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PERIODIC TRENDS IN PHYSICAL PROPERTIES

Most of the properties of the elements such as atomic volume, atomic size, ionization enthalpy, electron affinity and electronegativity are directly related to the electronic configuration of the atoms. These properties undergo periodic variation with the change in the atomic number within a period or a group. These properties indirectly control the physical properties such as melting point, boiling point, density, etc. Let us now proceed to study the variation of some of the atomic properties in the periodic table.

ATOMIC RADIUS

The atomic size is very important property of the atoms because it is related to many other chemical and physical properties. In dealing with atomic size, the atom is assumed to be a sphere and its radius determines the size. In general, atomic radius is defined as the distance of closest approach to another identical atom. However, it is not possible to find precisely the radius of the atoms because of the following reasons:

  1. Atom is too small to be isolated.
  2. The electron cloud of an atom has no well-defined boundary.

It is for this reason that atomic size is expressed in terms of different types of radii. Some of these are being discussed below:

  1. COVALENT RADIUS

The approximate radii of atoms can be determined by measuring the distance between the atoms in a covalent molecule by X-ray diffraction and other spectroscopic techniques. This radius of an atom is referred to as covalent radius. It may be defined as one-half of the distance between the centres of the nuclei of two similar atoms bonded by a single covalent bond.

In case of homonuclear bonds,

r covalent = 2 [Intern clear distance between two bonded atoms]

For example, the intern clear distance between two hydrogen atoms in H2 molecule is 74 pm (Fig. 6.4). Therefore, the covalent radius of hydrogen atom is 37 pm.

Similarly internuclear distance between two chlorine atoms in chlorine molecule is 198 pm. Therefore, covalent radius of chlorine is 19812 = 99 pm.

 

2 ‘JAN DER WAALS’ RADIUS

Van der Waals ‘ radius may be defined as half of the internuclear distance between two adjacent atoms of the same element belonging to two nearest neighboring molecules of the same substance in solid state.

Covalent radius of the elements is shorter than its van der Waal radius. The formation of covalent bond involves overlapping of atomic orbitals. As a result of this, the internuclear distance between the covalently bonded atoms is less than the internuclear distance between the non-bonded atoms. This has been shown in Fig below. Thus, 

 

  1. METALLIC OR CRYSTAL RADIUS 

The term is applied for metals only. A metallic lattice is considered to be consisting of Closely packed atoms which are spherical in shape. Metallic radius may be defined as half of the internuclear distance between two adjacent atoms in the metallic lattice. It is measured in angstrom units. Internuclear distances are measured by X-ray studies. The metallic radius of sodium is found to be 186 pm while that of copper is 128 pm.

The metallic radius of an atom is always larger than its covalent radius because metallic bond is weaker than covalent bond. Therefore, two atoms held by covalent bond are closer to each other. For example, the metallic radius of sodium is 186 pm, whereas its covalent radius as determined from its vapours which exist as Na2 molecules is 154 pm. Similarly, metallic radius of potassium is 231 pm whereas covalent radius is 203 pm.

 

VARIATION OF ATOMIC RADII IN THE PERIODIC TABLE

Atomic radii usually depend upon nuclear charge and number of main energy levels of an atom. The periodic trends in atomic radii have been described as follows:

Table 6.5. Atomic Radii of Elements of Second Period

(a) Variation in a Period

In general, the atomic radii decrease with the increase in the atomic number in a period. For example, atomic radii decrease from lithium to fluorine in second period.

The decrease of atomic radii along a period can be explained on the basis of nuclear charge. In moving from left to right across the period, the nuclear charge increases progressively by one unit but the additional electron goes to the same principal shell. As a result, the electron cloud is pulled closer to the nucleus by the increased effective nuclear charge. This causes the decrease of atomic size. The atomic radii of elements of second period are given in Table 6.5.

(b) Variation in a Group

In general, the atomic radii increase from top to bottom within a group of the periodic table. For example, atomic radii increase from lithium to cesium among alkali metals and the similar trend is followed by halogens from fluorine to iodine as shown in Table

 Variation of Atomic Radii of Group-1 and Group-17 elements

In moving down a group, the nuclear charge increases with increase in atomic number but at the same time, there is a progressive increase in the principal energy levels. The number of electrons in the outermost shell, however, remains the same. Since, the effect of additional energy level is more  pronounced than the effect of increased nuclear charge, therefore, the effective nuclear charge decreases. Consequently, the distance of the outermost electron from the nucleus increases on going down the group. In other words, the atomic size goes on increasing as we move down a group.

In short,

Atomic radii increase down the group.

Atomic radii decrease across the period

IONIC RADIUS 

Ions are formed when the neutral atoms lose or gain electrons. A positive ion or cation is formed by the loss of one or more electrons by the neutral atom whereas a negative ion or anion is formed by the gain of one or more electrons by the atom. The term ionic radii refers to the size of the ions in the ionic crystals. The ionic radius of an ion may vary from crystal to crystal because of change in surrounding ions.

The equilibrium distance between the nuclei of the two adjacent ions can be determined by X-ray analysis of ionic crystals. Assuming ions to be spheres, the internuclear distance can be taken as the sum of the ionic radii of the adjacent ions (Fig. 6.6). Knowing the ionic radius of one of the ions, the ionic radius of other can be calculated.

