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[SOLVED] Computation of bond angles and other angles in tetrahedral

Dhamnekar Winod

Active member
Nov 17, 2018
149
Hello,
I didn't understand the geometry of molecules in which central atom has no lone pairs of electrons. for example, in $CH_4, NH_4^+$ molecular shape is tetrahedral and bond angle is $109.5^\circ$. How is that bond angle computed? $CH_4$ stands for liquid methane and $NH_4^+$ is a polyatomic cation.


Now my other question involve mathematics as well.

If i want to compute other angles of this tetrahedral, how can i compute it?

If any member knows the answer to these question, may reply.
 

Dhamnekar Winod

Active member
Nov 17, 2018
149
Hello,
I didn't understand the geometry of molecules in which central atom has no lone pairs of electrons. for example, in $CH_4, NH_4^+$ molecular shape is tetrahedral and bond angle is $109.5^\circ$. How is that bond angle computed? $CH_4$ stands for liquid methane and $NH_4^+$ is a polyatomic cation.


Now my other question involve mathematics as well.

If i want to compute other angles of this tetrahedral, how can i compute it?

If any member knows the answer to these question, may reply.
Hi,

I got the answer to the question how is $109.5^\circ$ angle between all the bonds in tetrahedral structure computed.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,265
Hi,

I got the answer to the question how is $109.5^\circ$ angle between all the bonds in tetrahedral structure computed.
Here is one way to do it.

First consider that the points (+1,+1,+1), (+1,-1,-1), (-1,+1,-1), (-1,-1,+1) span a regular tetrahedron with its center at the origin.
\begin{tikzpicture}
%preamble \usepackage{tikz-3dplot}
\tdplotsetmaincoords{80}{110}
\begin{scope}[scale=3,tdplot_main_coords]
\coordinate[label=below:O] (O) at (0,0,0);
\coordinate[label=A] (A) at (+1,+1,+1);
\coordinate[label=left:B] (B) at (+1,-1,-1);
\coordinate[label=right:C] (C) at (-1,+1,-1);
\coordinate[label=D] (D) at (-1,-1,+1);

\draw[-latex] (O) -- (1,0,0) node[ left ] {x};
\draw[-latex] (O) -- (0,1,0) node[ right ] {y};
\draw[-latex] (O) -- (0,0,1) node[ above ] {z};

\draw[help lines] (-1,1,1) -- (-1,-1,1) -- (1,-1,1);
\draw[help lines] (1,1,-1) -- (-1,1,-1) -- (-1,1,1) -- (1,1,1);
\draw[help lines] (0,-1,1) -- (0,1,1) -- (0,1,-1);
\draw[help lines] (1,1,1) -- (1,-1,1) -- (1,-1,-1) -- (1,1,-1) -- cycle;
\draw[help lines] (1,-1,0) -- (1,1,0) -- (-1,1,0) (1,0,-1) -- (1,0,1) -- (-1,0,1);
\draw[dotted] (C) -- (D);
\draw[dashed] (O) -- (A);
\draw[dashed] (O) -- (B);
\draw[dashed] (O) -- (C);
\draw[dashed] (O) -- (D);
\draw[thick] (B) -- (C) -- (A) -- (D) -- (B);
\draw[ultra thick] (A) -- (B);
\end{scope}
\end{tikzpicture}

The angle between 2 bonds is $\phi=\angle AOB$.
We can calculate the angle $\phi$ from the definition of the dot product:
\[ \overrightarrow{OA} \cdot \overrightarrow{OB} = OA\cdot OB \cdot \cos\phi \\
\cos\phi = \frac{\overrightarrow{OA} \cdot \overrightarrow{OB}}{OA\cdot OB} = \frac{(+1,+1,+1)\cdot(+1,-1,-1)}{\|(+1,+1,+1)\|\cdot \|(+1,-1,-1)\|} = \frac{-1}{\sqrt 3\cdot \sqrt 3} = -\frac 13 \\
\phi=\arccos\left(-\frac 13\right)\approx 109.5^\circ \]
 
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