- ✓$1, 2-$ ethandiol
- B$2-$ butanone
- CNitrobenzene
- DEthanal
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$(I)$ $Be$ has smaller atomic radius compared to $Mg$
$(II)$ $Be$ has higher ionization enthalpy than $Al.$
$(III)$ Charge/radius ratio of $Be$ is greater than that of $Al$.
$(IV)$ Both Be and $Al$ form mainly covalent compounds.

Reason : $1-$ Butene is more stable than $2-$ butene According to Saytzeff's rule, $2-$ butene should be the product which is more branched or substituted compound and hence, more stable than butene $-1$
$(A)\, F^-, Na^+, Mg^{+2}$ $(B) \,Ni, Cu, Zn$
$(C)\, N^{-3}, Cs^+, H^-$ $(D)\, Li, He, Be^{+2}$
$\Delta_{\text {vap }} \mathrm{H}-\Delta_{\text {vap }} \mathrm{U}=...... \times 10^{2} \,\mathrm{~J}\, \mathrm{~mol}^{-1}$. (Round off to the NearestInteger)
$\left[\right.$ Use : $\left.R=8.31\, \mathrm{~J}\, \mathrm{~mol}^{-1}\, \mathrm{~K}^{-1}\right]$
[Assume volume of $\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$ is much smaller than volume of $\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$. Assume $\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ treated as an ideal gas]