Describe the state of hybridization, the shape and magnetic behavior of the following complex:  [Cr(H₂O)₂(C₂O₄)₂]^_
(At no's: Cr = 24)        

Dear Student,

Oxidation state of Cr in [Cr(H₂O)₂(C₂O₄)₂]^- :
$$\begin{gathered} \mathrm{x}+2\times 0+2\times (-2)=-1 \\ \mathrm{or}, \mathrm{x}-4=-1 \\ \mathrm{or} \mathrm{x}=+3 \end{gathered}$$

Now, Valence shell electronic configuration of Cr = 3d^​5 4s¹
Valence shell electronic configuration of Cr³⁺= 3d³ 4s⁰ 4p⁰
H₂O is a monodentate ligand while C₂O₄ is a bidentate ligand
So, 2 H₂O ligands will donate 2 lone pair and 2C₂O₄ will donate 4 lone pairs
Overall 6 lone pair are donated. These 6 pairs will go to the two empty d-orbital, one s orbital and three p orbital. 
Hence, the hybridisation will be d²sp³. 
d²sp³ means octahedral shape
and, here number of unpaired electrons = 3
Hence, the compound will be paramagentic as it contains unpaired electron.