Question

Consider the coordination compound Caz[Co(CN)6]: i. ii. Give the systematic name of the compound. What is the magnetic property of the complex (diamagnetic, strong or weakly paramagnetic)? (Use the appropriate d-orbital splitting diagram from the formula sheet to explain your answer) iii. Will the complex be a high-spin or low-spin complex?

Answer 1

Name of the complex is Calcium hexacyanocobaltate(2)

Detailed solution is given below in images
• i) Naming of coordination compound 1: followend are the Rules used in naming a coordination compound : 1) Always name the caution before the anion. 1 • 2) Naming the complen ion- → Name the ligands first, in alphabetic Order, and watter that mame the metal atom orion :-). Yox vanionic ligands, end the name . in 66-0" 1: eg CNo cyano 1. Greek prefines di, tri-, tetra- etc are used to • designate the number of each type of ligand : in complex ion. - Metals faesent in anionic complenes end E in 66-ate" eg. Iron =) ferrate ' Cobalt =) Cobaltate - Onidation state of the metal in the complen i is given as Roman numerals in Parentheses . after metal. te -

Na 10 DO caz [cocenda] Name of the above Coordination compaund is Calcium heracijama.cobaltote (II.) całt so total +ve charge wit 2(42) =+4 1. So, [co (CN);]" Hemer Ovelation number of co wi +2 i) Magnetic property of complexe [cocon)]4 het oudation State of co be u . so ' so, we have Co2in complex Configuration of co is [Ar]3d%45% 1 602+, [AY] 347 Naw there are six ligands present around a the central metal, so there will be octa- -hedral splitting of d- orbitals, and fine d- -orbitals will offelit in ful sels. One twill be tag (day, daz, dzn) and other one will be eg (droht and dz?). Splitting of d-orbituls in presence of octahedral o

field is shown below I crystal field splitting ma4111/11 < • degenerate d -orbitaly when ligands approach • central metal Symmetrically . ( Mypothetical) o in the presence of a Octahedral field of Ligands around Metal. Note o filling of es ase done in a manner tzer eg To because CNo is a strong Ligand and I the complen will be low spin. To Que to the presence of only one impaired electron, the compound ü Paramagnetic. Since due to presenes of Strong fricted ligand o le CNO, elections pair up against elec- 1-trostatic repulsion b/w them. This results in theack to pairing up of és un tay level. • Hence due to presence of one impaired electron o compound is weakly Paramagnétic o RE OR

ii) complen will be low-spin complex : Because of the presevice of strong field kigeudo for Stoong field ligands, Crystal field splitting * ieso is higher than pairing energy. : live for CNo A P Due to which electrons bretter to paik-up o rather go to higher level. This results • in pairing of electrons in Ang level. Because of which spin decreases.. Hence compleared is low spin complex