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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