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The correct order of bond angles (smallest first) in \(\mathrm{H}_{2} \mathrm{~S}, \mathrm{NH}_{3}, \mathrm{BF}_{3}\) and \(\mathrm{SiH}_{4}\) is: (a) \(\mathrm{H}_{2} \mathrm{~S}<\mathrm{SiH}_{4}<\mathrm{NH}_{3}<\mathrm{BF}_{3}\) (b) \(\mathrm{NH}_{3}<\mathrm{H}_{2} \mathrm{~S}<\mathrm{SiH}_{4}<\mathrm{BF}_{3}\) (c) \(\mathrm{H}_{2} \mathrm{~S}<\mathrm{NH}_{3}<\mathrm{SiH}_{4}<\mathrm{BF}_{3}\) (d) \(\mathrm{H}_{2} \mathrm{~S}<\mathrm{NH}_{3}<\mathrm{BF}_{3}<\mathrm{SiH}_{4}\)

Short Answer

Expert verified
Correct order is \(\mathrm{H}_{2}\mathrm{S}<\mathrm{NH}_{3}<\mathrm{SiH}_{4}<\mathrm{BF}_{3}\).

Step by step solution

01

Determine the Bond Angle of H2S

\(\mathrm{H_2S}\) is a bent molecule with a bond angle less than \(109.5^\circ\) due to two lone pairs on Sulphur, so the bond angle is approximately \(92.1^\circ\).
02

Determine the Bond Angle of NH3

\(\mathrm{NH_3}\) is a trigonal pyramidal molecule with a bond angle slightly less than tetrahedral \(109.5^\circ\) due to one lone pair, resulting in an angle of about \(107^\circ\).
03

Determine the Bond Angle of SiH4

\(\mathrm{SiH_4}\) is a tetrahedral molecule, having bond angles of approximately \(109.5^\circ\) due to no lone pairs on the central Silicon atom.
04

Determine the Bond Angle of BF3

\(\mathrm{BF_3}\) is a planar trigonal molecule with bond angles of exactly \(120^\circ\) due to its sp\(^2\) hybridization and no lone pairs.
05

Arrange the Compounds by Increasing Bond Angles

Order the molecules based on their bond angles from smallest to largest: \(\mathrm{H_2S}\) \(<\) \(\mathrm{NH_3}\) \(<\) \(\mathrm{SiH_4}\) \(<\) \(\mathrm{BF_3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms in a molecule. This concept is crucial for understanding various properties of substances. Knowing the molecular geometry helps predict the behavior of molecules during chemical reactions and their physical properties like boiling and melting points.
Molecules can adopt different shapes based on the number of bonds and lone pairs of electrons around the central atom. The common geometries include linear, bent, trigonal planar, tetrahedral, and more. The VSEPR theory (Valence Shell Electron Pair Repulsion) predicts molecular shapes by considering the repulsion between electron pairs around the central atom.
The arrangement of electron pairs is determined by minimizing repulsions to achieve the lowest energy state. For example, tetrahedral geometry results from four equally spaced bonds around the central atom, minimizing repulsion.
Lone Pairs
Lone pairs of electrons are the pairs of valence electrons that are not shared with other atoms in a bond. They reside solely on the central atom and significantly influence the molecular geometry and bond angles.
Lone pairs occupy more space around the central atom than bonding electrons because they are under the influence of only one nucleus rather than two. This effect often results in reduced bond angles when compared to molecules without lone pairs.
For instance, in the case of \(\mathrm{NH_3}\), the presence of one lone pair pushes the hydrogen atoms closer together, reducing the traditional tetrahedral bond angle from \(109.5^\circ\) to approximately \(107^\circ\). In contrast, \(\mathrm{SiH_4}\) exhibits the perfect tetrahedral angle due to a lack of lone pairs.
Trigonal Pyramidal
A trigonal pyramidal shape is a type of molecular geometry where the central atom is surrounded by three bond pairs and one lone pair. This configuration creates a pyramid-like shape with the lone pair at the apex.
This molecular shape is derived from a tetrahedral arrangement if one of the vertices is occupied by a lone pair instead of a bonded atom. The lone pair's repulsion causes the structure to slightly distort.
An example of trigonal pyramidal geometry can be seen in ammonia \(\mathrm{NH_3}\), where the nitrogen atom has three hydrogen atoms and one lone pair. The lone pair pushes the three hydrogen atoms closer, slightly contracting the bond angle to \(107^\circ\). This deviation from the ideal tetrahedral angle illustrates how lone pairs affect molecular shapes.
Tetrahedral
A tetrahedral geometry is an arrangement where four atoms or groups are located at the vertices of a tetrahedron, and all the bond angles are equal, approximately \(109.5^\circ\). It is one of the most common and important geometries in molecular chemistry.
This shape is characterized by four substituents symmetrically arranged around a central atom, which minimizes electron pair repulsion. The equal distribution results in maximum stability for molecules like \(\mathrm{SiH_4}\) and methane \(\mathrm{CH_4}\).
In tetrahedral molecules, the symmetry and equal bond angles are due to the absence of lone pairs on the central atom. This allows each bond to spread out evenly, creating the shape that is often visualized as a perfect three-dimensional triangle, ensuring molecule stability and balance.

