Give the number of (valence) \(d\) electrons associated with the central metal ion in each of the following complexes: (a) \(\mathrm{K}_{3}\left[\mathrm{TiCl}_{6}\right]\) (b) \(\mathrm{Na}_{3}\left[\mathrm{Co}\left(\mathrm{NO}_{2}\right)_{6}\right],\) (c) \(\left[\mathrm{Ru}(\mathrm{en})_{3}\right] \mathrm{Br}_{3},\) (d) \([\mathrm{Mo}(\mathrm{EDTA})] \mathrm{ClO}_{4},(\mathrm{e}) \mathrm{K}_{3}\left[\mathrm{ReCl}_{6}\right] .\)

For each of the given complexes, identify the central metal ion: (a) K3[TiCl6]: Ti (b) Na3[Co(NO2)6]: Co (c) [Ru(en)3]Br3: Ru (d) [Mo(EDTA)]ClO4: Mo (e) K3[ReCl6]: Re

For each complex, write out the charge balance equation and solve for the oxidation state of the central metal ion. (a) K3[TiCl6]: 3(+1) + x + 6(-1) = 0 -> x = +3 (Ti) (b) Na3[Co(NO2)6]: 3(+1) + x + 6(-1) = 0 -> x = +3 (Co) (c) [Ru(en)3]Br3: x + 3(-1) = 0 -> x = +3 (Ru) (d) [Mo(EDTA)]ClO4: x + (-1) = 0 -> x = +1 (Mo) (e) K3[ReCl6]: 3(+1) + x + 6(-1) = 0 -> x = +3 (Re)

Using the oxidation state from step 2, we can now find the number of valence d electrons: (a) Ti(III): There are 4 valence electrons for Ti in its ground state, so losing 3 will result in 1 d electron left. (b) Co(III): There are 9 valence electrons for Co in its ground state, so losing 3 will result in 6 d electrons left. (c) Ru(III): There are 8 valence electrons for Ru in its ground state, so losing 3 will result in 5 d electrons left. (d) Mo(I): There are 6 valence electrons for Mo in its ground state, so losing 1 will result in 5 d electrons left. (e) Re(III): There are 7 valence electrons for Re in its ground state, so losing 3 will result in 4 d electrons left. In conclusion, the number of valence d electrons associated with the central metal ion in each complex is: (a) 1 (b) 6 (c) 5 (d) 5 (e) 4