JEE Challenger
More from Chemical Bonding and Molecular Structure
ChemMedAdv 2015 • Paper 1

Linear Triatomic Molecules Hybridization without d Orbitals

Among the given triatomic species: BeCl2\text{BeCl}_2, N3−\text{N}_3^-, N2O\text{N}_2\text{O}, NO2+\text{NO}_2^+, O3\text{O}_3, SCl2\text{SCl}_2, ICl2−\text{ICl}_2^-, I3−\text{I}_3^-, and XeF2\text{XeF}_2, determine the total count of linear molecules or ions in which the central atom's hybridization does not involve any dd-orbital contribution. [Atomic numbers given: S=16\text{S} = 16, Cl=17\text{Cl} = 17, I=53\text{I} = 53, and Xe=54\text{Xe} = 54]

Official Numerical Answer4

Topics & Concepts

Chemical Bonding and Molecular StructureHybridizationVSEPR Theory

Step-by-Step Solution

To determine the total count of triatomic species that are linear and whose central atom's hybridization does not involve any dd-orbitals, we analyze each given species based on VSEPR theory and hybridization:

  1. BeCl2\text{BeCl}_2:

    • Central atom: Be\text{Be} (2 valence electrons)
    • Bonding pairs = 2, Lone pairs = 0
    • Steric Number = 2
    • Hybridization: spsp (no dd-orbitals involved)
    • Shape: Linear

  2. N3−\text{N}_3^- (Azide ion):

    • Central atom: N\text{N} (5 valence electrons +1+ 1 negative charge contribution shared, forming [N=N=N]−[\text{N}=\text{N}=\text{N}]^-)
    • Bonding pairs = 2 (σ\sigma-bonds), Lone pairs = 0 on central atom
    • Steric Number = 2
    • Hybridization: spsp (no dd-orbitals involved)
    • Shape: Linear

  3. N2O\text{N}_2\text{O} (Nitrous oxide):

    • Central atom: N\text{N} (Structure: N=N=O\text{N}=\text{N}=\text{O})
    • Bonding pairs = 2 (σ\sigma-bonds), Lone pairs = 0 on central atom
    • Steric Number = 2
    • Hybridization: spsp (no dd-orbitals involved)
    • Shape: Linear

  4. NO2+\text{NO}_2^+ (Nitronium ion):

    • Central atom: N+\text{N}^+ (4 valence electrons)
    • Bonding pairs = 2 (σ\sigma-bonds), Lone pairs = 0
    • Steric Number = 2
    • Hybridization: spsp (no dd-orbitals involved)
    • Shape: Linear

  5. O3\text{O}_3 (Ozone):

    • Central atom: O\text{O}
    • Bonding pairs = 2 (σ\sigma-bonds), Lone pairs = 1
    • Steric Number = 3
    • Hybridization: sp2sp^2
    • Shape: Bent / Angular (Not linear)

  6. SCl2\text{SCl}_2:

    • Central atom: S\text{S} (6 valence electrons)
    • Bonding pairs = 2, Lone pairs = 2
    • Steric Number = 4
    • Hybridization: sp3sp^3
    • Shape: Bent / Angular (Not linear)

  7. ICl2−\text{ICl}_2^-:

    • Central atom: I\text{I} (7 valence electrons +1+ 1 charge = 8)
    • Bonding pairs = 2, Lone pairs = 3
    • Steric Number = 5
    • Hybridization: sp3dsp^3d (Involves dd-orbital)
    • Shape: Linear (but contains dd-orbital contribution)

  8. I3−\text{I}_3^-:

    • Central atom: I\text{I} (7 valence electrons +1+ 1 charge = 8)
    • Bonding pairs = 2, Lone pairs = 3
    • Steric Number = 5
    • Hybridization: sp3dsp^3d (Involves dd-orbital)
    • Shape: Linear (but contains dd-orbital contribution)

  9. XeF2\text{XeF}_2:

    • Central atom: Xe\text{Xe} (8 valence electrons)
    • Bonding pairs = 2, Lone pairs = 3
    • Steric Number = 5
    • Hybridization: sp3dsp^3d (Involves dd-orbital)
    • Shape: Linear (but contains dd-orbital contribution)

Conclusion:

The linear species whose central atom hybridization does not involve any dd-orbital contribution (spsp hybridization) are: BeCl2,N3−,N2O,and NO2+\text{BeCl}_2, \text{N}_3^-, \text{N}_2\text{O}, \text{and } \text{NO}_2^+

Thus, the total count is 4.

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