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The species that does not contain peroxide ions is:
1. PbO₂
2. H₂O₂
3. SrO₂
4. BaO₂

The formation of ${\mathrm{PH}}_4^{+}$ is difficult as compared to ${\mathrm{NH}}_4^{+}$ because
1. Lone pair of phosphorous is optically inert.
2. Lone pair of phosphorous resides in almost pure p - orbital.
3. Lone pair of phosphorous resides at sp³ orbital.
4. Lone pair of phosphorous resides in almost pure s - orbital.

Iodine on reaction with concentrated nitric acid forms:
1. HOI
2. HI
3. HIO₃
4. HIO₄

Which of the following is the correct statement?
1. F₂ has higher dissociation energy than Cl₂
2. F has higher electron affinity than Cl
3. HF is stronger acid than HCl
4. Boiling point increases down the group in halogens

 An oxyacid among the following that contains both P-H and P-P bonds simultaneously is:
1. H₄P₂O₅
2. H₄P₂O₇
3. H₄P₂O₆
4. None of the above

Which of the following has been arranged in order of decreasing bond dissociation energy:
1. P-O>Cl-O>S-O
2. P-O>S-O>Cl-O
3. S-O>Cl-O>P-O
4. Cl-O>S-O>P-O

\(\mathrm{PCl}_3\) upon hydrolysis gives an oxyacid of phosphorous.
The basicity of that oxyacid is:

Which set of oxide of nitrogen is paramagnetic in nature?
(1) NO, N₂O
(2) NO₂, NO, N₂O
(3) NO,NO₂
(4) N₂O,NO₂

Correct order of boiling point of group 16 hydrides
(1) H₂O<H₂S<H₂Se<H₂Te
(2) H₂Te<H₂Se<H₂S<H₂O
(3) H₂S<H₂Se<H₂Te<H₂O
(4) H₂O<H₂Te<H₂Se<H₂S

Which is incorrectly given according to the order indicated?
(1)F₂>Cl₂>Br₂>I (Oxidizing power)
(2)HI>HBr>HCl>HF (Acidic Strength)
(3)F₂>Cl₂>Br₂>I₂ (Bond Energy)
(4) HF>HI>HBr>HCl (Boiling point)

The bleaching action of SO₂ is due to its
1. Acidic property
2. Basic property
3. Reducing property
4. Oxidising property

Which is not hydrolysed with water?
(1) PCl₃
(2) NF₃
(3) NCl₃
(4) BCl₃

The following solid exists in the form of ions:-
(1) BI₃
(2) BCl₃
(3) BF₃
(4) CaBO₂

Which is correctly given according to other indicated?
1. F₂<Cl₂<Br₂<I₂ (oxidising Power)
2. HI>HBr>HCl>HF (Acidic strength)
3. Cl₂<Br₂<F₂<I₂ (Bond Energy)
4. HF< HI<HCl< HBr (Boiling point)​

AlF₃ is soluble in HF only in presence of KF. It is due to the formation of
(1) K₃[AIF₃H₃]
(2) K₃[AIF₆]
(3) AlH₃
(4) K[AIF₃H]

Copper on heating with concentrated HNO₃ produces:
1. Cu(NO₃)₂ and NO       
2. Cu(NO₃)₂, NO and NO₂
3. Cu(NO₃)₂ and N₂O       
4. Cu(NO₃)₂ and NO₂

$\mathrm{A}+{\mathrm{H}}_2\mathrm{O}\stackrel{}{\to }\mathrm{B}+\mathrm{HCl}$
$\mathrm{B}+{\mathrm{H}}_2\mathrm{O}\stackrel{}{\to }\mathrm{C}+\mathrm{HCl}$
Compound (A),(B) and (C) will be respectively :
(a) ${\mathrm{PCl}}_5,<mspace\; width="1em"></mspace>{\mathrm{POCl}}_3,<mspace\; width="1em"></mspace>{\mathrm{H}}_3{\mathrm{PO}}_3$                 (b) ${\mathrm{PCl}}_5,<mspace\; width="1em"></mspace>{\mathrm{POCl}}_3,<mspace\; width="1em"></mspace>{\mathrm{H}}_3{\mathrm{PO}}_4$
(c) ${\mathrm{SOCl}}_2,<mspace\; width="1em"></mspace>{\mathrm{POCl}}_3,<mspace\; width="1em"></mspace>{\mathrm{H}}_3{\mathrm{PO}}_3$               (d) ${\mathrm{PCl}}_3,<mspace\; width="1em"></mspace>{\mathrm{POCl}}_3,<mspace\; width="1em"></mspace>{\mathrm{H}}_3{\mathrm{PO}}_4$