quantum theory and structure

Question

The following values are the only energy levels of a hypothetical one-electron atom:

E6 = -2 x 10-19 J E5 = -7 x 10-19 J E4 = -11 x 10-19 J E3 = -15 x 10-19 J E2 = -17 x 10-19 J E1 = -20 x 10-19 J

(i) If the electron is in the n = 5 level, calculate the highest frequency of radiation that could be emitted.

(ii) Calculate the ionization energy (in kJ mol-1 ) of the atom in its ground state.

(iii) If the electron is in the n = 3 level, calculate the shortest wavelength (in nm) of radiation that could be absorbed without causing ionization. (6 marks)

(d) The ionisation energy generally increases across a period. However, there are two exceptions to the otherwise smooth increase in the first ionisation energy across periods. In Periods 2, there are dips at the Group 3A (13) elements, boron and aluminium, and at the Group 6A (16) elements, oxygen and sulfur. Explain this trend using suitable examples. (7 marks)

Answer 1

Hi Prieyankka E.,

Since this is unrelated to your above chemistry problem, guessing you have a comprehensive review test here? If you're really unsure across many areas, I do recommend you find a live tutor, to get comprehension rather than spot help. That being said:

Your atom has been restricted to the 6 energy levels given. So, from n=5 it can ONLY go down as far as n=1. Figure the ΔΕ and then use E = hν .

Ionization energy from n=1 is the opposite of the given E1 level energy. Convert from J to kJ mol^-1 using Avogadro's number.

From n=3 the electron can ONLY go up to n=6 without ionizing. Figure the ΔΕ and then use E = hν .

Just for the record, this is NOT a hydrogenlike (one-electron) atom, for which you probably know the energy equations.

I'm not quite sure about (d): the usual explanation for the 3A's is that the 2p orbital is higher energy than was the 2s orbital was even with 2 electrons in it. And for the 6A's, following Hund's Rule, the electrons prefer to distribute over the suborbitals, so when the 4th one is obliged to pair up, it's sitting at higher energy. That's just the tidiest way of thinking about what electrons do, and there's no particular reason to invoke any other forces or trends. That reasoning is true throughout a lot of science, you'll find!

-- Cheers, -- Mr. d.