Problem 23 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Element \(X\) is strongly electropositive and element \(Y\) is strongly electronegative. Both are univalent. The compound formed would be [1980] (a) \(X^{+} Y^{-}\) (b) \(X^{-} X^{+}\) (c) \(X-Y\) (d) \(\quad X \rightarrow Y\)

Step-by-Step Solution

Answer

The compound formed is \(X^{+} Y^{-}\), so option (a) is correct.

Step 1: Understand Univalent Elements

Univalent elements have a valency of one, which means they can either lose or gain a single electron. Element X is strong electropositive, so it tends to lose one electron. Element Y is strongly electronegative, so it tends to gain one electron.

Step 2: Identify the Charge of Ions

Element X, being strongly electropositive and univalent, will lose an electron to form a cation with a charge of +1, indicated as \(X^{+}\). Element Y, being strongly electronegative, will gain that electron to become an anion with a charge of -1, indicated as \(Y^{-}\).

Step 3: Determine the Compound Formula

Since X loses one electron and Y gains one electron, they combine to form a neutral ionic compound \(X^{+}Y^{-}\). The compound is formed by ionic bonding between the oppositely charged ions.

Step 4: Select the Correct Option

Among the given options, option (a), \(X^{+} Y^{-}\), correctly represents the ionic compound formed between X and Y, as it reflects the ion combination resulting from their respective electropositive and electronegative natures.

Key Concepts

Electropositivity

Electropositivity refers to an element's tendency to lose electrons and form positive ions, also known as cations. In the context of our exercise, element \(X\) is noted as being strongly electropositive. This means that \(X\) readily loses an electron. But why would \(X\) want to lose an electron?

It's all about stability. Many electropositive elements prefer to shed their outermost electron to achieve a more stable electronic configuration, often resembling the nearest noble gas structure. Electropositivity is a characteristic feature of metals, which are typically found on the left side of the periodic table. The more electropositive an element is, the more it desires to lose electrons.

• Element \(X\) loses electrons.
• Forms cation \(X^{+}\).
• Contributes to ionic bonding.

Such strong tendencies can lead to the formation of an ionic compound, like the one in the exercise, when paired with an electronegative counterpart.

Electronegativity

Electronegativity is the propensity of an atom to attract and hold onto electrons. In our example, element \(Y\) is described as strongly electronegative. This inclination makes \(Y\) eager to gain electrons from another atom. But how does this work?

Electronegativity is a key factor in determining how an atom will interact with others. Atoms like \(Y\), often nonmetals, are usually found on the right of the periodic table. They have a high attraction for electrons, allowing them to complete their outer electron shells. Strongly electronegative elements tend to form anions, or negatively charged ions.

• Element \(Y\) gains electrons.
• Forms anion \(Y^{-}\).
• Facilitates ionic bonds with positive ions.

This results in the formation of ionic bonds, as seen in the compound \(X^{+}Y^{-}\) from the exercise.

Univalent Elements

Univalent elements have a valency of one, meaning they can either lose or gain just one electron. Both elements \(X\) and \(Y\) in the exercise are described as univalent. This characteristic plays a crucial role in the formation of ionic compounds. So how do univalent elements behave in reactions?

Their simple electron transfer processes make them perfect candidates for forming ionic bonds. For example:

• Element \(X\), losing an electron, becomes \(X^{+}\).
• Element \(Y\), gaining an electron, turns into \(Y^{-}\).

As univalent, they effortlessly engage in the electron transfer needed to stabilize both ions, leading to the creation of balanced, neutral compounds. The straightforward nature of univalent ions allows them to bond easily, giving rise to simple yet stable ionic compounds, like the one symbolized by \(X^{+}Y^{-}\) in the given scenario.