
- A$l_1 = l_2$
- B$l_1 > l_2$
- ✓$l_2 > l_1$
- D$l_1$ and $l_2$ cannot be compared


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|
LIST $I$ (Compound / Species) |
LIST $II$ (Shape / Geometry) |
| $A$ $\mathrm{SF}_4$ | $I$ Tetrahedral |
| $B$ $\mathrm{BrF}_3$ | $II$ Pyramidal |
| $C$ $\mathrm{BrO}_3^{-}$ | $III$ See saw |
| $D$ $\mathrm{NH}_4^{+}$ | $IV$ Bent $T$-shape |
Choose the correct answer from the options given below :
$(A)$ The decreasing order of atomic radii of group $13$ elements is $\mathrm{Tl}>\mathrm{In}>\mathrm{Ga}>\mathrm{Al}>\mathrm{B}$.
$(B)$ Down the group $13$ electronegativity decreases from top to bottom.
$(C)$ $\mathrm{Al}$ dissolves in dil. $\mathrm{HCl}$ and liberate $\mathrm{H}_2$ but conc. $\mathrm{HNO}_3$ renders Al passive by forming a protective oxide layer on the surface.
$(D)$ All elements of group 13 exhibits highly stable +1 oxidation state.
$(E)$ Hybridisation of $\mathrm{Al}$ in $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ ion is $\mathrm{sp}^3 \mathrm{~d}^2$.
Choose the correct answer from the options given below:
$(i)\,\,C\,({\rm{graphite}})\, + \,{O_2}{\kern 1pt} (g)\, \to \,C{O_2}\,(g);\,\Delta r{H^\circleddash} = x\,\,kJ\,mo{l^{ - 1}}$
$(ii)\,\,C\,({\rm{graphite}})\, + \,\frac{1}{2}{O_2}{\kern 1pt} (g)\, \to \,CO\,(g);\,\Delta r{H^\circleddash} = y\,\,kJ\,mo{l^{ - 1}}$
$(iii)\,\,CO\,(g)\, + \,\frac{1}{2}{O_2}{\kern 1pt} (g)\, \to \,C{O_2}\,(g);\,\Delta r{H^\circleddash} = z\,\,kJ\,mo{l^{ - 1}}$
Based on the above thermochemical equations, find out which one of the following algebraic relationships is correct?