Fig. 6.6.  lllustration of ionic radius

The radius of cation is smaller than that of the parent atom. 

Fig. 6.7. Relative sizes of Na atom and Na+ion

Cation is formed by the loss of one or more electron from the gaseous atom. Now, in the cation the nuclear charge remains the same as that in the parent atom but the number of electrons becomes less. As a result of this, the nuclear hold on the remaining electrons increases because of the increase in the effective nuclear charge per electron. This causes a decrease in the size.

In many cases the formation of cation also involves the removal of the valence shell completely. For example, formation of Na+ ion from Na atom involves the removal of third shell completely. This also result in the decrease in the size of the ion.

The comparative sizes of certain atoms and their corresponding cations are given in Table 6.7. 

Table 6.7. Atomic and Ionic Radii of Some Elements 

The radius of anion is larger than that of parent atom. Anion is formed by the gain of one or more electrons by the gaseous atom. In the anion, the nuclear charge is the same as that in the parent atom but the number of electrons has increased. Since same nuclear charge now acts on increased number of electrons, the effective nuclear charge per electron decreases in the anion. The electron cloud is held less tightly by the nucleus. This causes increase in the size. The relative sizes of chlorine atom and chloride ion have been shown in Fig. 6.8.

Fig. 6.8. Relative sizes of Cl atom and Cl- ion.

The comparative sizes of some atoms and their corresponding anions are given in Table 6.8. 

Table 6.8. Atomic and Ionic Radii of Some Elements 

ISO-ELECTRONIC IONS

 The ions having same number of electrons but different magnitude of nuclear charge are called iso-electronic ions. In fact, these are the ions of different elements having same electronic arrangement. For example, each one of sulphide (S2-), chloride (Cl-) and potassium (K+) ion has eighteen electrons but they have different nuclear charge, +16, +17 and + 19 respectively.

Variation of size among iso-electronic ions. Within the series of iso-electronic ions, as the nuclear charge increases, the attractive force between the electrons and nucleus also increases. This results in the decrease of ionic radius. In other words, size of the iso-electronic ions decreases with the increase in the magnitude of nuclear charge. For example, N3-, ()2-, F”, Na+, Mg 2+, Al3+ are iso-electronic and have 10 electrons each. The sizes of these ions are in the order:

Mg2+ having the highest nuclear charge (12 units) bas the smallest size whereas N3- ion having the smallest nuclear charge (7 units) has the largest size. Variation of size among these ions has been shown in Table 6.9. 

Table 6.9. Variation of Size Among Iso-electronic Ions

 

SOLVED EXAMPLES

`Example 6.3 The following species are isoelectronic with the noble gas neon. Arrange them in order of increasing size: Na+, r-, Q2-, Mg2+, Al3+.

Solution . In Na+, p-, 02-, Mg2+, AJ3+, the nuclear charges are 11, 9, 8, 12 and 13 respectively. Among isoelectronic species, greater the nuclear charge smaller is the size. Therefore, the sizes of the above ionic species are in the order:

Al3+ < Mg2+ < Na+ < F- < O2.

 

Example 6.4. Select from each group the species which has the smallest radius stating appropriate reason.

(i) O,O-, O2-   (ii) K+, srl+ , Ar

(iii) Si, P, CI

Solution. (i) The species 0 has the smallest radius because the radius of anion is always larger than the radius of the atom from which it is formed. o- and o:z- are anions of oxygen.

(ii) K+ has the smallest radius. In K+ and AI the outermost shell is third whereas in sc2+ it is fourth. Out of K+ and AI, K+ has smaller size because it has greater nuclear charge.

(iii) Cl has the smallest radius. Si, P and Cl belong to same period. In a period atomic radius decreases with increase in atomic number due to increase in effective nuclear charge.

 

Example 6.5. Name a species that will be isoelectronic with each of the following atoms or ions:

(i) Ne (ii) Cl (iii) Ca2+ (iv) Rb

Solution. (i) Sodium ion, Na+

(ii) Potassium ion, K+

(iii) Sulphide ion, S2 (iv) sr+

 

Example 6.6. Out of Na+ and Na which has smaller size and why?

Solution. Na+ has smaller size than Na. Na+ is formed by removal of one electron from Na. Therefore, Na+ has one electron less than Na. However, Na and Na+ have same nuclear charge. Therefore, electrons in Na+ are more tightly held than in Na. Moreover, removal of one electron from Na leads to complete removal of the third shell so that in Na+, the outermost shell is second. Hence, Na+ has smaller size than Na.

 

Example 6.7. Give examples of three cations and three anions which are isoelectronic with argon.

Solution       Cations: K+, Ca2+, Sc3+

Anions: CI-,

Ionization Energy

It is a well-known fact that the electrons in an atom are  attracted by the positively charged nucleus. In order to remove electron from an atom, energy has to be supplied to it to

overcome the attractive force. This energy is referred to as ionization energy. Thus, Ionization energy may be defined as the amount of energy required to remove the most loosely bound electron from the isolated gaseous atom in its ground state.

A (g) + Energy (I.E.)                          A+ (g) + e-

 

Ionization energy is a very important property which gives an idea about the tendency of an atom to form a gaseous positive ion.

