Thread: Science Assignment - Confusing radicals

To avoid as much confusion as possible I have scanned the question so you can see for yourself.


We've been given a science assignment, and no one I ask seems to know what on earth this question is talking about, and my science teacher refuses to help because in his opinion our school is, 'spoon feeding' us.

Anyway, I managed to do parts A, B, and C, but D (the one I scanned) was just too confusing, if anyone could help me with it I would extremely appreciate it!

Thanks to everyone in advanced,
Andrea

well, its oxidation numbers? thats what i would start playing with. . .

a) H + 4 (given) and because the ON have to add up to the charge of the complex ion, that makes N -3 difference in number of electrons, is has three more electrons than normal (normal elc configuration N = 2,5) that leaves it with an electron config of 2,8 which is stalble in that ion.

b) O -6 (given -2 times 3) ON numbers have to add to the charge. that makes N +5 the difference is it now has 5 less electrons than normal, and gives it the configuration of N= 2,
the diference is, teh configs are both stable in the ions but both have a differnt number of electrons. reason being that the Hydrogen and Oxegen take them,

now (after editing) im pretty sure im right, but still, same habbit of makign it up if it looks right.

Okay, thanks for that anyway. Any one else know?

I suppose for "a)" it's asking for the oxidation number of N, and not it's configuration, because you'd configure the entire molecule and not just the central atom.

N has many oxidation states, and in this one in the ammonium ion it is -3, because the ammonium ion has a +1 charge and each H is +1.

For part "c)" the oxidation number of N is +5 because each O is worth -2 and the overall charge is -1.

The inconsistencies lie in the special place of the group 5 elemens have p3 in their valence shell, and the uniqueness principle regarding N. The oxidation number of the 5th period is very much dictated by the other atoms involved.

If they want configuration, just check VSEPR and you should be fine. Ammonium is tertrahedral while I think nitrate is planar triangular (because of the e-neg of O) and it has 3 resonance structures as the pi-bonding changes b/t the 3 Os.

For part 2, the 3 Mg+2 combine to the the negative charge between 3 Os of each of 2 Phosphates. The forumula is Mg(3)P(2)O( 8 ), that is 3 Mg per 2 Phosphates. I don't know if the structure has a name.

I hope that helps, because that's all the help I can offer until next Monday.