Mr Rayon Posted June 14, 2009 Share Posted June 14, 2009 ...out of curiosity, in a molecule, how does the polarity of a molecule determine the type of intermolecular force present? Link to comment Share on other sites More sharing options...
Mr Rayon Posted June 14, 2009 Author Share Posted June 14, 2009 In what ways do the bonding and molecular shape determine the polarity of a molecule? Link to comment Share on other sites More sharing options...
hermanntrude Posted June 15, 2009 Share Posted June 15, 2009 if a molecule is polar, it will exhibit dipole-dipole interactions, which are (usually) the weakest of the three commonly-quoted intermolecular forces. Check your textbook for a complete list, try a few questions on the subject and then come back with specific questions if you have any. Link to comment Share on other sites More sharing options...
Theophrastus Posted July 25, 2009 Share Posted July 25, 2009 (edited) Simply put: many. Very many. You see, the polarity of a bond, is based upon two things, electronegativity and the geometric configuration of atoms within the molecule. Electronegativity, can be simplified, as an atom's affinity (not quite sure if it's an adequate word, but I'll go for it) for electrons. When there is a difference in electronegativity, between two bonded atoms, within a compound, chrages are formed, as the electrons are not equally "shared," by the two nuclei, and this results in a polar covalent bond (however, usually, only differences of around 0.4 and greater, in electronegativity, allow this to have a noticeable effect). An ionic bond, I generally like to think of, as a private case of the polar covalent bond, as it simply means that in this tug- of- war, between the two atoms, by means of a significant difference in electronegativity, (generally a difference in electronegativity, greater than 1.7 is required IIRC) the atom with a lesser electronegativity loses an electron, while the other atom, with the greater electronegativity, gains it. As the number of protons and electrons is no longer proportional, within the atoms, one molecule has a negative charge, (the one that gains an electron) and the other a positive charge (that which loses the electron). These opposing charges then result in an attraction between the two atoms; thus the ionic bond. This then allows for the resultant ionic compound, to freely dissociate in an aqueous medium. Because of the nature of ionic bonds, ionic compounds, are by definition, polar. However, in the case of polar covalent bonds, while charges may exist between two bonded atoms, in a compound, this may not necessarily result in polarity, as geometric configuration, must be taken into account. In figuring out whether the resultant molecule is polar, it's often useful to think of this separation of charge along the bonds axis, induced by the difference in electronegativities, as a vector. The direction of this vector, can be seen as moving along the bond axis, from positive charge, to negative charge. This separation of charge, along a polar covalent bond, is known as a dipole, while the representative vector, a dipole moment. You then proceed to add all of these vectors (the dipole moments), to attain the final charge, of the molecule. This is where molecular geometry comes into play. For example, while the carbon to oxygen bonds, in [ce] CO2 [/ce] are polar, as the molecule is linear, in adding the vectors, moving in opposing directions, you end up with a zero net charge. In contrast, in water, the bent structure of the molecule, means that the resultant vector will be of an adequately high positive value, to make the molecule polar. You could often perform such an analysis given molecular structure alone, estimating for magnitude, with the difference in electronegativity. Hope this helps. (As always, some lovely links...) http://en.wikipedia.org/wiki/Dipole#Molecular_dipoles http://en.wikipedia.org/wiki/Electronegativity http://www.green-planet-solar-energy.com/images/PT-small-electroneg.gif Edited July 25, 2009 by Theophrastus Link to comment Share on other sites More sharing options...
hermanntrude Posted July 27, 2009 Share Posted July 27, 2009 didn't you already ask this question and then abandon the thread? Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now