Q137P - Chemistry: Molecular Nature Of Matter And Change

Question

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

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

(b) Calculate the shortest distance between two  atoms.

Step-by-Step Solution

Answer

The required distances are

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

Step 1: 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.

 

 

Step 2: Pythagorean theorem

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

Step 3: 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.

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

Step 4: 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 two  atoms. 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

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