The effective atomic number for [R h(H2 O)6]3+ (at. no. for R h is 45 ) is

Question

The effective atomic number for [R h(H2 O)6]3+ (at. no. for R h is 45 ) is (A) 42 (B) 45 (C) 48 (D) 54

Tardigrade Admin · Accepted Answer

The effective atomic number for [R h(H2 O)6]3+ (atomic no. of R h=45 ) is 54 . Because rhodium is in oxidation state x+6 × 0=+3 x=+3 R h3+=45-3=42 42+12 electrons from 6 ligands (H2 O)=54 electrons.

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

42

45

48

54

The effective atomic number for $[R h (H_{2} O)_{6}]^{3 +}$
(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
$(H_{2} O) = 54$ electrons.

Q. The effective atomic number for [Rh(H2​O)6​]3+ (at. no. for Rh is 45 ) is

42

45

48

54

The effective atomic number for [Rh(H2​O)6​]3+ (atomic no. of Rh=45 ) is 54 . Because rhodium is in oxidation state x+6×0=+3 x=+3 Rh3+=45−3=42 42+12 electrons from 6 ligands (H2​O)=54 electrons.

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

The effective atomic number for $[R h (H_{2} O)_{6}]^{3 +}$
(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
$(H_{2} O) = 54$ electrons.