EFFECTIVE NUCLEAR CHARGE:
Between the outer most valence electrons and the nucleus of an atom, there exists finite number of shells containing electrons. Due to the presence of these intervening electrons, the valence electrons are unable to experience the attractive pull of the actual number of protons in the nucleus. These intervening electrons act as shield between the valence electrons and protons in the nucleus. Thus, the presence of intervening (shielding) electrons reduces the electrostatic attraction between the protons in the nucleus and the valence electrons because intervening electrons repel the valence electrons. The concept of effective nuclear charge allows us to account for the effects of shielding on periodic properties.
The effective nuclear charge (Zeff) is the charge felt by the valence electron. Zeff is given by Zeff = Z – s. Where Z is the actual nuclear charge (atomic number of the element) and s is the shielding (screening) constant.
Probability of finding the electron is never zero even at large distance from the nucleus. Based on probability concept, an atom does not have well-defined boundary. Hence, exact value of the atomic radius can't be evaluated. Atomic radius is taken as the effective size which is the distance of the closet approach of one atom to another atom in a given bonding state.
Atomic radius can be
(A) Covalent radius:
It is one-half of the distance between the centres of two nuclei (of like atoms) bonded by a single covalent bond.
Single Bond Covalent Radius, SBCR -
(a) For Homoatomic moleucles
where XA and XB electronegativity values of high electronegative element A and less electronegative element B, respectively. This formula is given by Stevenson & Schomaker.
Ex.2 Calculate the bond length of C – X bond, if C – C bond length is 1.54 Å, X – X bond length is 1.00 Å and electronegativity values of C and X are 2.0 and 3.0 respectively .
C – C bond length = 1.54 Å
rC = 154/100 = 0.77 Å
rX = 100/2 = 0.50 Å
(2) C – X bond length
dC–X = rC + rX – 0.09 (Xx – XC)
= 0.77 + 0.50 – 0.09 (3 – 2)
= 0.77 + 0.50 – 0.09 × 1
= 1.27 – 0.09 = 1.18 Å
Thus C – X bond length is 1.18 Å
(B) Van der Waals radius (Collision radius) :
It is one - half of the internuclear distance between two adjacent atoms in two nearest neighbouring molecules of the substance in solid state.
van der Waal's radius does not apply to metal. its magnitude depends upon the packing of the atoms when the element is in the solid state.
Comparision of convalent radius and van der Waal's radius
(i) The van der Waal's force of attractions are weak, therefore, their internuclear distances in case of atoms held by van der Waal's forces are much larger than those of between covalently bonded atoms.Therefore van der Waal's radii are always larger than covalent radii.
(ii) A covalent bond is formed by the overlaping of two half-filled atomic orbitals, a part of the orbital becomes common. Therefore, covalent radii are always smaller than the van der Waals radii. For example,
Covalent radius (A)
vander Waal's radius (A)
(C) Metallic radius (crystal radius) :
It is one -half of the distance between the nuclei of two adjacent metal atoms in the metallic crystal lattice. Metallic radius of an element is always greater than its covalent radius. It is due to the fact that metallic bond (electrical attraction between positive charge of an atom and mobile electrons) is weaker than covalent bond and hence the hence the internuclear distance between the two adjacent atoms in a metallic crystal is longer than the internuclear distance between the covalently bonded atom.
For example :
Variation In a Period
Variation In a Group
In a period left to right:
In a group top to bottom :
Z increases by one unit
Z increases by more than one unit
Zeff. also increases
Zeff. almost remains constant (due to increased screening effect of inner shells electrons)
n remains constant (no of orbits)
n increases (no. of orbits)
As a result of these electrons are pulled close to the nucleus by the increased Zeff.
rn ∝ 1/Z*
Thus atomic radii decreases with increase in atomic number in a period from left to right
The effect of increased number of atomic shells overweigh the effect of increased screening effect.
The effective distance from the centre of nucleus of the ion up to which it has an influence in the ionic bond is called ionic radius.
It is formed by the toss of one or more electrons from the valence shell of an atom of an element.
(ii) In a Cation, the number
It is formed by the gain of one or more electrons In the valence shell of an atom of an element.
