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Question

Match List I with List II.

List – IList – II
A. PCl₃I. See-saw shaped
B. SF₄II. Square planar
C. BrF₃III. Trigonal pyramidal
D. XeF₄IV. Bent T Shape

Choose the correct answer from the options given below:

The correct answer is

A-III, B-I, C-IV, D-II

Understanding Molecular Shapes and VSEPR Theory

Molecular shape is determined by the arrangement of electron pairs (both bonding and non-bonding or lone pairs) around the central atom. The Valence Shell Electron Pair Repulsion (VSEPR) theory helps predict these shapes by minimizing repulsion between electron pairs.

To determine the shape of a molecule, we typically follow these steps:

  1. Identify the central atom.
  2. Count the total number of valence electrons contributed by all atoms.
  3. Draw the Lewis structure to determine the number of bonding pairs and lone pairs around the central atom.
  4. Use the VSEPR theory to predict the electron geometry and molecular geometry based on the total number of electron domains (bond pairs + lone pairs).

Analyzing Molecular Shapes for Each Compound

A. PCl₃

Let's determine the shape of PCl₃:

  • Central atom: Phosphorus (P).
  • Valence electrons: P has 5, each Cl has 7. Total = \(5 + 3 \times 7 = 26\).
  • In PCl₃, P forms 3 single bonds with 3 Cl atoms. This uses \(3 \times 2 = 6\) electrons.
  • Remaining electrons: \(26 - 6 = 20\). \(3 \times 6 = 18\) electrons are placed as 3 lone pairs on each Cl atom. The remaining \(20 - 18 = 2\) electrons form 1 lone pair on the central P atom.
  • The central P atom has 3 bond pairs and 1 lone pair. Total electron domains = 4.
  • According to VSEPR, 4 electron domains lead to a tetrahedral electron geometry. With 3 bond pairs and 1 lone pair (AX₃E notation), the molecular geometry is Trigonal pyramidal.

So, PCl₃ has a Trigonal pyramidal shape.

B. SF₄

Let's determine the shape of SF₄:

  • Central atom: Sulfur (S).
  • Valence electrons: S has 6, each F has 7. Total = \(6 + 4 \times 7 = 34\).
  • In SF₄, S forms 4 single bonds with 4 F atoms. This uses \(4 \times 2 = 8\) electrons.
  • Remaining electrons: \(34 - 8 = 26\). \(4 \times 6 = 24\) electrons are placed as 3 lone pairs on each F atom. The remaining \(26 - 24 = 2\) electrons form 1 lone pair on the central S atom.
  • The central S atom has 4 bond pairs and 1 lone pair. Total electron domains = 5.
  • According to VSEPR, 5 electron domains lead to a trigonal bipyramidal electron geometry. With 4 bond pairs and 1 lone pair (AX₄E notation), the lone pair occupies an equatorial position to minimize repulsion, resulting in a See-saw shaped molecular geometry.

So, SF₄ has a See-saw shaped structure.

C. BrF₃

Let's determine the shape of BrF₃:

  • Central atom: Bromine (Br).
  • Valence electrons: Br has 7, each F has 7. Total = \(7 + 3 \times 7 = 28\).
  • In BrF₃, Br forms 3 single bonds with 3 F atoms. This uses \(3 \times 2 = 6\) electrons.
  • Remaining electrons: \(28 - 6 = 22\). \(3 \times 6 = 18\) electrons are placed as 3 lone pairs on each F atom. The remaining \(22 - 18 = 4\) electrons form 2 lone pairs on the central Br atom.
  • The central Br atom has 3 bond pairs and 2 lone pairs. Total electron domains = 5.
  • According to VSEPR, 5 electron domains lead to a trigonal bipyramidal electron geometry. With 3 bond pairs and 2 lone pairs (AX₃E₂ notation), the lone pairs occupy equatorial positions to minimize repulsion, resulting in a Bent T Shape molecular geometry.

So, BrF₃ has a Bent T Shape structure.

