The d - electron configuration of $\left[\mathrm{Ru}(\mathrm{en})_{3}\right] \mathrm{Cl}_{2}$ and $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$, respectively are:

(A) $t_{2 g}^{6} e_{g}^{0}$ and $t_{2 g}^{4} e_{g}^{2}$

(B) $\mathrm{t}_{2 \mathrm{~g}}^{4} \mathrm{e}_{\mathrm{g}}^{2}$ and $\mathrm{t}_{2 \mathrm{~g}}^{4} \mathrm{e}_{\mathrm{g}}^{2}$

(C) $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{e}}^{0}$ and $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$

(D) $\mathrm{t}_{2 \mathrm{~g}}^{4} \mathrm{e}_{\mathrm{g}}^{2}$ and $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Coordination Compounds

The d - electron configuration of

[Ru (en)_{3}] Cl_{2}

and

[Fe (H_{2} O)_{6}] Cl_{2}

, respectively are:

Answer: (A)

[R u (e n)_{3}] C l_{2}

ruthenium is in the

+ 2

state, a

4 d^{6}

ion, and ethylenediamine is a strong-field chelating ligand, so the electrons pair into the low-spin set

t_{2 g}^{6} e_{g}^{0}

. In

[F e (H_{2} O)_{6}] C l_{2}

iron is also

+ 2

, a

3 d^{6}

ion, but water is a weak-field ligand, so the electrons stay as spread out as possible in the high-spin set

t_{2 g}^{4} e_{g}^{2}

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Single correct option