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The oxidation state of K is +, and Ba is 2+.

When you start putting K2+, you are indicating that there are two potassium ions in the compound, but it would not be written this way.

2007-11-05 07:35:42 · answer #1 · answered by Anonymous · 0 0

For 'K+' and 'Ba2+' to become 'K2+' and 'Ba3+' requires a large amount of energy for these inner electrons to be ionised.

The electron configuration for K is:-
1s2, 2s2, 2p6, 3s2, 3p6, 4s1.
For the 4s1 electron to be ionised(K+) only requires 418 kJ mol^-1 of energy.
However, for a second electron to be ionised (K2+) requires 3070 kJ mol^-1 of energy, a huge jump. This is because this second electron is being removed from an inner energy shell (3p6), which requires much more energy because inner energy shells of electrons are held closer to the nucleus.
A similar argument applies to barium.
The electron configuration for Ba is:-
1s2,2s2,2p6,3s2,3p6,3d10,4s2,4p6,4d10, 5s2,5p6, 6s2.
For the 6s2 electrons to be ionised(Ba2+) only requires 502(1st) & 966(2nd) kJ mol^-1 of energy.
However, for a third electron to be ionised (Ba3+) requires 3390 kJ mol^-1 of energy, a huge jump. This is because this third electron is being removed from an inner energy shell (5p6), which requires much more energy because inner energy shells of electrons are held closer to the nucleus

2007-11-05 07:47:27 · answer #2 · answered by lenpol7 7 · 0 0

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