Periodicity In Atomic Properties

Periodicity In Atomic Properties

In a period from left to right there is a regular change in electronic configuration of elements. In a group from top to bottom the outermost shell electronic configuration is similar. The chemical properties of the elements depend upon their electronic configuration. So there is a regular change in chemical properties in a period while in a group element have similar chemical properties. In a period as well as in a group there is a regular gradation (gradual increase or decrease in a particular property) in physical and chemical properties of elements with the change in atomic number. This regular gradation in properties is called periodicity and all these properties which are directly or indirectly related to the atomic structure or electronic configuration

of the elements are called atomic properties. Periodicity in some atomic properties is discussed as under

ATOMIC RADIUS

The radius of an atom may be taken as the distance between atomic nucleus and the outermost shell of electrons of the atom. The size of atom is very important because many physical and chemical properties of the atom are related to it. According to the heisenberg's uncertainty principle the position of a moving electron can not be accurately determined. So the distance between the nucleus and the outermost electron is uncertain. Atomic radius can be determined indirectly from the inter-nuclear distance between the two atoms in a gaseous diatomic molecule. This inter-nuclear distance between the two atoms is called bond length. Different types of atomic radii are discussed below.

$\bullet$Covalent radius: One half of the distance between the nuclei (inter-nuclear distance) of two covalently bonded atoms in a homo-diatomic molecule is called the covalent radius of that atom. The covalent bond must be single covalent bond. The covalent radius (r_A) of atom A in a molecule A₂ may be given as –

$r_A=\frac{d_{A-A}}{2}$

i.e. the distance between the nuclei of the two single covalently bonded atoms in a homo-diatomic molecule is equal to the sum of covalent radii of both the atoms.

d_A-A = r_A+r_A

$\bullet$Metallic radius: Metal atoms are assumed to be closely packed spheres in the metallic crystal. These metal atom spheres are considered to touch one another in the crystal. One half of the internuclear distance between the two closest metal atoms in the metallic crystal is called metallic radius.

Metallic radius > Covalent radius

For example – Metallic radius and covalent radius of potassium are 2.3 Å and 2.03Å respectively.

$\bullet$Van der Waal's Radius or Collision radius: The molecules of non metal atoms are generally gaseous. On cooling, the gaseous state changes to liquid which is followed solid state on further cooling. In the solid state, the non metallic elements usually exist as aggregations of molecules are held together by Van-der Wall forces. One half of the distance between the nuclei of two adjacent atoms belonging to two neighbouring molecules of an element in the solid state is called Van der Waal's radius. It may also be defined as half of the inter-nuclear distance of two non bonded neighbouring atoms of two adjacent molecules.

Vander Waal's radius > Metallic radius> Covalent radius

The Vander Waal's radius and covalent radius of chlorine atom are 1.80Å and 0.99Å respectively

$\bullet$Ionic Radius: A neutral atom changes to a cation by the loss of one or more electrons and to an anion by the gain of one or more electrons. The number of charge on cation and anion is equal to the number of electrons lost or gained respectively. The ionic radii of the ions present in an ionic crystal may be calculated from the internuclear distance between the two ions

(a) Radius of a Cation–Radius of a cation is invariably smaller than that of the corresponding neutral atom

Na Na⁺

Number of e^_ = 11 10

Number of p = 11 11

1s²2s²2p⁶3s¹ 1s²2s²2p⁶3s⁰

Reasons

The effective nuclear charge increases. For example in Na atom 11 electrons are attracted by 11 protons and in Na⁺ 10 electrons are attracted by 11 protons. Thus in the formation of cation number of electrons decreases and nuclear charge remains the same.

