Question

The three hydrogen atoms and one nitrogen atom in an ammonia molecule form a tetrahedron—a pyramid with four triangular faces. Three of these faces are isosceles triangles with the nitrogen atom at the vertex. If the distance between the hydrogen...?

The three hydrogen atoms and one nitrogen atom in an ammonia molecule form a tetrahedron—a pyramid with four triangular faces. Three of these faces are isosceles triangles with the nitrogen atom at the vertex. If the distance between the hydrogen atoms is 1.64 angstroms, and the nitrogen atom is 1.02 angstroms from each hydrogen atom, determine the H-N-H bond angle. Round to the nearest tenth of a degree?

Answer 1

`theta=107.0125`

Explanation:

distance between hydrogen atoms of in base is 1.64 angstroms
distance between each hydrogen atoms of base and nitrogen atom at vertex is 1.02 angstroms

By cos rule

`1.64^2=1.02^2+1.02^2-2xx1.02xx1.02xxcostheta`

`2xx1.02xx1.02xxcostheta=1.02^2+1.02^2-1.64^2`

`costheta=(1.02^2+1.02^2-1.64^2)/(2xx1.02xx1.02)`
`=(-0.6088)/(2.0808)`
`costheta=-0.2926`

`theta=cos^-1(-0.2926)`
`theta=107.0125`

Answer 2

107.0°

Explanation:

Isosceles triangle with base 1.64 and other sides 1.02. Drop perpendicular from nitrogen to the H-H bond, bisecting that bond and the angle H-N-H.

Bond angle=`2*arcsin((1.64/2)/1.02)`
`~=107.0°` to 1 d.p.