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Question:

Match each set of hybrid orbitals from LIST-I with complex(es) given in LIST-II.
LIST - I
LIST - II
P. dsp²

  1. [FeF₆]⁴⁻
    Q. sp³
  2. [Ti(H₂O)₃Cl₃]
    R. sp³d²
  3. [Cr(NH₃)₆]³⁺
    S. d²sp³
  4. [FeCl₄]²⁻
  5. [Ni(CO)₄]
  6. [Ni(CN)₄]²⁻
    The correct option is:

P→5,6;Q→4;R→3;S→1,2

P→5;Q→4,6;R→2,3;S→1

P→6;Q→4,5;R→1;S→2,3

P→4,6;Q→5,6;R→1,2;S→3

Solution:

[1][FeF₆]⁴⁻
Fe²⁺=[Ar]3d⁶
[FeF₆]⁴⁻ is a high spin complex with sp³d² hybridization.
[2][Ti(H₂O)₃Cl₃]
Ti³⁺=[Ar]3d¹
[Ti(H₂O)₃Cl₃] is sp³d² hybridized.
[3][Cr(NH₃)₆]³⁺
Cr³⁺=[Ar]3d³
[Cr(NH₃)₆]³⁺ is d²sp³ hybridized.
[4][FeCl₄]²⁻
Fe²⁺=[Ar]3d⁶
[FeCl₄]²⁻ is sp³ hybridized.
[5][Ni(CO)₄]
Ni=[Ar]3d⁸4s²
[Ni(CO)₄] is sp³ hybridized.
[6][Ni(CN)₄]²⁻
Ni²⁺=[Ar]3d⁸
[Ni(CN)₄]²⁻ is dsp² hybridized.