do not contribute to the shielding effect constant.

iii.  The electrons in (ns, np) group contribute to an extent of 0.35 each to the screening effect constant (except the electrons in 1s which contribute to an extent of 0.30 each).

iv.  The electrons in (n-1) shells contribute to an extent of 0.85 each.

v.   All other electrons in the (n-2) and lower shells contribute to an extent of 1.0 each. 

2.      For the d- or f- orbital electrons:

Rules (I) to (iii) remain the same and (iv) and (v) get replaced by the rule (vi) as:

vi. The electrons present in the groups lying left to the (nd, nf) group contribute 1.0 each to the    screening effect constant.

Ex -1: Calculate the effective nuclear charge of the nucleus of zinc atom on the electrons in (i) 4s and (ii) 3d orbital.

  Solution: (i) for 4s- electron:

     (1s)²  (2s, 2p)⁸  (3s, 3p)⁸  (3d)¹⁰  (4s)²

     σ * = 1 x 0.35 + 18 x 0.85 + 10 x 1.0 = 25.65

     Z_eff  = Z - σ * = 30 - 25.65 = 4.35

             (ii) for d- electron:

     σ * = 9 x 0.35 + 18 x 1 = 21.15

      Z_eff  = Z - σ * = 30 - 21.15 = 8.85

c.       Penetration effect of the electrons (shape of orbitals): since s-orbital is spherical , electrons in s-orbital are more penetrated towards the nucleus. The order of penetration of various orbital is:

S > p > d > f.

      If penetration of an electron in an orbital is more, it is closer to the nucleus and attracted more by the nucleus. Thus an electron from a more penetrated orbital requires more energy to be removed. Ex: comparing Mg and Al, since last electron is being removed from the s orbital in Mg, it has higher ionization energy.

d.   Electronic configuration (half filled and full filled electronic configuration) : since exactly half and full filled electronic configuration are stable, higher energy is required  to remove an electron from such electronic configurations. For example Be (1s² 2s²) has higher ionization energy than B (1s² 2s²2p¹), also nitrogen has higher ionization energy than oxygen. All inert gaas elements have very high ionization energy.

Variation of ionization energy:

Along a period: in general, ionization energy increases as we go from left to right in a period. This is due to the increase in nuclear charge and decrease in atomic radius. The irregularities observed can be explained from any one of the factors influencing ionization energy (discussed above). E.g., Be and b; N and O; Mg and Al etc.

Note: in case of transition elements, some irregularities are observed along a period as the inner d-electrons screen the outer most s-electron. Increased screening effect is due to the increased number of (n-10 d-electrons. As a result the effective nuclear charge decreases and outer most electrons can be removed with comparatively smaller ionization energy. Ex: Fe, Co, Ni

Down a group: A regular decrease in the ionization energy is observed in case of representative elements as we go down a group. This is due to the increase in atomic size, increase in screening effect and decrease in effective nuclear charge.

Note: exceptionally, when we move from second to the third transition series, the corresponding member of the third transition series has a higher value than the member in the second in the same group. This is due to the lanthanide contraction, i.e., the atomic radii of corresponding elements in any group in these series (second and third transition series) are nearly the same but atomic number differs by 32. Thus the outer electrons are held firmly and higher energy required in removing such electrons.

    The following table of ionization energy of elements in kj mol^-1 confirms all the above mentioned points:

Li 520	Be 899	B 801	C 1086	N 1402	O 1314	F 1681	Ne 2080
Na 49	Mg 737.6	Al 577	Si 786	P 1011	S 999	Cl 1255	Ar 1520

 

Fe 762	Co 758	Ni 737
Ru 711	Rh 720	Pd 804
Os 840	Ir 900	Pt 870

Ex -2: The first (Δ_i H₁) and second (Δ_i H₂) ionization energies (kj mol^-1) of some elements are given below:

Element	Δi H1	Δi H2
I	2370	5250
II	520	7300
III	900	1760
IV	1680	3380

    Predict that who among them is likely a reactive metal, a metal that can form a stable binary compound with formula AX₂, a reactive non-metal and a noble gas.

5.         Electron affinity or electron gain enthalpy (Δ_eg H or E_A): it is defined as the energy change that occurs for the process of adding electron to an isolated neutral gaseous atom to convert it into a negative ion (to form a monovalent anion).

        In the above example 349 kj mol^-1 energy is released when one mole of isolated neutral gaseous atoms of chlorine are converted into chloride ions. The term ‘affinity’ refers to the measurement of the extent of attraction of a nucleus of an atom for the incoming electron. Greater the negative value of electron gain enthalpy, greater is the tendency to add one electron to the outer most level of an atom.

