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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: inferno_gogo on October 24, 2008, 05:43:36 PM

Title: IE and EA midpoint
Post by: inferno_gogo on October 24, 2008, 05:43:36 PM

Hello,

I need help on a homeowork problem. It is a little tough:

At large, interatomic separations, an alkali halide molecule MX has a lower energy as two neutral atoms, M + X; at short separations, the ionic form (M+)(X-) has a lower energy. At a certain distance R, the energies of the two forms become equal, and it is near that distance that and electron will jump. Because the distance is large we can ignore the variation in potential V(R) between R and infinity. In that range V is dominated by Coulomb attraction.
Express R in terms of first ionization energy of metal M and electron affinity of halogen X.

I am rather clueless on this question.
I know that the IE and AE are given numbers for each atom. So how to I set up a relationship? Coulomb's law might help, but i am not sure. I believe it is:

F = Q/4p*e*r

but what is the charge? and how can it relate to IE and AE?

Any help is appreciated.

Thanks,

-George

Title: Re: IE and EA midpoint
Post by: FeLiXe on October 25, 2008, 05:05:54 PM

I think the idea is the following
you set the energy of the two neutral atoms to zero as they are far away

the energy of the ionic molecule is:

$$ E_{ion} = I_p - E_{aff} - \frac{1}{R} /$$

at large distances it will be positive because Ip is larger than Eaff. at small distances it will be negative because the 1/R term is important. the crossing is what you are looking for