Arrange the elements tellurium, rubidium, potassium & bromine in order of increasing second energy (ie2) and in decreasing atomic size.

IE2: Te < Br < Rb < K

Atomic size: Rb > K > Te > Br

Explanation:

2nd Ionization energy

Ions with only one valence electron (K and Rb) should have the highest second ionization energies. Once the first electron has been removed, the next one would have to come from a filled valence shell.

Atoms lower down in the Periodic Table should have lower ionization energies, because their valence electrons are further from the nucleus and are less tightly held.

Br and Te each have a nearly complete subshell, so both electrons can come from the same level. Te is further down the Periodic Table, so Te < Br

The order of second ionization energies is Te < Br < Rb < K.

Atomic Size

Atomic size increases from right to left and from top to bottom of the Periodic Table. Thus, the biggest atoms are at the lower left and the smallest atoms are at the upper left.

Rb: Left of Row 5. ∴ Biggest atom.

Te: Right of Row 5. ∴ Rb > Te. Also, Br is in Row 4, so Te > Br

K: Left of Row 4. ∴ Rb > K

Br: Right of Row 4 ∴ K > Br

So far, we have Rb > K and K > Br, so Rb > K > Br .

We also have Rb > Te and Te > Br, but is Te > K or is K > Te?

In going from K to Te, the effect of adding 33 more protons to the nucleus outweighs the effect of adding one more shell, so K > Te.

The final order is Rb > K > Te > Br.

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