Note: since electron affinity is defined at absolute zero temperature, therefore, at any other temperature, heat capacity instead of electron affinity is to be considered. For this reason the two terms electron gain enthalpy and electron affinity are only same at absolute zero and at any other temperature the two terms are related as:

        The value of 5/2RT at 298k is just 2.477 kj mol^-1 which is very small and can be ignored. For this reason both the terms are considered to be same

        While for majority of the elements, energy is released (exothermic) when electron is added to the outer shell of atoms like halogens (attain stability by gaining electrons), for some other type of elements like noble gas elements energy is gained (endothermic). In other words noble gas elements have positive electron gain enthalpies because they attain a highly unstable state by adding electrons.

        Like second and higher ionization enthalpies, second and higher electron gain enthalpies are also possible. However addition of a second electron to a monovalent anion so as to make it X^2- is difficult as the new electron now experiences a repulsion from the electron cloud of the atom. Thus energy must be supplied (or gained by the atom) to add one more electron against the repulsion, and the second electron gain enthalpy becomes positive. the following example clarifies the above said concept: [Single click on android to see image clearly]

Ex-3: The electron gain enthalpy of bromine is 3.36 ev. How much energy in kcal is released when 8 gram of bromine is completely converted into Bromide ion in the gaseous state?

Solution: number of moles of bromine = 8/80 = 0.1 mole

Required energy = 0.1 x 3.36 x 23.06 kcal mol^-1 = 7.748 kcal mol^-1

Factors influencing electron gain enthalpy:

a.      Atomic Size: As previously explained under ionization energy, the distance between the nucleus and the last shell receiving the electron increases as atomic size increases. As a result the force of attraction between the nucleus and the incoming electron decreases. The electron is added less easily and electron gain enthalpy becomes less negative.

b.      Nuclear charge or effective nuclear charge: as the nuclear charge increases, the force of attraction between the nucleus and the incoming electron increases and hence the electron gain enthalpy increases, i.e., becomes more negative.

c.       Electronic configuration: elements having exactly half filled and fulfilled electronic configuration are very stable. Adding electron to such elements require energy since they do not accept electrons easily and on the addition of the electron they attain an unstable state. Thus electron gain enthalpies of such elements are positive. for example  nitrogen (1s² 2s² 2p³) has +31 kj mol^-1 of electron gain enthalpy.

 Variation of electron gain enthalpy: Both in period and group the values of electron gain enthalpies are not regular, still a generalization is being done as follows:

a.      Along a period: in general, the electron gain enthalpy becomes more and more negative as we move from left to right in a period. This is due to the decreasing atomic size and increasing nuclear charge. Thus the electron gain enthalpies of the halogen elements are the highest in a period and those of the noble gas has positive values.

b.      Down a group: In general, the electron gain enthalpies of elements in a group become less negative as we down. This is due to the increasing atomic size and decreasing nuclear charge.

Note: The electron gain enthalpies of some elements in the second period (O, F) are less negative than the corresponding elements in the third period (S, Cl).

      Due to the compact size of the O and F atoms, considerable electron-electron repulsion is already there inside such atoms. Thus further addition of electron is not as easier as it is the case with corresponding elements of the group such as S and Cl. As a result electron gain enthalpies of O and F atoms are less negative than S and chlorine respectively.

     The following table explains the above said points (electron affinities in kj mol^-1):

Li -60	Be +66	B -83	C -122	N +31	O -141	F -328	Ne +116
Na -53	Mg +67	Al -50	Si -119	P -74	S -200	Cl -349	Ar +96
K -48	Ca -	Ga -36	Ge -116	As -77	Se -195	Br -325	Kr +96

 

6.         Electronegativity: This is the property of a bonded atom. The relative tendency of an atom to attract the shared pair of electron towards itself is called the electronegativity of that atom.

