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Question 9C.9 in the textbook homework is related to determining a metal ion's coordination number. For part (c), en is bidentate, causing the coordination number of the complex to be 6. For part (d), edta is hexadentate, causing the coordination number of the complex to be 6. How and why does a ligand being mono, bi, or polydentate effect the coordination number of a metal ion in a complex?
Coordination number is the number of parts at which ligands are attached to the central metal atom. mono dentate has one binding site (an example would be chloride ions), bidentate has two donor atoms, and polydentate ranges in the bonding sites. For 9C part c en is bidentate therefore assume there could be the possibility of two binding sites. Since there are two en and two chloride ions there would be a total of six points to which ligands can attach. For part d edit is hexadentate therefore it has a coordination number of six because there six possible sites to bind to.
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