[funboy]

Question

Explain why the first ionization energy for sodium is much smaller than the second ionization energery for sodium.

Possible answers

when Na (element 11) has an electron removed from it this reduces the total electron repulsion and results in drawing the electron cloud closer to the nucleus.  The nucleus pulls the electrons closer to it and it takes more energy to seperate electrons that are close to each other than those that are farther apart.

The problem is that this is true for most (if not all elements) but there is a noticibly large gap between the first and second ionization energies of sodium.  I know that when you remove 1 electron from sodium that it has the same electron configuration of neon. 

I dont know what to write as an answer, I have a feeling it has something to do with electron configuration

I know that with sodium that outermost electron is unpaired and if you remove it the 2p level is full with paired electrons.  Does this have something to do with the answer??  Is it harder to seperate paried electrons than single??

Any help would be great

Thanks

[tamim83]

Ionization energies are all about the nuclear charge.  When you remove an electron you have more positive charge than negative charge.  As a result, the nucleus pulls the elevtrons in closer and harder.  As a result the ionization energy will increase, i.e. it will take a lot more energy to pull an electron away from a positively charged species. If you look at all of the ionization energies you will see that the energies increase; that is because as you remove electrons the atom becomes more posative. 

[funboy]

awsome, I answerd that and then some



Thanks

[Yggdrasil]

I think the point of the question is not to say that ionization energies depend on the charge of the species being ionized.  You are right when you say that the gap between the 1st and 2nd ionization energies is significantly larger than the gaps between the other ionization energies.  You are also right that it has to do with electron configurations.

Basically, there is a special stability associated with a full shell of electrons.  In Na⁺, the 2nd shell of electrons is completely full which makes Na⁺, and other species with ten electrons (such as Ne and F^-), more stable.  Because of this increased stability, it is especially difficult to add or remove an electron from Na⁺.

At the high school level, that's probably a good enough explanation.  There is a special stability associated with a full shell of electrons (i.e. a noble gas configuration) and also having a set of orbitals half full -- for example, N ([He]2s²2p³) has a half-full set of 2p orbitals and Mn ([Ar]4s²3d⁵) has a half-full set of 3d orbitals) which causes some irregularities in periodic trends.  For example, the first ionization energy of N is higher than that of O even though ionization energies generally increase as one moves to the right on the periodic table.

[xiankai]

as we're dealing with successive ionization energies of an element here, i thought i'll add that since there are several shells that hold electrons, whenever you remove from an inner shell as opposed to the outer shell (valence electrons), there is a huge leap in energy because basically, each shell has an average energy level that is quite similar for all the electrons in it.

the energy level generally increases the further the shell is from the nucleus. so, the electrons in the furthest shell are easiest to remove (need less energy to become fast enough to escape). like Yggdrasil mentioned, the 2nd shell of Na is more stable (less energy, so it cant escape so easily).

when Na (element 11) has an electron removed from it this reduces the total electron repulsion and results in drawing the electron cloud closer to the nucleus.

this is not really a factor, beacuse electron repulsion only has significant effects in the same shell. here we are dealing with different shells.

[kcn_10081991]

I think this can be solved with a simple explanation:
When an atom lose their electrons, they become a stronger cation, it mean the force of gravity became stronger for the other electrons. So, the following IE is alway higher than the previous IE.

[Borek]

the force of gravity became stronger

Gravity?

Besides, your explanation - while correct with regard to the fact the the second IE is always greater than the first - doesn't explain why these differences (between first and second IE) vary greatly for different elements. As the question was asked specifically for sodium answer should take it into account.

[xiankai]

the force of gravity became stronger

the force of gravity pales in comparison to the electrostatic force (gravitational constant = 6.67 X 10^-11 N m² kg^-2 vs. electrostatic constant = 8.988×¹⁰  N m² C^-2)

[kayamusty]

if sodium loses one electron,there will be 10 electrons on shells and protons(11) in nucleus pull them (electrons 10) more powerpul. that is why 2nd IE>1st IE

[Donaldson Tan]

How difficult is it to extract a negative charge from a postively charged cluster?

[english]

Well since sodium only has one s electron, when it is lost there is only core electrons, which is the electron configuration of neon.  These core electrons feel a greater effective nuclear charge (Z) than does the outer subshell, whose electron has been lost by absorption of the minimum energy required, IE.

So we see a considerable jump in IE energy as we go from IE1 to IE2.  But this only applies to sodium.  There are more core electrons as we move down the periodic table, and more outer electrons as we move across, so every element is variable, as Borek said.

[Mitch]

The core electron argument is sort of a slight of hand. The second ionization energy will always be higher than the first for any element on the PT. The reason is simply due to an increase in Z_eff.