- A$sp^3 + p_x$
- B$sp^3 + p_y$
- C$d_{xy} + p_x$
- ✓$sp^3 + s$
Ground state of electronic configuration $X e=[K r] 4 d^{10} 5 s^2 5 p^6$
Excited state of electronic configuration $Xe =[ Kr ] 4 d ^{10} 5 s ^2 5 p ^3 5 d ^3$
The electronic configuration of oxygen $=1 s^2 2 s^2 2 p^4$
In $XeO _3$, the hybridisation of $Xe$ is $sp ^3$, It form $3$ $\sigma$ bond and three $\pi$ bonds with three oxygen atoms and contain one lone pair of electron.
The oxygen atom form bond with $Xe$ like $s p^3+p_x, s p^3+p_y, s p^3+p_z$, but never form $s p^3+s$
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$\mathrm{Sn}(\mathrm{s})\left|\mathrm{Sn}^{2+}(\mathrm{aq}, 1 \mathrm{M}) \| \mathrm{Pb}^{2+}(\mathrm{aq}, 1 \mathrm{M})\right| \mathrm{Pb}(\mathrm{s})$
the ratio $\frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}$ when this cell attains equilibrium is
(Given $\mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{0}=-0.14 \mathrm{\;V}$ $\left.\mathrm{E}_{\mathrm{Pb}^{+2}/{\mathrm{Pb}}}^{0}=-0.13 \;\mathrm{V}, \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06\right)$