Ionization energy is expressed in terms of kilo Joules per mole of atoms (kJ moz-1)

 

SUCCESSIVE  IONIZATION  ENERGIES

Once the first electron has been removed from the gaseous atom, it is possible to remove second and successive electrons from positive ions one after the other. For example,

The amounts of energies required to remove most loosely bound electron from unipositive, dipositive, tripositive  ions of the element in gaseous state are called second, third, fourth  ionization energies respectively.

 

The second, third, fourth, etc. ionization energies are collectively known as successive ionization energies. It may be noted that:

The variation in the values of successive ionization energies can be explained, in general, as follows:

 

After the removal of first electron, the atom changes into monopositive ion. In the ion, the number of electrons decreases but the nuclear charge remains the same as in the parent atom. As a result of this, effective nuclear charge per electron increases. The remaining electrons are, therefore, held more tightly by the nucleus. Thus, more energy is required to remove the second electron. Hence, the value of second ionization energy is greater than the first.

 

Similarly, the removal of second electron results in the formation of divalent positive ion and the attraction between the nucleus and the remaining electrons increases further. This

accounts for the progressive increase in the values of successive ionization energies.

 

If the removal of first, second or third electron also results in the removal of the valence shell, the next electron (which , belongs to lower energy shell) will require very large amount

of energy for its removal. For example, in sodium, the removal of first electron leads to the removal of third shell completely. Therefore, for sodium, I.E.2 >> LET

The first, second and third ionization energies of some elements are given in Table 6.10.

 

Table 6.10. I.E.l‘ I.E.2 and I.E.3 Values

of Some Elements

FACTORS ON WHICH ION IZATION EN ERGY DEPENDS

The ionization energy depends upon the following factors:

1 . Size of the atom;

  1. Magnitude of nuclear charge;
  2. Screening effect of the inner electron;
  3. Penetration effect of the electrons; and
  4. Electronic configuration.

 

  1. Size of the Atom. The attractive force between the electron and the nucleus is inversely proportional to the distance between them. Therefore, as the size of the atom increases, the outermost electrons are less tightly held by the nucleus. As a result, it becomes easier to remove the electron. Therefore, ionization energy decreases with increase in  atomic size.

 

  1. Magnitude of Nuclear Charge. The attractive force between the nucleus and the electrons increases with the increase in nuclear charge provided their main energy shell remains the same. This is because, the force of attraction is directly proportional to the product of charges on the nucleus  and that on the electron. Therefore, with the increase in I nuclear charge, it becomes more difficult to remove an electron and, therefore, ionization energy increases.

 

  1. Screening Effect of the Inner Electrons. In multi electron atoms, the electrons present in the outermost shell do not experience the complete nuclear charge because of

repulsive interaction of the intervening electrons. Thus, the outermost electrons are shielded or screened from the nucleus by the inner electrons. This is known as screening effect.

 

If the number of electrons in the inner shells is large, the screening effect will be large. As a result, the attractive interactions between the nucleus and outermost electrons will be less. Consequently, ionization energy will decrease. Thus, if other factors do not change, an increase in the number of inner electrons tends to decrease the ionization energy.

See also  FARADAY’S LAWS OF ELECTROLYSIS AND CALCULATIONS

 

  1. Penetration Effect of the Electrons. It is a well-known fact that in case of multi-electron atoms, the electrons in the s-orbital have the maximum probability of being found near the nucleus and this probability goes on decreasing in case of p, d and f orbitals. In other words, s-electrons are more penetrating towards the nucleus than p-electrons. The penetration power decreases in a given shell (same value of n) in the order: s > p > d > f

 

Now, if the penetration of the electron is more, it experiences less shielding effect by the inner electrons and will be held firmly. Consequently, ionization energy will be high. This means· that ionization energy increases with increase in penetration power of the electrons. Thus, for the same shell, it is easier to remove the p-electrons in comparison to the s-electrons.

 

  1. Electronic Configuration. It has been noticed that certain electronic configurations are more stable than the other. The atom having a more stable configuration has less tendency to lose the electron and consequently, has high value of ionization energy.. For example:

 

  1. The noble gases have stable configuration (ns2np6). They have highest ionization energies within their respective periods.

 

  1. The elements like N (ls2, 2s2, 2p1, 2py, 2p1) and P ( l s 2 2s 2 2p 6 3 s2 3p 1 3p 1 3p 1) have  configurations in which orbitals belonging to same sub-shell are exactly half-filled. Such  configurations are quite stable and consequently, require more energy for the removal of electron. Hence, their ionization energies are relatively high.

 

 VARIATION OF ION IZATION ENERGY IN THE PERIODIC TABLE

Let us now, study the variation of ionization energy across the period and along the group of representative elements on :be basis of above factors.
 Variation Across the Period

In general, the value of ionization energy increases with e increase in atomic number across the period. This can be attributed to the fact that moving across the period from left  right,

(i) nuclear charge increases regularly;

(ii) addition of electrons occurs in the same energy level;

(iii) atomic size decreases.

 

Thus, due to the gradual increase in nuclear charge and simultaneous decrease in atomic size, the attractive force between the nucleus and the electron cloud increases. Consequently  the electrons are more and more tightly bound to the nucleus. This results in the gradual increase in ionization across the period. The ionization energies of the ments of 2nd periods are given in Fig. 6.9.