Anions are larger than the parent atoms because (1) Anion Is formed by gain of one or more electrons in the neutral atom and thus number of electron
For example :
Number of Protons
1s2 2s2 2p6 3s1
Number of Electrons
Number of Proton
Within a series of isoelectronic species as the nuclear charge increases, the force of attraction by the nucleus on the electrons also increases. As a result, the ionic radii of isoelectronic species decrease with increases in the magnitude of nuclear charges.
lonisation enthalpy/energy (IE) , sometimes also called ionisation potential (IP) , of an element is defined as the amount of energy required to remove an electron from an isolated gaseous atom of that element resulting in the formation of positive ion.
IE2 & IE3 are the IInd & IIlrd ionization energies to remove electron from monovalent and divalent cations respectively.
In general: (IE)1 < (IE)2 < (IE)3 < .............
because, as the number of electrons decreases, the attraction between the nucleus'and the remaining electrons increases considerably and hence subsequent 1.E.(s) increase.
(A) Size of the Atom : lonisation energy decreases with increase in atomic size. As the distance between the outermost electrons and the nucleus increases, the force of attraction between the valence shell electrons and the nucleus decreases. As a result, outer most electrons are held less firmly and lesser amount of energy is required to knock them out.
For example, ionisation energy decreases in a group from top to bottom with increase in atomic size.
(B) Nuclear Charge: The ionisation energy increases with increase in the nuclear charge. This is due to the fact that with increase in the nuclear charge, the electrons of the outer most shell are more firmly held by the nucleus and thus greater amount of energy is required to pull out an electron from the atom. For example, ionisation energy increases as we move from left to right along a period due to increase in nuclear charge.
(C) Shielding effect: The electrons in the inner shells act as a screen or shield between the nucleus and the electrons in the outermost shell. This is called shielding effect. The larger the number of electrons in the inner shells, greater is the screening effect and smaller the force of attraction and thus (IE) decreases.
(D) Penetration Effect of the Electron : The ionisation energy increases as the penentration effect of the electrons increases.
It is a well known fact that the electrons of the s-orbital has the maximum probability of being found near the nucleus and this probability goes on decreasing in case of p, d and f orbitals of the same energy level. Within the same energy level, the penetration effect decreases in the order
s > p > d > f
Greater the penetration effect of electron more firmly the electron will be held by the nucleus and thus higher will be the ionisation energy of the atom.
For example, ionisation energy of aluminium is comparatively less than magnesium as outer most electron is to be removed from p-orbital (having less penetration effect) in aluminium where as in magnesium it will be removed from s-orbital (having large penetration effect) of same energy level.
(E) Electronic Configuration : If an atom has exactly half-filled or completely filled orbitals, then such an arrangement has extrastability.
The removal of an electron from such an atom requires more energy than expected. For example,
Be IE1 > B IE1
As noble gases have completely filled electronic configuration, they have highest ionisation energies in their respective periods.
Ex.3 First and second ionisation energies of Mg(g) are 740 and 1450 kJ mol–1. Calculate percentage of Mg+(g) and Mg2+(g), if 1 g of Mg(g) absorbs 50 kJ of energy.
Sol. Number of moles of 1g of Mg = 1/24 = 0.0417
Energy required to convert Mg(g) to Mg+(g) = 0.0417 x 740 = 30.83 kJ
Remaining energy = 50 – 30.83 = 19.17 kJ
Number of moles of Mg2+ formed = 17.19/1450 = 0.0132
Thus, remaining Mg+ will be = 0.0417 – 0.0132 = 0.0285
% Mg+ = (0.0285/ 0.0417) × 100 = 68.35%
% Mg+ = 100 – 68.35 = 31.65%
ELECTRON GAIN ENTHALPY (ELECTRON AFFINITY) :
When an electron is added to a neutral gaseous atom (X) to convert it into a negative ion, the enthalpy change accompanying the process is defined as the electron gain enthalpy.
Electron gain enthalpy provides a measure of the ease with which an atom adds an electron to form anion.
