Chemistry Help!

[nehaoscar]

Check out the attached question
The answer is A (1,2 and 3 are correct)
But V3+ means that it has 20 electrons
so that means it has a configuration of 1s2 2s2 2p6 3s2 3p6 4s2
this shows that all the orbitals are filled right?
Then how come it still has unpaired electrons?

 

[SebastianM]

Vn3+'s electronic configuration is 1s2 2s2 2p6 3s2 3p6 3d2. So Vn3+ has 2 unpaired electrons. That is why Option 3 is CORRECT.

 

[nehaoscar]

Vn3+'s electronic configuration is 1s2 2s2 2p6 3s2 3p6 3d2. So Vn3+ has 2 unpaired electrons. That is why Option 3 is CORRECT.

But then I was taught that the 4s shell fills before the 3d shell...
So for CIE AS do i consider 3d to be filled before 4s?

 

[SebastianM]

You were taught correctly. But not quite completely.
-The D shell fills before the S shell.
-When the electrons leave, however, it is the other way round. The electrons leave from the S shell first. After the S shell's electrons are lost, the electrons proceed to leave from D shell.

 

[nehaoscar]

You were taught correctly. But not quite completely.
-The D shell fills before the S shell.
-When the electrons leave, however, it is the other way round. The electrons leave from the S shell first. After the S shell's electrons are lost, the electrons proceed to leave from D shell.

Thankyou so much!

 

[SebastianM]

I think it will. Even though acid is in excess, acid is a reactant, which will be affected by the concentration. Excess only ensures that the reaction goes to completion.

 

[Xylferion]

View attachment 51702
Check out the attached question
The answer is A (1,2 and 3 are correct)
But V3+ means that it has 20 electrons
so that means it has a configuration of 1s2 2s2 2p6 3s2 3p6 4s2
this shows that all the orbitals are filled right?
Then how come it still has unpaired electrons?

Electronic configuration of V ---> 1s2.2s2.2p6.3s2.3p6.4s2.3d5

During ionization, 4s is repelled to a higher level than 3d, making it take priority during ionization.

When you ionize Vanadium to [V]3+, you remove 3 electrons starting with 2 from the 4s orbital and the remaining electron from the 3d orbital.

Giving you, 1s2.2s2.2p6.3s2.3p6.3d4

There are 5 d-subshells, each occupying 2 electrons at most, which is why 3d10 is the max it go before moving onto f.

When filling the subshells, you fill them singly first.
In [V]3+, it says 3d4 at the end. This tells you that there are only 4 electrons present, so you place them singly. This gives you 4 3d subshells each with 1 electron, and an empty 5th with no electrons. That is 4 unpaired electrons, and one subshell with no electrons at all.

Which is practically why [V]3+ is also a correct option.

Hope that made sense ^_^

Note:- It's for transition elements that you remove an electron from the 4s orbital.