0
$\begingroup$

What are all the possible quantum numbers ml and ms of the electrons in the outermost s orbital of silver? I'm new to this concept. Can anyone walk me through this?

$\endgroup$
1
$\begingroup$

You have four quantum numbers: n, l, ml and ms.

The relationship between them is

n = 1, 2, 3... (theoretically speaking, up to infinite)

l = 0, 1, 2... n-1

-l <= ml <= l (that means, ml can assume all integer values between -l and l, including 0).

ms can be either 1/2 or -1/2.

l gives the "shape" of the orbital, and chemists give nicknames to the orbitals:

l = 0 is called s, l = 1 is called p, l = 2 is called d, l = 3 is called f, and so on (l = 4 is g, l = 5 is h, you got the idea)

Now, Silver is in the fifth period, therefore your n = 5. Since we are only interested in the s orbitals, we have n = 5, l = 0 (for the reason given above). Since l = 0, then also ml is 0. You have two electrons in the s orbital, therefore you have both 1/2 and -1/2 for ms.

Therefore, your quantum numbers are n = 5, l = 0, ml = 0, ms = +1/2 and - 1/2.

I hope this helps!

$\endgroup$
1
$\begingroup$

There are 4 quantum numbers.

n - Principal quantum number with values 1,2,3,4... - in simple words look at the row in periodic table and this is the max value of n for a particular atom.

l - Azimuthal quantum number, has always values from 0 to (n-1). In simple words it representsthe type of orbital (s, p, d, f). Look at the following table.

enter image description here

ml - magnetic quantum number distinguishes the orbitals available within a subshell and has values from -l to +l. In simple words, each orbital wavefunction has a specific shape (photo above). There is 1 s orbital, 5 p orbitals, 10 d orbitals and so on. This numbers points to one of the orbitals.

ms - spin quantum number, represents the spin of an electron. Values +1/2 and -1/2.


Back to your question - quantum number of outermost s orbital of silver. Silver has an electron configuration [Kr] 4d10 5s1.

n = 5 as silver is in the 5th row (row is called period).

l = 0 for s orbital

ms = 0 as there is only 1 s orbital

ml = +1/2 or -1/2

I hope this explains your problem.

$\endgroup$
1
$\begingroup$

Silver has electronic configuration as [Kr] 4d10 5s1 .
The s orbital ( 5s ) has only one electron. And the s orbital here has n = 5 (as it is in the fifth period) and l = 0 .
The s orbital is symmetric in space (sphere) which has only one orientation and hence has no "different" orientations each of which is represented by a particular value of ml. So ml can take only one value and that is 0.( If this was a p orbital which could take 3 different orientations, ml would take 3 values -> -1,0,1. In general an orbital with azimuthal quantum number l, ml would take 2l+1 values of ml ranging from -l to +l )
The quantum number ms represents the spin. It can either be +1/2 or -1/2.
Hence , ml = 0 amd ms = +1/2 or -1/2.
Hope this helps!

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.