Question

Among the following, find out number of $sp^3$ hybridized polar as well as planar species, having all bond lengths are identical.

$H_2S, I_3^{\ominus}, O_3, ClF_3, H_2O_2, H_2O, SOCl_2, ClO_2^{\oplus}, CCl_4, SO_4^{2-}, NH_4^+, SCl_2, NH_3, I_3^{\oplus}, ClO_2^{\ominus}$

Answer 1

5

Answer 2

To determine the number of $sp^3$ hybridized, polar, planar species with identical bond lengths from the given list, we will analyze each species based on these four criteria.

1. Hybridization (of central atom) and Geometry: We use VSEPR theory to determine the steric number (SN = number of bond pairs + number of lone pairs).

• SN = 4 $\implies sp^3$ hybridization.
  • If 4 BP, 0 LP (SN=4): Tetrahedral
  • If 3 BP, 1 LP (SN=4): Trigonal pyramidal
  • If 2 BP, 2 LP (SN=4): Bent (V-shaped)
• SN = 3 $\implies sp^2$ hybridization.
  • If 3 BP, 0 LP (SN=3): Trigonal planar
  • If 2 BP, 1 LP (SN=3): Bent (V-shaped)
• SN = 5 $\implies sp^3d$ hybridization.
  • If 5 BP, 0 LP (SN=5): Trigonal bipyramidal
  • If 4 BP, 1 LP (SN=5): See-saw
  • If 3 BP, 2 LP (SN=5): T-shaped
  • If 2 BP, 3 LP (SN=5): Linear

2. Polarity: A molecule is polar if it has a net dipole moment. This occurs when there are polar bonds and the molecular geometry is asymmetrical, preventing the cancellation of individual bond dipoles. Molecules with lone pairs on the central atom are often polar unless highly symmetrical.

3. Planarity: A molecule is planar if all its atoms lie in the same plane.

• Bent molecules (e.g., $H_2O$) are planar as three points define a plane.
• Trigonal pyramidal (e.g., $NH_3$) and tetrahedral (e.g., $CCl_4$) molecules are non-planar.

4. Identical Bond Lengths: All bonds of the same type must have identical lengths. This is often true for symmetrical molecules or those exhibiting resonance.

Let's analyze each species:

1. $H_2S$:

   • Central atom S: 2 bond pairs (BP) + 2 lone pairs (LP) = 4. $sp^3$ hybridized.
   • Geometry: Bent.
   • Polarity: S-H bonds are polar, and the bent geometry ensures a net dipole moment. Polar.
   • Planarity: Bent molecules are planar. Planar.
   • Bond lengths: Both S-H bond lengths are identical. Identical.
   • Conclusion: Fits all criteria.

2. $I_3^{\ominus}$:

   • Central atom I: 2 BP + 3 LP = 5. $sp^3d$ hybridized.
   • Conclusion: Not $sp^3$ hybridized.

3. $O_3$:

   • Central atom O: 2 BP + 1 LP = 3. $sp^2$ hybridized.
   • Conclusion: Not $sp^3$ hybridized.

4. $ClF_3$:

   • Central atom Cl: 3 BP + 2 LP = 5. $sp^3d$ hybridized.
   • Conclusion: Not $sp^3$ hybridized.

5. $H_2O_2$:

   • Each O atom: 2 BP + 2 LP = 4. $sp^3$ hybridized.
   • Geometry: Open book structure (non-planar).
   • Conclusion: Not planar.

6. $H_2O$:

   • Central atom O: 2 BP + 2 LP = 4. $sp^3$ hybridized.
   • Geometry: Bent.
   • Polarity: O-H bonds are polar, and the bent geometry ensures a net dipole moment. Polar.
   • Planarity: Bent molecules are planar. Planar.
   • Bond lengths: Both O-H bond lengths are identical. Identical.
   • Conclusion: Fits all criteria.

7. $SOCl_2$:

   • Central atom S: 3 BP (1 S=O, 2 S-Cl) + 1 LP = 4. $sp^3$ hybridized.
   • Geometry: Trigonal pyramidal.
   • Polarity: S-O and S-Cl bonds are polar, and the trigonal pyramidal geometry ensures a net dipole moment. Polar.
   • Planarity: Trigonal pyramidal molecules are non-planar.
   • Conclusion: Not planar.

8. $ClO_2^{\oplus}$:

   • Central atom Cl: 2 BP + 1 LP = 3. $sp^2$ hybridized.
   • Conclusion: Not $sp^3$ hybridized.