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Most popular questions from this chapter

The sequence that correctly describes the relative bond strength pertaining to oxygen molecule and its cation or anion is: (a) \(\mathrm{O}_{2}^{2-}>\mathrm{O}_{2}^{-}>\mathrm{O}_{2}>\mathrm{O}_{2}^{+}\) (b) \(\mathrm{O}_{2}>\mathrm{O}_{2}^{+}>\mathrm{O}_{2}^{-}>\mathrm{O}_{2}^{2}\) (c) \(\mathrm{O}_{2}^{2}>\mathrm{O}_{2}>\mathrm{O}_{2}^{2-}>\mathrm{O}_{2}^{2}\) (d) \(\mathrm{O}_{2}^{+}>\mathrm{O}_{2}>\mathrm{O}_{2}^{-}>\mathrm{O}_{2}^{2}\)

The charge/size ratio of a cation determines its polarizing power. Which one of the following sequences represents the increasing order of the polarizing power of the cationic species, \(\mathrm{K}^{+}, \mathrm{Ca}^{2+}\), \(\mathrm{Mg}^{2+}, \mathrm{Be}^{2+} ?\) (a) \(\mathrm{Be}^{2+}<\mathrm{K}^{+}<\mathrm{Ca}^{2}+<\mathrm{Mg}^{2+}\) (b) \(\mathrm{K}^{+}<\mathrm{Ca}^{2+}<\mathrm{Mg}^{2+}<\mathrm{Be}^{2+}\) (c) \(\mathrm{Ca}^{2+}<\mathrm{Mg}^{2+}<\mathrm{Be}^{2+}<\mathrm{K}^{+}\) (d) \(\mathrm{Mg}^{2+}<\mathrm{Be}^{2+}<\mathrm{K}^{+}<\mathrm{Ca}^{2+}\)

The decreasing order of the boiling points of the following hydrides is: 1\. \(\mathrm{NH}_{3}\) 2\. \(\mathrm{PH}_{3}\) 3\. \(\mathrm{AsH}_{3}\) 4\. \(\mathrm{SbH}_{3}\) 5\. \(\mathrm{H}_{2} \mathrm{O}\) (a) \(5>4>1>3>2\) (b) \(5>1>2>3>4\) (c) \(2>4>3>1>5\) (d) \(4>3>1>2>5\)

The most polar bond is: (a) \(\mathrm{O}-\mathrm{H}\) (b) \(\mathrm{C}-\mathrm{O}\) (c) \(\mathrm{C}-\mathrm{Br}\) (d) \(\mathrm{C}-\mathrm{C}\)

Acetylene has ......... bonds: (a) \(2 \sigma, 2 \pi\) (b) \(2 \sigma, 3 \pi\) (c) \(3 \sigma, 2 \pi\) (d) \(3 \sigma, 3 \pi\)

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