Generally the formation of cation results in the removal of the whole outer shell due to which interelectronic repulsion decreases. The interelectronic repulsion in Na is among $11e^{-}$and in Na⁺ among $10e^{-}$

(b) Radius of an anion – Radius of an anion is invariably bigger than that of the corresponding atom

Cl Cl^-

Number of = $e^{-}$ 17 18

Number of p = 17 17

Reasons

The effective nuclear charge decreases in the formation of anion. Thus the electrostatic force of attraction between the nucleus and the outer electrons decreases and the size of the anion increases.

Interelectronic repulsion increases due to which expansion of electron cloud takes place

Isoelectronic species:

A series of atoms, ions and molecules in which each species contains same number of electrons but different nuclear charge is called isoelectronic series

In isoelectronic series atomic radii decreases, nuclear charge increases as

(i) Number of electrons is same.

(ii) Number of protons is increasing

(iii) So the effective nuclear charge is increasing and atomic size is decreasing. In an isoelectronic series atomic size decreases with the increase of effective nuclear charge.

Some of the examples of isoelectronic series are as under

$S^{2-},C{l}^{-},K^{+},C{a}^{2+},S{c}^{3+}$

So₂, $N{O}_3^{-}$, $C{O}_3^{2-}$

N₂, CO , $C{N}^{-}$

NH₃, H₃O⁺

PERIODICITY IN ATOMIC RADIUS AND IONIC RADIUS

$\bullet$For normal elements

(a) In a period from left to right effective nuclear charge increases because the next electron fills in the same shell. So the atomic size decreases. For example the covalent radii of second period elements in Å are as follows –

Li Be B C N O F

1.23 0.89 0.80 0.77 0.74 0.74 0.72

(b) In a group, from top to bottom the number of shells increases. So the atomic size increases. Although the nuclear charge increases but its effect is negligible in comparison to the effect of increasing number of shells. For example the covalent radii of IA group elements in Å are as follows –

$\bullet$The atomic radius of inert gas (zero group) is shown largest in a period because of its Vander Waal's radius which is generally larger than the covalent radius. The Vander Waal's radius of inert gases also increases on moving from top to bottom in a group.

$\bullet$For transition elements – There are four series of transition elements

3d – Sc (21) to Zn (30)

4d – Y (39) to Cd (48)

5d – La (57), Hf (72) to Hg (80)

6d – Ac(89), Rf(104) …………. Unb (ununbium) 112 (incomplete)

(a) From left to right in a period

The atomic size first decreases due to the increase in effective nuclear charge and then becomes constant and then increases. In transition elements, electrons are filled in the (n-1)d orbitals. These (n-1)d electrons screen the ns electrons from the nucleus. So the force of attraction between the ns electrons and the nucleus decreases. This effect of (n-1)d electrons over ns electrons is called shielding effect or screening effect. The atomic size increases due to shielding effect and balances the decrease in size due to increase in nuclear charge to about 80%. Thus moving from left to right in a period, there is a very small decrease in size and it may be considered that size almost remains the same. In the first transition series the atomic size slightly decreases from Sc to Mn because effect of effective nuclear charge is stronger than the shielding effect. The atomic size from the Fe to Ni almost remains the same because both the effects balance each other. The atomic size from Cu to Zn slightly increases because shielding effect is more than the effective nuclear charge due to d¹⁰ structure of Cu and Zn. The atomic radii of the elements of 3d transition series are as under.

Inner transition elements - As we move along the lanthanide series, there is a decrease in atomic as well as ionic radius. The decrease in size is regular in ions but not so regular in atoms. This is called lanthanide contraction. The atomic radii in Å are as under

There are two peaks one at Eu (63) and other at Yb (70). This is due to the difference in metallic bonding. Except Eu and Yb other lanthanides contribute three electrons in metallic bond formation. These two atoms contribute two electrons in the bond formation leaving behind half filled and completely filled 4¦-orbitals respectively.