Depending on the elements, the process of adding an electron to the atom can be either endothermic or exothermic. When an electron is added to the atom and the energy is released, the electron gain enthalpy is negative and when energy is needed to add an electron to the atom, the electron gain enthalpy is positive. The addition of second electron to an anion is opposed by electrostatic repulsion and hence the energy has to be supplied for the addition of second electron.
EA (i) is exothermic whereas EA(ii) is endothermic.
(i) Electron affinity
(ii) Electron affinity ∞ Effective nuclear charge (zeff)
(iii) Electron affinity
(iv) Stability of half filled and completely filled orbitals of a subshell is comparatively more and the addition of an extra electron to such an system is difficult and hence the electron affinity value decreases.
Ex.4 How many CI atoms can you ionise in the process he energy liberated for the process or one Avogadro number of atoms.
Given IP = 13.0 eVand EA= 3.60 eV
Sol. Let n atoms be ionised. 6.02 × 1023 × EA = n × IP
Ex.5 The first ionisation potential of Li is 5.4 eV and the electron affinity of CI is 3.6 eV Calculate Δ H in kcal mol–1 for the reaction.
Sol. The overall reaction is written into two partial equations
= 1.8 × 23.06 kcal mol–1
= 41.508 kcal mol–1
Ex.6 For the gaseous reaction, was calculated to be 19 kcal under conditions where the cations and anions were prevented by electrosatic separation from combining with each other. The ionisation potential of K is 4.3 eV. What is the electron effinity of F ?
Ex.7 The electron affinity of chlorine is 3.7 eV. How much energy in kcal is released when 2 g of chlorine is completely converted to CI– ion in a gaseous state? (1 eV = 23.06 kcal mol-1)
35.5 3.7 × 23.06 kcal
l .'. Energy released for conversion of 2 g gaseous chlorine into CI– ions
× 2 = 4.8 kcal
Electronegativity is a measure of the tendency of an element to attract electrons towards itself in a covalently bonded molecules .
The magnitude of electronegativity of an element depends upon its ionisation potential & electron affinity. Higher ionisation potential & electron affinity values indicate higher electronegativity value.
There is no direct method to measure the value of electronegativity, however, there are some scales to measure its value .
(a) Pauling's Scale: Linus Pauling developed a method for calculating relative electronegativities of most elements. According to Pauling
(b) Mulliken's scale
Electronegativity (EN) can be regarded as the average ofthe ionisation energy (IE) and the electron affinity (EA) of an atom.
If both (EA) and (IE) are determined in eV units then paulings's electronegativity (EN)p is related to Mulliken's electronegativity. Mulliken's values were about 2.8 times larger than the Pauling's values.
(c) Allred-Rochow's Electronegatlvlty Allred and Rochow defined electronegativity as the force exerted by the nucleus of an atom on its valence electrons:
where Zelleclive is the effective nuclear charge and r the covalent radius (in Å ).
Ex.8 lonisation potential and electron affinity of fluorine are 17.42 and 3.45 eV respectively. Calculate the electronegativity of fluorine.
Sol. According to Mulliken equation
when both IP and EA are taken in eV..
APPLICATIONS OF ELECTRONEGATIVITY :
(I) Nomenclature: Compounds formed from two nonmetals are called binary compounds. Name of more electronegative element is written at the end and 'ide' is suffixed to it. The name of less electronegative element is written before the name of more electronegative element of the formula.
Ex.9 Write the correct formula and name of the following (a) ICI or CIl (b) FCI or CIF (c) BrCI or CIBr (d) BrI or IBr (e)OF2 or F2O (f)Cl2O or OCI2
Sol. Correct formula Name
(a) I+ Cl– Iodine chloride
(b) CI+ F– Chlorine fluoride
(c) Br+ CI– Bromine chloride
(d) IBr Iodine bromide
(e) OF2 Oxygen difluoride
(f) Cl2O Dichlorine oxide
(II) Nature of Bond: If difference of electronegativities of the two elements is 1.7 or more, then ionic bond is formed between them whereas if it is less than 1.7, then covalent bond is formed. (HF is exception in which bond is covalent although difference of electronegativity is 1.9)
(iii) Metallic and Nonmetallic Nature : Generally values of electronegativity of metallic elements are low, whereas electronegativity values of nonnmetals are high.
