ELECTRONEGATIVITY

Electronegativity and its applications
Pauling Scale and Mulliken Scale

1. Electro - negativity

2. Sincere efforts made by :-

3. Electro-negativity is a chemical property that describes the tendency of an atom or a functional group to attract electrons towards itself. Definition :-

4. In 1932 Pauling was first to purpose scale of electro negativity Pauling Scale

5. Principle on which Pauling Scale is based : Difference between the measure energy of AB bond and expected for purely covalent AB bond . ∆ E=E measure - E expected covalent

6. Conditions given by Pauling If two atom A and B have same electro negativity value than the molecules AB is bonded by purely covalent bond .Then energy of AB covalent bond would be the mean of energy of A2 and B2 molecules. E A-B = (E A-A +E B-B ) 1/2 ∆ E=E A-B - [E A-A * E B-B ] 1/2 E A-B = [E A-A * E B-B ] 1/2 ∆ E= 0

7. Second Conditions If two atoms A and B have different electro negativity . The bond AB will no longer be purely covalent and energy would be greater than mean of energy A2 and B2 molecule. E A-B >[E A-A * E B-B ] ∆ E >0 This excess bond energy ∆ E is known ionic covalent resonance energy.

8. Expression to co-relate ∆ E with electro negativity difference 0.18 √ ∆ E = χ A – χ B ,S.I unit for E=Kcal/mol 0.208√∆E= χ A - χ B, S.I unit for E= ev /mol 0.1017√ ∆E= χ A - χ B, S.I unit for E=-KJ/mol

9. Advantages And Disadvantages Advantages This method has simple theoretical bases Different values can be obtained for different oxidation state of same element . Disadvantages Only if electro affinities are known.

10. MULLIKEN’S SCALE

11. IN 1934 MULLIKEN SUGGESTED AN APPROACH TO ELECTRONEGATIVITY BASED ON IONISATION ENTHALPY & ELECTRON AFFINITY ACCORDING TO MULLIKEN ELECTRONEGATIVITY OF AN ELEMENT IS AIRTHMETIC MEAN OF ITS I.E. & E.A .

12. A B e- E = I(A) - E(B) If B A e- E = I(B) - E(A) If A B are formed then this process required less energy. ( I(A) –E(B) ) < ( I (B) - E(A) )

13. The correlation between Mulliken electronegativities (x- axis in KJ / mol) and Pauling electronegativities ( Y axis) MULLIKEN VALUE = 2.8 PAULING VALUE

14. ADVANTAGES Simple theoretical basis Different values can be obtained for different oxidation states of the same element disadvantages Only few electron affinities are known

15. Charge On The Atom Hybridization Ionization Energy & Electron Affinity Effective Nuclear Charge Effect of the Substituent FACTORS AFFECTING ELECTRONEGATIVITY 1. 2. 3. 4. 5.

16. The overall tendency of any bonded atom to attract electrons is clearly influenced by its bonding environment. As such electronegativity is more specifically a property of the orbital of an atom. It is now reasonable to expect that electronegativity should be influenced by such factors like state of hybridization , oxidation state and partial charge of the atom.

17. 1. CHARGE ON THE ATOM Oxidation state may be defined as the charge left on the central atom when all the other atoms of the compound have been removed in their usual oxidation states. An atom which acquires a positive charge would tend to attract electrons more strongly than a neutral atom. As the oxidation number of element increases, electronegativity increases.

18. For example In PCl5 & PCl3, P (+ V ) > P (+ III ) In SnCl4 & SnCl2, Sn (+ IV ) > Sn (+ II ) In TlCl4 & TlCl , Tl (+ IV ) > Tl (+ I ) Also HClO ₃ > HCl⁺O Thus, a cation will be more electronegative than parent atom ( M+ > M ). And a parent atom will be more electronegative than an anion ( M+ > M- )

19. So, the decreasing order of electronegativity will be: M+ > M > M- Higher the oxidation states, greater will be the force of attraction for the electrons and higher will be the Electronegativity. The following values of electronegativity (Allred- Rochow ) are illustrative:

20. 2. HYBRIDIZATION Hybrid orbitals are composed of orbitals of different character to a varying degree of mixing. Hybridization affects electronegativity values because s-orbital due to its higher penetration effect and thereby higher effective nuclear charge, is more tightly held than other orbitals. It also has lower energy. Consequently, it will have greater tendency to attract electrons and hence show higher electronegativity values.

21. This conclusion agrees with the observed variations of acidity among the compounds CH₄ , C₂H₄ & C₂H₂. Compound: Hybridization: sp³ sp² sp s-character : 25% 33% 50% Electronegativity of C atom increases as we move from CH₄ to C₂H₂ . Due to greater Electronegativity of carbon in C₂H₄ & C₂H₂ , the electron pair of C-H bond is pulled more towards carbon atom, releasing H+ ion. The H atoms in CH₄ are neutral while in C₂H₄ is slightly acidic & C₂H₂ is more acidic. CH₄ C₂H₄ C₂H₂

22. R NH₂ Molecule: sp³ sp² sp fairly basic feebly basic No basic character Similarly, the availability of the lone pair of electrons on the nitrogen atom in a number of compounds is determined by the state of hybridization involved. Here, the basic character increases with decreasing s-character in the hybrid orbitals .

23. 3. IONISATION ENERGY AND ELECTRON AFFINITY:: More is the ionization energy and electron affinity, more is the electron negativity

24. 4. EFFECTIVE NUCLEAR CHARGE:: As the effective nuclear charge increases , the electronegativity increases

25. APPLICATIONS OF ELECTRONEGATIVITY ::IN CALCULATION OF PERCENTAGE OF IONIC CHARACTER :: If two atoms have similar electronegativity then the bond between them would be covalent whereas if two atoms have different electronegativities then then bond will be predominantly ionic. Percentage ionic character= 18( Ӽ A - Ӽ B ) 1.4 (this is the suggested and empirical equation for calculating % ionic character .) On the basis of this equation 50% ionic character occurs when electronegativity difference 1.7 This equation was modified by Hanny and Smith Percentage ionic character     = 16 ( c A  -  c B ) + 3.5 ( c A  -  c B ) 2   When electronegativity difference is 2.1 then the bond is 50% ionic.

26. Larger the difference in electronegativity of two combining atoms more stable will be the bond. It is due to increase in ionic character . HF HCl HBr HI Xa - Xb 1.9 0.9 0.7 0.4 Stability = HF > HCl > HBr > HI 1.STABILITY OF BOND IN A MOLECULE

27. The bond angle decreases in the series NH3 > PH3 > AsH3 > SbH3 Lesser the electronegativity of the central atom in the poly atomic molecule , the lesser would be the magnitude of the bond angle. If electronegativity of central atom is less, it would not be able to hold the bonded electron towards itself . Therefore , bonding electron pair will shift more towards bonding electron . This would result in decrease in Bond pair-bond pair repulsion and decrease in bond angle. 2. BOND ANGLES IN MOLECUES

28. 3. Acidic and Basic Character of Molecule in Aqueous Solution Electro-negativity can help determine whether compound of the type M-OH will behave as an acid or base in aq. s ol.

29. Let Ӽ O-H > Ӽ O - Ӽ M i.e E.N. difference of O-H > E.N. difference of O & M Hence OH bond will be more polar than O-M bond And Ionization of molecule in aq sol. Will take place at O-H bond So H + ions will be released Nickel Hydroxide Eg :-

30. If Ӽ O-H < Ӽ O - Ӽ H i.e. E.N. difference of O-H > E.N. difference of O & M Then OH bond will be less polar than O-M bond i.e. M-O bond will be more polar .: Ionization of molecule in aq. sol. will take place at M-O bond So OH- ions will be released

31. Eg :-

32. 4. Calculation of bond length 2 Atoms A & B bonded through a covalent bond differ in E.N. Greater the polarity Shorter the bond length b/w A & B Covalent bond acquires ionic character or polarity δ + δ -

33. Bond length of A & B [d (A-B) ] = r A + r B or r + + r - or 2r A Shoemaker & Stevenson’s numerical equation for calculating bond length [d (A-B) ] = r A + r B - 0.09( Ӽ A - Ӽ B ) Where r A & r B are atomic or molecular covalent radii Normal length of covalent bond b/w A & B = r A + r B Reduction in bond length leads to more stability

34. 5. Heat Of formation For a diatomic molecule like A-B Δ f H= Δ E Where Δ E= Energy difference between measured energy of bond A-B & energy expected for the purely covalent bond A-B. For AB n type compounds Δ f H = n x 96.49 ( Ӽ A - Ӽ B ) 2 Or ( Ӽ A - Ӽ B ) = 0.102

35. Assumption Reactants and products have same no. of covalent bonds For compounds containing N & O - Δ f H = n x 96.49 ( Ӽ A - Ӽ B ) 2 – 231 n N – 105 n O n N = no. of nitrogen atoms n O = no. of oxygen atoms

36. What did we study?

37. Thank You