Cause of Lanthanide contraction – In lanthanides the additional electrons enters the (n-2)¦ orbital. The mutual shielding effect of (n-2)¦ electrons is very little because the shape of ¦-subshell is very much diffused. Thus the effective nuclear charge increases then the mutual shielding effect of (n-2) ¦ electrons. The outer electrons are attracted more by the nucleus. Consequently the atomic and ionic radii decreases from La (57) to Lu (71)

This type of contraction also occurs in actinides. The jump in contraction between the consecutive elements in the actinides is greater than lanthanides. This is due to the lesser shielding of 5¦-electrons which are therefore pulled more strongly by the nucleus.

(b) In a group

The atomic radius of elements increases moving from first transition series (3d) to second transition series (4d). This is due to the increase in number of shells with the increase in atomic number.

The atomic radii of second (4d) and third (5d) transition series in a group is almost same except Y(39) and La (57). In third transition series, there are fourteen lanthanides in between La (57) of III B and Hf (72) of IV B groups, so the atomic radius of Hf(72) decreases much due to lanthanide contraction in lanthanides. The difference in the nuclear charge in the elements of a group in first and second transition series is + 18 units while this difference in second and third transition series is + 32 units except Y (39)$\to$ La(57). Due to the increase of + 32 units in the nuclear charge there is a sizable decrease in the atomic radius which balances the increase in size due to the increase in number of shells.

So in a group moving from second to third transition series, the atomic radii of the elements almost remain the same except IIIB. The difference is about 0.02Å.

Illustration 1. Compare the size of Cl, Cl^–, and Fe⁺⁺ ion

Solution: $\frac{Z}{e}$.ratio for Cl = $\frac{17}{17}$ = 1.00

Cl^– = $\frac{17}{18}$ = 0.944 and Fe⁺⁺ = $\frac{26}{24}$ = 1.08

Cl^– > Cl > Fe⁺⁺

IONIZATION POTENTIAL OR IONIZATION ENERGY

The amount of energy required to remove the most loosely bound electron of the outermost shell (i.e. the outermost electron) from one mole of an isolated gaseous atom of an element in its ground state to produce a cation is known as ionization energy of that element. Ionization energy is generally expressed in electron volts, so it is also known as ionization potential. Energy required for the removal of first, second and the third electron from the gaseous atom is called first, second and third ionization energy respectively.

${\text{A}}_{\text{(g)}}\text{+I}{\text{E}}_{\text{1}}\to {{\text{A}}_{\text{(g)}}}^{\text{+}}\text{+}{\text{e}}^{\text{-}}$

${\mathrm{A}}_{(\mathrm{g})}^{+}+\mathrm{I}{\mathrm{E}}_2\to {\mathrm{A}}_{(\mathrm{g})}^{2+}+{\mathrm{e}}^{-}$

${\mathrm{A}}_{(\mathrm{g})}^{2+}+\mathrm{I}{\mathrm{E}}_3\to {\mathrm{A}}_{(\mathrm{g})}^{3+}+{\mathrm{e}}^{-}$

The order of first, second and third ionization energies may be given as, $\text{I}{\text{E}}_{\text{1}}\text{I}{\text{E}}_{\text{2}}\text{I}{\text{E}}_{\text{3}}$

This is because second and third electron is removed from unipositive and di-positive cations respectively. Effective nuclear charge increases with the increase of positive charge. So the attraction between the nucleus and the outermost electron increases and more energy is required for the removal of electron.

Factors Affecting Ionization Potential:

$\bullet$Atomic radius: The values of ionization potential of an element decreases as its atomic radius increases. This is because the electrostatic force of attraction between the nucleus and the outermost electron decreases as the distance between them increases. So the energy required for the removal of electron will comparatively be less.

$Ionisation\mathrm{potential}\alpha \frac{1}{Atomicradius}$.

$\bullet$Effective nuclear charge: The greater the effective charge on the nucleus of an atom, the more difficult it would be to remove an electron from the atom because electrostatic forces of attraction between the nucleus and the outermost electron increases. So the greater energy will be required to remove the electron.