9. $CCl_4$:

   • Central atom C: 4 BP + 0 LP = 4. $sp^3$ hybridized.
   • Geometry: Tetrahedral.
   • Polarity: C-Cl bonds are polar, but the symmetrical tetrahedral geometry results in cancellation of bond dipoles. Non-polar.
   • Conclusion: Not polar.

10. $SO_4^{2-}$:

    • Central atom S: 4 BP + 0 LP = 4. $sp^3$ hybridized.
    • Geometry: Tetrahedral.
    • Polarity: S-O bonds are polar, but the symmetrical tetrahedral geometry results in cancellation of bond dipoles. Non-polar.
    • Conclusion: Not polar.

11. $NH_4^+$:

    • Central atom N: 4 BP + 0 LP = 4. $sp^3$ hybridized.
    • Geometry: Tetrahedral.
    • Polarity: N-H bonds are polar, but the symmetrical tetrahedral geometry results in cancellation of bond dipoles. Non-polar.
    • Conclusion: Not polar.

12. $SCl_2$:

    • Central atom S: 2 BP + 2 LP = 4. $sp^3$ hybridized.
    • Geometry: Bent.
    • Polarity: S-Cl bonds are polar, and the bent geometry ensures a net dipole moment. Polar.
    • Planarity: Bent molecules are planar. Planar.
    • Bond lengths: Both S-Cl bond lengths are identical. Identical.
    • Conclusion: Fits all criteria.

13. $NH_3$:

    • Central atom N: 3 BP + 1 LP = 4. $sp^3$ hybridized.
    • Geometry: Trigonal pyramidal.
    • Polarity: N-H bonds are polar, and the trigonal pyramidal geometry ensures a net dipole moment. Polar.
    • Planarity: Trigonal pyramidal molecules are non-planar.
    • Conclusion: Not planar.

14. $I_3^{\oplus}$:

    • Central atom I: 2 BP + 2 LP = 4. $sp^3$ hybridized.
    • Geometry: Bent.
    • Polarity: Although I-I bonds are non-polar, the bent geometry with lone pairs on the central atom creates an uneven electron distribution, making it polar. Polar.
    • Planarity: Bent molecules are planar. Planar.
    • Bond lengths: Both I-I bond lengths are identical due to symmetry. Identical.
    • Conclusion: Fits all criteria.

15. $ClO_2^{\ominus}$:

    • Central atom Cl: 2 BP + 2 LP = 4. $sp^3$ hybridized.
    • Geometry: Bent.
    • Polarity: Cl-O bonds are polar, and the bent geometry ensures a net dipole moment. Polar.
    • Planarity: Bent molecules are planar. Planar.
    • Bond lengths: Due to resonance (O=Cl-O$^-$ $\leftrightarrow$ $^-$O-Cl=O), both Cl-O bond lengths are identical and intermediate between single and double bonds. Identical.
    • Conclusion: Fits all criteria.

The species that satisfy all four conditions are: $H_2S, H_2O, SCl_2, I_3^{\oplus}, ClO_2^{\ominus}$. There are 5 such species.

Core Solution:

1. Identify $sp^3$ hybridization: Calculate steric number (SN = BP + LP). SN=4 means $sp^3$.
   • $H_2S$ (SN=4), $H_2O_2$ (SN=4 per O), $H_2O$ (SN=4), $SOCl_2$ (SN=4), $CCl_4$ (SN=4), $SO_4^{2-}$ (SN=4), $NH_4^+$ (SN=4), $SCl_2$ (SN=4), $NH_3$ (SN=4), $I_3^{\oplus}$ (SN=4), $ClO_2^{\ominus}$ (SN=4).
   • Eliminate: $I_3^{\ominus}$ ($sp^3d$), $O_3$ ($sp^2$), $ClF_3$ ($sp^3d$), $ClO_2^{\oplus}$ ($sp^2$).
2. Check for Polarity:
   • Eliminate non-polar: $CCl_4$, $SO_4^{2-}$, $NH_4^+$ (all tetrahedral and symmetrical).
3. Check for Planarity:
   • Eliminate non-planar: $H_2O_2$ (open book), $SOCl_2$ (trigonal pyramidal), $NH_3$ (trigonal pyramidal).
4. Check for Identical Bond Lengths:
   • All remaining species ($H_2S$, $H_2O$, $SCl_2$, $I_3^{\oplus}$, $ClO_2^{\ominus}$) are bent. The two bonds in each of these molecules are either inherently identical (e.g., S-H in $H_2S$) or made identical by resonance (e.g., Cl-O in $ClO_2^{\ominus}$).

The species satisfying all criteria are $H_2S, H_2O, SCl_2, I_3^{\oplus}, ClO_2^{\ominus}$. There are 5 such species.