The thing I am having trouble with: why does $\small\ce{PCl4}$ form a +1 ion and $\small\ce{Cl}$ a -1 ion?
There is no ionic bonding that occurs, the phosphorus is able to make a hybrid orbital out of its $\small3s$ and $\small3p$ electrons, along with an empty $\small d$ orbital to form 5 $sp^3d$ molecular orbitals, of equal energy, which yields 5 spots for bonding electrons.
This is a known exception to the octet rule.
(graciously borrowed from here)
-
$\begingroup$ Thanks, my information came from a 1957 Pauling lecture, so it was bound to be a little off. $\endgroup$ – Meow Sep 27 '12 at 20:44
-
$\begingroup$ @Alyosha No problem, I had to refresh my own memory. $\endgroup$ – jonsca♦ Sep 27 '12 at 20:46
-
$\begingroup$ Also- from the link you supplied: 'The 3-level electrons now rearrange (hybridise) themselves to give 5 hybrid orbitals, all of equal energy. ' How do they actually do this? Is the one electron from 3s pulled towards the 5th Chlorine very strongly so to make it a lower potential energy to 'jump out' towards it? $\endgroup$ – Meow Sep 27 '12 at 20:51
-
-
1$\begingroup$ The concept of spd hybrid orbitals should be completely removed from the textbooks, esp. for non-transition metal complexes. Hybridisation is a mathematical concept, that allows neither energy gain nor loss. The so-called hypervalancy is also a concept, that should be treated very critically. A mathematical explanation can be found with more-centre bonds. This concept can also explain the geometry of the molecule: $\ce{Cl\bond{-}PCl3\bond{-}Cl}$. $\endgroup$ – Martin - マーチン♦ Apr 8 '14 at 11:33
phosphorus has 5 valence electrons.each chlorine has 7 valence electrons.phosphorus give 1 electron to each chlorine and hence all the 6 attain 8 electrons in the outer shell thus forming an ionic bond.
-
-
2$\begingroup$ They mean all the six atoms end up with a closed valence shell. pappu is suggesting $\ce{PCl5}$ is an ionic substance composed of $\ce{P^{5+}}$ and $\ce{Cl^{-}}$ ions, which is a very poor description of the bonding. Discrete +5 ions are essentially impossible, because their resulting high charge density would force the molecules/counterions around them to donate electrons to the cation and either decrease their positive charge by electrons transfer or at least establish a covalent bond, making a polyatomic cation. $\endgroup$ – Nicolau Saker Neto Mar 22 '15 at 11:16