On carefully examining the values given in Fig. 6.9, wed some exceptions within the period. These can be explained on the basis of other factors governing ionization energy .

(i) There is an increase in ionization energy from Li to Be. This is due to increased nuclear charge and smaller atomic size.

Fig. 6.9. Variation of ionization energies

     among elements of 2nd period.

 

(ii) There is a decrease in the value of ionization energy from Be to B inspite of increased nuclear charge. The effect of increased nuclear charge is cancelled by (a) greater penetration of 2s electron as compared to 2p electron: (b) better shielding of 2p electrons by the inner electrons. Thus. 2p electron of boron is relatively less tightly held by its nucleus in comparison to 2s electrons of Be .

 

(iii) There is a regular increase in ionization energy from B to C to N. It is again due to gradual increase of nuclear charge and decrease of atomic size.

 

(iv) There is slight decrease in ionization energy from N to O. It is attributed to the relatively stable configuration of the nitrogen due to a half filled 2p-orbital. In the nitrogen atom three 2p-electrons are present in different atomic orbitals (Hund’s rule) whereas in the oxygen atom, two of the four 2p-electrons are present in the same 2p-orbital resulting in an increased electron-electron repulsion. Therefore, it is easier to remove one of the 2p-electron from oxygen than it is, to remove one of the 2p-electrons from nitrogen.

 

(v) There is an expected increase in ionization energy from O to F to Ne.

 

Variation in a Group

The values of ionization energies of elements decrease regularly with the increase in atomic number within a group.

The values of ionization energies of the elements of group 1 have been represented graphically in

Variation of ionization energy among elements of group 1.

The decrease in the value of ionization energy within the group can be explained on the basis of net effect of the following factors:

 

As we move down the group there -is:

(i) A gradual increase in the atomic size due to progressive addition of new energy shells;

(ii) Increase in the shielding effect on the outermost electron due to in’2rease in the   number of inner electrons.

 

The nuclear charge also increases but the effect of increased nuclear charge is cancelled by the increase in atomic size and the shielding effect. Consequently, the nuclear hold on the valence electron decreases gradually and ionization energy also decreases. The variation of first ionization energy as a function of atomic number for elements with atomic number up to 60 have been shown in Fig. 6. 1 1 . The maxima in the curve represents noble gases which means that the

ionization energies of noble gases are highest- within their periods. The minima in the curve represent alkali metals which implies that ionization energies of alkali metals are the lowest

within their respective periods.

 

SOLVED EXAMPLES

Example 6.8  From each set, choose the atom which has the largest ionization energy and explain your answer

(i) F, 0, N.       (ii) Mg, P, Ar.

(iii) B, Al, Ga.

 

Solution . (i) F has the highest ionization energy among F, 0 and N because it has smallest size and highest nuclear charge. In general, ionization energy increases as we go from left to right in a period.

 

(ii) Ar (a noble gas) has the highest ionization energy among the elements Mg, P and Ar because it has stable electronic configuration and maximum nuclear charge.

 

(iii) B, AI and Ga belong to group- 13. Among these elements B has the largest ionization energy because on moving down a group, from top to bottom, ionization energy decreases. B is the first element of group 13.

 

Example 6.9. Out of Na+ and Ne which has higher ionization energy ? Explain why.

Solution. Na+ has higher ionization energy than Ne. Na+ and Ne are isoelectronic species. However, the nuclear charge in Na+ is more than in Ne. Hence, the electrons are more tightly held in Na+ and it has higher ionization energy.

 

Example 6. 10. How do you explain that 31 Ga has slightly higher ionization energy than 13Al, although it occupies lower position in the group ?

Solution.      13AJ : 1s2, 2s2, 2p6, 3s2, 3pl

13Ga : ls2, 2?, 2p6, 3s2, 3p6, 3d10, 4s2, 4p1

 

In Ga, the 10 electrons present in 3d-sub-shell do not shield the outer electrons from the nucleus effectively. As a result, effective 1uclear charge in Ga increases. This explains why ionization energy of Ga is slightly more than that of 13Al.

Electron Affinity, Eea

We have already learnt that ionization energy is a measure of the tendency of the atom to form cation. In the similar way, the tendency of a gaseous atom to form anion is expressed in terms of electron affinity.

Electron affinity may be defined as the energy change taking place when an isolated gaseous atom accepts an electron to form a monovalent gaseous anion.

The values of electron affinity are expressed in kilo joules per mole of atoms. For example, electron affinity of chlorine is – 348 kJ mol-1, i.e.,

Cl(g) + e- à    CI- (g) Eea = – 348 kJ mot-1

 

Depending on the element, the process of adding an electron can be either exothermic or endothermic.

The magnitude of electron affinity measures the tightness with which the atom can hold the additional electron. The large negative value of electron affinity reflects the greater tendency of an atom to accept the electron. For example, the elements of group-17 (the halogens) have  very high negative electron affinities because they can attain stable noble gas electronic configuration by gaining an electron. Thus, halogens have great tendency to accept an electron. On the other hand, the elements of group-18 (the noble gases) have large positive electron affinities because the additional electron goes to the next principal shell resulting in a very unstable electronic configuration.

 

FACTORS AFFECTING ELECTRON AFFINITY

Some of the important factors which affect electron affinity are discussed below:

  1. Nuclear Charge. Greater the magnitude of nuclear charge greater will be the attraction for the incoming electron and as a result, larger will be the negative value of electron affinity.