Ionization potential$\alpha$Effective nuclear charge (Z_eff)

$\bullet$Penetration effect of the orbitals: The order of energy required to remove electron from s,p,d-and¦-orbitals of a shell is s>p>d>¦ because the distance of the electron from the nucleus increases. For example – The value of ionization potential of Be(Z=4, 1s²2s²) and Mg(Z=12, 1s²2s²2p⁶3s²) are more than the I.P.s of B (Z=5, 1s²2s²2px¹) and Al (Z= 13, 1s²2s²2p⁶3s²3px¹) because the penetration power of 2s and 3s electrons is more than 2p and 3p electrons respectively. More energy will be required to separate the electrons from 2s and 3s orbitals.

$\bullet$Shielding or screening effect: The shielding or screening effect increases if the number of electrons in the inner shells between the nucleus and the outermost electrons increases. This results in decreases of force of attraction between the nucleus and the outermost electron and lesser energy is required to separate the electron. Thus the value of I.P. decreases.

Ionization potential $\alpha$$\frac{1}{Shielding\mathrm{effect}}$$\bullet$Stability of half-filled and fully-filled orbitals: According to Hund&aposs rule the stability of half filled and completely filled degenerate orbitals is comparatively high. So comparatively more energy is required to separate the electron from the atoms having half filled and full filled electronic configuration.

For example:

$\bullet$Removal of electron is comparatively difficult from the half filled configuration of N (Z=7, Is²2s²2px¹2py¹pz¹).$\bullet$The ionization potential of inert gases is very high due to stable ns²np⁶ electronic configurations.PERIODICITY IN IONIZATION POTENTIAL $\bullet$For normal elements: On moving from left to right in a period, value of ionization potential of elements increases because effective nuclear charge also increases.Exceptions: $\bullet$In a period, the ionization energy of IIA group elements is more than the elements of IIIA because penetration power of s-electrons is more than that of p- electrons. The value of ionization energy of Be(1s²2s²) is more than B (1s²2s²px¹) because the penetration power of 2s-electrons of Be is more than the 2px electrons of B. Moreover the electrons are paired i.e. completely filled s-subshell.$\bullet$In a period, the ionization energy of VA group elements is more than the elements of VI A group because the half filled p³ configuration of VA group elements is comparatively of higher stability. VIA group elements (p⁴) have the tendency to acquire comparatively more stable (p³) configuration by the loss of one electron? Ionization energy of N(1s²2s²2px¹py¹pz¹)>O $1s^{2}2s^{2}2p_x^2{p}_y^1{p}_z^1$Thus P>S, As>Se.But the value of I.P. of Sb (VA) & Te (VIA) and Bi (VA) & Po(VIA) are according to general rule i.e.Sb (VA) Exceptions: $\bullet$In group IIIA the ionization potential of Al (13) is equal to the ionization potential of Ga(31). Before Ga (31) the electrons are filled in 3d – orbitals of ten transition elements. These 3d orbital electrons do not completely shield the 4p – electron. So the increase of +18 units in nuclear charge results in the greater increase of effective nuclear charge. Due to increase in nuclear charge the I.P. increases which counter balance the decrease in I.P. due to the increase in number of shells.$\bullet$The values of I.P. of Tl (81) and Pb (82) of sixth period is more than that of I.P. values of In (49) and Sn (50) of same groups in period fifth.This is because of the electrons are filled in 4¦-orbitals before Tl (81) and Pb (82) which do not completely shield the outer electrons. Thus increase in + 32 units in nuclear charge results in the increase of ionization potential values For transition elements: On moving from left to right in a transition series – $\bullet$As the atomic number increases the effective nuclear charge also increases. Hence the I.P. increases.$\bullet$The shielding effect of (n – 1)d electrons over ns electrons increases with the addition of electrons in (n-1)d orbitals. Hence the I.P. decreases.$\bullet$The increased values of I.P due to the increase of effective nuclear charge almost balances the decreased value of I.P. due to increase in shielding effect. There is a very small increase in the values of I.P. or it may be said that I.P. almost remains the same.$\bullet$In first transition series from Sc to Cr the value of I.P. increases because effect of increase in effective nuclear charge is more than the shielding effect. I.P. values of Fe, Co, Ni and Cu are almost same. Due to (n-1)d¹⁰ns² configuration of Zn, the first I.P. increases.On moving from top to bottom in a group in transition series: $\bullet$In a group on moving from first to second transition series, the values of I.P decreases because atomic size increases.$\bullet$When moving from second to third transition series the value of I.P some what increases except IIIB group [Y(39) $\to$ La(57)]. This is because of 14 electrons are filled in 4¦-orbitals of lanthanides which do not shield the 5d electrons effectively. Thus the increase in +32 units in nuclear charge results in the increase of I.P., on moving from left to right this effect decreases and becomes negligible in the later part.Illustration 2.Magnesium has the first and second ionization potential 7.646 and 15.035 eV respectively. What is the amount of energy required to convert all the magnesium atoms to Mg²⁺ ions present in 24 mg to magnesium vapour.Solution: Mg(g) + IE₁ $$ Mg⁺(g) + e^– IE₁ = 7.646 eV Mg⁺(g) + IE₂ $$ Mg⁺⁺ (g)+ e^–IE₂ = 15.035 eVSo, for the reaction Mg(g) $$ Mg⁺⁺ (g) + 2e^–, energy required = 7.646 eV+ 15.035 eV = 22.681 eV = 22.681 x 96.49 kJ mol^–1 = 2188.48.9 K mol^–1

