Question at position 18 Predict the actual bond angles in N2F+ using the VSEPR theory. less than 109.5° 120° 90° 109.5°180° less than 120°

Question

BlackTom AI · Accepted Answer

To approach this question, first we restate the scenario: predicting bond angles in N2F+ using VSEPR theory and selecting from the given angle options.
Option 1: less than 109.5°. This would correspond to a highly distorted tetrahedral environment or strong lone-pair repulsion beyond what is typical for a trigonal arrangement; in N2F+ the local geometry around involved atoms is not expected to be tetrahedral, so this is unlikely.
Option 2: 120°. This is the ideal angle for a perfect trigonal planar arrangement (three regions of electron density around a central atom). However, in a realistic molecule with lone pairs present, the observed angle is typically reduced from 120°, so this exact value would be too large if lone-pair effects are significant.
Option 3: 90°. A 90° angle is characteristic of octahedral or certain square-planar/other highly constrained environments. It does not describe a simple N–N–F fragment in a small cation like N2F+, so this is not the expected angle.
Option 4: 109.5°. This is the ideal tetrahedral angle. If the central atom had four electron domains arranged tetrahedrally, you would expect ~109.5°. But in N2F+, the electron-domain geometry is not tetrahedral around the relevant center due to having fewer than four substituents and the presence of lone pairs, so this exact angle is unlikely.
Option 5: 180°. This corresponds to a perfectly linear arrangement. While some diatomic or simple fragments can be linear, the presence of a fluorine substituent and the overall electron-pair distribution in N2F+ makes a strictly linear N–N–F interior angle unlikely.
Option 6: less than 120°. This option aligns with the idea that a bond angle deviates downward from the ideal trigonal planar value (120°) due to lone-pair repulsion and limited central-atom substitution, yielding a bent geometry with angles smaller than 120° but not as small as 109.5° in most simple cases.
In this context, VSEPR reasoning points to a scenario where the bond angle is constrained by two bonding regions plus at least one lone pair around the relevant center, producing a bond angle smaller than 120° but not necessarily as small as 109.5°; hence a result like "less than 120°" is the most consistent choice. It accounts for the typical reduction from 120° due to lone-pair effects and steric interactions without forcing it to the tetrahedral or linear extremes.
Overall, options asserting exact 120°, 90°, 109.5°, or 180° are not consistent with the expected electron-domain geometry for N2F+ under VSEPR, whereas the statement "less than 120°" captures the common downward deviation from 120° due to lone-pair repulsion and the specific bonding situation. Therefore, the reasoning supports selecting that the actual bond angles are less than 120°.

Question at position 18 Predict the actual bond angles in N2F+ using the VSEPR theory. less than 109.5° 120° 90° 109.5°180° less than 120°

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