 Solution

ID: past-exam-of-the-chemistry-b-course-of-the-university-of-cambridge/2022/paper-2/b7/a/i/solution

d^{0}

N_{e} = 0 + 4 (2) = 8.

tetrahedral molecular geometry separates the four donors without placing them at

9 0^{\circ}

d^{0}

centre has no occupied d orbitals providing a square-planar stabilization; the 18-electron rule is not a requirement for this early-metal compound.

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