Give the systematic name for this coordination compound. K

Question

Give the systematic name for this coordination compound. K₃ Cr(CN)₆

Accepted Answer

Hey everyone, we're asked to name the following compounds. Starting off with our first one, we have sodium as our metal Cagayan. Now let's go ahead and take this step by step first. So we have sodium and for our complex ion we have iron which is a transition metal. So we'll have to sulfur its charge since it's going to affect our name. We have CNN which is going to be our cyanide. And we have no which is our nitric oxide. So let's go ahead and solve for our charge first. Before we name this now, we know that sodium has a plus one charge. And since we have two of them, we're going to multiply this by two And Cyanide has a - charge. And since we have five, we're going to multiply this by five. And we know that nitric oxide is a neutral Liggan, so have a charge of zero. Now looking at our iron, we're going to label this as X. So when we sulfur its charge, we got two from our sodium plus X minus five from our cyanide And this is going to equal zero. And since we're solving for X, We're going to end up with X -3 equals zero and X is going to be plus three. So our iron has a charge of plus three. Now riding out its name, we know that sodium is going to be placed before our an ionic complex and it's going to be kept as is now for our complex, we're going to name this in an alphabetical fashion. So we know that cyanide is going to be changed to Siano. And since we have five, we're going to input the prefix of Penta. So we have Penta Siano. Now moving on to our nitric oxide, we know that this is going to change into nitro sil. So we have Penta Siano nitro sil And lastly we have our iron. So the metal is going to keep its base name and remember iron has changed as fair and it's going to have an ending of A T E. And to indicate its charge, we're going to have that roman numeral three. So the name of our compound is going to be sodium Pentothal, Siano nitro sill fairy. Now moving on to part B. We have platinum, which is also a transition metal. So we'll have to settle for that as well. And we have NH three which is a neutral ligand. So that will have a charge of zero. And lastly we have chlorine. So solving for platinum's charge, we're going to label this as X again And we know that this charge is zero since it's neutral And Chlorine has a -1 charge. And since we have to, we're going to multiply that by two. So we get X minus two equals zero, So x equals positive two. So our charge on our platinum is going to be a positive too. Now when we name this, we're going to start off with our amine. And the reason why is because our complex ion is in the caddy on portion now and our medal has to be named at the end of our complex ion. So ammonia is going to be changed to a mean. And since we have to we're going to write die for two, I mean for ammonia and then we're going to write platinum and to indicate our platinum's charge, we're going to put roman numeral two. And lastly we're going to add our chlorine and that's going to keep its base name, but it's going to change its ending into ID. So the name of our compound is going to become dia mean platinum to chloride. Let's go ahead and look at sea. So for C we have cobalt, which is also a transition metal. So we'll have to solve for that Again. We have NH three and we know that this is going to become, I mean when we name it and we have chlorine again in our metal complex. And lastly we have roaming So since COBOL is a transition metal, let's go ahead and find out its charge. Now we know that NH three is a neutral ligand. So it's going to have a charge of zero And Chlorine has a charge of -1. And we're going to multiply that by two. Since we have to cobalt will label as X And bro mean we know has a charge of -1. So solving for cobalt charge, we have X minus two minus one equals zero. So we end up with x equals positive three. Now when we name this, we're going to start it off with our amine since it's alphabetized but we need to remember that we have that subscript of four, So the prefix for four is going to be tetra and then we're going to write our amine and for chlorine, we had two of chlorine. So we're going to write tetra amine, di chlor Oh and we're going to end it with our metal. So we're going to write cobalt And since it's a transition metal we can't forget the Roman numeral three to denote its charge. And lastly we're going to end it with our bruning and it's going to keep its space but it's going to change the ending into ID. So the full name is going to be tetra amine dyke. Loro, cobalt three bromide. So I hope this made sense. And let us know if you have any questions

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Give the systematic name for this coordination compound. K₃ Cr(CN)₆