MCQ
$XeF_2$ reacts with $PF_5$ to give
- A$XeF_6$
- ✓$[XeF]^+ [PF_6]^-$
- C$XeF_4$
- D$[PF_4]^+ [XeF_3]^-$
✓
Answer
Correct option: B.
$[XeF]^+ [PF_6]^-$
b
$PF _5$ is a fluoride ion acceptor so it yields cationic species with xenon fluorides.
$PF _5$ is a fluoride ion acceptor so it yields cationic species with xenon fluorides.
$\left[ XeF _2\right]+\left[ PF _5\right] \rightarrow[ XeF ]^{+}\left[ PF _6\right]^{-}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If the radius of the $3^{\text {rd }}$ Bohr's orbit of hydrogen atom is $r_{3}$ and the radius of $4^{\text {th }}$ Bohr's orbit is $r_{4}$. Then ....
→Which statement is false for the balanced equation given below ?
→$CS_2 + 3O_2 \,\,\to \,\, 2SO_2 + CO_2$
Which mixture forms a buffer when dissolved in $1\, lit$. of water
→Which of the following reactions does not yield an amine
$XeF_6$ on partial hydrolysis gives
→is prepared best by the reaction
For the galvanic cell,
→$\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(0.02 \mathrm{M}) \rightarrow \mathrm{Zn}^{2+}(0.04 \mathrm{M})+\mathrm{Cu}(\mathrm{s})$
$\mathrm{E}_{\text {cell }}=...... \,\times 10^{-2} \,\mathrm{~V} { (Nearest integer) }$
${\left[\text { Use }: \mathrm{E}_{\mathrm{Cu} / \mathrm{Cu}^{2+}}^{0}=-0.34\, \mathrm{~V}, \mathrm{E}_{2 \mathrm{n} / \mathrm{Zn}^{2+}}^{0}=+0.76 \,\mathrm{~V}\right.}$
$\left.\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059\, \mathrm{~V}\right]$
The energy required to break one mole of $Cl- Cl$ bonds in $Cl_2$ is $242\, kJ \,mol^{- 1}.$ The longest wavelength of light capable of breaking a single $Cl - Cl$ bond is ............ $\mathrm{nm}$
→$(C= 3 \times 10^8 \,ms^{-1}$ and $N_A =6.02 \times mol^{-1}).$
On the basis of the electrochemical theory of aqueous corrosion, the reaction occurring at the cathode is
Electron affinity depends on