24 mg of magnesium = 1 x 10^–3 moles of magnesium

$\therefore$ Total energy required = 2188.489 x 10^–3 J= 2.188 kJ

ELECTRON AFFINITY

The amount of energy released when an electron is added to the outermost shell of one mole of an isolated gaseous atom in its lower energy state is called electron affinity. Electron affinity just defined is actually first electron affinity since it corresponds to the addition of one electron only. In the process of adding further electron, the second electron will be added to gaseous anion against the electrostatic repulsion between the electron being added and the gaseous anion. Sometimes energy instead of being released is supplied for the addition of an electron to an anion.

$A_{\left(g\right)}+e^{-}\to {A_{\left(g\right)}}^{-}+E_1$

${A_{\left(g\right)}}^{-}+e^{-}+\text{ Energy supplied }\to {{\text{A}}_{\left(\text{g}\right)}}^{2-}$

Factors Affecting the Magnitude of Electron Affinity:

$\bullet$Atomic size: In general electron affinity value decreases with the increasing atomic radius because electrostatic force of attraction decreases between the electron being added and the atomic nucleus due to increase of distance between them.

Electron affinity $\alpha$$\frac{1}{Atomic\text{ size }}$$\bullet$Effective nuclear charge: Electron affinity value of the element increase as the effective nuclear charge on the atomic nucleus increases because electrostatic force of attraction between the electrons being added and the nucleus increases. As the electrostatic force of attraction increases, amount of energy released is more.Electron affinity a Effective nuclear charge (Z_eff)$\bullet$Screening or Shielding effect: Electron affinity value of the elements decreases with the increasing shielding or screening effect. The shielding effect between the outer electrons and the nucleus increases as the number of electrons increases in the inner shells. Electron affinity $\alpha$$\frac{1}{Shielding\text{ effect}}$

$\bullet$Stability of half filled and completely filled orbitals: The stability of half filled and completely filled degenerate orbitals of a sub shell is comparatively more, so it is difficult to add electron in such orbitals and lesser energy is released on addition of electrons hence the electron affinity value will decrease.

PERIODICITY IN ELECTRON AFFINITY

$\bullet$In general electron affinity value increases in moving from left to right in a period because effective nuclear charge increases.

