Chapter 22, Problem 7QP

Interpretation Introduction

Interpretation:

The highest oxidation states from Sc to Cu are to be determined.

Concept introduction:

The transition metals are those that form at least one stable ion with incompletely filled d sub shell.

The transition metals show variable oxidation states.

The transition metals have partially filled d-sub shell.

Answer to Problem 7QP

Solution: The metals with highest oxidation states as follows:

Sc $+3$, Ti $+4,$ V $+5$, Cr $+6$, Mn $+7$, Fe $+6$, Co $+4$, Ni $+3$, and Cu $+3$

Explanation of Solution

Transition metals have a greater number of oxidation states than any other element due to their ability to lose multiple electrons from the d subshell. The highest oxidation state of a transition element is determined by adding a number of unpaired electrons in a d subshell and number of electrons present in 4s orbital electrons.

The electron configurations of transition metals are as follows:

$\text{Sc - 4}{s}^{2}3d^{1}oneunpairedelectronoxidationstate=+3$

$\text{Ti - 4}{s}^{2}3d^{2}twounpairedelectronsoxidationstate=+4$

$\text{V - 4}{s}^{2}3d^{3}threeunpairedelectronsoxidationstate=+5$

$\text{Cr - 4}{s}^{1}3d^{5}sixunpairedelectronsoxidationstate=+6$

$\text{Mn - 4}{s}^{2}3d^{5}fiveunpairedelectronsoxidationstate=+7$

$\text{Fe - 4}{s}^{2}3d^{6}fourunpairedelectronsoxidationstate=+6$

$\text{Co - 4}{s}^{2}3d^{7}threeunpairedelectronsoxidationstate=+4$

$\text{Ni - 4}{s}^{2}3d^{8}twounpairedelectronsoxidationstate=+3$

$\text{Cu - 4}{s}^{1}3d^{10}oneunpairedelectronoxidationstate=+3$

Sc, Ti, V, Cr, Mn, Fe, Co, Ni, and Cu have $1,2,3,4,5$, 4, 3, 2, and 1 unpaired electrons in a d subshell, respectively.

Conclusion

The highest oxidation states of Sc, Ti, V, Cr, Mn, Fe, Co, Ni, and Cu are

$+3,+4,+5,+6,+7,+6,+4,+3,\text{ and }+3$ respectively.