  Consider covalently bonded molecules like H₂ or F₂ .  Since the two bonded atoms are identical, the shared pair of electrons is equally attracted by nuclei of the two atoms and the electron distribution around the two nuclei is similar. But when the bonded atoms in a molecule are different like HF, they attract the shared pair of electron to different extent and the electron distribution around the two nuclei does not remain symmetrical. The atom attracting the shared pair more gets a partial negative charge and the other gats a partial positive charge, as shown below:

Factors influencing electronegativity:

a.      Size of the atom: The smaller the size of the atom, the greater the attraction of bonding electrons, greater is the electroneggativity.

b.      Types of ion: Cations are more electronegative then the corresponding atom from which it is formed as the cation has smaller size (for example Li⁺ has E.N of 2.5 and Li has 1.0). Similarly anion has less electronegativity than the corresponding atom from which it is formed as anion larger in size (for example F^- has E.N of 0.78 and F has 4).

c.       Hybridization: The greater the s-character of the hybrid orbital of an atom greater is the electronegativity of that atom. Let us compare the basicity of three molecules:

    Since greater the s-character greater is the electronegativity the nitrogen atom, it tends to attract the lone pair of electron more towards it and availability of lone pair becomes less and basicity becomes less.

d.      Electron affinity and ionization energy: Mulliken says, the electronegativity of an atom is one half of the sum total of its E.A and I.E. Thus an element having higher E.A and I.E should have higher electronegativity.

e.      Effective nuclear charge: As shown by Allred and Rochow, smaller the effective nuclear charge (or greater the screening effect) smaller is the electronegativity. For example the electronegativities of the halogen atoms are F(4.0), Cl(3.0), Br(2.8), I(2.5).

 Measurement of electronegativity:

A.     Mullikens’ Scale (from electron affinities and ionization energy): According to Mulliken, the electronegativity of an atom is one half of the sum total of its electron affinity, E.A and ionization energy, I.E.

B.      Pauling method (from bond energy): According to Pauling, [Single click on android to see image clearly]

E_A-B   = bond energy of covalent bond A-B and accordingly are E_A-A and E_B-B.

The following example makes it clear:

Ex: 4: calculate the electronegativity of carbon in C-H bond if E_C-H, E_H-H and E_C-C are 98.8, 104 and 83 kcal mol^‑1 respectively. E.N of hydrogen = 2.05 [Single Click to see image solution clearly]

C.         Allred – Rochow scale: The force of attraction on the electron in the valence shell by the nucleus can be calculated using coulombs’ law:

 

Where Z*e = effective nuclear charge felt by the electron

Z* can be calculated by the Slaters’ rule.

e = charge on the electron, r = mean radius of the orbital (covalent radius).

Accordingly, the force of attraction of the atom on the shared pair of electron in a molecule which is known as electronegativity can be calculated by the formula:

 

 Ex: 5: Calculate the electronegativity of arsenic atom (z=33) having covalent radius 1.21 A° .

Solution: the effective nuclear charge can be calculated from Slaters’ rule and then the formula of Allred Rochow scale can be used to calculate the E.N of the said atom taking r = 1.21 A° . 

Trends in electeronegativity:

Along a period: Since the effective nuclear charge increases from left to right, the E.N also increases as  we move from left to right in a period. For example if we will consider the second period then the lowest value of E.N is of Li (0.98) where as the highest is that fluorine (3.98).

Down a group: Since the effective nuclear charge decreases from top to bottom in a group, the E.N also decreases accordingly. For example if we will consider the alkali metals then the highest value of E.N are of Lithium (0.98) where as the lowest is that cesium (0.79).

Application of electronegativities:

1.    Calculation of percent ionic character: A homonuclear diatomic molecule due to its similar electronegativities of atoms is nonpolar. Its bond can be considered as purely covalent. Whereas a partial ionic character is developed in a covalent bond joining any two different atoms differing in their electronegativities. This can be calculated by using Pauling method:

Percent ionic character = 18 (X_A - X_B) ^1.4   ;  X_A - X_B = E.N difference between two atoms A and B.

Note: It has been observed, that two atoms having an electronegativity difference greater than 1.7 (% ionic character = 50%) develop an ionic character in their bond predominantly. In a similar way E.N diff. less than 1.7 gives rise to a covalent bond predominantly. 

2.    Calculation of Bond length:

     In case a molecule (AB) is purely covalent (non polar) then Bond length (A-B) =

                          Covalent radius (r_A) of atom A in A° + cov. Rad. (r_B) of  atom B in A°

And in case atoms A and B differ in their electronegativities, bond A-B becomes polar, increased polarity in the bond results in the shortening of the bond.

Bond length of a polar bond A-B can be calculated using Shoemaker and Stevenson’s’ formula:

Bond length of a polar bond A-B = r_A + r_B – 0.09 (X_A - X_B) 

3.   Reactivity: Greater electronegativity difference also indicates greater reactivity. Let us compare the reactivity of Cl₂ and ICl molecule. Even a small electronegative difference between I and Cl atom in ICl molecule makes it more reactive than Cl₂ which has a covalent bond.

