Q1.
i) Kr **1s**^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6}
ii) F^{- }**1s**^{2} 2s^{2} 2p^{6}
iii) N^{3- }**1s**^{2} 2s^{2} 2p^{6}
iv) K^{+ }**1s**^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}
v) S^{2- }**1s**^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}
vi) Li^{+ }**1s**^{2
}

Q2.
**Al**^{2+}_{(g)} → Al^{3+}_{(g)} + e^{-} this is denoted ΔH_{i3}

Q3.
i) There is a larger step in ionisation energy after the 4^{th} 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 3p^{2} 3s^{2} 2p^{6} 2s^{2} 1s^{2}, so the 7^{th} ionisation is removal
of the 2p^{4} electron. It will therefore have an ionisation energy similar
to, and a little greater than, the 6^{th} ionisation energy.
This is because the 2p^{4} 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 2p^{5} electron. It is therefore
reasonable to suggest that the 7^{th} ionisation energy will be in the
region of **23000 kJmol**^{-1} (accept any value from 20000 to 26000 kJmol^{-1}).

### Like this:

Like Loading...