 

  1. Atomic Size. Larger the size of an atom is, more will be the distance between the nucleus and the additional electron and smaller will be the negative value of electron affinity.

 

  1. Electronic Configuration. Stable the electronic configuration of an atom is, lesser will be its tendency to accept the electron and larger will be the positive value of its electron affinity. For example, the elements having completely filled sub-levels of the valence shell have relatively stable configurations and consequently, possess large positive values of electron affinity.

 

VARIATION OF ELECTRON AFFINITY IN THE PERIODIC TABLE

Since experimental determination of electron affinity is not as easy as that of ionization energy, the sufficient data regarding electron affinities is not available. Consequently, the varying trends of electron affinities are not well-defined. The electron affinities of some elements are given in Table 6. 1 1 .
 

Table 6.11. Electron Affinities of Some Elements (kJ moi-1)

Variation in a Period

 

On moving across the period, the atomic size decreases and nuclear charge increases. Both these factors result into greater attraction for the incoming electron, therefore, electron

affinities tend to become more negative as we go from left to right across a period. However, some irregularities are observed among the elements of group 2, group 1 5 and group 18. The electronic configurations of these elements are relatively stable and hence, these elements have positive or very low negative electron gain enthalpies.

 

Variation Down a Group

On moving down a group, the atomic size as well as nuclear charge increases. But the effect of increase in atomic size is much more pronounced than that of nuclear charge and thus, the additional electron feels less attraction. Consequently, electron affinity becomes less negative on going down the group. Let us now examine the values of electron affinity for halogens as shown in Table 6. 1 1 .

 

It may be noted that the electron affinity becomes less negative as we go from chlorine to bromine to iodine. However, as we move from fluorine to chlorine the electron affinity becomes more negative whereas reverse was expected. This is because when an electron is added· to fluorine atom, it goes to the relatively compact second energy level. As a result, it experiences significant repulsion from the other electrons present in this shell. On the other hand, in chlorine atom, the added electron goes to the third energy shell which is relatively larger. Hence, it experiences Jess electron-electron repulsion. Therefore, electron affinity of fluorine is less negative as compared with chlorine. The unexpected trend is observed in case of many other elements of third period as their electron affinities are more negative than those of the elements of second period.

 

SUCCESSIVE ELECTRON AFFINITIES

When the first electron is added to the gaseous atom, it forms a uninegative ion and the energy change during the process is called .first electron affinity. Now, if an electron is added to the uninegative ion, it experiences a repulsive force from the anion. As a result, the energy has to be supplied to overcome the repulsive force. Thus, in order to add the second electron, the energy is required rather than released. Therefore, the value of second electron affinity is positive. Similarly, addition of third, fourth electrons, etc., also requires energy. Hence, the values of successive electron affinities are positive. For example, let us study the addition of electrons to oxygen atom

    SOLVED EXAMPLE

Example 6.11 . Which of the following will have the most negative electron affinity and which the least negative?

P, S, Cl, F.

Explain your answer.

Solution. Electron affinity generally becomes more negative across a period as we move from left to right. Therefore, among P, S and Cl the order of negative electron affinity is Cl > S > P. Within a group, electron affinity becomes less negative down a group. However, adding an electron to the 2p-orbital leads to greater repulsion than adding an electron to the larger 3p-orbital. Therefore, electron affinity of Cl is more negative than that of F. Hence, the element with most negative electron affinity is Cl the one with the least negative electron affinity is P.

Electronegativity

Electronegativity

Electronegativity may be defined as the tendency of an atom in a molecule to attract towards itself the shared pair of electrons.

It may be mentioned here that unlike ionization energy and electron affinity, electronegativity is not a measurable quantity. However, in order to compare the electronegativity values of different elements a number of numerical scales of electronegativity of elements have been proposed. The most common and widely use scale of electronegativity is the

 

Pauling scale

Pauling’s scale of electronegativity is based on excess bond energies. Electronegativity of fluorine, the most electronegative element, is arbitrarily taken as 4.0 in this scale.

The main factors on which the electronegativity depends are effective nuclear charge and atomic radius.

  • Greater the effective nuclear charge greater is the electronegativity.
  • Smaller the atomic radius greater is the electronegativity.

The electronegativity of any given element is not const3nt but varies depending on the element to which it is bound.

In a period electronegativity increases in moving from left to right. This is due to the reason that nuclear charge increases whereas atomic radius decreases as we move from left to right in a period. Halogens have the highest value of electronegativity in their respective periods. The electronegativity values (on Pauling scale) of elements of second and third periods are listed in Table 6.12.

Table 6.12. Electronegativity Values of Elements of Second and Third Periods

Elements of Second Period                  Li                Be

Electronegativity                                   1.0             1.5

Elements of Third Period                     Na             Mg

Electronegativity                                  0.9             1.2

       B           C         N         O           F

2.0          2.5     3.0         3.5      4.0

Al             Si      p           s          CI

1.5         1.8       2.1        2.5      3.0

 

In a group electronegativity decreases on moving down the group. This is due to the effect of increased atomic radius. for example, among halogens fluorine has the highest etectronegativity. In fact, fluorine has the highest value of etectronegativity among all the elements.

