a) As as in H3aSo4
b) Cr on HCr2O7-
2007-01-07 02:31:32 · 3 answers · asked by Joyce 4 in Science & Mathematics ➔ Chemistry
Using oxygen as -2 (Minus two).
Then with 4 oxygens in the molecule there is a total of 4 x -2 = -8 for the oxygen part of the species.
Using hydrogen as +1 (plus one)
The with 3 hydrogens in the molecule there is a total of 3 x 1 = +3 for the hydrogen part of the species.
Adding these numbers together -8 + 3 = -5
This -5 must be 'cancelled' as the molecule is neutral. So the third item (As) must be +5.
So As is in oxidation state 5 (V).
Written as As(V).
Similarly for the bi-chromate species.
Using oxygen as -2
Then with seven oxygens in the species there is a total of 7 x -2 = -14.for the oxygen part of the species.
Using hydrogen as +1 (plus one)
Then with one hydrogen there is a total of 1 x +1 = +1 for the hydrogen part of the species.
Adding these numbers together -14 + 1 = -13
Noting that this is an ion of charge minus one (-1) the overall addition must come to minus one (-1)
So the two x Cr must each be +6. (2 x +6 = +12
So making a full addition..
-14 + 1 + 12 = -1
So each Cr is in oxidation state +6.
Written as Cr(VI).
2007-01-07 08:05:11 · answer #1 · answered by lenpol7 7 · 0⤊ 0⤋
As is +5 Cr is +7
2007-01-07 07:20:50 · answer #2 · answered by lykovetos 5 · 0⤊ 0⤋
oxidn state of As is +5
and that of Cr is +6
2007-01-07 02:38:57 · answer #3 · answered by me myself 1 · 1⤊ 0⤋