Problem 14
Question
Boron when heated with carbon forms (a) \(\mathrm{B}_{4} \mathrm{C}\) (b) \(\mathrm{B}_{2} \mathrm{C}_{3}\) (c) \(\mathrm{BC}_{2}\) (d) \(\mathrm{B}_{4} \mathrm{C}_{3}\)
Step-by-Step Solution
Verified Answer
Boron and carbon form \\(\mathrm{B}_{4}\mathrm{C}\\).
1Step 1: Identify the Elements
The problem statement mentions that boron (B) is heated with carbon (C) to form a compound. It's important to focus on these two elements and their typical reactions.
2Step 2: Understand the Common Compound
Boron and carbon typically react to form compounds used in ceramics and materials science. One common compound formed by boron and carbon is \(\mathrm{B}_{4} \mathrm{C}\), known for its hardness and high melting point.
3Step 3: List the Possible Compounds
The provided options include different stoichiometric combinations of boron and carbon: \(\mathrm{B}_{4}\mathrm{C}\), \(\mathrm{B}_{2}\mathrm{C}_{3}\), \(\mathrm{BC}_{2}\), and \(\mathrm{B}_{4}\mathrm{C}_{3}\). Not all these combinations are common or stable for boron and carbon.
4Step 4: Choose the Correct Compound
Based on chemical knowledge and typical reactions, the most recognized and stable compound formed by heating boron with carbon is \(\mathrm{B}_{4}\mathrm{C}\). Other proportions are not typically seen in chemical reactions involving boron and carbon.
Key Concepts
Understanding Boron ReactionsExploring Carbon CompoundsApplying Stoichiometry to Chemical Compounds
Understanding Boron Reactions
Boron is a fascinating element that plays a key role in various chemical reactions. One notable reaction is its combination with carbon to form boron carbide, a compound known for its extreme hardness. When heated, boron reacts with carbon through a process that involves the breaking and forming of chemical bonds. This reaction leads to the creation of stable compounds. Typically, boron forms compounds that are robust and can withstand high temperature and stress. In the context of the exercise, the compound
- formed is Boron Carbide (\(\mathrm{B}_{4}\mathrm{C}\)), which is widely used in industries for its high hardness similar to diamond.
Exploring Carbon Compounds
Carbon is a remarkable element with the ability to form numerous compounds due to its tetravalency. This means carbon can form four bonds with other elements, creating a vast number of possible compounds. When carbon combines with elements like boron, it results in compounds with unique characteristics.
- For instance, carbon's ability to form long chains or networks contributes to the solid structures found in materials like boron carbide.
- In the compound \(\mathrm{B}_{4}\mathrm{C}\), carbon atoms play a critical role in stabilizing the structure, enhancing its hardness and thermal stability.
Applying Stoichiometry to Chemical Compounds
Stoichiometry is the branch of chemistry that deals with the quantitative relationships of the reactants and products in chemical reactions. It is an essential concept, especially in determining the proportions of elements in compounds. When considering the formation of boron carbide, stoichiometry helps in identifying the specific ratios in which boron and carbon combine to form stable compounds.
- For the compound \(\mathrm{B}_{4}\mathrm{C}\), stoichiometry confirms that it consists of four boron atoms and one carbon atom.
- This ratio is significant because it defines the properties and stability of the compound.
Other exercises in this chapter
Problem 12
\(\mathrm{H}_{3} \mathrm{BO}_{3}\) is (a) monobasic and a weak lewis acid (b) monobasic and a weak bronsted acid (c) monobasic and a strong lewis acid (d) triba
View solution Problem 13
Mark the wrong statement which of the following statements about/regarding \(\mathrm{BF}_{3}\) is incorrect? (a) it forms an addition compound with \(\mathrm{NH
View solution Problem 15
When strongly heated, orthoboric acid gives (a) \(\mathrm{H}_{2} \mathrm{~B}_{4} \mathrm{O}_{7}\) (b) \(\mathrm{B}_{2} \mathrm{O}_{3}\) (c) \(\mathrm{HBO}_{2}^{
View solution Problem 16
From \(\mathrm{B}_{2} \mathrm{H}_{6}\), all the following can be prepared except (a) \(\mathrm{B}_{2}\left(\mathrm{CH}_{3}\right)_{6}\) (b) \(\mathrm{NaBH}_{4}\
View solution