 

The electronegativity values (on Pauling scale) of group and group 17 elements are given in Table 6.13.

 

Relationship between Electronegativity and Non-metallic or Metallic) Character of an Element

Non-metallic elements have strong tendency to gain electrons. Therefore, electronegativity is directly related to he non-metallic character of elements. We can also say that the electronegativlty is inversely related to the metallic character of elements. Thus, the increase in electronegativities .across a period is accompanied by an increase in non-metallic character (or decrease in metallic character) of elements. Similarly, the decrease in electronegativity down a group is accompanied by a decrease in non-metallic character or increase in metallic character) of elements.

 

Electropositivity or Metallic Character

Tendency of atoms of an element to lose electrons and form positive ion is known as electropositivity.

A more electropositive element has more metallic character.

Whether an element behaves as a metal or a non-metal is directly related to its ionization energy. The elements having low values of ionization energies are metals whereas e1ements having high values of ionization energies are non-metals. The border line elements behave as metalloids.

 

Variation of Metallic Character Across a Period

Metallic character decreases across a period from left to right. On the other hand non-metallic character increases with increase in atomic number across a period. For example, let us consider elements of second and third periods.

In the second period, lithium and beryllium are metals, boron is a metalloid while carbon, nitrogen, oxygen, fluorine and neon are non-metals.

 

In the third period, sodium, magnesium and aluminium are metals, silicon is a metalloid while phosphorus, sulphur chlorine and argon are non-metals.

 

Variation of Metallic Character along a Group

On going along a group from top to bottom, the metallic character of elements increases.

In each group, the first element is least metallic while the last element is most metallic. For example, let us consider the elements of groups 14 and 15.

In group 14, the first element carbon is a non-metal, silicon and germanium are metalloids while tin and lead are metals.

In group 15, the first two elements, nitrogen and phosphorus, are non-metals while arsenic and antimony are metalloids and bismuth is a metal.

It may be mentioned here that metals generally form cations by losing electrons from the outermost shell while non-metals generally form anions by accepting one or more electrons. For example, alkali metals form M+ ions by losing

 

SOLVED EXAMPLES

Example 6.12 Arrange the following elements in the increasing order of metallic character:

8, At, Mg, K

 

Solution. Metallic character increases on moving down the group and decreases on going across a period from left to right. Hence, the· order of increasing metallic character is B <Al < Mg < K. one electron while alkaline earth metals form M2+ ions by losing two electrons from the outermost shell. The nonmetallic elements of group 17 form anions (X-) by accepting one electron.

 

Example 6.13 Arrange the following elements in the increasing order of non-metallic character:

B, C, Si , N, F

Solution .  Non-metallic character increases across a period from left to right and decreases on moving down the group from top to bottom. Hence, the order of increasing non-metallic character is

Si < B < C < N < F.

 

Example 6.14 Among the elements B, AI, C and Si

( i) Which has the highest first ionization energy?

(ii) Which has the most negative electron gain energy?

(iii) Which has the largest atomic radius?

(iv) Which has the most metallic character?

 

Solution . (i) Carbon (C) has the highest first ionization energy.

(ii) Carbon (C) has the most negative electron gain energy.

(iii) Aluminium (AI) has the largest atomic radius.

(iv) Aluminium (AI) has the most metallic character.

Valency

The valency of an element may be defined as the combining capacity of element.

The valency of an element may be expressed in terms of the number of electrons that an atom of the element makes available for bonding. The valency of an element is determined by the number of electrons in the outermost shell. These electrons are known as valence electrons.

In case of representative elements, the valency is generally  equal to either the number of valence electrons or eight minus the number of valence electrons. However,· the transition elements, exhibit variable valency.

 

VARIATION OF-VALENCY IN THE PERIODIC ABLE

variation in a Period

The number of valency electrons increases from 1 to 8 on moving across a period, the valency of the elements with respect to hydrogen and chlorine increases from 1 to 4 and then decreases from 4 to zero. Table 6.14 shows the valencies of the elements of 2nd and 3rd periods. The valencies of the elements have been shown in brackets.

 

Variation in a Group

On moving down a group, the number of valence electrons remains .same and, therefore, all the elements in a group exhibit same valency. For example, all the elements of group 1 have valency equal to 1 and those of group 2 have valency equal to 2.

 

SOLVED EXAMPLE

Example 6.15  Predict the formulae of the stable binary compounds that would be formed by the following pairs of elements:

(i) silicon and oxygen              (ii) aluminium and bromine

(iii) calcium and iodine

Solution. (i) Silicon belongs to group 14. Its valency is 4. The valency of oxygen is 2. Thus, the binary compound between silicon and oxygen would have formula Si02

(ii) Aluminium belongs to group 13. Its valency is 3. The valency of bromine is 1. Thus, the binary compound between  aluminium and bromine would have formula AlBr 3

(iii) Calcium belongs to group 2. Its valency is 2. The valency of iodine is 1. Thus, the compound between calcium and iodine would have formula CaI 2

Valency

The valency of an element may be defined as the combining capacity of element.

The valency of an element may be expressed in terms of the number of electrons that an atom of the element makes available for bonding. The valency of an element is determined by the number of electrons in the outermost shell. These electrons are known as valence electrons.

In case of representative elements, the valency is generally  equal to either the number of valence electrons or eight minus the number of valence electrons. However,· the transition elements, exhibit variable valency.

