I know outer electrons include (n-1)d...+ all n'ss.. But i don't understand, never been told what unpaired electron is supposed to be? is it the same thing.

so in the following electron configuration:

$4s^2\: 3d^{10}\: 4p^3$:

OuterElectrons: $15$

ValenceElectrons: $5$

from what I remember someone told me unpaired electrons are $15$ too for this...which is the same as outer electrons so I guess it's the same thing. But I am not sure so please help.


Outer electrons and unpaired electrons are not the same. Outer electrons are, as you said, $(n-1)d$ and all $n$'s. But unpaired electrons are different. To understand what they are, you must know what is electron spin (which, I hope you do). If you don't, read on.
Every electron in an atom has a quantum spin state (either clockwise or anticlockwise, nothing else) denoted by $+\frac{1}{2}$ and $-\frac{1}{2}$. These spin quantum numbers have no classical analogue, i.e., you can't relate them to any phenomenon in real life situations that you can visually perceive.
So, you see, every electron is paired with another electron with opposite spin. Together, the two of them fill one orbital, as for example, $2p_x$, $3d_{xy}$ etc. These orbitals are found out by solving the Schrodinger Equation.
Now, whenever an electron in an orbital does not have another electron with opposite spin, it is called an unpaired electron.

The electronic configuration that is given is $4s^2\: 3d^{10}\: 4p^3$. As, you can see, the $4p$ is filled with $3$ electrons whereas its maximum capacity is $6$. $4p$ has $3$ orbitals and so, each orbital is filled with one electron each, as given (according to Hund's Rule). So, these three are unpaired electrons. All other shells are fully filled and so, there are no more unpaired electrons.
So, number of unpaired electrons is 3 not 15.

  • $\begingroup$ how do you know 4p has three orbitals $\endgroup$ – Muhammad Umer Dec 12 '13 at 8:16
  • $\begingroup$ well, I just know. If you aren't convinced, you should know that s has 1 orbital, p has 3, d has 5, f has 7 etc. Learn atomic structure and you will know. $\endgroup$ – Ris97 Dec 12 '13 at 15:38
  • $\begingroup$ 1,3,5,7...hmm..nature sometimes make me wonder things...:D $\endgroup$ – Muhammad Umer Dec 12 '13 at 17:00
  • 1
    $\begingroup$ @MuhammadUmer To be more precise, a subshell with secondary quantum number $l$ is composed of $2l+1$ orbitals. These conditions arise while solving the angular part of the Schrödinger equation for a hydrogen-like atom. $\endgroup$ – Nicolau Saker Neto Dec 12 '13 at 21:19

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