4.   Prediction of acidic hydrogen and bond cleavage of a particular type: Consider an example of acetic acid. It contains four hydrogen atoms, yet it releases only one hydrogen ion (H⁺) to the solution (mono basic acid). This is because the electronegative difference between the carbon and hydrogen is only 0.35 (2.55 – 2.2 = 0.35), whereas this difference between hydrogen and oxygen is 1.24 (3.44 – 2.2 = 1.24).

     Now consider Sulphuric acid. Since the E.N difference between O and S (3.44 – 2.58 = 0.86) is less than that between O and H (3.44 – 2.2 = 1.24), the bond between O and H breaks (more reactive) and acts as an acid.

     Similarly we can explain why NaOH is a base (furnish OH‑ ion) in solution not an acid (furnishes H⁺ ion). The bond between Na and O breaks due to the high E.N difference.

5.   Explanation of why HCl (g) is polar covalent and where as HCl (aq) is ionic. In HCl (g) the E.N difference is 0.96 (3.16 – 2.2 = 0.96), which is small and hence it is polar covalent. But in HCl (aq) water plays a vital role in weakening the force of attraction between H and Cl atom by a factor 80, (dielectric const. of water, D = 80) given by the Coulombs’ formula; thus H⁺ and Cl^- get separated and HCl is ionic by nature.

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Expected Long Questions (7-10 marks):

1.What is electronegativity? Factors affecting it. Trends. Applications.

2.Ionisation energy, Electron affinity, Slater’s Rule.

Expected Short Questions (2-3 marks):

Questions may be asked from Atomic radii. Lanthanide contraction. Slater’s Rule. Effective nuclear charge.  General electronic configurations of various blocks. Diagonal relationship. Applications of E.N. isotons, isosters, isoelectronic species. Comparision between E.A of F and Cl.

·            Which among 3d and 4s subshell should be closer to the nucleus of scandium?

·            The first and second ionisaton energies of magnesium atom are 7.64 and 15.03 eV respectively. Calculate the amount of energies required in kj to convert all the atoms in 12g of magnesium.

·            The five successive ionization energies of an element are 800, 2427, 3658, 25024 and 32824 kj mol^-1 respectively. What is the number of valence electrons in this element?

·            The second electron gain enthalpy of oxygen is positive. Explain.

·            The values of E.N of atoms A and B are 1.2 and 4.0 respectively. What is the percent ionic character in A—B bond?

·            Calculate the effective nuclear charge on a 3d electron of Zn atom.

·            What is the formula for determining the ionic radii of cation and anion in an ionic compound according to Pauli?

Hints:  r⁺ (radius of cation) = (Effective nuclear charge of anion in the compound / sum of eff. Nuclear charges of anion and cation) x internuclear distance in the compound,

Similarly    r ^--  (radius of anion) = (Effective nuclear charge of cation in the compound / sum of eff. Nuclear charges of anion and cation) x internuclear distance in the compound

·            Calculate the E.N of arsenic (Z = 33) having covalent radius 1.20 A⁰.

Expected Very Short Questions (1 mark):

1.Diagonal relationship is not shown by the pair:  Mg,Ba; Mg,Na; Mg,Cu; Mg,Cl

2.What is the characteristic electronic configuration of transition elements?

3.What is the number of elements in the fifth period of modern periodic table?

4.Ce (Z=58) belongs to which block?

5.Write the electronic configuration of Eu (Z= 63).

6.Which among Ag⁺ , K⁺ , Fe²⁺and Mg²⁺ is paramagnetic?

7.Which among oxygen, nitrogen, carbon and boron has highest ionization potential?

8.The most predominantly electrovalent compounds are formed between which groups?

9.What are isosters?

10.     What happens to the basic nature of oxides of elements in going from left to right in a period?

11.      Write four characteristics of transition metals.

12.     An element ‘E’ forming its highest oxide as E₂O₅ should belong to which group?

13.     Two elements A and B have 3 and 6 valence electrons respectively. What should be the formula of the compound made from them?

14.     Which among HF, HCl, HBr and HI is a strong acid and why?

15.     What is the name and atomic number of the last element of the modern periodic table?

16.      What is the relationship between pauling and Mulliken’s scale of E.N?

17.     What are the atomic numbers of the 14 elements of lanthanides?

18.     First member of each group shows anomalous behavior. Explain.

19.     The first ionization enthalpy of sodium is lower than that of magnesium but the second ionization energy of sodium is greater than that of magnesium. Explain.

20.     How can you summarise the nature of acidic strength of oxides of elements along a period and down the group? Hints: Acidic strength increases left to right in a period and decreases down a group.

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