 

VARIATION OF-VALENCY IN THE PERIODIC ABLE

variation in a Period

The number of valency electrons increases from 1 to 8 on moving across a period, the valency of the elements with respect to hydrogen and chlorine increases from 1 to 4 and then decreases from 4 to zero. Table 6.14 shows the valencies of the elements of 2nd and 3rd periods. The valencies of the elements have been shown in brackets.

 

Variation in a Group

On moving down a group, the number of valence electrons remains .same and, therefore, all the elements in a group exhibit same valency. For example, all the elements of group 1 have valency equal to 1 and those of group 2 have valency equal to 2.

 

SOLVED EXAMPLE

Example 6.15  Predict the formulae of the stable binary compounds that would be formed by the following pairs of elements:

(i) silicon and oxygen              (ii) aluminium and bromine

(iii) calcium and iodine

Solution. (i) Silicon belongs to group 14. Its valency is 4. The valency of oxygen is 2. Thus, the binary compound between silicon and oxygen would have formula Si02

(ii) Aluminium belongs to group 13. Its valency is 3. The valency of bromine is 1. Thus, the binary compound between  aluminium and bromine would have formula AlBr 3

(iii) Calcium belongs to group 2. Its valency is 2. The valency of iodine is 1. Thus, the compound between calcium and iodine would have formula CaI 2

Periodic Trends and Chemical Reactivity

We have already studied the periodic trends in the various fundamental properties such as atomic and ionic radii, ionization energy, electron affinity, electronegativity and valence. The periodic trends in these properties can be explained on the basis of electronic configuration of the elements. The chemical reactivity of elements can be related to the fundamental properties of elements. As already discussed the ionization energy is least for the element at the extreme left of the period and the electron affinity is most negative for the element at the extreme right of the period (For group-17 elements). The elements of group-18 have positive electron affinity due to their stable electronic configurations. As a result, the chemical reactivity is maximum at the two extremes and lowest in the centre. The extreme reactivity of group-1 elements is due to the ease with which these elements can lose an electron leading. to the formation of corresponding cation. On the other hand, the reactivity of halogens is due to the ease with which these elements can gain an electron to form the corresponding anion. Thus, elements at the extreme left exhibit strong reducing behaviour whereas the elements at the extreme right exhibitstrong oxidizing behaviour.

 

Since ionization energy of alkali metals decreases on moving down the group their reactivity increases from lithium to caesium. For example, lithium reacts with water slowly, sodium reacts vigorously whereas potassium reacts almost violently. Reaction of caesium with water is explosive

2M + 2H2O                2MOH + H2

(M = Li, Na or K)

 

On the other hand among halogens, the reactivity decreases on going down the group because tendency to gain electron becomes less on descending the group.

Noble gases are almost inert due to their stable electronic configuration.

 

EVALUATION

  1. The correct order of size among Cl, Cl+ and CI- is

(a) CI+ < CI- < Cl (b) Cl+ > CI- > Cl

(c) Cl+ < Cl < CI- (d) CI- < Cl < CJ+.

 

  1. Considering the elements B, Al, Mg, and K, the correct order of their metallic character is

(a) B > Al > Mg > K               (b) Al > Mg > B > K

(c) Mg > Al > K > B                (d) K > Mg > Al > B.

 

  1. Which of the following represents the correct order of ionic radii?

(a) s2- > Cl- > K+ > Ca2+      (b) Ca2+ > K+ > CI- > s2-

(c) CI- > S2- > K+ > Ca2+      (d) S2- > Ca2+ > K+ > CI-.

 

  1. The first ionization energies of Na, Mg, Al and Si are in the order

(a) Na < Mg > Al < Si             (b) Na > Mg > Al > Si

(c) Na < Mg < Al > Si              (d) Na > Mg > AI < Si.

 

  1. Keeping in view, the periodic law and periodic table, suggest which of the following elements should have maximum electronegative character?

(a) Oxygen      (b) Nitrogen

(c) Fluorine      (d) Astatine.

 

  1. Which of the following structures is associated with the biggest jump between the second and the third ionization energies?

(a) 1s22s22p6              (b) 1s22s22p63 s1

(c) ls22s22p63s2         (d) 1s22s22p1.

  1. The statement that is not correct for periodic classification of elements is

(a)    the properties of elements are a periodic function of their atomic numbers.

(b)   non-metallic elements are less in number than metallic elements.

(c)    the first ionization energies of elements along a period do not vary in a regular manner   with increase in atomic number.

(d)   for transition elements, the d-subshells are filled with electrons monotonically with increase in atomic number.

 

  1. Which one of the following statements is incorrect in relation to ionization energy?

(a)    Ionization energy increases for each successive electron.

(b)   The greatest increase in ionization energy is experienced on removal of electron from core noble gas configuration.

(c)    End of valence electrons is marked by a big jump in ionization energy.

(d)   Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.

