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And what are the orbitals in the 5th and 6th energy levels?

Coz how come Iodine is in period 5 of the periodic table of elements when its greatest energy level is only in the 4th...erm, correct me if I'm wrong.

2006-07-19 04:49:33 · 3 answers · asked by Anonymous in Science & Mathematics Chemistry

3 answers

As indicated earlier for iodine, [Kr] 4d10 5s2 5p5

The orbitals *possible* in the n=5 level are l=0, 1, 2, 3, 4 or s-, p-, d-, and f. The largest atoms discovered to date have some electrons while in their ground state configuration in the 5f levels.

The orbitals *possible* in the n=6 level are l=0, 1, 2, 3, 4, and 5 or s-, p-, d-, f-, and (theoretically) g-. No elements discovered to date have enough electrons to populate the g-orbitals in the ground state.

The periods correspond to the highest principal quantum number in the electron configuration. The energy of an electron is a function of its principle quantum number (n).

The order of filling (Aufbau) of electrons into an atom isn't the same as the order of ionization (removal of electrons). It's a fine point that intro chem courses tend to gloss over.

An analogy is to suppose that you are on a pier, stepping down onto a raft made of multiple segments that can move up and down independently. You are going to step down onto the highest point (hey, you're lazy) since it's the easiest to access. But by doing so, you (and the raft and its other occupants) will change heights. Maybe now another part of the raft is now higher. So when the time comes to get off the raft and climb back up onto tthe pier, even though you went in last, that other higher (closer) part gets off first.

It's not a great analogy, but it's meant to point out that the energy states of an atom aren't fixed or stationary, and depend on the occupancy of the atom.

2006-07-19 05:41:34 · answer #1 · answered by ChemDoc 3 · 2 0

[Kr] 4d10 5s2 5p5

all orbitals above are empty and will stay that way, most of the time

2006-07-19 06:44:03 · answer #2 · answered by shiara_blade 6 · 0 0

the electronic configuration of I is: [Kr] 4d10 5s2 5p5

2006-07-19 04:55:44 · answer #3 · answered by Noel 4 · 0 0

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