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Chemical Bonding and Molecular Structure question

2023 · Shift 2 · Q9
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Chemical Bonding and Molecular Structure question

2023 · Shift 2 · Q9

JEE AdvancedChemistryChemical Bonding and Molecular StructureNumerical+4 / −1
Among [I3]+,[SiO4]4−,SO2Cl2,XeF2,SF4,ClF3,Ni(CO)4,XeO2 F2,[PtCl4]2−,XeF4\left[\mathrm{I}_3\right]^{+},\left[\mathrm{SiO}_4\right]^{4-}, \mathrm{SO}_2 \mathrm{Cl}_2, \mathrm{XeF}_2, \mathrm{SF}_4, \mathrm{ClF}_3, \mathrm{Ni}(\mathrm{CO})_4, \mathrm{XeO}_2 \mathrm{~F}_2,\left[\mathrm{PtCl}_4\right]^{2-}, \mathrm{XeF}_4[I3​]+,[SiO4​]4−,SO2​Cl2​,XeF2​,SF4​,ClF3​,Ni(CO)4​,XeO2​ F2​,[PtCl4​]2−,XeF4​, and SOCl2\mathrm{SOCl}_2SOCl2​, the total number of species having sp3s p^3sp3 hybridised central atom is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 5

To determine the number of species with an sp3sp^3sp3 hybridized central atom, we need to find the hybridization of the central atom for each given species. We can use the concept of steric number (SN) for p-block elements and Valence Bond Theory (VBT) for coordination compounds.

Steric Number (SN) = (Number of sigma bonds) + (Number of lone pairs on the central atom).

  • If SN = 2, hybridization is spspsp.
  • If SN = 3, hybridization is sp2sp^2sp2.
  • If SN = 4, hybridization is sp3sp^3sp3.
  • If SN = 5, hybridization is sp3dsp^3dsp3d.
  • If SN = 6, hybridization is sp3d2sp^3d^2sp3d2.

Let's analyze each species:

  1. [I3]+\left[\mathrm{I}_3\right]^{+}[I3​]+: The central atom is Iodine (I).

    • Valence electrons of I = 7.
    • The cation has a +1 charge, so the central I atom has 7−1=67 - 1 = 67−1=6 valence electrons to consider.
    • It forms 2 single bonds (sigma bonds) with the other two I atoms.
    • Number of lone pairs on the central I = 6−22=2\frac{6 - 2}{2} = 226−2​=2.
    • Steric Number (SN) = 2 (sigma bonds) + 2 (lone pairs) = 4.
    • A steric number of 4 corresponds to sp3sp^3sp3 hybridization.

  2. [SiO4]4−\left[\mathrm{SiO}_4\right]^{4-}[SiO4​]4−: The central atom is Silicon (Si).

    • Valence electrons of Si = 4.
    • It forms 4 sigma bonds with the 4 oxygen atoms.
    • Number of lone pairs on Si = 4−42=0\frac{4 - 4}{2} = 024−4​=0. (Assuming single bonds with O⁻ ions)
    • Steric Number (SN) = 4 (sigma bonds) + 0 (lone pairs) = 4.
    • A steric number of 4 corresponds to sp3sp^3sp3 hybridization.

  3. SO2Cl2\mathrm{SO}_2 \mathrm{Cl}_2SO2​Cl2​ (Sulfuryl chloride): The central atom is Sulfur (S).

    • Valence electrons of S = 6.
    • It forms 2 single bonds with Cl and 2 double bonds with O.
    • Number of sigma bonds = 4 (one for each attached atom).
    • Electrons used by S in bonding = 2×1(for Cl)+2×2(for O)=62 \times 1 (\text{for Cl}) + 2 \times 2 (\text{for O}) = 62×1(for Cl)+2×2(for O)=6.
    • Number of lone pairs on S = 6−62=0\frac{6 - 6}{2} = 026−6​=0.
    • Steric Number (SN) = 4 (sigma bonds) + 0 (lone pairs) = 4.
    • A steric number of 4 corresponds to sp3sp^3sp3 hybridization.

  4. XeF2\mathrm{XeF}_2XeF2​: The central atom is Xenon (Xe).

    • Valence electrons of Xe = 8.
    • It forms 2 sigma bonds with F atoms.
    • Number of lone pairs on Xe = 8−22=3\frac{8 - 2}{2} = 328−2​=3.
    • Steric Number (SN) = 2 (sigma bonds) + 3 (lone pairs) = 5.
    • A steric number of 5 corresponds to sp3dsp^3dsp3d hybridization.

  5. SF4\mathrm{SF}_4SF4​: The central atom is Sulfur (S).

    • Valence electrons of S = 6.
    • It forms 4 sigma bonds with F atoms.
    • Number of lone pairs on S = 6−42=1\frac{6 - 4}{2} = 126−4​=1.
    • Steric Number (SN) = 4 (sigma bonds) + 1 (lone pair) = 5.
    • A steric number of 5 corresponds to sp3dsp^3dsp3d hybridization.

