Following is the structural representation of \(N_2O_3\) molecule in which \(x, y, z\) are bond angles and \(p, q, r\) are bond lengths. The correct orders of bond angles and bond lengths respectively are
center
center
Show Hint
Remember:
\[
\text{Bond length} \propto \frac{1}{\text{Bond order}}
\]
Thus:
\[
\text{Triple bond} < \text{Double bond} < \text{Single bond}
\]
in terms of bond length.
Step 1: Picture the N2O3 molecule.
$N_2O_3$ has two nitrogen atoms joined by a single $N-N$ bond. One nitrogen carries two oxygens with double bond character, and the other nitrogen carries an oxygen by a longer single bond. The labels $x,y,z$ are bond angles and $p,q,r$ are bond lengths.
Step 2: Recall the bond length rule.
A higher bond order means a shorter bond. So a double bond is shorter than a single bond. This single idea decides the order of $p,q,r$.
Step 3: Order the bond lengths.
The $N-N$ single bond is the longest, so $q$ is the largest. The $N=O$ double bond is the shortest, so $p$ is the smallest. The remaining bond $r$ is in between. Hence \[ p < r < q. \]
Step 4: Recall the bond angle rule.
A double bond and lone pairs both push on nearby bonds. Strong repulsion opens an angle wider; a lone pair squeezes an angle smaller.
Step 5: Order the bond angles.
The angle near the doubly bonded oxygens, $x$, is opened widest by the double bond repulsion, so $x$ is the largest. The angle $y$ is squeezed smallest by lone pair repulsion. The angle $z$ is medium. Hence \[ y < z < x. \]
Step 6: Combine both orders.
Putting the two results together gives the matching pair of orders. \[ \boxed{y < z < x \ \text{and} \ p < r < q} \]