Exceptions:

$\bullet$The electron affinity value of alkaline earth metals of IIA group is zero. (only Be and mg)

$\bullet$Electron affinity value of alkali metals of IA group is also approximately zero because these elements have the tendency of losing the electron instead of gaining the electron.

$\bullet$Electron affinity values of nitrogen and phosphorus (VA) are lesser than the electron affinity values of carbon and silicon respectively. It is due to the comparatively stable half filled configuration (np³) of nitrogen and phosphorus and the tendency to acquire the stable np³ configuration by the gain of one electron in carbon and silicon np² orbitals.

$\bullet$The theoretical value of the electron affinity of zero group inert gas elements is zero due to stable s²p⁶ configuration.

$\bullet$In a group moving from top to bottom the electron affinity value of elements decreases because the atomic size increases

Illustration 3. Why in lithium more powerful reductant than sodium in aqueous medium?

Solution: Lithium ion having smaller size possesses higher heat of hydration and thus net reaction becomes more exothermic0.

ELECTRONEGATIVITY

It may be defined as the tendency of an atom to attract shared pair of electrons towards itself in a covalently bonded molecules. The numerical value of the electronegativity of an element depends upon its ionization potential and electron affinity values. Higher ionization potential and higher electron affinity values implies higher electronegativity value.

ELECTRONEGATIVITY SCALES

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

Some arbitrary scales for the quantitative measurement of electronegativities are as under

$\bullet$Pauling's scale: Pauling related the resonance energy($\Delta$)_AB of a molecule AB with the electronegativities of the atoms A and B. If x_A and x_B are the electronegativities of atoms A and B respectively then

0.208$\sqrt{{\Delta }_{AB}}$ = x_A – x_B if x_A > x_B

or 0.043 $\Delta$_AB = 23.06 (x_A – x_B)²

${\Delta }_{AB}=\left[E_{A-B}-\sqrt{E_{A-A}\times E_{B-B}}\right]$

Where E_A-B, E_A-B, and E_B-B are the bond energy of A-B, A-A and B-B in kcalmol^-1 respectively.

$\bullet$Mulliken's electronegativity: According to Mulliken, the electronegativity of an element is the average value of its ionization potential and electron affinity. $=\frac{Electron\text{ affinity }+\text{ Ionisation potential}}{2}$(where both IE and EA are expressed in eV)

Mulliken's electronegativity (x_M) – 0.615)

or x_p = 0.336 x_m – 0.2

Mulliken's values are

$\approx$ 2.8 times greater than that of Pauling value.

$\bullet$Alfred Rochow's electronegativity: The electronegativity of an element is the electrostatic force of attraction between the electron present on the circumference of the outermost shell of its atom and the atomic nucleus. If the distance between the circumference of outermost shell and the nucleus is r and the effective nuclear charge Z_eff then electronegativity of an element:

$=\frac{Z_{eff}{e}^2}{r^2}=\frac{0.359Z_{eff}}{r^2}+0.744$, Z_eff = Z - $\sigma$

Z = the actual number of charge present on the nucleus i.e. number of protons and

$\sigma$ = Shielding constant

Factors Affecting the Magnitude of Electronegativity:

$\bullet$Atomic radius: As the atomic radius of the element increases the electronegativity value decrease.

$Electronegativity\alpha \frac{1}{Atomic\text{ radius }}$

$\bullet$Effective nuclear charge: The electronegativity value increases as the effective nuclear charge on the atomic nucleus increases.

Electronegativity $\alpha$ Effective nuclear charge (Z_eff)

$\bullet$Oxidation state of the atom: The electronegativity value increases as the oxidation state (i.e. the number of positive charge) of the atom increases.

$\bullet$Hybridization state of an atom in a molecule: if the s- character in the hybridization state of the atom increases electronegativity also increases because s-electrons are having more penetration effect. For example the electronegativity values of C-atom in various hybridization states are as under —

s-character is increasing

So the electronegativity value is increasing

PERIODICITY IN ELECTRONEGATIVITY

$\bullet$In a period moving from left to right, the electronegativity increases due to the increase in effective nuclear charge.

