A) 42
B) 45
C) 48
D) 54
Correct Answer: D
Solution :
The effective atomic number for\[{{[Rh{{({{H}_{2}}O)}_{6}}]}^{3+}}\](atomic no. of\[Rh=45\]) is 54. Because rhodium is in oxidation state \[x+6\times 0=+3\] \[x=+3\] \[R{{h}^{3+}}=45-3=42\] 42+12 electrons from 6 ligands\[({{H}_{2}}O)\] =54 electrons.You need to login to perform this action.
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