The correct increasing order for bond angles among \( \text{BF}_3, \, \text{PF}_3, \, \text{and} \, \text{CF}_3 \) is:
\( \text{PF}_3 \, < \, \text{BF}_3 \, < \, \text{CF}_3 \)
\( \text{BF}_3 \, < \, \text{PF}_3 \, < \, \text{CF}_3 \)
\( \text{CF}_3 \, < \, \text{PF}_3 \, < \, \text{BF}_3 \)
\( \text{BF}_3 \, = \, \text{PF}_3 \, < \, \text{CF}_3 \)
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:
Based on these observations, we conclude:
Thus, the increasing order of bond angles is:
\(\text{CF}_3 \, < \, \text{PF}_3 \, < \, \text{BF}_3\)
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$_3$.
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 : \(A_2 \rightleftharpoons 2A\)