Question

The graph compares the 1s orbital energies for the F atom (Z = 9), the Ne+ ion (Z = 10), and the Na++ ion (Z = 11).
a. How many electrons does each species have?
b. According to the graph, which species is most stable?
c. Use Coulomb's law to explain the energy measurements shown in the graph and how this affects the distribution of electrons.

Answer 1

a) all the species have the same number of electrons. 9 electrons
b) I dont understand that graph what its represent. The electron in the 1s are core electron and the energy usually are negative ( are electrons)

Answer 2

thank you!

Answer 3

\[F = \frac{ kQ _{1}q _{2} }{ r ^{2}}\]

To me this is a tough question

Above is columbs law which states that the columbia force is inversely proportional to the distance squared. so as the charges get closer together (r)^2 becomes big and your force goes up i think it's by a factor of r^2. as the charges get farther apart in distance the columbic force decreases.

the issue here is that we're dealing with the 1s orbital. (the graph is showing us the energies of the 1s orbital) but i've never encountered a question talking about 1s orbital energies b.c usually we deal with valence electrons.

I can probably guess/infer that removal of electrons, creates a higher positive charge so the inner electrons most notably the ones in the one s orbital will be held much more tightly once you remove an electron from the outer shell.

Now to explain the effect, I would guess that the more electrons you remove the atomic radius (r) decreases and the force of attraction goes up and would (occupy lower energy) don't quote me on that last part I'm not sure. But I don't think this is the same as the amount of energy you would need to put in more energy to remove electrons as that + charge increases.

I would guess Fc would be the greatest for Na++ followed by Neon+ and then F.


another thing in the graph, I guess we'll base it off of this also *Higher energy implies less stable*

Now based off of this i would guess that the most stable would be Na++ followed by Ne+ then F as least stable. (not sure completely)

Neon+ Z 9 similar configuration to fluorine
Flourine z = 9
Na++ Z = 9

All of these ions have a similar electron configuration to fluorine and thus have 9, electrons.

Answer 4

typo** Columb force,

Answer 5

also in the first part as the charges get closer as (r)^2 becomes smaller not bigger, and the force increases** sorry was writing that quickly.

Answer 6

thanks again!