Solution

(A) $$[Fe(CO)_5]$$ involves $$dsp^3$$ hybridization. The orbitals involved are $$d_{z^2}, s, p_x, p_y and p_z $$ It is low spin complex. Thus option A is incorrect. (B) $$[Pt(NH_3)_2Cl_2]$$ undergoes $$dsp^2$$ hybridization to form high spin complex. The orbitals involved in the hybridization are $$s, p_x, p_y, d_{x^2-y^2}$$. (C) In $$[Cr(H_2O)_6]^{3+}$$, the $$Cr^{3+} (d^3)$$ ion undergoes $$sp^3d^2$$ hybridization to form outer orbital complex with 3 unpaired electrons. The orbitals involved in the hybridization are $$s, p_x, p_y, p_z, d_{x^2}, d_{x^2-y^2}$$ Thus option C is incorrect. (D) $$Ni(CO)_4$$ involves $$sp^3$$ hybridization to form low spin complex. The orbitals involved in the hybridization are $$s, p_x, p_y, p_z$$. Thus option D is correct.