6.
To determine this the CrO4 ion must be 2- as it is balanced by 2 potassium ions which can only have a +1 charge.
The oxygen atoms are considered to be present as O2- so all four contribute 8-. As the net charge is only 2- then chromium must have charge (oxidation number or state) of 6+.
K2CrO4 has 2x K(potassium) atoms per 1xCr (chromium) atom per 4xO (oxygen) atoms. The oxidation number of oxygen is always -2 (except when chemically bound to fluorine, which is the only element higher in electronegativity/ionization energy than oxygen) K is an alkali metal, (in column 1A of the periodic table) and henceforth will almost always have an oxidation number of +1. With this information, we can deduct that... 2 K atoms will each have +1 as an oxidation number (totaling +2) 4 O atoms will each have -2 as an oxidation number (totaling -8) We set the equation to zero (because the compound has no charge) so. 2K+Cr+4O=0 or 2(+1)+Cr+4(-2)=0 or +2+cr-8=0 simple algebra from here. cr=+6 each, individually, would be, k=+1, cr=+6, o=-2. Generally, you'll only be looking to figure out the oxidation number of cr in this problem.
Well, without being given an equation, we presume you mean the o.n. of chromium. H in this is +1, so H+H=+2 4 x O= -2 x 4 = -8, therefore Cr = +6 as the whole element has to equal zero. and i do believe in this form it normally goes to Cr +3
+2 for Ca, +6 for Cr, -2 for each O
In CrBr (chromium monobromide) Cr would have an oxidation number of +1. This compound is not known perhaps you meant CrBr3, where chromium has an oxidation number of +3
Cr is a transition metal and has a variation of oxidation number from 0 in the elemental state to +6 in the dichromate ion.
K2CrO4 has 2x K(potassium) atoms per 1xCr (chromium) atom per 4xO (oxygen) atoms. The oxidation number of oxygen is always -2 (except when chemically bound to fluorine, which is the only element higher in electronegativity/ionization energy than oxygen) K is an alkali metal, (in column 1A of the periodic table) and henceforth will almost always have an oxidation number of +1. With this information, we can deduct that... 2 K atoms will each have +1 as an oxidation number (totaling +2) 4 O atoms will each have -2 as an oxidation number (totaling -8) We set the equation to zero (because the compound has no charge) so. 2K+Cr+4O=0 or 2(+1)+Cr+4(-2)=0 or +2+cr-8=0 simple algebra from here. cr=+6 each, individually, would be, k=+1, cr=+6, o=-2. Generally, you'll only be looking to figure out the oxidation number of cr in this problem.
Well, without being given an equation, we presume you mean the o.n. of chromium. H in this is +1, so H+H=+2 4 x O= -2 x 4 = -8, therefore Cr = +6 as the whole element has to equal zero. and i do believe in this form it normally goes to Cr +3
+2 for Ca, +6 for Cr, -2 for each O
In CrBr (chromium monobromide) Cr would have an oxidation number of +1. This compound is not known perhaps you meant CrBr3, where chromium has an oxidation number of +3
Cr is a transition metal and has a variation of oxidation number from 0 in the elemental state to +6 in the dichromate ion.
In Cr2O72- chromium (Cr) has an oxidation number of 6+ while oxygen has an oxidation number of 2-.
As with any other element, the oxidation number of Cr depends on whether and how it is chemically bonded. The oxidation number of pure elements is arbitrarily defined to be 0. In compounds, Cr has oxidation numbers of +2, +3, and +6, depending on the compound.
+6
The answer is +6
+6
+6 for Cr
O.S. of Cr = +6