Question

Eka-aluminum forms chloride of the type:
(A) ${E_2}Cl$
(B) $EC{l_3}$
(C) ${E_2}C{l_3}$
(D) ${E_3}C{l_4}$

Answer 1

When Mendeleev proposed his periodic table he predicted the existence of undiscovered element he named them eka-aluminum, eka-Boron, eka-silicon and eka-manganese they existed with the properties of Gallium, scandium, germanium and technetium.

Complete answer:
The elements in the group contain the same properties and the same atomic number. Mendeleev placed eka-aluminum under the aluminum because the element was undiscovered. When gallium was discovered it contained the same properties as eka-aluminum, thus the eka-aluminum was replaced by gallium.
All the properties of eka-aluminum are of gallium, so the chloride type of eka-aluminum will be the same as of the chloride of gallium. Thus the chloride formed by gallium is gallium trichloride $(GaC{l_3})$ .
Gallium has three positive charge because gallium has three valence electrons as same of aluminum, so it tends to lose electrons then after losing the $3$ electrons, these three electrons are gained by the three atoms of chlorine because chlorine have the shortage of $1$ electrons.
Gallium trichloride is a Lewis acid catalyst such as in the field-crafts reaction, and is also used in the carboxylation reaction of compounds with a carbon-carbon triple bond.
Therefore, Option $(b), EC{l_3}$ is correct.

Note:
Mendeleev named the elements as eka boron because this element lies below boron, eka-aluminum as this element lies below aluminum, eka-silicon as this element lies below silicon. Mendeleev named these unknown elements using the term eka. Which means the first element of the same family beyond the one to whose name it is prefixed.