Indicate the coordination number for the central metal atom in each of the following coordination compounds:
(a) [pt(h2o)2br2]
(b) [pt(nh3)(py)(cl)(br)] (py = pyridine, c5h5n)
(c) [zn(nh3)2cl2]
(d) [zn(nh3)(py)(cl)(br)]
(e) [ni(h2o)4cl2]
(f) [fe(en)2(cn)2]+ (en = ethylenediamine, c2h8n2)

A. 4, since all 4 ligands (2 H2O&2Br) are monodentate hence 4x1=4

B. 4, (4x1=4)

C. 4(4x1=4)

D. 4(4x1=4)

E. 6(6x1=6)

F. 6(2(2)+2(1)=4+2=6) en is bidentate ligand

Explanation:

Coordination number is equal to number of ligands surrounded the metal atom.

Coordination number= No of ligands X denticity

Denticity is nothing but nature of ligands like mono, bi dentate, or tetrads rate etc.

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.