Question

Determine the chemical formula for the compound, diaquadicarbonylzinc tetrabromopalladate(iv).

Answer 1

Missing question:
A. [PdZn(H2O)2(CO)2]Br4. 
B. [Zn(H2O)2(CO)2]2[PdBr4]. 
C. [Pd(H2O)2][Zn(CO)2]Br4. 
D. [Pd(H2O)2]2[Zn(CO)2]3Br4. 
E. [Zn(H2O)2(CO)2][PdBr4].
Answer is: E. [Zn(H2O)2(CO)2][PdBr4]..
In this complex diaqua means two waters (H₂O), dicarbonyl means two carbonyl groups (CO), zinc(Zn) and palladium (Pd) are central atoms or metals, bromine has negative charge -1. Bromine, water and carbonyl are ligands.

Answer 2

Answer:

$[Zn(H_2O)_2(CO)_2][PdBr_4]$

Explanation:

Hello,

In this case, one could build-up the formula, via analyzing the name as follows:

- It says tetrabromopalladate (IV), it means that it is a mixed anion, composed by four bromine atoms (tetrabromo) and one palladium atom due suffix ate), therefore the anion is $[PdBr_4]$.

- By means of the cation, it starts by diaqua, so it has two $H_2O$ as an aquocation. Next, is says dicarbonyl, so two carbonyl groups $C=O$ are bonded as well. Finally, the metallic part is given by the presence of zinc, so the cation turns out as $[Zn(H_2O)_2(CO)_2]$ by complexity.

Therefore, the resulting formula is: $[Zn(H_2O)_2(CO)_2][PdBr_4]$

Best regards.