Number of Unpaired Electrons

Question

How many unpaired electrons are there in the following transition metal ions. Write your answer as a sequence of numbers, for example, 23542141.

A) V3

B) Cu2

C) Sc3

D) Cr3

E) Cr2

F) Mn2

G) Ni4

H) Ni3

Could someone please help me! Thanks!!

Answer 1

Would you believe I had a perodic table layng on my desk for days .... this must be fate :P

27134567

1)find the element

2)count 'back' (like reading) the number of positive charges

3)including the element your finger is on, count left until the row has ended

4)that's usually your answer, but once the number reaches half the rw length, th number of spaces reduce :) it's a little connfusing

Answer 2

The first 5 3d electrons fill unpaired orbitals before the filled.  Usually the 4s electrons are lower energy, but sometimes this isn't the case, leading to the occasional exception to Aufbau (not a rule, just a general principle)

For V, the neutral atom would have 2 4s and 3 3d electrons.  V3+ has 3 electrons stripped from the V neutral atom.  Leaving 2 electron in 4s and removing the 3 3d electrons leave the ion in the lowest (and most stable) possible energy state.  Note that the 2 electron in 4s are PAIRED; there is only one s orbital, the 2 electrons have to be paired by Pauli's Exclusion Law.

Between this, your class notes, and the Wiki article, you should be able to piece out the rest for yourself.

Answer 3

We're assuming that when you write "V3" what you really mean is a [V]3+ cation.

These are all first-row transition metals, so the neutral atoms all have valence electrons in the 4s and 3d subshells.  By finding the Group each metal is in, you can determine the electronic configuration of the neutral atom.  e.g. V is Group 5, so it has a (4s)2(3d)3 configuration.

But you have cations, so you have to remove a number of electrons.  [V]3+ has 5 - 3 = 2 valence electrons, [Cu]2+ has 11 - 2 = 9, and so on.  So, what orbitals are they in?

For main group elements, an ion has the same configuration as its isoelectronic neutral atom: [O]– has 9 electrons like neutral F, and both [O]– and F have the same configurations.  One of the previous posters has forgotten that this isn't true with transition metals.  Although aufbau tells you that the neutral atom fills the 4s before the 3d, that's not true for the cations: TM atoms lose the 4s electrons before the 3d electrons, so that [V]3+ is (3d)2, *not* (4s)2.  That actually makes it easier to figure out, because it means that none of your ions has any s electrons left, all the valence electrons are d electrons, and you just take Group Number – Oxidation State to find their number: [V]3+ is 5-3=(3d)2, [Cu]2+ is 11-2=(3d)9, and so on.

So now we know how many electrons and in what orbital subshell.  How many are unpaired?  Hund's rule says you populate a set of degenerate orbitals with one electron at a time, until each orbital carries one, and only then do you go back and pair them up.  The first poster has forgotten that there are only five d orbitals, so it's impossible to have more than 5 unpaired electrons in a d subshell.  Take your number of electrons, add them one at a time to each of the five 3d orbitals, when you get to a sixth you pair it, and so on.  Which means [V]3+ is (3d)2, two unpaired, [Cu]2+ is (3d)9, one unpaired, etc.