According to Valence Bond theory, in the square planar Ni(CN)₄²⁻ complex ion, the orbital hybridization pattern is A)sp ³. B)dsp ². C)d ² sp. D)d ² sp ³. E)None of these choices are correct.

Question

Quizplus · Accepted Answer

In the square planar Ni(CN)₄²⁻ complex ion, the nickel center is surrounded by two cyanide ligands (CN), one bipyridyl ligand (B14), one sulfur ligand (S), and one phosphorus ligand (P). The coordination number of nickel is 4, and the electronic configuration of nickel is (d8). In order to form four bonds, the nickel ion must hybridize its 4d and 4p orbitals. The two pairs of electrons in the bipyridyl ligand occupy the sp2 hybridized orbitals of nickel, and the two cyanide ligands occupy the remaining two hybridized orbitals. Therefore, the hybridization pattern of nickel is dsp ².

According to Valence Bond Theory, in the Square Planar Ni(CN)42−

According to Valence Bond Theory, in the Square Planar Ni(CN)₄²⁻