$\bullet$In a period the electronegativity value of IA alkali metal is minimum and that of VIIA halogen is maximum.

$\bullet$In a group moving from top to bottom, the electronegativity decreases because atomic radius increases.

$\bullet$The electronegativity value of F is maximum and that of Cs is minimum in the periodic table.

$\bullet$The electronegativity of Cs(55) should be more than Fr(87) but it is less. This is due to the increase of +32 units in nuclear charge of Fr which makes the effective nuclear charge comparatively high.

$\bullet$On moving from second to third transition series in a group [except Y(39) $\to$ La(57)] electronegativity increases due to the increase of +32 units in nuclear atoms.

$\bullet$The electronegativity of inert gas elements (zero group) is zero due to stable ns²np⁶ configuration. Inert gases are monoatomic molecules and the electronegativity is the property of bonded atoms.

Applications of Electronegativity:

$\bullet$Partial ionic character in covalent Bond: The ionic character of a covalent bond increases as the electronegativity difference of bonded atoms increases. According to Haney and Smith if the electronegativity difference of bonded atoms is $\Delta$x then.

Ionic character percentage of the bond =16$\Delta$x+3.5$\Delta$x²

If the value of $\Delta$x is 1.7 then ionic character percentage is about 50. For example the order of ionic character in H–X bond is as follows–

H–F > H–Cl > H–Br > H–I

Because the electronegativity difference of bonded atoms ($\Delta$x) decreases.

$\bullet$Bond strength If the electronegativity difference of covalently bonded atoms ($\Delta$x) increases, the bond energy of the covalent bond also increases. For example – the order of the H–X bond strength is H– F > H–Cl > H–Br > H–I

As the bond strength is decreasing the acid strength is increasing. So order of increasing acid strength is HF < HCl < HBr < HI

$\bullet$Acidic and basic nature of oxides of normal elements in a period: The acidic nature of the oxides of normal elements increases as we move from left to right in a period. In a period from left to right the electronegativity of the elements increases. So the difference of the electronegativities of Oxygen and the elements (x₀ –x_E) decreases. If the (x₀ – x_E) values is about 2.3 or more then oxide will be basic. If (x₀ – x_E) values is less than 2.3 the oxide will be acidic. The oxides of the IIIA elements are amphoteric.

The order of acidic or basic nature of the oxides of third period elements may be given as under

Illustration 4. Which two elements would you expect to combine in the most violent fashion?

Solution: Cs and F; Cs is most electropositive and F is most electronegative.

$\bullet$Metallic and non metallic properties of elements:

(a) The metallic character decreases as the electronegativity of the element increases

(b) On moving from left to right in a periods, the electronegativity of the elements increases. So the metallic character decreases.

(c) On moving down a group, the electronegativity of the elements decreases. So the metallic character increases.

$\bullet$Basic nature of the hydroxides of elements:

A hydroxide MOH of an element M may ionize in two ways in water.

M–O–H + H₂O $$ MO^– + H₃O⁺ …….(1)M–O–H +H₂O$$ MOH₂⁺ + OH^– …… (2)

If the ionization is according to eqn (1) then it is acidic. It is possible when ionic character of O–H bond is more than the ionic character of M-O bond i.e. x₀ – x_H > x₀–x_M where x₀, x_H and x_M are the electronegativities of oxygen, hydrogen and element respectively. If the ionization is according to eqn. (2) then it is basic. This is possible only when ionic character of O –H bond is less than M–O bond i.e x₀ – x_H <x₀ –x_M

Illustration 5. Arrange the following compounds in the order of their decreasing stability if the electro negative values of elements are as follows

