Which one of the orders is correctly matched with the property mentioned against it?
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General trends to remember:
- Boiling points of hydrides increase down a group, with exceptions for NH$_3$, H$_2$O, and HF due to H-bonding.
- Acidic character of oxides increases with the oxidation state of the central atom.
- Acidic strength of binary hydrides increases down a group in p-block (e.g., HF<HCl).
- Bond angles in similar hydrides (like H$_2$O, H$_2$S) decrease down the group as the central atom's electronegativity decreases.
(A) Boiling point of Group 16 hydrides: The boiling points of hydrides generally increase down the group due to increasing van der Waals forces. However, H$_2$O has an exceptionally high boiling point due to extensive hydrogen bonding. The correct order is H$_2$S<H$_2$Se<H$_2$Te<H$_2$O. The given order is incorrect. (B) Acidic nature of nitrogen oxides: The acidic nature of oxides increases with the increasing oxidation state of the central atom. Oxidation states of N in each oxide: N$_2$O: +1 NO: +2 N$_2$O$_3$: +3 N$_2$O$_4$: +4 N$_2$O$_5$: +5 N$_2$O and NO are neutral oxides, while N$_2$O$_3$, N$_2$O$_4$, and N$_2$O$_5$ are acidic. The acidic character increases with oxidation state, giving the correct trend: N$_2$O<NO<N$_2$O$_3$<N$_2$O$_4$<N$_2$O$_5$. This statement is correct. (C) Acidic nature of hydrogen halides: Acidic strength depends on bond dissociation enthalpy. As we go down the group from F to I, the H-X bond length increases, making it easier to release H$^+$. The correct order is HF<HCl<HBr<HI. The given order is incorrect. (D) Bond angle of Group 16 hydrides: All have a bent shape with two lone pairs on the central atom. Electronegativity decreases down the group, reducing lone pair-bond pair repulsion and decreasing bond angle. Correct order: H$_2$O (>104.5$^\circ$)>H$_2$S>H$_2$Se>H$_2$Te ($\sim$90$^\circ$). The given order is incorrect.
