I\'m trying to figure out why the complex ion Pt(NH3)4(2+) is considered square

Question

I'm trying to figure out why the complex ion Pt(NH3)4(2+) is considered square planar. I understand the whole concept of coordination number and that I have to distinguish between square planar or tetrahedral.
Pt's configuration: [Ar] 6s^(2)5d^(8) Pt^ (2+) config: [Ar]3d^10
I concluded that the configuration for the complex ion should be sp^3 (tetrahedral) since there is no orbital that is empty where a coordinate bond can form. Also, the d-orbitals are not used to get to Pt^(2+).
I was wondering what is wrong in my reasoning. I'm trying to figure out why the complex ion Pt(NH3)4(2+) is considered square planar. I understand the whole concept of coordination number and that I have to distinguish between square planar or tetrahedral.
Pt's configuration: [Ar] 6s^(2)5d^(8) Pt^ (2+) config: [Ar]3d^10
I concluded that the configuration for the complex ion should be sp^3 (tetrahedral) since there is no orbital that is empty where a coordinate bond can form. Also, the d-orbitals are not used to get to Pt^(2+).
I was wondering what is wrong in my reasoning.

Explanation / Answer

all 4 atom compound of pt are square planar its an exception to general theory and every book states that only

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