Question

An element ${}_{11}{X^{23}}$ forms an ionic compound with another element $Y$ . Then the charge on the ion formed by $X$ is:
A) $+ 1$
B) $+ 2$
C) $- 1$
D) $- 2$

Answer 1

We know that the ionic bond is formed between the cation and the anion. We also know that cation is formed by the donation of electrons and anion is formed by accepting the electron.

Complete step-by-step answer:
We know that the number of electrons in which atom loses or gains while forming an ionic bond is called its electrovalency.
Here, electrovalency is nothing but the unit charge on the ions used in the formation of the ionic bond.
We know that while the formation of ionic bonds complete transfer of electrons takes place and this complete transfer of electrons occurs to attain stability of the atoms taking part in the formation of an ionic bond.
We also know that stability can be attained by the formation of the octet.
We are given the element having atomic number 11 and atomic mass number 23.
Let us write the electronic configuration of the element $X$ :
$1{s^2}2{s^2}2{p^6}3{s^1}$
Now, the completion of the octet element $X$ will tend to donate 1 electron from the $3s$ orbital to form the ionic bond. As the element $X$ is donating 1 electron so it will gain $+ 1$ a positive charge.
As the element $X$ is donating the electrons so the element $Y$ will accept the electron to form an ionic bond.
So, the ionic bond will be formed in between ${X^ + }{Y^ - }$.

Therefore, we can conclude that the correct answer to this question is option A.

Note: We have to focus on the electronic configuration of the element as the atomic number is provided in the question. In this type of question formation of octet decides whether the element will accept the electron or donate the electron. According to the question, the element X has an atomic number of 11, i.e. it is sodium. It is an s-block element and loses an electron to form $N{a^ + }$ ion. It forms ionic bonds with most elements, especially halogens like chlorine, bromine etc.