How come PH3 molecules have permanent dipole dipole attraction arh? Dont P and H have the same electronegativity?
Like ammonia (N and P are both Grp V), phosphorus hydride or phosphane (IUPAC names) aka phosphine (common name) has VSEPER geometry of trigonal pyramidal with 3 bond pairs and one lone pair. Phosphorus is more electronegative than hydrogen. Thus permanent dipole-dipole interactions exist between PH3 molecules.
Electronegativity increases from left to right, and bottom to top, of the periodic table. Most electronegative element is flourine, followed by oxygen, then third place is a draw between nitrogen and chlorine.
Originally posted by UltimaOnline:Like ammonia (N and P are both Grp V), phosphorus hydride or phosphane (IUPAC names) aka phosphine (common name) has VSEPER geometry of trigonal pyramidal with 3 bond pairs and one lone pair. Phosphorus is more electronegative than hydrogen. Thus permanent dipole-dipole interactions exist between PH3 molecules.
Electronegativity increases from left to right, and bottom to top, of the periodic table. Most electronegative element is flourine, followed by oxygen, then third place is a draw between nitrogen and chlorine.
erm no.. For PH3, the electronegativity difference is 0 between P and H (2.1 for each)..
Ah yes, you're correct. P and H both have the same electronegativity of 2.1.
Hold a moment...
Ok, there is more to this matter, but here is a brief summary. If anyone has anything further to add, please do so.
As the electron geometry is tetrahedral (the molecular geometry is trigonal pyramidal), you might expect the P bonding orbitals to be sp3 hybridized, in accordance to the bond angles.
However, note that because of the following (from Wikipedia) :
>>> PH3 is a trigonal pyramidal molecule with C3v molecular symmetry. The length of the P-H bond 1.42 Å, the H-P-H bond angles are 93.5°. The dipole moment is 0.58 D, which increases with substitution of methyl groups in the series: CH3PH2, 1.10 D; (CH3)2PH, 1.23 D; (CH3)3P, 1.19 D. In contrast, the dipole moments of amines decrease with substitution, starting with ammonia, which has a dipole moment of 1.47 D. The low dipole moment and almost orthogonal bond angles lead to the conclusion that in PH3 the P-H bonds are almost entirely pσ(P) – sσ(H) and the lone pair contributes only a little to the molecular orbitals. The high positive chemical shift of the P atom in31P NMR spectrum accords with the conclusion that the lone pair electrons occupy the 3s orbital and so are close to the P atom. This electronic structure leads to a lack of nucleophilicity and an inability to form hydrogen bonds. <<<
(edited : For those who might be confused, I will give a simplified explanation of the statement "the almost orthogonal bond angles lead to the conclusion that in PH3 the P-H bonds are almost entirely pσ(P) – sσ(H) and the lone pair contributes only a little to the molecular orbitals". Recall that p orbitals are px, py and pz, 90 deg angles. Whilst sp3 hybridized orbitals have 109.5 deg angles. Since in PH3 the angles are much closer to 90 deg than 109.5 deg, you would expect the P-H bonds to be mainly sigma bonds from the overlap of p electrons (from P) and s electrons (from H). Notice then, that the s orbitals are largely not involved in bonding, ie. the lone pair occupies the s orbital. Recall that the electrons in the s orbital, on average, are closer (ie. spend more time nearer) to the nucleus, compared to the electrons in the p orbitals. In conclusion, the lone pair is close to the P nucleus, and hence P has slightly increased electron density, which has an effect on polarity of the molecule.)
Hence, due to the particular nature of the PH3 molecular orbitals and the trigonal pyramidal molecular geometry, specifically that the lone pair occupying the 3s orbital and thus being relatively closer to the P atom than the bond pairs between P and H, there is consequently a slightly greater electron density and a slightly partial negative charge on the P atom (even as P and H have the same electronegativity), and hence a slight dipole moment, making PH3 a slightly polar molecule.
Therefore, you would expect the permanent dipole-dipole interactions to be weak, unlike the stronger hydrogen bonding in NH3. Induced dipole-dipole van der Waals interactions, that are present between all molecules (whether polar or non-polar), would play a significant role in the intermolecular attractive forces for PH3.
If anyone has anything further to add, please do so.
Thanks! =D
A student of mine came across something similar (and in fact related, to this thread's discussion... you should be able to figure out the connection, by the time you read the end of this post), in her SAJC 2007 prelim paper practice.
The SAJC question required that the ('A' level) student be aware of the fact that the bond angle in H2S is smaller than the bond angle in H2O. (Check for 'A' level students : Did you know this?). Since it was a Paper 1 qn, it is uncertain if SAJC expected their students to be able to explain why.
Read the following url for the answer (several suggestions by different people are made, you should be able to pick up the most relevant and helpful one from the list... how about that? another MCQ question for you right there!).
"Why is the bond angle smaller in H2S compared to in H2O?"
http://www.newton.dep.anl.gov/askasci/chem03/chem03447.htm
(When you're done with the above, ask yourself if you see the link to the PH3 vs NH3 discussion... see the connection? Yes, the helpful reply by Roberto Gregorius does add on an important point to my explanation (see previous post in this thread) on PH3, namely the link between atomic radii (or number of electron shells) and the *need* for differing extents of hybridization.)