Arrange the following compounds in increasing order of their dipole moment:
HBr, H\(_2\)S, NF\(_3\), and CCl\(_3\)
HBr, H\(_2\)S, NF\(_3\), and CCl\(_3\)
Show Hint
- CCl\(_3\)<NF\(_3\)<HBr<H\(_2\)S
- NF\(_3\)<HBr<H\(_2\)S<CCl\(_3\)
- H\(_2\)S<HBr<NF\(_3\)<CCl\(_3\)
- HBr<H\(_2\)S<NF\(_3\)<CCl\(_3\)
The Correct Option is C
Solution and Explanation
The dipole moment of a molecule is determined by both the electronegativity difference between atoms and the molecular geometry.
\(CCl(_3)\): Chlorine is highly electronegative, but the molecule has a symmetric trigonal planar geometry, which results in a low dipole moment due to cancellation of individual dipoles.
\( NF_3\): Nitrogen is more electronegative than fluorine, but due to the geometry of \(NF_3\) (a trigonal pyramidal shape), the dipole moment is moderate.
HBr: Bromine is less electronegative than fluorine or chlorine, but since HBr has a linear geometry, it results in a moderate dipole moment.
\( H_2S\): Due to the bent geometry of \(H_2S\) and the significant electronegativity difference between sulfur and hydrogen, \(H_2S\) has the highest dipole moment among the given compounds.
Thus, the increasing order of dipole moments is: \[ \text{H}_2\text{S} < \text{HBr} < \text{NF}_3 < \text{CCl}_3 \]
Top JEE Main Chemistry Questions
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 : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
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
- The correct order of atomic radii is:
Top JEE Main Dipole moment Questions
- Number of compounds/species from the following with non-zero dipole moment is ______
\(\text{BeCl}_2, \, \text{BCl}_3, \, \text{NF}_3, \, \text{XeF}_4, \, \text{CCl}_4, \, \text{H}_2\text{O}, \, \text{H}_2\text{S}, \, \text{HBr}, \, \text{CO}_2, \, \text{H}_2, \, \text{HCl}\) - Arrange the following compounds in increasing order of their dipole moment:
HBr, H\(_2\)S, NF\(_3\), and CCl\(_3\)
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
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 : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.