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Q.

The degenerate orbitals of [Cr(H2O)6]3+ are

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a

dx2-y2 and dxy

b

dx2 and dxz

c

dxz and dyz

d

dyz and dx2

answer is C.

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Detailed Solution

  • Cr has an oxidation state of +3 in this compound since water is a neutral ligand, thus has a d3 configuration. 
  • The hybridization of the chemical so discovered is d2sp3, meaning that dx2-y2 and dz2 are the two-dimensional orbits involved, while the other three are degenerate orbitals, or orbitals with the same energy levels.
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