In acidic aqueous solution, the complex trans-[Co(en)₂Cl₁]²⁺(aq) undergoes the following substitution reaction:

trans-[Co(en)₁Cl₂]⁺(aq) + H₂O(l) → trans-[Co(en)₂(H₂O)Cl]²⁺(aq) + Cl^–(aq)

The reaction is first order in trans-[Co(en)₂Cl₂]⁺(aq), and the rate constant at 25°C is 3.2×10^–5 s^–1.

(d) Is the reaction product chiral or achiral? Explain.

Video transcript

All right. Hi, everyone. So for this question, let's consider the acid hydrolysis reaction of the complex ion cis Coen two N CS cl positive. The reaction is a pseudo first order reaction with a rate constant of 1.1 multiplied by 10 to the power of negative five inverse seconds at 25 °C determine whether the product is Cairo or a Cairo and justify your answer. And here we have four different answer choices proposing whether or not the product is Cairo or a Cairo depending on whether or not its sis structures are identical or not. So first, let's start by writing out the name of the product form. It is sis co en two N CS h2o close brackets to positive. So here based on this information, the central metal ion is going to be Cobalt to positive. And in this case, we happen to have a few different Liggins. The first of which is me actually move this, the first of which is en which is the abbreviation of Celine Diamine. And recall that Ethylene diamine is a B dentate ligand that has two donor atoms capable of connecting to the central metal ion in two different locations. In this case, our nitrogen atoms are the the donor atoms. On the other hand, our next legend is N CS minus which is or iso tho cyano ligand, which happens to be mono dentate. Therefore, only having one donor atom. Next is water referred to as aqua as a ligand, which is also monogenic. Now recall that bent ligands contribute a co-ordination number of two or as mono dentate ligands contribute a co-ordination number of one. So considering the fact that I have two B dentate ligands in this molecule and two mono dentate leggings, that means that my co ordination number is equal to six, which means that the geometry of this compound is going to be octahedral. Now recall also that the prefix sis indicates that two ligands are located to adjacent corners of the octahedron, right. So let's start by drawing the structure here is Cobalt in the center and I will start off by drawing by ethylene diamine ligands. So the curve that's combining my two nitrogens together represents the two carbons in between both nitrogens and ethylene diamine. And now I will be adding my molecule of water as well as N CS. So the first step is to imagine a mirror image of the compound I just drew. So I will withdraw my Cobalt center and then I'm going to add the same leggings. However, on the opposite sides to reflect that they are supposed to be mirror images of each other. So here are the mirror images of each other. So now that we've generated the mirror images of our product, the question is whether or not these mirror images are super imposable to each other. So to visualize this better, I'm going to actually rotate the first structure. So that now my ethylene diamine are on the left side. And also I'm going to go ahead and rotate the first structure horizontally should be N CS, right. So now when I compared these structures together, after rotating, the first one, I recognize that they are actually non super imposable mirror images of each other, they are non superimposable, which means that they are not the same molecule. Therefore, our product here is actually a Cairo complex. So our answer is going to be option B in the multiple choice, right. The product is a chiral molecule because it has non identical cyst structures. And with that being said, thank you so very much for watching. And I hope you found this helpful.