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:

Always analyze the number of lone pairs on the central atom (VSEPR theory) to predict bond angles. Lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion, compressing bond angles. For radicals, an unpaired electron exerts less repulsion than a lone pair.

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$_2^+$.