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P Block Elements question

2007 · Shift 1 · Q15
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P Block Elements question

2007 · Shift 1 · Q15

JEE AdvancedChemistryP Block ElementsMCQ+3 / −1
The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions. The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers +2, +4 and +6. XeF 4{}_44​ reacts violently with water to give XeO 3{}_33​. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.The structure of XeO 3{}_33​
  1. A
    linear
  2. B
    planar
  3. C
    pyramidal
  4. D
    T-shaped
View written solutionFree

Correct answer: C

  1. Identify the central atom and valence electrons: The central atom in XeO₃ is Xenon (Xe). Xenon is a noble gas and belongs to Group 18 of the periodic table, so it has 8 valence electrons. Oxygen (O) is in Group 16 and has 6 valence electrons.

  2. Determine the number of bond pairs and lone pairs around the central atom (Xe) using VSEPR theory:

    • The Xenon atom is bonded to three Oxygen atoms.
    • Oxygen is divalent, so each oxygen atom typically forms two bonds to complete its octet. In XeO₃, each oxygen atom forms a double bond with the central xenon atom.
    • A double bond consists of one sigma (σ) bond and one pi (π) bond.
    • The number of sigma bonds formed by Xe is 3 (one with each oxygen atom).
    • The number of electrons from Xe used to form these three double bonds is 2 electrons per oxygen atom, totaling 3×2=63 \times 2 = 63×2=6 electrons.
    • The number of valence electrons remaining on the Xe atom as lone pair electrons is calculated as: (Total valence electrons of Xe) - (Electrons used in bonding) = 8−6=28 - 6 = 28−6=2 electrons.
    • These 2 electrons constitute one lone pair.

  3. Determine the steric number and electron geometry:

    • The total number of electron domains (steric number) around the central Xe atom is the sum of the number of sigma bonds and the number of lone pairs.
    • Steric Number = (Number of σ bonds) + (Number of lone pairs) = 3+1=43 + 1 = 43+1=4.
    • A steric number of 4 corresponds to a tetrahedral electron geometry.

  4. Determine the molecular geometry (shape):

    • The arrangement of atoms in the molecule is determined by the number of bonding pairs and lone pairs. The molecule has the general VSEPR formula AX₃E₁, where A is the central atom, X are the bonded atoms, and E is the lone pair.
    • For an AX₃E₁ system, the four electron pairs are arranged tetrahedrally. One position is occupied by a lone pair, and the other three are occupied by the oxygen atoms.
    • Due to the presence of the lone pair, which exerts greater repulsion than bonding pairs, the shape of the molecule is not tetrahedral. The resulting molecular geometry is trigonal pyramidal.

  5. Evaluate the given options:

    • A: linear - Incorrect. This geometry corresponds to AX₂ or AX₂E₃ systems.
    • B: planar - Incorrect. The lone pair on xenon pushes the three oxygen atoms down, making the molecule non-planar. Trigonal planar geometry (AX₃) has no lone pairs.
    • C: pyramidal - Correct. This matches the trigonal pyramidal geometry predicted for an AX₃E₁ molecule.
    • D: T-shaped - Incorrect. This geometry corresponds to an AX₃E₂ system.
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