Problem 56 - A Complete Resource Book in Chemistry for JEE Main

Question

Which one of the following elements does not form the compound, \(\mathrm{M}_{4} \mathrm{O}_{10}(\mathrm{M}=\) element \() ?\) (a) As (b) \(\mathrm{Bi}\) (c) \(\mathrm{Sb}\) (d) \(\mathrm{Pb}\)

Step-by-Step Solution

Answer

The element that does not form \(\mathrm{M}_{4}\mathrm{O}_{10}\) is \(\mathrm{Pb}\) (d).

Step 1: Understand the Given Compound Formula

The compound given is \(\mathrm{M}_{4} \mathrm{O}_{10}\). This indicates that for every 4 atoms of element M, there are 10 atoms of oxygen. The compound is likely an oxide of a group named element.

Step 2: Analyze the Valency Requirement

The formula \(\mathrm{M}_{4}\mathrm{O}_{10}\) can be derived from the simplest oxide formula \(\mathrm{M}_{2}\mathrm{O}_{5}\). Therefore, it suggests that the element M must have a valency of +5 to balance with two oxygen atoms (each with a valency of -2).

Step 3: Evaluate Given Elements

Determine which elements are likely to form a +5 oxidation state:1. (a) As (Arsenic) forms \(\text{As}_2\text{O}_5\).2. (b) Bi (Bismuth) commonly forms +3 or +5 oxides.3. (c) Sb (Antimony) forms \(\text{Sb}_2\text{O}_5\). 4. (d) Pb (Lead) generally forms oxides as \(\text{PbO}\) and \(\text{PbO}_2\), with typical oxidation states of +2 and +4, rather than +5.

Step 4: Identify the Element that Doesn't Fit

From the evaluations, elements As, Bi, and Sb can theoretically reach the +5 oxidation state to form \(\mathrm{M}_{4}\mathrm{O}_{10}\), whereas Pb primarily forms lower oxidation state compounds like \(\text{PbO}\) and \(\text{PbO}_2\). Thus, Pb is the element that does not fit the required oxidation state for \(\mathrm{M}_{4}\mathrm{O}_{10}\).

Key Concepts

Valency

Valency is the concept used to describe the combining power of an element, particularly in forming compounds. It usually corresponds to the number of electrons an atom can lose, gain, or share during a chemical reaction. Understanding valency is crucial because it helps predict how elements interact with one another.

In the context of oxides like \(\text{M}_4\text{O}_{10}\), the valency of the element \(M\) becomes a pivotal point to consider. Oxygen typically has a valency of \(-2\). This means that for every oxygen in the compound, it should form two bonds. Hence, to balance with \(5\times\left(-2\right)=-10\) oxidation state required by the oxygen atoms, the combined valency of the 4 M atoms in \(\text{M}_4\text{O}_{10}\) must be \(+10\).

• Hints to balance the charges of components in the compound.
• Gives insight on which elements can form stable compounds at respective valencies.

Group 15 Elements

The elements in Group 15 of the periodic table are known as the pnictogens. These include nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb), and bismuth (Bi). These elements have 5 electrons in their outer shell and often show a variety of oxidation states, mainly due to the availability of vacant d orbitals in heavier pnictogens such as As, Sb, and Bi.

Their general oxidation states are -3, +3, and +5. While nitrogen uses oxidation states ranging from -3 to +5, arsenic, antimony, and bismuth exhibit +3 and +5, making them fits for forming the compound \(\text{M}_4\text{O}_{10}\):

• Arsenic (As) forms stable pentoxide compounds, highlighting its +5 state.
• Antimony (Sb) is similar, also forming \( \text{Sb}_2\text{O}_5 \) with oxygen.
• Bismuth (Bi) can form +5 states though less stable compared to its +3.
• Only lead (Pb), a Group 14 element, lacks the common ability to reach a stable +5 oxidation state required for \(\text{M}_4\text{O}_{10}\).

Oxide Compounds

Oxide compounds are formed when elements combine with oxygen. They play a crucial role in various chemical reactions and industrial processes. Understanding how elements form oxides requires knowledge of valency and oxidation states.

The general formula for an oxide compound is \(\text{M}_x\text{O}_y\), where M is typically a metal or non-metal and O represents oxygen. Oxidation states help in determining how many oxygen atoms will be needed to create a stable compound with M. In the case of \(\text{M}_4\text{O}_{10}\), this implies a balanced ratio where oxygen balances out the element's total possible oxidation state:

• Gives insights on requirements for forming stable compound structures.
• Shows why elements choose specific oxidation states, influencing practical application in industries.

Understanding these relationships is essential to chemists in predicting compound formation and evaluating reaction outcomes.