- A$13, 22$
- ✓$3, 11$
- C$4, 24$
- D$2, 4$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| List $I$ | List $II$ | ||
| $A$ | $\alpha$-Glucose and $\alpha-$Galactose | $I$ | Functional isomers |
| $B$ | $\alpha$-Glucose and $\beta-$-Galactose | $II$ | Homologous |
| $C$ | $\alpha$-Glucose and $\alpha$-Galactose | $III$ | Anomers |
| $D$ | $\alpha$-Glucose and $\alpha$-Galactose | $IV$ | Epimers |
Choose the correct answer from the options given below:
$\left( i \right)\,2F{e_2}{O_3}\left( s \right) \to 4Fe\left( s \right) + 3{O_2}\left( g \right)$
${\Delta _r}{G^o} = + 1487.0\,kJ\,mo{l^{ - 1}}$
$\left( {ii} \right)\,2CO\left( g \right) + {O_2}(g) \to 2C{O_2}\left( g \right)$
${\Delta _r}{G^o} = - 514.4\,kJ\,mo{l^{ - 1}}$
Free energy change, $\Delta_rG^o$ for the reaction
$\,2F{e_2}{O_3}\left( s \right) + 6CO\left( g \right) \to 4Fe\left( s \right) + 6C{O_2}\left( g \right)$ will be .....$kJ\, mol^{-1}$
|
LIST $I$ (Hybridization) |
LIST $II$ (Orientation in Space) |
| $A$ $\mathrm{sp}^3$ | $I$ Trigonal bipyramidal |
| $B$ $\mathrm{dsp}^2$ | $II$ Octahedral |
| $C$ $\mathrm{sp}^3 \mathrm{~d}$ | $III$ Tetrahedral |
| $D$ $s p^3 d^2$ | $IV$ Square planar |
Choose the correct answer from the options given below :

$(i)$ $\left[ M ( NCS )_{6}\right]^{(-6+ n )}$
$(ii)$ $\left[ MF _{6}\right]^{(-6+ n )}$
$(iii)$ $\left[ M \left( NH _{3}\right)_{6}\right]^{ n ^{+}}$