The correct increasing order for bond angles among \( \text{BF}_3, \, \text{PF}_3, \, \text{and} \, \text{CF}_3 \) is:
The correct increasing order for bond angles among \( \text{BF}_3, \, \text{PF}_3, \, \text{and} \, \text{CF}_3 \) is:
\( \text{PF}_3 \, < \, \text{BF}_3 \, < \, \text{CF}_3 \)
\( \text{BF}_3 \, < \, \text{PF}_3 \, < \, \text{CF}_3 \)
\( \text{CF}_3 \, < \, \text{PF}_3 \, < \, \text{BF}_3 \)
\( \text{BF}_3 \, = \, \text{PF}_3 \, < \, \text{CF}_3 \)
The Correct Option is C
Approach Solution - 1
To determine the increasing order of bond angles among \( \text{BF}_3, \, \text{PF}_3, \text{and} \, \text{CF}_3 \), we need to understand the molecular geometry and electronic effects influencing these compounds:
- \(\text{BF}_3\) (Boron Trifluoride) has a trigonal planar structure due to the \( \text{sp}^2 \) hybridization of the boron atom. This leads to bond angles of 120°.
- \(\text{PF}_3\) (Phosphorus Trifluoride) exhibits a trigonal pyramidal shape, with phosphorus undergoing \( \text{sp}^3 \) hybridization. The presence of a lone pair causes the bond angle to decrease from the ideal tetrahedral angle (109.5°) to approximately 97°.
- \(\text{CF}_3\) (represented as part of a radical such as \( \text{CF}_3^+ \) in chemistry) typically has a trigonal planar arrangement similar to \( \text{BF}_3 \), but due to the presence of differing electronic factors (such as resonance or inductive effects involving fluorine), the bond angle can be slightly different from 120°.
Based on these observations, we conclude:
- The bond angle in \( \text{BF}_3 \) is the largest because there are no lone pair repulsions.
- The bond angle in \( \text{PF}_3 \) is smaller due to the lone pair repulsion.
- The bond angle in \( \text{CF}_3 \) is influenced by both the resonance and inductive effects, which make it smaller than the ideal angles in \( \text{BF}_3 \) and \( \text{PF}_3 \) structures.
Thus, the increasing order of bond angles is:
\(\text{CF}_3 \, < \, \text{PF}_3 \, < \, \text{BF}_3\)
Approach Solution -2
BF$_3$: Planar structure with 120$^\circ$ bond angles ($sp^2$ hybridization).
PF$_3$: Tetrahedral geometry distorted by lone pair on phosphorus, bond angle $<$ 109.5$^\circ$.
CF$_3$: Tetrahedral geometry with strong electron-withdrawing fluorine atoms, bond angle $\sim$ 104$^\circ$.
The order of bond angles is CF$_3$ $<$ PF$_3$ $<$ BF$_3$.
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 The Valence Shell Electron Pair Repulsion (VSEPR) Theory Questions
Top JEE Main Questions
- 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 ______.
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,