Question

Most transition metals form multiple cations, that is, they have more than one possible amount of positive charge. Virtually all of the transition metals form dipositive $(2+)$ cations, along with one or more other forms. For example, titanium forms $T{i}^{2+}$ and $T{i}^{4+}$, and cobalt forms $C{o}^{2+}$ and $C{o}^{3+}$
The principal exceptions to this predominance of cations with a $2^{+}$ charge are scandium and yttrium, which form only tripositive cations, $S{c}^{3+}$ and $Y^{3+}$, and copper and silver, which form cations with a single positive charge: $C{u}^{+}$ and $A{g}^{+}$.
[While copper also forms dispositive $C{u}^{2+}$, silver exhibits the single positive state essentially exclusively.]
Which of the following offers a possible explanation for the existence of $C{u}^{+}$ and $A{g}^{+}$?

• A. Mendeleev placed copper and silver in Group IB on his original periodic table because they formed compounds with the same combining ratios as elements in Group $1A$
• B. Elemental copper and silver have a single valence electron, which is lost in cation formation.
• C. Copper and silver have higher ionization energies, than the other transition metals in their respective periods, so are unable to lose more than one electron.
• D. The loss of one electron gives copper and silver noble gas configurations.

Answer 1

The correct option is A Mendeleev placed copper and silver in Group IB on his original periodic table because they formed compounds with the same combining ratios as elements in Group $1A$

The elements Copper ($Cu$) and Silver ($Ag$) have only one valence electron in their outermost shell and have fully filled d sub shells. Thus the single valence electron is lost in formation of cation.

Hence answer is option $A$.