Question
Given below are two statements:
Statement-I : Presence of large number of unpaired electrons in transition metal atoms results in higher enthalpies of their atomization.
Statement-II : $d_{xy} = d_{xz} = d_{yz} < d_{x^{2} - y^{2}} = d_{z^{2}}$ and $d_{x^{2} - y^{2}} = d_{z^{2}} < d_{xy} = d_{xz} = d_{yz}$ are the d-orbital splittings in $\left\lbrack Fe\left( H_{2}O \right)_{6} \right\rbrack^{3 +}$ and $\left\lbrack Ni(Cl)_{4} \right\rbrack^{2 -}$ complex ions respectively.
In the light of the above statements, choose the correct answer from the options given below:
Options
Solution
Sol. [Key Concept: More unpaired electrons give stronger metal-metal bonding and higher atomisation enthalpy; $\lbrack Fe(H_{2}O)_{6}\rbrack^{3 +}$ is octahedral ($t_{2g} < e_{g}$) while $\lbrack Ni\ {Cl}_{4}\rbrack^{2 -}$ is tetrahedral ($e < t_{2}$).]
Statemet I :
More are the number of unpaired electrons , stronger is metal-metal boding thus higher enthalpy of atomisation.
Statement II :
$\left\lbrack Fe\left( H_{2}O \right)_{6} \right\rbrack^{3 +}$ is octahedral complex.
Thus, $\left\lbrack t_{2\text{ }g}\left( {\text{ }d}_{xy} = d_{yz} = d_{zx} \right) < e_{g}\left( d_{x^{2} - y^{2}} = d_{z^{2}} \right) \right\rbrack$
$\left\lbrack {NiCl}_{4} \right\rbrack^{2 -}$ is tetrahedral complex.
Thus, $\left\lbrack e\left( d_{x^{2} - y^{2}} = d_{z^{2}} \right) < t_{2}\left( {\text{ }d}_{xy} = d_{yz} = d_{zx} \right) \right\rbrack$
Create a free account to view solution
View Solution Free