Bond Angles and Molecular Geometry

The bond angle in a molecule is determined by the arrangement of electron pairs around the central atom. According to the Valence Shell Electron Pair Repulsion (VSEPR) theory, electron pairs (both bonding and non-bonding) repel each other and arrange themselves as far apart as possible to minimize repulsion. This arrangement dictates the molecular geometry and thus the bond angles.

Geometry of $\text{NF}_3$, $\text{SbF}_3$, and $\text{SbCl}_3$

Let's analyze the structure of each molecule:

• $\text{NF}_3$: Nitrogen (N) is the central atom. It has 5 valence electrons. It forms 3 single bonds with 3 Fluorine (F) atoms. This leaves 5 - 3 = 2 electrons, forming one lone pair on N. Thus, N has 3 bond pairs and 1 lone pair. The steric number (bond pairs + lone pairs) is 3 + 1 = 4. The electron geometry is tetrahedral, and the molecular geometry is trigonal pyramidal.
• $\text{SbF}_3$: Antimony (Sb) is the central atom. It has 5 valence electrons. It forms 3 single bonds with 3 F atoms. This leaves 5 - 3 = 2 electrons, forming one lone pair on Sb. Thus, Sb has 3 bond pairs and 1 lone pair. The steric number is 3 + 1 = 4. The electron geometry is tetrahedral, and the molecular geometry is trigonal pyramidal.
• $\text{SbCl}_3$: Antimony (Sb) is the central atom. It has 5 valence electrons. It forms 3 single bonds with 3 Chlorine (Cl) atoms. This leaves 5 - 3 = 2 electrons, forming one lone pair on Sb. Thus, Sb has 3 bond pairs and 1 lone pair. The steric number is 3 + 1 = 4. The electron geometry is tetrahedral, and the molecular geometry is trigonal pyramidal.

All three molecules have a similar trigonal pyramidal structure with one lone pair on the central atom. The ideal bond angle for a tetrahedral arrangement is 109.5 degrees. However, the presence of a lone pair causes greater repulsion than bond pairs, compressing the bond angles.

Factors Affecting Bond Angles

For molecules with the same number of bond pairs and lone pairs around the central atom, the bond angle is influenced by two main factors:

• Electronegativity of the central atom: If the electronegativity of the central atom increases, the bonding electron pairs are pulled closer to the central atom. This increases the repulsion between the bonding pairs, leading to a larger bond angle.
• Electronegativity of the surrounding atoms: If the electronegativity of the surrounding atoms increases, the bonding electron pairs are pulled further away from the central atom towards the surrounding atoms. This reduces the repulsion between the bonding pairs, leading to a smaller bond angle.

Comparing Bond Angles

Let's compare the bond angles based on these factors:

1. Comparing $\text{NF}_3$ and $\text{SbF}_3$: The surrounding atoms are the same (F), but the central atoms are different (N vs Sb). The electronegativity of N is higher than that of Sb (N > Sb). A more electronegative central atom pulls the bonding electrons closer, increasing bond pair-bond pair repulsion. Therefore, the bond angle in $\text{NF}_3$ is expected to be larger than in $\text{SbF}_3$.

   Bond angle($\text{NF}_3$) > Bond angle($\text{SbF}_3$).

2. Comparing $\text{SbF}_3$ and $\text{SbCl}_3$: The central atom is the same (Sb), but the surrounding atoms are different (F vs Cl). The electronegativity of F is higher than that of Cl (F > Cl). More electronegative surrounding atoms pull bonding electrons further away from the central atom, reducing bond pair-bond pair repulsion. Therefore, the bond angle in $\text{SbF}_3$ is expected to be smaller than in $\text{SbCl}_3$.

   Bond angle($\text{SbCl}_3$) > Bond angle($\text{SbF}_3$).

Combining these observations, we have Bond angle($\text{NF}_3$) > Bond angle($\text{SbF}_3$) and Bond angle($\text{SbCl}_3$) > Bond angle($\text{SbF}_3$). Now we need to compare $\text{NF}_3$ and $\text{SbCl}_3$. Comparing $\text{NF}_3$ (N-F bonds) and $\text{SbCl}_3$ (Sb-Cl bonds), we see differences in both central and surrounding atoms. The electronegativity difference between N and Sb is significant (N is much more electronegative). The effect of the central atom's electronegativity is generally more pronounced in determining the relative bond angles when comparing molecules with central atoms from different periods like N and Sb. The higher electronegativity of N compared to Sb leads to stronger repulsion between N-F bond pairs than between Sb-Cl bond pairs (despite F being more electronegative than Cl). This results in a larger bond angle in $\text{NF}_3$ compared to $\text{SbCl}_3$.

Order of Bond Angles

Based on the analysis, the order of bond angles is:

Bond angle($\text{NF}_3$) > Bond angle($\text{SbCl}_3$) > Bond angle($\text{SbF}_3$)

Typical approximate values are $\text{NF}_3$ (102.3°), $\text{SbCl}_3$ (97.2°), and $\text{SbF}_3$ (92°).