Question

How to find no. of hybrid orbitals Na2 CrO4

Answer 1

4

Answer 2

• Identify the central atom and its oxidation state: In Na₂CrO₄, the central atom is Chromium (Cr) within the chromate ion (CrO₄²⁻).

  • Let the oxidation state of Cr be 'x'.
  • The oxidation state of Oxygen (O) is -2.
  • The overall charge of the chromate ion is -2.
  • So, x + 4(-2) = -2
  • x - 8 = -2
  • x = +6.
  • Therefore, Chromium is in the +6 oxidation state (Cr(VI)).

• Determine the electronic configuration of Cr(VI):

  • Ground state electronic configuration of Cr (atomic number 24): [Ar] 3d⁵ 4s¹
  • To form Cr(VI), 6 electrons are removed (1 from 4s, then 5 from 3d).
  • Electronic configuration of Cr(VI): [Ar] 3d⁰ 4s⁰. This indicates that both the 3d and 4s orbitals are empty.

• Determine the number of sigma bonds and lone pairs around the central atom:

  • In CrO₄²⁻, there are four oxygen atoms bonded to the central chromium atom. Each Cr-O bond involves a sigma bond.
  • Number of sigma bonds = 4.
  • Since Cr(VI) has no valence electrons (3d⁰ 4s⁰), there are no lone pairs on the central Cr atom.
  • Number of lone pairs = 0.

• Calculate the steric number and determine the number of hybrid orbitals:

  • Steric Number (SN) = (Number of sigma bonds) + (Number of lone pairs on the central atom)
  • SN = 4 + 0 = 4.
  • The number of hybrid orbitals formed is equal to the steric number.
  • Therefore, there are 4 hybrid orbitals.
  • For transition metals in high oxidation states with vacant d-orbitals, these 4 hybrid orbitals are formed by the mixing of one 4s orbital and three 3d orbitals, leading to d³s hybridization.

The number of hybrid orbitals in Na₂CrO₄ (specifically on the central Cr atom in CrO₄²⁻) is 4.