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d1i1m1o1n [39]
3 years ago
12

Determine the oxidation number of each element in these compounds or ions. (a) au2(seo4)3 (gold(iii) selenate) au = se = o = (b)

ni(cn)2 (nickel(ii) cyanide) ni = c = n =
Chemistry
1 answer:
Stels [109]3 years ago
6 0
<span>                                                    Au</span>₂(SeO₄)₃

                                         O = -2 × 4 = -8
                                             Se  =  + 6
So,
                                            (+6 - 8) = -2

Means (SeO₄) contains -2 charge, Now multiply -2 by 3
                                             
                                             -2 ₓ 3 = -6
Means,
                             Au₂ + (-6) = 0
               
                            Au₂  = +6
Or,
                            Au  =  6 / 2

                            Au  = +3
Result:
                            Au  =  +3
                            Se  =  +6
                            O   =  -2

                                                      Ni(CN)₂


Cyanide (CN⁻) contains -1 charge,
So,
                              N  =  -3
                              C  =  +2
Then,
                                         Ni + (-1)₂  =  0

                                               Ni - 2  =  0
Or,
                                                     Ni =  +2
Result:
                            N  =  -3
                            C  =  +2
                           Ni  =  +2




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