Question

Most stable in aq. solution
A.$M{n^{ + 3}}$
B.$C{r^{2 + }}$
C.\[{V^{2 + }}\]
D.$T{i^{2 + }}$

Answer 1

Hint: In the given question we have to find out that what is the set of elements that are mentioned as they all belong to a specific group of elements. Now we can relate to the trend of the metallic character by the help of the general trend of the series. Thenwe know that it would increase down the group so the one at the top is weakest.

Complete Step by step Solution:
The given question is asked about the metallic character of the given elements. All these elements belong to the group $1$ of the modern periodic table and are considered generally most metallic among the entire series of elements in the world into existence.
Now there is a general trend regarding the metallic character of any group in the table which is:
The metallic character would decrease on the descend into any of the groups of the table.
This is because of a reason, because when we will move down any group, the Ionization Energy (IE) of the element will decrease and therefore that would result in the increase of their tendency to lose electrons.
This in result would provide a significant over the metallic character as their metallic character would also increase.
Hence among given options, the weakest metallic character is of $Li$ as it is at the top of the group.

Therefore, the right option would be option B, $C{r^{2 + }}$ .

Note: The trend in carbocations is that the more substituents on the carbocation, the greater the stability. Tertiary cations, with three substituents on the carbocation, are more stable than secondary cations, with two substituents on the carbocation. Primary cations are more stable than methyl cations