Charges of anions...

[FreeThinker]

I am having a bit of trouble understanding the charges of certain atoms.

 

 

For example, Potassium reacts with Phosphorus producing Potassium phosphide.

 

The balanced equasion is: 3K+P--->K3P

 

Phosphorus has an atomic number of 15 and if I put 2 electrons in the first shell, 8 in the second, and 5 in the third I can see that it would gain 3 electrons from the cation.

 

But as soon as the atomic number gets bigger , I can not use this method to figure out the charge of certain anions. What I am doing wrong? Or is there no method and I simply have to remeber each charge?

 

 

Thanks in advance,

 

Free Thinker

[Tannin]

Generally we are looking only on the last non-complete shell:

phosphorus has electronic configuration of 1s2 2s2 2p6 3s2 3p3, but the complete shells play no part in the chemical bonding, so we are interested only in the p-electrons from 3p shell: we have 3 and we can take 3 more. Potassium is 1s2 2s2 2p6 3s1, but again we are looking only on the last non-complete shell and thus it has only one electron to donate.

Another way to say the same thing is "completing octet - stable 8 electrons configuration".

So everything you need to know is the number of column where the element is: P is in the 5th column, K is in the first. And then take the atoms in such a proportion that the sum of the electrons in their outer shells would equal 8.

But, unfortunately, there is no such thing like a rule without the exceptions of the rule.

The notable one is the "transition elements" (like Fe, Cr etc.). Here we are interested in the second from the last shell - e.g. 3d. And also another stable configurations (and not just octet) are possible.

The reason is that the energies of 4s and 3d orbitals become very close, and thus the order of their filling is not so easy to figure.

Also, the charge on the cations/anions may be defined in the different ways: the electrons are shared between the atoms and the charge (electron) transfer is never complete. So, it's rather a crude approximation to assume that K is 3+ cation.

But this is our way - moving from one approximation to another and more answers are available in the fascinating field of quantum chemistry.

[Ilja]

Just figure out the charge , like this:

 

K can only be in +1 so K3P there must the P be in -3 and so on.

 

Other examples are Oxygen , in compounds its in the majority of cases -2 charged.Or the earth-alkali metals ar always +2...

[FreeThinker]

Ilja ,I am not provided with balanced equation, that I have to work out my self. There fore I dont know that K3P. If I did I would not have a problem.

 

Tannin, I did not realise that the group number was actually the number of electrons in the outer shell. That helps a lot. Thanks you.

 

Everything is clear now, once again thanks for you help!

[rthmjohn]

So, it's rather a crude approximation to assume that K is 3+ cation.

 

What exactly did you mean by this? I'm guessing you just made a mistake and meant to say that P forms -3 anions because K only forms +1 cations. Or did I misunderstand you?...

[jowrose]

Freethinker, not only does the group number tell the number of electrons in the outer shell, but it also tells you the element's outer subshell. The skinny rectangle at the far left of the periodic table (with the alkali and alkaline earth metals) has elements that end in the s subshell. Their period tells which level the subshell belongs to. e.g. magnesium ends in 3s2.

 

Cross over the transition metals, and you get to the square box containing the metalloids and nonmetals. You'll notice that this section is 6 spaces wide, and it corresponds with all the p orbitals (with the exception of helium, which is 1s2). Again, the period of the element gives its element, so you can tell that chlorine ends in 3p5.

 

The transition metals contain the d subshell, but the d subshell doesn't occur in level 3, as it should. This is because the 4s subshell is slightly less energetic than the 3d, so the electrons fill it first. You must remember this detail, and infer that Iron ends in 3d6 (not 4d6, as the period of iron would imply), but is preceded by 4s2.

 

The f subshells don't begin in the periodic table until the 6th level, when they, in theory, should begin in the 4th. You can use the same method as in observing the d subshells of the transition metals.

 

What is odd about determining the charges of the transition metals is that electrons are not most easily removed from their last energy shell (the d shell). Rather, they are drawn from the shell with the highest level number, which would be the preceding s subshell. So Iron doesn't have a charge of +6, but rather one of + 2 (as it loses the 4s electrons). Of course, most of the transition metals have several possible charges, but as a general rule the +2 is most common.

 

I hope I got all that right, somebody correct me if I'm wrong.