Question
Given below are two statements
Statements-I : The number of paramagnetic species among $\left\lbrack {CoF}_{6} \right\rbrack^{3 -},\left\lbrack {TiF}_{6} \right\rbrack^{3 -},V_{2}O_{5}$ and $\left\lbrack Fe(CN)_{6} \right\rbrack^{3 -}$ is 3 .
Statement-II : $K_{4}\left\lbrack Fe(CN)_{6} \right\rbrack < K_{3}\left\lbrack Fe(CN)_{6} \right\rbrack < \left\lbrack Fe\left( H_{2}O \right)_{6} \right\rbrack{SO}_{4} \cdot H_{2}O < \left\lbrack Fe\left( H_{2}O \right)_{6} \right\rbrack{Cl}_{3}$ is the correct order in terms of number of unpaired electron(s) in the complexes.
In the light of the above statements, choose the correct answer from the options given below.
Options
Solution
Paramagnetic species :
$$\left\lbrack {CoF}_{6} \right\rbrack^{3 -},\left\lbrack {TiF}_{6} \right\rbrack^{3 -},\left\lbrack Fe(CN)_{6} \right\rbrack^{3 -} $$Diamagnetic species : $V_{2}O_{5}$
In $K_{4}\left\lbrack Fe(CN)_{6} \right\rbrack \Rightarrow$ No. of unpaired electron $= 0$
$K_{3}\left\lbrack Fe(CN)_{6} \right\rbrack \Rightarrow$ No. of unpaired electron $= 1$
$\left\lbrack Fe\left( H_{2}O \right)_{6} \right\rbrack{SO}_{4} \cdot H_{2}O \Rightarrow$ No. of unpaired electron $= 4$
$\left\lbrack Fe\left( H_{2}O \right)_{6} \right\rbrack{Cl}_{3} \Rightarrow$ No. of unpaired electron $= 5$
Create a free account to view solution
View Solution Free