Question at position 9 Predict the actual bond angles in N2F+ using the VSEPR theory. less than 109.5° 90° 120° 180° less than 120° 109.5°Single choice
A
less than 109.5°
B
90°
C
120°
D
180°
E
less than 120°
F
109.5°

Question

Question at position 9 Predict the actual bond angles in N2F+ using the VSEPR theory. less than 109.5° 90° 120° 180° less than 120° 109.5°

BlackTom AI · Accepted Answer

We start by restating the question and listing all answer options to ensure clarity.
Question: Predict the actual bond angles in N2F+ using VSEPR theory.
Answer options: "less than 109.5°", "90°", "120°", "180°", "less than 120°", "109.5°".
Now, evaluate each option in turn.
Option 1: less than 109.5°. This would be unusually small for a simple diatomic-natured or small polyatomic cation like N2F+. In VSEPR terms, bond angles are typically set by the steric arrangement around the central atoms; while some distortions can occur, suggesting a bond angle far below 109.5° requires strong repulsions or unconventional geometry. Without evidence for such extreme crowding, this option is unlikely.
Option 2: 90°. An angle of 90° is characteristic of square-planar or certain octahedral-derived interactions, or linear arrangements in specific cyclizations, but it is not a common default for a neutral or positively charged diatomic-nonmetal framework like N2F+. In VSEPR, 90° occurs in certain antibonding or coordination contexts rather than typical covalent bonding between light elements, so this angle is improbable for N2F+.
Option 3: 120°. An angle of 120° corresponds to a trigonal planar arrangement, which arises when there are three regions of electron density around a central atom with sp2 hybridization and minimal lone-pair repulsion. For N2F+ (a nitrogen-nitrogen framework with fluorine substituents and a positive charge), this exact planar arrangement is not the standard expectation; some degree of three-dimensional distortion is more typical, depending on lone-pair activity and bond pair repulsions.
Option 4: 180°. A linear arrangement yields 180° angles, seen in diatomic molecules or linear polyatomic ions with minimal steric crowding and no angular strain. In the case of N2F+, the presence of multiple substituents and potential lone pairs generally prevents a perfectly linear N–N–F or N–N–N linkage, making a strict 180° angle unlikely.
Option 5: less than 120°. This option allows for bond angles that are somewhat acute but not as small as 109.5° or 90°, reflecting a geometry where there is some lone-pair or bond-pair repulsion that compresses angles below a typical 120° trigonal benchmark, yet not to extreme lows. It accommodates common deviations seen in molecules with lone pairs or multiple substituents that reduce angles from the idealized trigonal planarity.
Option 6: 109.5°. This is the classic tetrahedral angle, arising when a central atom has approximately four regions of electron density with minimal lone-pair distortion. For N2F+, achieving a perfect 109.5° would require a near-tetrahedral arrangement without significant distortion, which is unlikely given the likely presence of lone-pair repulsions or asymmetrical substituent effects in this cation.
In evaluating these options, one looks for the description that best aligns with VSEPR predictions for a small nitrogen-containing cation with multiple substituents and potential lone-pair interactions. The reasoning favors an angle regime that is compressed relative to 120° but not forced into extreme values, reflecting typical deviations caused by lone-pair and substituent repulsions while avoiding the unrealistic extremes.
Thus, the most consistent option with the expected VSEPR-derived geometry is the choice that specifies angles somewhat smaller than 120° but not as small as 109.5° or other extreme values, acknowledging the common distortions in such species.

Question at position 9 Predict the actual bond angles in N2F+ using the VSEPR theory. less than 109.5° 90° 120° 180° less than 120° 109.5°Single choice

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