  1. The incorrect statement among the following is

(a) The first ionization energy of Al is less than the first ionization energy of Mg

(b) The second ionization energy of Mg is greater than the second ionization energy of Na

(c) The first ionization energy of Na is less than the first ionization energy of Mg

(d) The third ionization energy of Mg is greater than the third ionization energy of AL

 

  1.  Which of the following ions has the highest value of ionic radius?

(a) Li+ (b) BJ+

(c) oz- (d) p-

 

  1.  Which one of the following sets of ions represents the collection of isoelectronic species?

(a) K+, Ca2+, Sc3+, Cl-          (b) Na+, Ca2+, Sc3+, p-

(c) K+, Cl-, Mg2+, Sc3+         (d) Na+, Mg2+, Al3+, Cl-

(Atomic numbers: F= 9, Cl = 17, Na = 11, Mg = 12, Al = 13,

K = 19, Ca = 20, Sc = 21).

 

  1. Fill in the blanks.

12 . Complete the following sentences by supplying appropriate words.

(i) The most electropositive element in second period is …… whereas the most electronegative element is …….

(ii) The ionization energy of nitrogen is …… than that of oxygen.

(iii) There are …… periods in long form of periodic table and …… groups.

(v) The d-block elements are known as …… metals.

(vi) Ca2+ has a smaller ionic radius than K+ because it has …… .

(vii) Among halogens, the most negative electron affinity is of

(viii) The element with highest value of first ionization energy is …… .

(ix) Metallic radius is …… than the covalent radius.

(x) Moderm periodic law was proposed by ……

(xi) Metallic character …… from top to bottom in a group.

(xii) Amongst the elements Na, Mg and Al the highest value of second ionization energy is of …… .

 

  1. Out of metallic radius and covalent radius of an element, which is larger and why?
  2. Why van der Waals’ radius of an element is always larger than the covalent radius?
  3. Out of the following ions which has the smallest ionic size?

Li+, Na+, K+

  1. Arrange the following ions in the increasing order of their sizes.

CI-, P3-, s2-, F –

  1. Arrange the following sets of ions in the decreasing order of their sizes

(i) Al3-, Mg2+, Na+,O2-, F(

(ii) Na+, Mg2+, K+.

  1. Out of I and I+ which has larger size and why?
  2. Which element has the highest ionization energy?
  3. Among the alkali metals which element has the highest ionization energy?
  4. Arrange each of the following sets of elements in the increasing order of their ionization energies:

(i) 0, N, S                    (ii) C, N, 0

(iii) Li, Be, Na             (iv) Ne, He, Ar.

  1. Out of Na (Z = 11) and Mg (Z = 12), which has higher second ionization energy and why?
  2. Explain why

(i) Be has higher ionization energy than B

(ii) 0 has lower ionization energy than N and F?

  1. Why halogens have highest negative electron affinities in their respective periods?
  2. Why noble gases have largest positive electron affinity in their respective periods?
  3.  Comment on the statement that “all elements having high ionization energies also have high negative electron affinities.”
  4. Out of oxygen and sulphur, which has greater negative electron affinity and why?
  5. Consider the electronic configurations of the elements X, Y and Z and answer.

X : ls2, 2s2, 2px1, 2py1

Y : ls2 2 s2 2p 2 2p 1 2p 1

Z: 1s2, 2s2, 2px1, 2py1, 2pz1

(i) Which element has highest negative electron affinity?

(ii) Which element has lowest negative electron affinity?

  1. In each of the following sets, arrange the elements in the increasing order of their negative electron affinities:

(i)C,N,O          (ii) 0, N, S

(iii) S, Cl, Ar    (iv) F, Cl, Br.

  1. What is the trend of metallic character on going down from top to bottom in a group?
  2. Define electronegativity. How does it vary along a period and along a group?
  3. What is the basic difference between the terms electron affinity and electronegativity?
  4. Why second ionization energy is always higher than first ionization energy?
  5. A, B, C are three elements with atomic numbers, Z- 1, Z, Z + 1 respectively. B is an inert gas.

Answer the following questions:

(i) Predict the group of A and C.

(ii) Which out of the three has positive electron gain energy and why?

(iii) Which of the three has least value of ionization energy?

  1. The electronic configurations of some elements are given as:

(a) [Ne] 3s2 3p3          (b) [Ne] 3s2 3p4

(c) [Ne] 3s2 3r             (d) [Ne] 3s2 3p63tf’ 4s1

(i) Which element will be most metallic ?

(ii) Which element will have most negative electron affinity?

(iii) Which element belongs to d-block?

(iv) Which element belongs to group 17?

(v) Out of a, b and c which will have least ionization energy?

  1. Account for the following:

(i) Be has slightly higher value of ionization energy than that of boron (B).

(ii) The ionization energy of Na+ is more than that of Ne although they have same configuration.

(iii) Electron affinity of Cl is more negative than that of F. Among the elements of second period Li to Ne pick out the element

(i) with the highest ionization energy

(ii) with highest negative electron affinity

(iii) with largest covalent radius

(iv) that is most reactive non-metal

(v) that is most reactive metal.

 

  1. The first (IE1) and the second (IE2) ionization energies (in kJ moJ-1) and the electron affinity (in kJ mol-1) of a few elements are a given as follows:

Which of the above element is likely to be:

(i) the least reactive element.

(ii) the most reactive metal.

(iii) the most reactive non-metal.

(iv) the least reactive non-metal.

(v) the metal which can form a stable binary halide of the formulae .

 

See also

PERIODICITY

AMINES AND AMIDES

CARBOHYDRATES

NATURAL AND SYNTHETIC POLYMERSFats and Oils As Higher Esters

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