Electron Arrangements Practice Answers

Q1.
i) Kr     1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6
ii) F-     1s2 2s2 2p6
iii) N3-   1s2 2s2 2p6
iv) K+     1s2 2s2 2p6 3s2 3p6
v) S2-     1s2 2s2 2p6 3s2 3p6
vi) Li+    1s2
Q2.
Al2+(g) → Al3+(g) + e-   this is denoted ΔHi3 
Q3.
i) There is a larger step in ionisation energy after the 4th ionisation
⇒ the element has four outer shell electrons so is in group 4. 
The element in period 3 group 4 is silicon.

ii) Silicon is 3p2 3s2 2p6 2s2 1s2, so the 7th ionisation is removal 
of the 2p4 electron. It will therefore have an ionisation energy similar 
to, and a little greater than, the 6th ionisation energy. 
This is because the 2p4 electron experiences the same shielding and 
nuclear charge, but is slightly closer to the nucleus and therefore 
more strongly attracted due to the decrease in repulsion between 
electrons in n=2 with removal of the 2p5 electron. It is therefore 
reasonable to suggest that the 7th ionisation energy will be in the 
region of 23000 kJmol-1 (accept any value from 20000 to 26000 kJmol-1).