H = 2.1 F = 4, Cl = 3.0, Br = 2.8

I = 2.3, N = 3.0

HF, NCl₃, HBr, HI, HCl

Solution: Electronegativity difference for

HF = 4.0 – 2.1 = 1.9

NCl₃ = 3.0 – 3.0 = 0.0

HBr = 2.8 – 2.1 = 0.7

HI = 2.3 – 2.1 = 0.2

HCl = 3.0 – 2.1 = 0.9

So their order of their decreasing stability

HF > HCl > HBr > HI > NCl₃

1.9 0.9 0.7 0.4 0.0

SOME OTHER PERIODIC PROPERTIES

$\bullet$Atomic Volume: Atomic volume increases in a group from top to bottom. The increase is due to the increases in the number of energy shells.

In period from left to right, atomic volume varies cyclically, i.e., it decreases at first for some elements, becomes minimum in the middle and then increases.

$\bullet$Density: The density of the elements in solid state varies periodically with their atomic numbers. At first, the density increases gradually in a period and becomes maximum somewhere for the central members and then start decreasing afterwards gradually.

In solids Osmium has the highest density (22.6) & in liquid Hg has the highest density (13.6)

$\bullet$Melting and boiling points: The melting points of the elements exhibit some periodicity with rise of atomic number. It is observed that elements with low values of atomic volumes have high melting point while elements with high values of atomic volumes have low melting points. In general, melting points of elements in any period at first increases and become maximum somewhere in the centre and thereafter begins to decreases

$\bullet$Nature of oxides: In a period, from left or right the basic nature of the oxides decreases while acidic nature of the oxides increases. Down the group, basic nature increases or acidic nature decreases. Oxides of the metal are generally basic and oxides of the non-metals are acidic or neutral. The oxide of the metalloids are generally amphoteric, The oxide of Al, Zn, Sn, As and Sb are amphoteric. We can summarize that as the electronegativity of element increases, acidic character of oxides increases. In general oxide of higher oxidation states of same elements are acidic and lower oxidation states are basic.

$\bullet$Nature of oxy – acids: In a period, the strength of the oxy – acids formed by non – metals increases from left to right

$\bullet$Nature of hydrides: The nature of the hydrides changes from basic to acidic in a period from left to right.

In a group, the acidic nature of the hydrides of non metals increases. The reducing nature also increases but stability decreases from top to bottom.

HF HCl HBr HI

$\to$strength of the acid increases

$\to$stability decreases

$\to$ reducing nature increases

Effective Nuclear Charge and Shielding Effect:

$\bullet$In a multielectronic atom, the effect of nuclear charge experienced by the outermost electron is less than the theoretical value of the nuclear charge (Z).

$\bullet$If the nuclear charge of an atom is Z, effective nuclear charge is Z_eff and shielding constant is s then according to Slater

$Z_{eff}=Z--\sigma$

DIAGONAL RELATIONSHIP

Elements of second period Li, Be and B resemble closely with the elements Mg, Al and Si of third period in the next higher group.

Diagonal relationship is due to similar size of ions and almost similar electronegativities of the elements or more correctly is due to similar polarizing power of ions (ionic charge/ionic radius²).

Diagonal relationship is predominant only in lighter non-transition elements.

Illustration 6. What would be the atomic number of alkaline earth element if it is discovered and belong to 8^th period?

Solution: 120

Illustration 7. Arrange the Na⁺, Mg⁺² and Al³⁺ ions in order of increasing;

(a) extent of hydration (d) ionic mobility

(b) hydration energy (e) $E_{RP}^{{}^{\circ }}$

(c) size of hydrated ions (f) size of ions

Solution: (a) Na⁺ < Mg⁺² < Al³⁺ (d) Al⁺³ < Mg⁺² < Na⁺ (b) Na⁺ < Mg⁺² < Al³⁺ (e) Na⁺ < Mg⁺² < Al³⁺

(c) Na⁺ < Mg⁺² < Al³⁺ (f) Al⁺³ < Mg⁺² < Na⁺