Chemistry, asked by jeannie2814, 11 months ago

The hybridization involved in complex [Ni(CN)₄]²⁻ is
(At. No. Ni = 28) [2015 RS]
(a) dsp² (b) sp³
(c) d²sp² (d) d²sp³

Answers

Answered by yogichaudhary
1

Answer:

The hybridization involved in complex [Ni(CN)₄]²⁻ is

(At. No. Ni = 28) [2015 RS]

(a) dsp²

(b) sp³

(c) d²sp²✔

(d) d²sp³

Answered by Anonymous
0

The hybridization involved in complex [Ni(CN)₄]²⁻ is (a) dsp².

  • The outer electronic configuration of Ni  2+  ion is 3D`84S`0
  • dsp² is a square planar, while others are octahedral and tetradedral
  • One 3d, one 4s, and two 4p orbitals undergo dsp² hybridization.
  • Four identical dsp² hybrid orbitals are occupied by four pairs of four cyanide ligands of electrons, thus forming [Ni(CN)₄]²⁻.

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