Question
Given below are two statements:
Statement I: Crystal Field Stabilization Energy (CFSE) of $\left\lbrack Cr\left( H_{2}O \right)_{6} \right\rbrack^{2 +}$ is greater than that of $\left\lbrack Mn\left( H_{2}O \right)_{6} \right\rbrack^{2 +}$.
Statement II: Potassium ferricyanide has a greater spin-only magnetic moment than sodium Ferrocyanide.
In the light of the above statements, choose the correct answer from the options given below:
Options
Solution
$\left\lbrack Mn\left( H_{2}O \right)_{6} \right\rbrack^{2 +} \Rightarrow$ CFSE value is zero because of $d^{5}$ configuraiton with WFL in coordination number 6 $\left\lbrack Cr\left( H_{2}O \right)_{6} \right\rbrack^{2 +}$.
$\left\lbrack Cr\left( H_{2}O \right)_{6} \right\rbrack^{2 +} \Rightarrow$ CFSE value is $- 0.6\Delta_{0}$ because of $d^{4}$ configuraiton with WFL in coordination number 6.
For : $K_{3}\left\lbrack Fe(CN)_{6} \right\rbrack,\mu = \sqrt{1(1 + 2)} = \sqrt{3}$ B.M.
For : ${Na}_{4}\left\lbrack Fe(CN)_{6} \right\rbrack,\mu = \sqrt{0}$ B.M.
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