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2007-01-14 21:44:39 · 2 answers · asked by Anonymous in Science & Mathematics Chemistry

2 answers

The cation is hexaqua nickel(II) cation

Because Nickel is 4s2 3d8( I mean Electronic configuration)

In this complex Ni loses 2 electrons ......so the +2 charge.
's' electrons easiest to lose for transition elements .....2 electrons lost in this way!

the 'd' orbital remains unchanged so 3d8 remains 3d8

Thus Ni(H2O)6]2+ is d8 for Ni

2007-01-14 21:52:17 · answer #1 · answered by Som™ 6 · 1 0

Lancenigo di Villorba (TV), Italy

Your complex is hexa-aquo-nickel(II) ion.
Since you know ion's electrical charge (as two) and water's electrical charge (as zero), you retrieve that this complex belongs to bi-valent nickel compounds.
You return to a suggestions which a great scientist (e.g. SOM) sent us : transition's metals lose "s electrons" before "d electrons". "S electrons" are two, no more, so zinc atoms maintain all its "d electrons".
As any bi-valent zinc compounds, [Ni(H2O)6]++ results "3d8".

I hope this helps you.

2007-01-15 06:13:08 · answer #2 · answered by Zor Prime 7 · 0 0

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