Which one of the following compounds on strong heating evolves ammonia gas ?

1. ${\left({\mathrm{NH}}_4\right)}_{2}S{O}_4$

2. ${\mathrm{HNO}}_3$

3. ${\left({\mathrm{NH}}_4\right)}_{2}C{r}_2{O}_7$

4. ${\mathrm{NH}}_4{\mathrm{NO}}_3$

Match List - I with List - II

1. I-a, II-b, III-d, IV-c

2. I-b, II-c, III-a, IV-d

3. I-a, II-d, III-c, IV-b

4. I-a, II-c, III-b, IV-d 

Ammonia can be dried by

1. conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$

2. ${\mathrm{P}}_4{\mathrm{O}}_{10}$

3. CaO

4. anhydrous ${\mathrm{CaCl}}_2$

There is no S-S bond in

1. ${\mathrm{S}}_2{\mathrm{O}}_4^{2-}$

2. ${\mathrm{S}}_2{\mathrm{O}}_5^{2-}$

3. ${\mathrm{S}}_2{\mathrm{O}}_3^{2-}$

4. ${\mathrm{S}}_2{\mathrm{O}}_7^{2-}$

Which of the following leaves no residue on heating ?

1. $\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2$

2. ${\mathrm{NH}}_4{\mathrm{NO}}_3$

3. $\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2$

4. ${\mathrm{NaNO}}_3$

Which amongst the following reactions cannot be used for the preparation of the halogen acid?

1. $2\mathrm{KBr}+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{conc}.\right)\to {\mathrm{K}}_2{\mathrm{SO}}_4+2\mathrm{HBr}$

2. $2\mathrm{NaCl}+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{conc}.\right)\to {\mathrm{NaHSO}}_4+\mathrm{HCl}$

3. ${\mathrm{NaHSO}}_4+\mathrm{NaCl}\to {\mathrm{Na}}_2{\mathrm{SO}}_4+\mathrm{HCl}$

4. ${\mathrm{CaF}}_2+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{conc}.\right)\to {\mathrm{CaSO}}_4+2\mathrm{HF}$

Which of the following pair has bleaching property ?

1. ${\mathrm{O}}_3 \mathrm{and} {\mathrm{NO}}_2$

2. ${\mathrm{O}}_3 \mathrm{and} {\mathrm{H}}_2\mathrm{S}$

3. ${\mathrm{SO}}_2 \mathrm{and} {\mathrm{Cl}}_2$

4. ${\mathrm{Cl}}_2 \mathrm{and} {\mathrm{NO}}_2$

Nitrogen is liberated by the thermal decomposition of only

1. ${\mathrm{NH}}_4{\mathrm{NO}}_2$

2. ${\mathrm{NaN}}_3$

3. ${\left({\mathrm{NH}}_4\right)}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$

4. All the three

In which of the following arrangement, the order is not according to the property indicated against it

The variation of the boiling point of the hydrogen halides is in the order HF > HI >HBr>HCl, What explains the higher boiling point of hydrogen fluoride?

1. The electro negativity of flourine is much higherthan for other elements in the group

2. There is strong hydrogen bonding between HF molecules

3. The bond energy of HF molecules is greater than in other hydrogen halides

4. The effect of nuclear shielding is much reducedin flourine which polarizes the HF molecule