Question

Compound	No. of lone pairs on 1 central atom	No. of d-orbitals used for hybridisation
$XeF_3^+$	$\omega_1$	$\omega_2$
$I_2Cl_6$	$\omega_3$	$\omega_4$

Find $[\frac{\omega_1 + \omega_2}{\omega_3 + \omega_4}] \times 100$

Answer 1

75

Answer 2

To determine the values of $\omega_1, \omega_2, \omega_3, \omega_4$, we need to analyze the hybridization and lone pairs on the central atom for each compound.

1. For $XeF_3^+$:

• Central atom: Xe (Xenon)
• Valence electrons of Xe: 8 (Group 18)
• Charge: +1. This means one electron is removed from Xe.
  Effective valence electrons on Xe = 8 - 1 = 7.
• Surrounding atoms: 3 F atoms. Each F forms a single bond with Xe.
• Number of bond pairs (BP): 3 (for 3 Xe-F bonds)
• Electrons used in bonding: 3 bonds $\times$ 2 electrons/bond = 6 electrons.
• Remaining electrons on Xe (for lone pairs): 7 (effective valence electrons) - 3 (electrons contributed by Xe to bonds) = 4 electrons.
• Number of lone pairs (LP): 4 electrons / 2 electrons/pair = 2 lone pairs.
  So, $\omega_1 = 2$.
• Steric Number (SN): BP + LP = 3 + 2 = 5.
• Hybridization for SN = 5: $sp^3d$.
• Number of d-orbitals used for hybridization: 1.
  So, $\omega_2 = 1$.

2. For $I_2Cl_6$:

$I_2Cl_6$ is a dimer of $ICl_3$. Each iodine atom is bonded to 4 chlorine atoms (2 terminal, 2 bridging). The geometry around each iodine is square planar.

Let's consider one central Iodine atom in $I_2Cl_6$.

• Central atom: I (Iodine)
• Valence electrons of I: 7 (Group 17)
• Number of atoms bonded to one I: 4 (2 terminal Cl, 2 bridging Cl).
  So, Number of bond pairs (BP): 4.
• Geometry around each I: Square planar.
  For a square planar geometry, the steric number is 6 (4 bond pairs + 2 lone pairs).
• Number of lone pairs (LP): SN - BP = 6 - 4 = 2 lone pairs.
  So, $\omega_3 = 2$.
• Steric Number (SN): 6.
• Hybridization for SN = 6: $sp^3d^2$.
• Number of d-orbitals used for hybridization: 2.
  So, $\omega_4 = 2$.

Calculation:

We need to find $[\frac{\omega_1 + \omega_2}{\omega_3 + \omega_4}] \times 100$.

• $\omega_1 = 2$
• $\omega_2 = 1$
• $\omega_3 = 2$
• $\omega_4 = 2$

$[\frac{2 + 1}{2 + 2}] \times 100 = [\frac{3}{4}] \times 100 = 0.75 \times 100 = 75$.