Chapter 23: Q23.90P (page 1061)

Give the number of d electrons ( n of $d^{n}$ ) for the central metal ion in (a) ${[{TiCl}_{6}]}^{2 -}$ ; (b) $K [{AuCl}_{4}]$ ; (c) ${[{RhCl}_{6}]}^{3 -}$ .

Expert verified

Step by step solution

The electron configuration of an element describes how electrons are distributed in its atomic orbitals.

Electron configuration of Ti: $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{2} 4 s^{2} \ln {[{TCl}_{6}]}^{2 -}$ the oxidation state of Ti is $+ 4 ({Ti}^{4 +})$ , so it means we have to remove 2 electrons from the 4 s subshell and 2 electrons from the 3d subshell, which leaves ${Ti}^{4 +}$ with $0 3 d$ electrons.
Electron configuration on localid="1663328440485" $Au : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{1} 04 s^{2} 4 p^{6} 4 d^{10} 5 s^{2} 5 p^{6} 4 f^{14} 5 d^{10} 6 s^{1} InK [AuCl 4$ the oxidation state of Au is $+ 3 ({Au}^{3 +})$ , so it means we have to remove 1 electron from the 6s subshell and 2 electrons from the 5d subshell, which leaves ${Au}^{3 +}$ with 8, 5d electrons.
Electron configuration of $Rh : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{8} 5 s^{1} \ln {[{RhCl}_{6}]}^{3 -}$ the oxidation state of Rh is $+ 3 ({Rh}^{3 +})$ , so we have to remove 1 electron from the 5 s subshell and 2 electrons from the 4d subshell, which leaves ${Rh}^{3 +}$ with 6, 4d electrons.

Unlock Step-by-Step Solutions & Ace Your Exams!

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.