PCl5 acts as
1. Reducing agent
2. Oxidising agent
3. Dehydrating Agent
4. None of the above.
The electron gain enthalpy values for O → O– and O → O2– are –141 and 702 kJ mol-1 , respectively. The correct statement about the formation of oxides is-
1. | Higher the electron gain enthalpy lower is the lattice energy. |
2. | Higher the electron gain enthalpy higher is the lattice energy. |
3. | Higher the electron gain enthalpy higher is the stability of oxide. |
4. | O- is more stable than O2- |
The group which contains the strongest oxidizing agent is-
1. Group 15
2. Group 16
3. Group 17
4. Group 18
Halogen that forms only one oxoacid is :
1. Chlorine
2. Fluorine
3. Bromine
4. Iodine
\(\mathrm{Xe}+\mathrm{PtF}_6 \rightarrow \text { ? }\)
The product of the above reaction is:
1. | \(\mathrm{O}_2{ }^{+}\left[\mathrm{PtF}_6\right]^{-}. \) | 2. | \(\mathrm{XeF}_4 \) |
3. | \(\mathrm{XeF}_6 \) | 4. | \(\mathrm{Xe}^{+}\left[\mathrm{PtF}_6\right]^{-}\) |
In which compound does phosphorous have the lowest oxidation state?
1. H3PO3
2. PCl3
3. Ca3P2
4. POF3
A xenon fluoride that does not exist is:
1. XeF4
2. XeF2
3. XeF3
4. XeF6
(a) XeF6 + 3H2O →
(b) XeF6 + H2O
The products formed in the above reactions, respectively, are:
1. XeO3 , XeOF4
2. XeO6 , XeOF4
3. XeO3, XeOF6
4. XeO6 , XeOF6
Arrange the following compounds in increasing order of their acidity:
HF, HCl, HBr, HI
1. HCl < HBr < HI<HF
2. HF < HCl < HI< HBr
3. HF < HCl < HBr < HI
4. HF< HBr < HI< HCl
XeF4 is isostructural with :
1.
2.
3.
4.