(iv) Partial Ionic
Character in Covalent bonds Partial ionic characters are generated in covalent compounds by the difference of electronegativities.
Hanny and smith calculated percentage of ionic character from the difference of electronegativity.
Percentage of ionic character =
XA is electronegativity of element A
XB is electronegativity of element B
Δ = XA – XB
(v) Bond length
When difference of electronegativities of atoms present in a molecule is increased, then bond length decreases. Shoemaker and stephensen determined.
(vi) Bond Strength & Stability
Bond strength and stability of A – B increases on increase in difference of electronegativities of atoms A and B bonded A – B. Therefore H – F > H – Cl > H – Br > H – I
Ex.10 Electronegativity of which of the following is high ?
(2) H2C = CH2(sp2)
(3) CH ≡ CH(sp)
(4) Equal in all
Ex.11 CF3NH2 is not a base, whereas CH3NH2 is a base. What is the reason ?
Sol. Due to high electronegativity of F tendency of donating the lone pair of electrons present on N will be less
Ex.12 OF2 is called oxygen difluoride, whereas Cl2O is called dichlorine monoxide. Why ?
Sol. Electronegativity of O in OF2 is less than F. Therefore, there will be positive charge on oxygen and negative charge on fluorine. Whereas in Cl and O, electronegativity of Cl is less than that of O therefore there will be positive charge on Cl and negative charge on O. Positive charge is written first followed by negative charge.
Ex.13 Calculate the electronegativity of fluorine from the following data :
EH – H = 104.2 kcal mol–1,
EF–F = 36.6 kcal mol–1
EH–F = 134.6 kcal mol–1,
XH = 2.1
Sol. Let the electronegativity of fluorine be XF.
Applying Pauling's equation.
In this equation, dissociation energies are taken in kcal mol–1.
Ex.14 The electron affinity of chlorine is 3.7 eV. How much energy in kcal is released when 2 g of chlorine is completely converted to Cl– ion in a gaseous state ? (1 eV = 23.06 kcal mol–1)
35.5 3.7 × 23.06 kcal
∴ Energy released for conversion of 2 g gaseous chlorine into Cl– ions
× 2 = 4.8 kcal
Ex.15 Calculate the electronegativity of fluorine from following data :
EH–H = 104.2 kcal mol–1
EF–F = 36.6 kcal mol–1
EH–F = 134.6 kcal mol–1
Electronegativity of H is 2.05.
Sol. On Paulling scale :
(using B.E. in kcal mol–1)
= = 1.5534
xF = xH + 1.4434 = 2.05 + 1.5534 = 3.6034
(VII) METALLIC PROPERTY
Metals have the tendency to form cations by loss of electrons and this property makes the elements as electropositive elements or metals.
Oxygen react with all elements except noble gases, Au, Pd and Pt to form oxides. In general, metallic oxides (O2–), peroxides and super oxides are ionic solids.
The tendency of group IA metals (alkali metals) to form oxygen rich compounds increases from top to bottom i.e. with increasing cation radii and decreasing charge density on the metal ion.
IIA metals also show the similar trend. Except Be, the IIA metals react with oxygen at normal conditions to form normal ionic oxides and at high pressure of O2, they form peroxides (CaO2, SrO2, BaO2). Oxides of metals are called as basic anhydries as most of them combine with water forming hydroxides with no change in oxidation state of metals.
Oxides of IA and IIA dissolve in water forming basic solution where as other oxides do not dissolve in water.
Oxygen combines with many non-metals to form covalent oxides such as CO, CO2, SO2, P4O10, Cl2O7 etc.
Non-metals with limited supply of oxygen usually form oxides in which non-metals are present in lower oxidation states where as with excess of oxygen, oxides with higher oxidatin state are formed. Oxides of non-metals are called as acid anhydrides as most of them dissolve in water forming acids of oxy-acids.
P4O10 + 6H2O → 4H3 PO4 ;
SO3 + H2O → H2SO4 : Cl2O7 + H2O → 2HClO4