MCQ
The $'d'$ orbital with the orientation along $X$ and $Y$ axis is called.
- A$d_{z^2}$
- B$d_{yz}$
- C$d_{xy}$
- ✓$d_{x^2 -y^2}$
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$(a)\, \Delta U = 0$
$(b)\, q = 0$
$(c)\, \Delta T = 0$
$(d)\,q = 2.303\,nRT\,{\log _{10}}\,\left( {\frac{{{V_2}}}{{{V_1}}}} \right)$
$ \mathrm{N}_2=3.0 \times 10^{-3} \mathrm{M}, \mathrm{O}_2=4.2 \times 10^{-3} \mathrm{M} \text { and } \mathrm{NO}=2.8 \times 10^{-3} \mathrm{M} . $
$ 2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$
If $0.1 \mathrm{~mol} \mathrm{~L} \mathrm{~L}^{-1}$ of $\mathrm{NO}_{(\mathrm{g})}$ is taken in a closed vessel, what will be degree of dissociation ( $\alpha$ ) of $\mathrm{NO}_{(\mathrm{g})}$ at equilibrium?
$PCl _{5}, BrF _{5}, \& IF _{7}, BrF _{5} \& IF _{7}$