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The d-electron configuration of \mathrm{\left [ Ru(en)_{3} \right ]Cl_{2}} and \mathrm{\left [ Fe(H_{2}O)_{6} \right ]Cl_{2}}, respectively are :
Option: 1 \mathrm{t_{2 g}^{6} e_{g}^{0} \text { and } t_{2 g}^{6} e_{g}^{0}}
Option: 2 \mathrm{t_{2 g}^{4} e_{g}^{2} \text { and } t_{2 g}^{6} e_{g}^{0}}
Option: 3 \mathrm{t_{2 g}^{6} e_{g}^{0} \text { and } t_{2 g}^{4} e_{g}^{2}}
Option: 4 \mathrm{t_{2 g}^{4} e_{g}^{2} \text { and } t_{2 g}^{4} e_{g}^{2}}

Answers (1)

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\mathrm{[Ru(en)_3]Cl_2}

Ru belongs to the 4d series of elements and ethylene diamine(en) is a chelating ligand which causes a larger splitting in the 4d orbitals. This leads to a pairing of electrons in the \mathrm{t_{2g}} orbital due to which the electronic configuration is \mathrm{t_{2g}^6e_g^0}

\mathrm{[Fe(H_2O)_6]Cl_2}

\mathrm{H_2O} is a weak field ligand which is unable to cause pairing in \mathrm{Fe^{2+}} and hence the \mathrm{d^{6}} configuration is accomodated as \mathrm{t_{2g}^{4}e_g^2}

Therefore, Option 3 is correct.

Posted by

Kuldeep Maurya

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