Copper(I) complexes tend to be tetrahedral rather than octahedral or square planar. Why is this the case? Explain your reasoning.

The ion is d^10

Explanation:

Tetrahedral geometry is one in which the central atom/ion is located at the center of four ligands positioned at the corners of a tetrahedron.

It is different from a square planar complex wherein the ligands are positioned at the corners of a square with the central atom/ion at the center of the square.

Tetrahedral complexes are more common with d^10 species.The electronic configuration of Cu I is [Ar]3d104s1. Hence, a tetrahedral geometry is preferred over a square planar geometry.

може ли да зареждате повече информация, моля

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally