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).
Identify the correct orders against the property mentioned:
A. H$_2$O $>$ NH$_3$ $>$ CHCl$_3$ - dipole moment
B. XeF$_4$ $>$ XeO$_3$ $>$ XeF$_2$ - number of lone pairs on central atom
C. O–H $>$ C–H $>$ N–O - bond length
D. N$_2$>O$_2$>H$_2$ - bond enthalpy
Choose the correct answer from the options given below:
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to: