The analysis of the molecules \(\text{NH}_3\) and \(\text{NF}_3\) is as follows:
Step 1: Structure and dipole moment of \(\text{NH}_3\)
\(\text{NH}_3\) has a pyramidal shape due to the presence of one lone pair on the nitrogen atom.
- The dipole moments of the \(\text{N–H}\) bonds and the lone pair point in the same direction, leading to a higher resultant dipole moment.
Step 2: Structure and dipole moment of \(\text{NF}_3\)
\(\text{NF}_3\) also has a pyramidal shape, but the \(\text{N–F}\) bonds are highly electronegative.
- The dipole moment of the lone pair on nitrogen is opposite to the resultant dipole moment of the \(\text{N–F}\) bonds, which reduces the overall dipole moment.
Step 3: Comparison of dipole moments
- The dipole moment of \(\text{NH}_3\) is approximately \(1.47 \, \text{D}\), while that of \(\text{NF}_3\) is approximately \(0.80 \, \text{D}\).
- This confirms that \(\text{NH}_3\) has a greater dipole moment than \(\text{NF}_3\).
Step 4: Validating the statements
- Assertion (A): True, because \(\text{NH}_3\) has a higher dipole moment than \(\text{NF}_3\).
- Reason (R): True, as the lone pair’s dipole in \(\text{NH}_3\) aligns with the bond dipoles, while in \(\text{NF}_3\), it opposes them.
- \((R)\) is the correct explanation of \((A)\).
Final Answer: (1).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,