Problem 15
Question
An element \(M\) reacts with chlorine to form a compound \(\mathrm{X}\). The bond angle in \(\mathrm{X}\) is \(120^{\circ}\). What is \(\mathrm{M}\) ? (a) \(\underline{B}\) (b) \(\mathrm{N}\) (c) \(\mathrm{Be}\) (d) \(\mathrm{Mg}\)
Step-by-Step Solution
Verified Answer
The element \( M \) is boron (\( \underline{B} \)).
1Step 1: Understanding the Problem
The problem asks us to determine the element \( M \) that forms a compound \( X \) with chlorine and has a bond angle of \( 120^{\circ} \). We need to identify which element among the given options can form such a compound.
2Step 2: Understanding Bond Angles
A bond angle of \( 120^{\circ} \) suggests a trigonal planar molecular geometry, which is common for compounds where the central atom forms three bonds and has no lone pairs on it.
3Step 3: Analyzing Each Element
Let's consider each element:- \( \underline{B} \): Boron can form \( \text{BCl}_3 \), which is trigonal planar and has a bond angle of \( 120^{\circ} \).- \( \text{N} \): Nitrogen typically forms \( \text{NCl}_3 \), which has a trigonal pyramidal shape due to a lone pair, with bond angles less than \( 109.5^{\circ} \).- \( \text{Be} \): Beryllium forms \( \text{BeCl}_2 \), which is linear with a bond angle of \( 180^{\circ} \).- \( \text{Mg} \): Magnesium forms \( \text{MgCl}_2 \), which also leads to a linear structure.
4Step 4: Conclusion
Boron, \( \text{B} \), forms the compound \( \text{BCl}_3 \) with the required bond angle of \( 120^{\circ} \). Therefore, the correct element \( M \) is boron.
Key Concepts
Trigonometry of compoundsMolecular geometryBond angles
Trigonometry of compounds
When it comes to determining the structure of molecules, trigonometry plays a crucial role in understanding how atoms are arranged in a compound. The term "trigonometry of compounds" refers to the use of geometric principles to predict and describe the spatial arrangement of atoms. In compounds like \( \text{BCl}_3 \), this concept helps to recognize the angles between chemical bonds. These angles are vital for defining the shape of the molecule and how it interacts with other molecules. By applying trigonometric principles, chemists can predict patterns and behaviors in molecular structures, helping in understanding reactivity and properties. For instance, in the case of boron trifluoride, the bonding is even around the central atom, forming equal bond angles of \(120^{\circ}\). Understanding these principles helps chemists infer structures with ease and precision.
Molecular geometry
Molecular geometry is a fundamental aspect of inorganic chemistry that describes the three-dimensional arrangement of atoms in a molecule. It's determined by the number of bonds and lone pairs surrounding the central atom. For instance, the compound \(\text{BCl}_3\) forms a trigonal planar geometry. This means that the boron atom in the center creates three equal bond angles of \(120^{\circ}\) with chlorine atoms placed at the vertices of an equilateral triangle. Key points to remember about molecular geometry:
- It affects the physical and chemical properties of the compound, such as polarity, color, and reactivity.
- The shape is predicted using the VSEPR (Valence Shell Electron Pair Repulsion) theory.
- The geometry influences how the molecule interacts with light, determines melting and boiling points, and affects its reactivity.
Bond angles
Bond angles refer to the angles between adjacent chemical bonds within a molecule. These angles are influenced by the types of bonds and the presence of lone pairs on atoms. The bond angle is a key factor that determines the molecular shape and, in turn, the molecule's properties. In \(\text{BCl}_3\), the bond angle is exactly \(120^{\circ}\), typical of a trigonal planar geometry. This angle is a result of the even distribution of bonding pairs around the central boron atom. Important insights on bond angles:
- Bond angles can vary due to the presence of lone pairs, which tend to occupy more space and thus decrease the angles between bonded atoms.
- They are essential for predicting the spatial distribution of atoms in a molecule.
- Consistent bond angles are crucial for maintaining molecular symmetry.
Other exercises in this chapter
Problem 13
Alkalimetals have high oxidation potential and hence, they behave as (a) electrolytes (b) Lewis bases (c) oxidizing agents (d) reducing agents
View solution Problem 14
Alkali metals present in their compounds are always (a) monovalent (b) bivalent (c) zerovalent (d) none of these
View solution Problem 18
Which of the following has the least ionization potential? (a) \(\mathrm{He}\) (b) \(\mathrm{Li}\) (c) \(\mathrm{Zn}\) (d) \(\mathrm{N}\)
View solution Problem 19
As the alkaline earth metals (except Be) tend to lose their valence electrons readily, they act as (a) bases (b) strong reducing agents (c) weak oxidizing agent
View solution