Question

The effective atomic number for $\left[R h\left(H_{2} O\right)_{6}\right]^{3+}$ (at. no. for $R h$ is $45$ ) is

• A. 42
• B. 45
• C. 48
• D. 54

Answer 1

Answer: (D)

54

Answer 2

The effective atomic number for $\left[R h\left(H_{2} O\right)_{6}\right]^{3+}$
(atomic no. of $R h=45$ ) 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
$\left(H_{2} O\right)=54$ electrons.