Is this statement true or false and why?

If one prepared a gas-phase Be3+ ion with its electron in a 4f orbital, it would take exactly as much energy to remove that electron as it does to ionize a ground state H atom.

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Is this statement true or false and why?

If one prepared a gas-phase Be3+ ion with its electron in a 4f orbital, it would take exactly as much energy to remove that electron as it does to ionize a ground state H atom.

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## Answers (1)

Sorry I don't have time to do the calculations, BUT

Be 3+ is a single electron atom.

You should know that you can calculate the exact energies of the electron orbitals in the hydrogen atom.

You should also be able to calculate the energies of the Be 3+ ion, though you may need to modify the formula.

I may edit this answer if I get around to revising this stuff.

edit: This question is easier than I expected. A single electron atom is called a Bohr atom.

The energy levels of a Bohr atom are given by:

E = -Z²×RH/n²

Z is nuclear charge (atomic number)

RH is a constant 2.180 x 10^-18 J

n is orbital principle quantum number.

In the ground state the electron should be in the 1s orbital. ie n = 1 for hydrogen. For Be3+ electron is in 4f orbital. that is, n = 4.

The ionisation energy is that required to move the electron to n = ∞. At n = ∞, E = 0, which makes sense. So the energy calculated by the equation is the ionisation energy.

because I CBF doing calcs, and I know other stuff is constant, I just compare Z²/n²

H: 1s: 1²/1² = 1

Be3+: 4f: 4²/4² = 1

So yes, ionisation energies are identical.

A larger nuclear charge would be expected to more strongly attract the electron (greater electrostatic force), thus requiring more energy to liberate electron. But when the electron is further away it is less strongly attracted (force decays with distance squared). Of course the orbital isn't an exact distance from centre (it's a function) but I guess the average election density gets further away for higher energies.