Question

Some octahedral complexes have dissented shapes. In some, two metal-ligand bonds that areapart are shorter than the other four. In, for example, twobonds arelong, and the other four arelong.

(a) Calculate the longest distance between twoatoms in this complex.

(b) Calculate the shortest distance between twoatoms.

Short answer

The required distances are

$$\begin{gathered} \left(a\right)524\text{pm} \\ \text{(b)}334\text{pm} \end{gathered}$$

Step-by-step solution

Given information

In the${\left[\text{Cu}{\left({\text{NH}}_3\right)}_6\right]}^{2+}$compound, two${\left[\text{Cu}{\left({\text{NH}}_3\right)}_6\right]}^{2+}$bonds are$207\text{pm}$long, while the 4 other$\text{Cu}-\text{N}$bonds are$262pm$long.

Pythagorean theorem

Pythagorean theorem: ${\mathit{a}}^{\mathbf{2}}\mathbf{+}{\mathit{b}}^{\mathbf{2}}\mathbf{=}{\mathit{c}}^{\mathbf{2}}$

Longest distance

(a)

The longest distance between two$\text{Nac}$atoms can be found along the plane.

The distance between the two atoms with Cu-N bond length $262\text{pm}$is the longest distance.

$\text{Distance}=262\text{pm}+262\text{pm}=524\text{pm}$

Shortest distance

(b)

The shortest distance can be found between two atoms which are${90}^{\circ }$apart involving both lengths of$\text{Cu}-\text{N}$bond present.

The two bonds can form two sides of a right triangle, with the third side being the distance between twoatoms. Using the Pythagorean theorem, the distance can be computed.

Pythagorean theorem:$a^2+b^2=c^2$, with$\text{a}=262\text{pm},\text{b}=207\text{pm}$, and $\mathbf{\text{c}}$the distance

$$\begin{gathered} c=\sqrt{a^2+b^2} \\ c=\sqrt{{262}^2+{207}^2} \\ c=\sqrt{{262}^2+{207}^2} \\ \text{Distance}=334pm \end{gathered}$$