characteristics of covalent bond n.
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  2. 1. Bond polarity

  3. Uneven sharing of electrons produces partial negative charges on the atom having greater than half share, leaving partial positive charge on the atom having less than half share. There will be surplus of negative charge around one nucleus and deficiency around the other.

  4. Bond polarity depends on electronegativity difference. The greater the electronegativity difference, the greater the unevenness of sharing, the greater share is acquired by the initially more electronegative atom, thus electron density will be displaced in the direction of the more electron attracting atom. The greater the difference between the electron attracting power of the two atoms, the greater is the ionic character of the covalent bonds.

  5. Electronegativity: F > O > Cl, N > Br > C, H

  6. Question: The electronegativity difference of C - I bond is zero but the bond is polarizable. Why? • In polar environment this covalent bond can acquire an appreciable degree of ionic character. This is due to the polarizability of the outermost electrons of iodine as a consequence of the big size of iodine atom. The distance between the nucleus and the outermost shell is great so that the outermost electrons are not as strongly drawn towards the nucleus as are the electrons of small atoms.

  7. Formal positive charge effectively increases the electronegativity, thus increases polarity. Calculation of Formal charge: FC= group # - ½ ( # of shared electrons) – # of unshared electrons

  8. Example: 1. N in ammonia FC= 5- 3- 2 = 0 • Example: 2. N in ammonium FC = 5 – 4 = +1 more polar compared to ammonia

  9. The formal positive charge effectively increases the electronegativity of nitrogen. There will be a tendency to pull the electron closer to nitrogen resulting in increased partial positive charges on each of the four hydrogen atoms

  10. a. Determine which C – O is more polar and explain. or ( FC = 0) (FC = +1) The excess positive charge increases its electron attracting power so that there will be a greater displacement of electron density towards oxygen.

  11. b. Which C – H is more polar? or The C-H bond of chloroform is more polar than the C-H bond in methane because in chloroform adjacent atoms are chlorine atoms which possess greater electron-attracting capacity than hydrogen atoms in methane. The 3 chlorine will reduce the electron density around C, increasing the partial positive charge on C which in turn will attract electron from hydrogen increasing the partial positive charge on this atom.

  12. c. Which C – H is more polar? or The C-H bond in acetylene is more polar than C-H bond in ethylene because C in acetylene is an sp ( ½ s and ½ p character) while C in ethylene is an sp2 ( 1/3 s and 2/3 p). The s character of hybrid orbital is a measure of the electron-attracting capacity of the atom. The electron-attracting capacity of C in acetylene is greater than in that of ethylene. The electron attracting power as a result of s character of the hybrid orbital, responsible for the polarity of C-H bond in acetylene is also known as ORBITAL ELECTRONEGATIVITY.

  13. d. Which C – H is more polar? or The C-H bond in ethylene is more polar than the C-H bond in ethane because of orbital electronegativity. Importance of bond Polarity:1. It contributes to the characteristic of bond that is it influenced bond length and energy.2. It influences molecular polarity, thus in the end determines the physical properties like melting point, boiling point, solubility. 3. It influences chemical reactivity.

  14. 2. BOND LENGTH

  15. The bond length is the distance between those two atoms. The greater the number of electrons between two atoms, the closer the atoms can be brought towards one another, and the shorter the bond. This is measured in Angstrom units which 1A is equal to 10-8 cm. It has been shown that bond length is correlated with bond polarity, hybridization of orbital and delocalization of electrons.

  16. When two atoms approach each other, their interaction is influenced by: a. repulsion between two electron, clouds b. repulsion between the two nuclei c. attraction between the nucleus of each, and the electrons of the other.

  17. Factors influencing bond length a. Bond Polarity Increasing polarity decreasing bond length

  18. b. Hybridization Increasing s ( near the nucleus) character of hybrid orbital decreasing bond length or orbital electronegativity leads to reduction of bond distance. Ethane CH3-CH3 sp3 – sp3 1.538 Propylene CH3-CH=CH2 sp3 – sp2 1.501 Methylacetylene sp3-sp 1.459 Ethylene CH2=CH2 sp2-sp2; p-p 1.339 Acetylene sp-sp; p-p; p-p 1.207

  19. c.atomic size small atoms will form shorter bond. Example: H-H < CH4-CH4

  20. d. bond order decreasing bond order increasing bond length. Triple bond< double bond< single bond sp sp2 sp3

  21. e. π and σ electron delocalization illustrated by system containing atoms in the trigonal state of hybridization. Example: a. CH2 = CH - Cl b. O = C c. CH3 – CH = CH2


  23. energy required to break the bond and at the same time the energy release when the bond is formed. • This is express in terms of kilo calories per mole • Bond energy is variable depending on length, the shorter the bond the stronger the bond.

  24. Bond strengthening are attributed to: a. orbital hybridization – when the s character of the bonding orbitals is high, there will be closer interaction between the bonding electrons and the nuclei thus covalent bond is stronger. sp3 – sp3sp3- sp2sp3 – spsp2 – sp2Bond energy increases bond length decreasessp2 – spsp - sp

  25. b. bond polarity - when bond polarity increases bond energy also increases - C – C sp3 –sp3 0 = C – C = sp2 – sp2 0 - C = C - sp2 – sp2 1 increasing pi bond increasing = C – C= sp – sp 0 bond energy - C = C- sp-sp 2

  26. c. pi bonding - C – C sp3 –sp3 0 = C – C = sp2 – sp2 0- C = C - sp2 – sp2 1 increasing pi bond = C – C= sp – sp 0 increasing bond-C = C- sp-sp 2 energy

  27. d.Reduction of bond strength is observed when there is loss of overlap of hybrid orbitals. The loss of overlap is a consequence of forced bending of orbitals. A vivid example of loss of overlap as a result of forced bending of sp3orbitals is seen in cyclopropane. Forced bending results in angular strain due to compression of the tetrahedral bond angle.

  28. 4. BOND order

  29. The bond order is equal to the number of bonds between two atoms.

  30. The BO is an indication of the bond length, the greater the bond order, the shorter the bond and the greater the bond strength.

  31. 5. BOND ANGLE

  32. a. governed by hybridization of central atomsp3 - 109.5osp2 – 120 osp - 180 o

  33. b. influenced by the presence of lone pairsCH4 sp3 109o 28’NH3 sp3 106o 47’H2O sp3 104o 31’As the number of lone pair increases, the bond angle decreases.