![1.\,\,{{[Ir{{({{C}_{2}}{{O}_{4}})}_{2}}C{{l}_{2}}]}^{3-}}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-692e4aaaca8376b322d57a9135b32d09_l3.png "Rendered by QuickLaTeX.com")
![2.\,\,{{[CoC{{l}_{2}}{{(en)}_{2}}]}^{+}}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-dd1620a388a2143f22e2867088a90788_l3.png "Rendered by QuickLaTeX.com")
![3.\,\,{{\left[ Co{{(en)}_{2}}{{(N{{O}_{3}})}_{2}} \right]}^{+}}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c24550a7f86653e173c395b71a61990c_l3.png "Rendered by QuickLaTeX.com")
![4.\,\,\left[ Pt{{(N{{H}_{3}})}_{2}}C{{l}_{2}} \right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cf3b4bcb7cb8c9599927883a92170271_l3.png "Rendered by QuickLaTeX.com")
Solution:
Because Pt(II) is a transition metal, Pt(NH3)2Cl2 is a square planar. For transition metals with a d8 electron configuration, square planar is the favoured shape.
The oxidation sate of Pt in the given complex is +2.
![P{{t}^{2+}}:{{[Xe]}^{54}}4{{f}^{14}}5{{d}^{8}}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b4cd9243e1f36450377c1915dbbb6e11_l3.png "Rendered by QuickLaTeX.com")
The eight electrons in the 5d-subshell pair up in four d-orbitals, followed by 4sp2-hybridization, resulting in a square planar complex.