In the Lewis dot structure for NH_3, the central atom of the molecule has three atoms bonded to it and one lone pair of electrons. What is the shape of an NH_3 molecule?

Jul 7, 2017

Trigonal pyramidal

Explanation:

An ammonia molecule has nitrogen as the central atom with three hydrogen bonds and one lone electron pair. This gives the molecule a steric number of four. Therefore, the molecular shape is trigonal pyramidal. Jul 8, 2017

Pyrimidal $A {X}_{3} {E}_{1}$ => 3 Bonded Prs & 1 Nonbonded Pr

Explanation:

A method for determining geometry of binary compounds.

Given a binary compound $A {X}_{n}$ where A is the central element, X the attached substrate element and n the number of substrate elements attached. If the structure has non-bonded electron pairs, then the objective is $A {X}_{n} {E}_{n}$ where ${E}_{n}$ is the number of non-bonded electron pairs.

Needed is the number of 'Bonded electron pairs' (BPrs) and the number of 'Nonbonded electron pairs' (NBPrs). The total number of bonded pairs and nonbonded pairs => the Parent Structure. The Geometry is defined by the 'bonded pairs' attached to the central element of the Parent Structure. The nonbonded pairs occupy the remaining orbitals of the parent structure but are not used to define the final geometry. The table posted in the 1st answer illustrates this concept.

To Determine the Geometry of a Binary Compound ...

Determine the Number of Bonded Pairs $\left(B P r s\right)$:
=> number of Substrate elements attached to the central element.

$N {H}_{3}$ has 3 bonded electron pairs b/c there are 3 Hydrogens bonded to the central element, Nitrogen.

$B P r s = 3$

Determine the Number of Nonbonded Pairs ($N B P r s$):
a. Determine the 'Valence Number'
=> $\left({V}_{e}\right)$ = number of valence electrons

${V}_{e}$ = Valence electrons of A + n(Valence electrons of X)

$\text{For}$ $N {H}_{3}$ => ${V}_{e}$ = $3 H + 1 N$ = $3 \left(1\right) + 1 \left(5\right)$ = $8$

b. Determine the 'Substrate Number'
=> $\left({S}_{e}\right)$= total electrons in valence shell of substrate elements when bonded to the central element.

$\text{For} N {H}_{3}$ => ${S}_{e}$ = $3 H = 3 \left(2\right) = 6$

c. Determine the number of non-bonded electron pairs.

$N B P r s = \left(\frac{{V}_{e} - {S}_{e}}{2}\right) = {E}_{n}$

$\text{For} N {H}_{3}$ => $N B P r s = \left(\frac{8 - 6}{2}\right) = \frac{2}{2} = 1 N B P r$

Determine Parent Structure:
$B P r s + N B P r s$ = Parent structure electron pairs

$\text{For} N {H}_{3}$ = BPRs + NBPRs = 3 + 1 = 4 > Electron Pairs around central element of a Parent Structure => $A {X}_{4}$ => Tetrahedral parent configuration.

Determine Geometry of Structure of Interest:

$N {H}_{3}$ Geometry => $A {X}_{3} {E}_{1}$ => 3 Bonded Pair & 1 Nonbonded Pair about the central element Nitrogen. The electron pairs are arranged in a 'Parent' Tetrahedral Configuration. (Note: In the 'Parent' configuration one of the legs of the Tetrahedron is occupied by the nonbonded pair while the other 3 legs are occupied by bonded pairs of electons.) The geometry is then Pyrimidal Geometry and related to the Tetrahedral parent configuration 