The correct stability order of the following species/molecules is:
To determine the correct stability order of the given species/molecules, we need to examine factors that contribute to their stability. These factors can include electron distribution, resonance structures, and inductive effects, among others.
Step 1: Analyze Species \( q \)
Species \( q \) has favorable characteristics such as extensive resonance stabilization, which can delocalize charge across the molecule, enhancing its stability.
Step 2: Analyze Species \( r \)
Species \( r \) has some degree of resonance or an inductive effect that provides moderate stabilization, placing it in an intermediate position between \( q \) and \( p \).
Step 3: Analyze Species \( p \)
Species \( p \) lacks sufficient resonance structures or may even have some electron-withdrawing elements destabilizing the molecule, making it the least stable.
Final Stability Order:
After examining the key factors influencing stability, the correct order is \( q>r>p \), as species \( q \) benefits most from stabilizing effects, followed by \( r \), and then \( p \).
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
A hydrocarbon which does not belong to the same homologous series of carbon compounds is
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: