168k views
4 votes
What is the type of hybridization and bond angle in [Ni(CN)4]−2?

a) sp3, 109.5°

b) dsp2, 90°

c) sp3d2, 120°

d) d2sp3, 180°

User Rkd
by
7.1k points

1 Answer

6 votes

Final answer:

The hybridization of [Ni(CN)4]−2 is dsp2, and the bond angles are 90°, corresponding to a square planar geometry.

Step-by-step explanation:

The correct answer to the question 'What is the type of hybridization and bond angle in [Ni(CN)4]−2?' is (b) dsp2, 90°. In this complex, nickel typically forms a square planar geometry which is consistent with dsp2 hybridization. The square planar geometry indicates that the ligands are arranged in the same plane, with 90° angles between them. Unlike the sp3 hybridization which leads to a tetrahedral geometry and 109.5° bond angles, dsp2 hybridization results in a planar molecule with 90° bond angles. This outcome correlates with the characteristics of nickel complexes with strong field ligands such as cyanide (CN−).

User Ken Lyon
by
7.5k points