Question

The correct order of ONO bond angle in the given species is

• A. NO$_2^+$ \textless NO$_2$ \textless NO$_3^-$ \textless NO$_2^-$
• B. NO$_2^-$ \textless NO$_3^-$ \textless NO$_2$ \textless NO$_2^+$
• C. NO$_3^-$ \textless NO$_2$ \textless NO$_2^-$ \textless NO$_2^+$
• D. NO$_2^-$ \textless NO$_3^-$ \textless NO$_2^+$ \textless NO$_2$

Hint:

First sort the species by how many electron-dense regions surround nitrogen: two regions gives a strictly linear sp arrangement, three regions gives an sp2 arrangement with an ideal reference angle of 120 degrees. For the three-region species, identify what the extra region actually is, an ordinary bonding pair, a full lone pair, or a single unpaired electron, since only a full lone pair squeezes the angle below 120 degrees while an unpaired electron actually lets it relax above 120 degrees.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks to arrange four nitrogen-oxygen species (NO$_2^+$, NO$_2$, NO$_3^-$, NO$_2^-$) in the correct order of their ONO bond angles. This involves applying VSEPR (Valence Shell Electron Pair Repulsion) theory and considering resonance/lone pair effects.

Step 2: Key Formula or Approach:
1. Draw the Lewis structure for each species.
2. Determine the hybridization and electron pair geometry of the central nitrogen atom.
3. Count the number of lone pairs on the central nitrogen atom.
4. Apply VSEPR theory: Lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion. This will affect the bond angles.

Step 3: Detailed Explanation:
Let's analyze each species:
1. NO$_2^+$ (Nitronium ion):
Lewis structure: O=N$^+$=O.
Central N has 2 sigma bonds and 0 lone pairs.
Steric Number = 2.
Hybridization: sp.
Geometry: Linear.
ONO bond angle = 180\(^{\circ}\).
2. NO$_2$ (Nitrogen dioxide radical):
Lewis structure: O=N-O. Central N has 2 sigma bonds, 1 pi bond, and 1 unpaired electron (a half lone pair, or radical electron). Due to resonance, it is O-N-O with partial double bond character.
Steric Number = 2 (sigma bonds) + 1 (radical electron, counted as half a lone pair for VSEPR purposes, but still causes repulsion). Effective electron groups = 3.
Hybridization: sp$^2$.
Geometry: Bent.
Repulsion from the unpaired electron is less than that of a lone pair but still present.
ONO bond angle \(\approx\) 134.1\(^{\circ}\).
3. NO$_3^-$ (Nitrate ion):
Lewis structure (one resonance form): O=N(O$^-$)-O$^-$. Central N is bonded to three O atoms.
Central N has 3 sigma bonds and 0 lone pairs.
Steric Number = 3.
Hybridization: sp$^2$.
Geometry: Trigonal planar.
ONO bond angle = 120\(^{\circ}\) (due to resonance, all bonds are equivalent, and no lone pairs on central N).
4. NO$_2^-$ (Nitrite ion):
Lewis structure: O=N-O$^-$. Central N is bonded to two O atoms and has 1 lone pair.
Central N has 2 sigma bonds and 1 lone pair.
Steric Number = 3.
Hybridization: sp$^2$.
Geometry: Bent.
Repulsion from the lone pair is significant, pushing the bond angle to be smaller than 120\(^{\circ}\).
ONO bond angle \(\approx\) 115\(^{\circ}\).
Order of Bond Angles:
NO$_2^+$ (180\(^{\circ}\)) \textgreater NO$_2$ (134.1\(^{\circ}\)) \textgreater NO$_3^-$ (120\(^{\circ}\)) \textgreater NO$_2^-$ (115\(^{\circ}\)).
So, the increasing order is: NO$_2^-$ \textless NO$_3^-$ \textless NO$_2$ \textless NO$_2^+$.
This matches option (B).

Step 4: Final Answer:
The correct order of ONO bond angle is NO$_2^-$ \textless NO$_3^-$ \textless NO$_2$ \textless NO$_2^+$.

Answer 2

Concept:

• Grouping the species by the steric number (number of electron-dense regions) of nitrogen separates them into two hybridization types: two regions gives a strictly linear sp arrangement, three regions gives an sp2 arrangement with an ideal reference angle of $120^{\circ}$.
• Within the three-region group, the actual O-N-O angle depends only on what the third region is. An ordinary bonding pair keeps the ideal $120^{\circ}$. A full lone pair repels more strongly than a bonding pair and squeezes the angle below $120^{\circ}$. A single unpaired electron, as in a radical, repels less than a full pair and lets the angle relax above $120^{\circ}$.

Step 1: Sort the species by the steric number of nitrogen.
NO$_2^+$: nitrogen forms two sigma bonds with no lone pair and no odd electron, structure O=N$^+$=O, steric number $2$.
NO$_3^-$, NO$_2$, NO$_2^-$: nitrogen has two N-O bonds plus one extra region in each case, steric number $3$.

Step 2: Identify what occupies the third region in each steric-number-$3$ species.
NO$_3^-$: all three N-O bonds are equivalent by resonance, so the third region is just another ordinary bonding pair.
NO$_2$: two N-O bonds plus one single unpaired electron, since NO$_2$ is a radical rather than a closed-shell anion.
NO$_2^-$: two N-O bonds plus one full lone pair.

Step 3: Convert the identity of the third region directly into a bond-angle order.
Repulsion strength order for the third region: lone pair $>$ bonding pair $>$ unpaired electron.
Stronger repulsion at the third region pushes the two N-O bonds closer together, giving a smaller angle. Weaker repulsion lets them relax further apart, giving a larger angle.
So among the three sp2 species: NO$_2^-$ (lone pair, most compressed) $<$ NO$_3^-$ (bonding pair, stays at $120^{\circ}$) $<$ NO$_2$ (unpaired electron, least compressed, relaxes above $120^{\circ}$).

Step 4: Place the linear species.
NO$_2^+$ has no competing third region at all. Being strictly linear, its angle is fixed at the maximum possible value, $180^{\circ}$, larger than any of the sp2 species.

Step 5: Combine into the full order.
NO$_2^-$ $<$ NO$_3^-$ $<$ NO$_2$ $<$ NO$_2^+$

Final Answer: NO$_2^-$ $<$ NO$_3^-$ $<$ NO$_2$ $<$ NO$_2^+$