K[Cu(en)2 (CN)2 ]
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Re: K[Cu(en)2 (CN)2 ]
This would have to do with it being bonded to the K+, which "cancels out" the second negative charge from the CN-
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Re: K[Cu(en)2 (CN)2 ]
You can look at this entire compound as an ionic compound, where K+ is the cation and the coordination compound is the anion. Since the whole compound is neutral, the coordination compound must have a charge of -1. So, since each CN has a charge of -1, Cu has a charge of +1.
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Re: K[Cu(en)2 (CN)2 ]
K has a +1 charge, which makes the coordination compound have a -1 charge. CN has a -1 charge and en is neutral, and since there are 2 CN, the charge would be -2. Everything in the brackets should add up to the total charge, which is -1, so Cu would have a charge of +1. Hope this helps!
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Re: K[Cu(en)2 (CN)2 ]
K[Cu(en)2(CN)2] as a neutral compound would have an overall charge of 0. Because K has a charge of +1, then the coordination compound in the brackets would have a charge of -1 that cancels out with K. In the brackets (en)2 is neutral, while (CN)2 has a charge of -2; meaning Cu needs to be +1 for the compound to be -1.
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Re: K[Cu(en)2 (CN)2 ]
each Cn has a negative one charge for a total of negative two, K has one positive charge. Therefore for the overall compound to be nuetral it should be Cu(I)
Re: K[Cu(en)2 (CN)2 ]
K has a positive charge of 1 and en is neutral, CN has a negative charge of 1 and there are two of them, so Cu has to be +1
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Re: K[Cu(en)2 (CN)2 ]
Because it's bonded with a K+, the necessary Cu charge for it to be a non charged compound is Cu(I)
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