New answers tagged

1

The short answer is that we'd like to do the first, but solving the multi-electron Schrodinger equation is impossible, so we use variations of your second option (ie considering many separate one-electron wavefunctions) to get something that matches experimental data as closely as possible, even though we don't ever get an explicit form of the true multi-...


3

Spin multiplicity = $2S+1$, where $S$ is the total spin angular momentum. Now $S = \frac{n}{2}$ where $n$ represents total number of unpaired electrons. So now we can write spin multiplicity = $n+1$. Now coming to your question, $\ce{Mn}$ has 5 unpaired electrons in it. Therefore, its spin multiplicity $(S) = 5+1 = 6$. Similarly, $\ce{Mn^{2+}}$ has 5 ...


3

The answer of your problem lies in the existence of sub-shells, that you seem not to know. Let me explain. If you give a number $n$ to the successive shells, $K$ shell gets $1$, $L$ gets $2$, $M$ gets $3$, etc. The maximum amount of electrons per shell is $2n^2$ , which is $2$ for $K$ shell, $8$ for $L$ shell, $18$ for $M$ shell, etc. $n$ is called first ...


-1

It is based on the aufbau principle. As it is a multi-electron system it follows the order for the energies of atomic orbitals as 1s<2s<2p<3s<3p<4s,...and so on. For K shell n=1,for M shell n=2, and so on(where 'n' refers to principal quantum number). "S-sub shell" contains only one orbital whereas in "P" it contains three ...


9

What's wrong with the reasoning that there is now a larger negative charge and hence the positive nucleus and negative electron cloud have a stronger attraction so the ionic radius decreases. The problem is that you're treating it as if all the electrons can be collectively reasoned about as one single negative charge. This isn't true. The nucleus is ...


Top 50 recent answers are included