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Question

The correct shape & hybridisation for XeF2 are:

The correct answer is

linear, sp3d

Understanding the Structure of XeF$_2$

To determine the correct shape and hybridisation for the molecule XeF$_2$, we use the Valence Shell Electron Pair Repulsion (VSEPR) theory. VSEPR theory helps predict the geometry of molecules based on the repulsion between electron pairs around the central atom.

Steps to Determine Shape and Hybridisation of XeF$_2$

Let's analyze the XeF$_2$ molecule step-by-step:

  1. Identify the Central Atom: Xenon (Xe) is the central atom in XeF$_2$.
  2. Count Valence Electrons on the Central Atom: Xenon (Xe) is in Group 18, so it has 8 valence electrons.
  3. Count Surrounding Atoms and Bonds: There are two fluorine (F) atoms surrounding the central xenon atom. Fluorine is in Group 17 and typically forms one single bond. Thus, Xe forms two single bonds with the two F atoms.
  4. Calculate Electrons Used in Bonding: Two single bonds use $2 \times 1 = 2$ valence electrons from Xe.
  5. Calculate Remaining Valence Electrons on the Central Atom: Remaining electrons on Xe = Total valence electrons on Xe - Electrons used in bonding = $8 - 2 = 6$ electrons.
  6. Determine the Number of Lone Pairs: Six remaining electrons form lone pairs. Number of lone pairs = Remaining electrons / 2 = $6 / 2 = 3$ lone pairs.
  7. Calculate the Steric Number: The steric number is the total number of electron groups around the central atom. It is the sum of the number of bonding pairs and the number of lone pairs.
    Steric Number = Number of bonding pairs + Number of lone pairs
    Steric Number = $2$ (for the two Xe-F single bonds) + $3$ (lone pairs) = $5$.
  8. Determine Hybridisation: The steric number corresponds to the hybridisation of the central atom.
    • Steric Number 2 → sp
    • Steric Number 3 → sp$^2$
    • Steric Number 4 → sp$^3$
    • Steric Number 5 → sp$^3$d
    • Steric Number 6 → sp$^3$d$^2$

    Since the steric number for Xe in XeF$_2$ is 5, the hybridisation is sp$^3$d.

  9. Determine Electron Geometry: The electron geometry is determined by the total number of electron groups (steric number). For a steric number of 5, the electron geometry is trigonal bipyramidal. These 5 electron groups are arranged in a trigonal bipyramidal shape around the central atom.
  10. Determine Molecular Shape: The molecular shape is determined by the arrangement of *atoms* only, ignoring the lone pairs, although the lone pairs influence the positions of the bonding pairs due to repulsion. In a trigonal bipyramidal arrangement with 3 lone pairs and 2 bonding pairs, the lone pairs occupy the equatorial positions to minimize repulsion (LP-LP repulsion is greater than LP-BP and BP-BP repulsion). The two bonding pairs occupy the axial positions. The two fluorine atoms are at the ends of the axial positions with Xenon in the middle, resulting in a linear shape.

Summary for XeF$_2$

  • Number of bonding pairs: 2
  • Number of lone pairs: 3
  • Steric Number: 5
  • Hybridisation: sp$^3$d
  • Electron Geometry: Trigonal bipyramidal
  • Molecular Shape: Linear

Based on this analysis, the correct shape for XeF$_2$ is linear and the correct hybridisation is sp$^3$d.

Comparing with Options

Let's look at the provided options:

  • Option 1: linear, sp$^3$d
  • Option 2: linear, sp$^3$d$^2$
  • Option 3: trigonal bipyramidal, sp$^3$d
  • Option 4: trigonal bipyramidal, sp$^3$d$^2$

Our determination shows a linear shape and sp$^3$d hybridisation, which matches Option 1.


Revision Table: VSEPR Theory Basics

Steric Number Hybridisation Electron Geometry Examples (Shape)
2 sp Linear BeCl$_2$ (Linear)
3 sp$^2$ Trigonal Planar BF$_3$ (Trigonal Planar), SO$_2$ (Bent)
4 sp$^3$ Tetrahedral CH$_4$ (Tetrahedral), NH$_3$ (Trigonal Pyramidal), H$_2$O (Bent)
5 sp$^3$d Trigonal Bipyramidal PCl$_5$ (Trigonal Bipyramidal), SF$_4$ (See-saw), ClF$_3$ (T-shaped), XeF$_2$ (Linear)
6 sp$^3$d$^2$ Octahedral SF$_6$ (Octahedral), BrF$_5$ (Square Pyramidal), XeF$_4$ (Square Planar)

Additional Information on Noble Gas Compounds

Xenon is a noble gas, traditionally thought to be inert. However, under specific conditions, noble gases like Xenon can form compounds, especially with highly electronegative elements like fluorine and oxygen. XeF$_2$, XeF$_4$, XeF$_6$, XeO$_3$, and XeOF$_4$ are some examples. The formation of these compounds involves the expansion of the valence shell and participation of d orbitals in bonding and hybridisation, as seen in the sp$^3$d hybridisation of Xe in XeF$_2$. Studying the shapes and hybridisation of noble gas compounds helps us understand the bonding capabilities of elements beyond the traditional octet rule.

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Important Questions from Coordination Compounds

  1. Match List-I with List-II:

    List-IList-II
    (A) Diamagnetic solid(I) CrO₂
    (B) Ferromagnetic solid(II) Fe₃O₄
    (C) Antiferromagnetic solid(III) NaCl
    (D) Ferrimagnetic solid(IV) MnO

    Choose the correct answer from the options given below:

  2. [NiCl₂(PPh₃)₂] is named as:

  3. Inner orbital complex among the following is:

    (A) [Co(NH₃)₆]³⁺

    (B) [CoF₆]³⁻

    (C) [Ni(CN)4]²⁻

    (D) [MnCl₆]³⁻

    (E) [FeF₆]³⁻

    Choose the correct answer from the options given below:

  4. Which will form the most stable complex?

  5. How many Cr-O bonds in dichromate ions are of the same bond length and are in resonance?

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