D. XeF₄

Let's determine the shape of XeF₄:

  • Central atom: Xenon (Xe).
  • Valence electrons: Xe has 8, each F has 7. Total = \(8 + 4 \times 7 = 36\).
  • In XeF₄, Xe forms 4 single bonds with 4 F atoms. This uses \(4 \times 2 = 8\) electrons.
  • Remaining electrons: \(36 - 8 = 28\). \(4 \times 6 = 24\) electrons are placed as 3 lone pairs on each F atom. The remaining \(28 - 24 = 4\) electrons form 2 lone pairs on the central Xe atom.
  • The central Xe atom has 4 bond pairs and 2 lone pairs. Total electron domains = 6.
  • According to VSEPR, 6 electron domains lead to an octahedral electron geometry. With 4 bond pairs and 2 lone pairs (AX₄E₂ notation), the lone pairs occupy positions opposite to each other to minimize repulsion, resulting in a Square planar molecular geometry.

So, XeF₄ has a Square planar shape.

Matching List I with List II

Based on our analysis:

  • A. PCl₃: Trigonal pyramidal (III)
  • B. SF₄: See-saw shaped (I)
  • C. BrF₃: Bent T Shape (IV)
  • D. XeF₄: Square planar (II)

This matches the combination A-III, B-I, C-IV, D-II.

Molecule Bond Pairs (BP) Lone Pairs (LP) on Central Atom Total Electron Domains (BP + LP) Electron Geometry Molecular Geometry (Shape) Matching List II
PCl₃ 3 1 4 Tetrahedral Trigonal pyramidal III
SF₄ 4 1 5 Trigonal Bipyramidal See-saw shaped I
BrF₃ 3 2 5 Trigonal Bipyramidal Bent T Shape IV
XeF₄ 4 2 6 Octahedral Square planar II

Revision Table: Molecular Geometry & VSEPR

Total Electron Domains (BP+LP) Lone Pairs (LP) Molecular Geometry Examples
2 0 Linear \(CO_2\)
3 0 Trigonal Planar \(BF_3\)
3 1 Bent \(SO_2\)
4 0 Tetrahedral \(CH_4\)
4 1 Trigonal Pyramidal \(NH_3\), PCl₃
4 2 Bent \(H_2O\)
5 0 Trigonal Bipyramidal \(PCl_5\)
5 1 See-saw SF₄
5 2 T-shaped (Bent T) BrF₃
5 3 Linear \(XeF_2\)
6 0 Octahedral \(SF_6\)
6 1 Square Pyramidal \(IF_5\)
6 2 Square Planar XeF₄

Additional Information: VSEPR Theory Basics

VSEPR theory is a model used to predict the geometry of individual molecules from the number of electron pairs around their central atoms. The core principle is that valence shell electron pairs repel each other and will arrange themselves as far apart as possible to minimize repulsion. The repulsion strength generally follows this order:

Lone Pair - Lone Pair > Lone Pair - Bond Pair > Bond Pair - Bond Pair

The electron geometry describes the arrangement of all electron domains (both bonding and non-bonding) around the central atom, while the molecular geometry describes the arrangement of only the atoms (determined by the positions of the bonding pairs).

Understanding how to calculate bond pairs and lone pairs from the Lewis structure is crucial for applying VSEPR theory correctly and predicting the molecular shape.

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Important Questions from p-block Elements

  1. Second most abundant element in alloy misch metal is:

  2. Match List-I with List-II:

    List-IList-II
    (A) Gel(I) Hair cream
    (B) Foam(II) Dust
    (C) Emulsion(III) Cheese
    (D) Aerosol(IV) Whipped cream

    Choose the correct answer from the options given below:

  3. Rate of a reaction changes from 2.48 × 10⁻³ mol⁻¹ sec⁻¹ to 4.96 × 10⁻³ mol⁻¹ sec⁻¹ when concentration of reactant is changed from 0.6 M to 2.4 M respectively, the order of reaction is:

  4. Degree of dissociation, when molar conductivity of X at its concentration C is 24.14 and its limiting molar conductivity is 48.28 will be:

  5. A divalent ion of 'V' (Atomic no. 23) in aqueous solution is:

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