  6. ClF3\mathrm{ClF}_3ClF3​: The central atom is Chlorine (Cl).

    • Valence electrons of Cl = 7.
    • It forms 3 sigma bonds with F atoms.
    • Number of lone pairs on Cl = 7−32=2\frac{7 - 3}{2} = 227−3​=2.
    • Steric Number (SN) = 3 (sigma bonds) + 2 (lone pairs) = 5.
    • A steric number of 5 corresponds to sp3dsp^3dsp3d hybridization.

  7. Ni(CO)4\mathrm{Ni}(\mathrm{CO})_4Ni(CO)4​: This is a coordination compound. The central atom is Nickel (Ni).

    • The oxidation state of Ni is 0. Its electronic configuration is [Ar]3d84s2[\mathrm{Ar}] 3d^8 4s^2[Ar]3d84s2.
    • CO is a strong field ligand, which causes the pairing of the 4s4s4s electrons into the 3d3d3d orbitals. The configuration becomes [Ar]3d104s04p0[\mathrm{Ar}] 3d^{10} 4s^0 4p^0[Ar]3d104s04p0.
    • To form 4 coordinate bonds, Ni uses one vacant 4s4s4s and three vacant 4p4p4p orbitals.
    • Thus, the hybridization is sp3sp^3sp3.

  8. XeO2 F2\mathrm{XeO}_2 \mathrm{~F}_2XeO2​ F2​: The central atom is Xenon (Xe).

    • Valence electrons of Xe = 8.
    • It forms 2 single bonds with F and 2 double bonds with O.
    • Number of sigma bonds = 4.
    • Electrons used by Xe in bonding = 2×1(for F)+2×2(for O)=62 \times 1 (\text{for F}) + 2 \times 2 (\text{for O}) = 62×1(for F)+2×2(for O)=6.
    • Number of lone pairs on Xe = 8−62=1\frac{8 - 6}{2} = 128−6​=1.
    • Steric Number (SN) = 4 (sigma bonds) + 1 (lone pair) = 5.
    • A steric number of 5 corresponds to sp3dsp^3dsp3d hybridization.

  9. [PtCl4]2−\left[\mathrm{PtCl}_4\right]^{2-}[PtCl4​]2−: This is a coordination compound. The central atom is Platinum (Pt).

    • The oxidation state of Pt is +2. Its electronic configuration is [Xe]4f145d8[\mathrm{Xe}] 4f^{14} 5d^8[Xe]4f145d8.
    • For 5d series metals, all ligands are treated as strong field ligands. This causes pairing of the 8 d-electrons.
    • This leaves one 5d5d5d orbital empty. For bonding with 4 Cl⁻ ligands, it uses one 5d5d5d, one 6s6s6s, and two 6p6p6p orbitals.
    • The hybridization is dsp2dsp^2dsp2.

  10. XeF4\mathrm{XeF}_4XeF4​: The central atom is Xenon (Xe).

    • Valence electrons of Xe = 8.
    • It forms 4 sigma bonds with F atoms.
    • Number of lone pairs on Xe = 8−42=2\frac{8 - 4}{2} = 228−4​=2.
    • Steric Number (SN) = 4 (sigma bonds) + 2 (lone pairs) = 6.
    • A steric number of 6 corresponds to sp3d2sp^3d^2sp3d2 hybridization.

  11. SOCl2\mathrm{SOCl}_2SOCl2​ (Thionyl chloride): The central atom is Sulfur (S).

    • Valence electrons of S = 6.
    • It forms 2 single bonds with Cl and 1 double bond with O.
    • Number of sigma bonds = 3.
    • Electrons used by S in bonding = 2×1(for Cl)+1×2(for O)=42 \times 1 (\text{for Cl}) + 1 \times 2 (\text{for O}) = 42×1(for Cl)+1×2(for O)=4.
    • Number of lone pairs on S = 6−42=1\frac{6 - 4}{2} = 126−4​=1.
    • Steric Number (SN) = 3 (sigma bonds) + 1 (lone pair) = 4.
    • A steric number of 4 corresponds to sp3sp^3sp3 hybridization.

The species with sp3sp^3sp3 hybridized central atoms are: [I3]+\left[\mathrm{I}_3\right]^{+}[I3​]+, [SiO4]4−\left[\mathrm{SiO}_4\right]^{4-}[SiO4​]4−, SO2Cl2\mathrm{SO}_2 \mathrm{Cl}_2SO2​Cl2​, Ni(CO)4\mathrm{Ni}(\mathrm{CO})_4Ni(CO)4​, and SOCl2\mathrm{SOCl}_2SOCl2​.

Total number of such species is 5.

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