Chapter MS, Problem 1PQ

Interpretation Introduction

Interpretation:

The geometry around the nitrogen and boron centers in the addition compound, ${\text{H}}_{\text{3}}{\text{NBF}}_{\text{3}}$ has to be determined.

Concept Introduction:

VSEPR or valence shell electron pair repulsion theory predicts the shape of the molecules with the use of arrangement of electrons around the central metal atom. Its postulates are as follows:

1. The shape of the molecules is governed by the repulsion between electron pairs situated in the outermost or valence shell of the atoms.

2. A distortion is observed in the shape of the molecules if lone pairs are present in it. These are the electron pairs that do not participate in the chemical bonding between the atoms. The electron pairs that take part in bond formation are known as bond pairs.

3. The order of repulsion of electron pairs is as follows:

  $\text{lp}-\text{lp repulsion}>\text{lp}-\text{bp repulsion }>\text{bp}-\text{bp repulsion}$

4. Multiple bonds create more repulsion as compared to single bonds.

5. The following table shows the various molecular shapes that match to different number of electron pairs:

$\begin{array}{ccc}\begin{array}{l}\text{Steric number or}\\ \text{number of}\\ \text{electron groups}\end{array}& \begin{array}{l}\text{Number of }\\ \text{lone pairs}\end{array}& \text{Molecular geometry}\\ 2& 0& \text{Linear}\\ 3& 0& \text{Trigonal planar}\\ 3& 1& \text{Bent}\\ 4& 0& \text{Tetrahedral}\\ 4& 1& \text{Trigonal pyramid}\\ 4& 2& \text{Bent}\\ 5& 0& \text{Trigonal bipyramidal}\\ 5& 1& \text{See}-\text{saw}\\ 5& 2& \text{T}-\text{structure}\\ 5& 3& \text{Linear}\\ 6& 0& \text{Octahedral}\\ 6& 1& \text{Square pyramidal}\\ 6& 2& \text{Square planar}\end{array}$

The geometry of the molecule is predicted with the help of steric number. The formula to calculate the steric number is as follows:

  $\text{Steric number}=\left[\begin{array}{c}\left(\text{Atoms bonded to the central atom}\right)+\\ \left(\text{Number of lone pairs on the central atom}\right)\end{array}\right]$        (1)

The least electronegative atom is usually taken to be the central atom.

Answer to Problem 1PQ

Option (A) is the correct one.

Explanation of Solution

Reason for correct option:

The formation of ${\text{H}}_{\text{3}}{\text{NBF}}_{\text{3}}$ by the addition of ${\text{NH}}_{\text{3}}$ and ${\text{BF}}_{\text{3}}$ is shown as follows:

Substitute 4 for atoms bonded to the central atom and 1 for number of lone pairs on the central atom in equation (1) to calculate the steric number of $\text{N}$.

  $\begin{array}{c}\text{Steric number of N}=\text{4}+\text{0}\\ =\text{4}\end{array}$

So the number of electron pairs around nitrogen atom is 4 and no lone pairs are present on it. Therefore the geometry around this atom is tetrahedral according to the VSEPR theory.

Substitute 4 for atoms bonded to the central atom and 1 for number of lone pairs on the central atom in equation (1) to calculate the steric number of $\text{B}$.

  $\begin{array}{c}\text{Steric number of B}=\text{4}+\text{0}\\ =\text{4}\end{array}$

So the number of electron pairs around boron atom is 4 and no lone pairs are present on it. Therefore the geometry around this atom is also tetrahedral according to the VSEPR theory.

Therefore the geometry around both the nitrogen and boron atoms in the addition compound ${\text{H}}_{\text{3}}{\text{NBF}}_{\text{3}}$ is tetrahedral.

Reason for incorrect option:

Since geometry around both $\text{N}$ and $\text{B}$ atoms in addition compound ${\text{H}}_{\text{3}}{\text{NBF}}_{\text{3}}$ is tetrahedral. Hence option (B), (C) and (D) are incorrect.

Conclusion

The geometry around the nitrogen and boron centers in the addition compound ${\text{H}}_{\text{3}}{\text{NBF}}_{\text{3}}$ is tetrahedral. Hence, the correct option is (A).