1

### JEE Main 2017 (Online) 9th April Morning Slot

XeF6 on partial hydrolysis with water produces a compound ‘X’. The same compound ‘X’ is formed when XeF6 reacts with silica. The compound ‘X’ is :
A
XeF2
B
XeF4
C
XeOF4
D
XeO3

## Explanation

XeF6 on hydrolysis with water can produces 3 compounds XeOF4, XeO2F2 and XeO3.

Here XeOF4 and XeO2F2 are produced on partial hydrolysis and XeO3 is produced on complete hydrolysis.

XeF6  +  H2O  $\buildrel {Partial} \over \longrightarrow$  XeOF4  +  2HF

XeF6   +   2H2O  $\buildrel {Partial} \over \longrightarrow$  XeO2F2   +  4HF

XeF6   +   3H2O  $\buildrel {Complete} \over \longrightarrow$  XeO3  +  6HF

So, the compound X can be XeOF4   or   XeO2F2

Reaction with silica (SiO2) of XeF6 :

When ration of XeF6 and SiO2 is 2 : 1 then produce XeOF4.

2XeF6   +   SiO2   $\to$   2XeF4   +   SiF4

When ratio of XeF6 and SiO2 is 1 : 1 , then produce XeO2F2.

XeF6   +   SiO2   $\to$   XeO2F2   +   SiF4

So, the compound X can be XeOF4   or  XeO2F2.

Here in option only XeOF4 is given so this will be right answer.
2

### JEE Main 2017 (Online) 9th April Morning Slot

The number of P−OH bonds and the oxidation state of phosphorus atom in pyrophosphoric acid (H4P2O7) respectively are :
A
four and four
B
five and four
C
five and five
D
four and five

## Explanation Number of P - OH bonds = 4

Let the oxidation number of P = x

$\therefore$   In H4 P2 O7

for 7 oxygen = 7 $\times$ ($-$ 2) = $-$ 14

for 4 Hydrogen = 4 $\times$ (+1) = 4

For 2 Phosphorus = 2x

$\therefore$   2x + 4 $-$ 14 = 0

$\Rightarrow$   x = + 5
3

### JEE Main 2017 (Online) 9th April Morning Slot

The correct sequence of decreasing number of $\pi$-bonds in the structures of H2SO3, H2SO4 and H2S2O7 is :
A
H2SO3 > H2SO4 > H2S2O7
B
H2SO4 > H2S2O7 > H2SO3
C
H2S2O7 > H2SO4 > H2SO3
D
H2S2O7 > H2SO3 > H2SO4

$\pi$

## Explanation   4

### JEE Main 2018 (Offline)

The compound that does not produce nitrogen gas by the thermal decomposition is
A
(NH4)2SO4
B
Ba(N3)2
C
(NH4)2Cr2O7
D
NH4NO2

## Explanation

Thermal decomposition reaction of the given compounds are . . .

(NH4)2SO4 $\buildrel \Delta \over \longrightarrow$ 2 NH3(g) + H2SO4

Ba (N3)2 $\buildrel \Delta \over \longrightarrow$ Ba + 3N2(g)

NH4NO2 $\buildrel \Delta \over \longrightarrow$ N2(g) + 2H2O

(NH4)2Cr2O7 $\buildrel \Delta \over \longrightarrow$ N2(g) + 4H2O + Cr2O3

So, here you can see only (NH4)2SO4 does not give N2 by thermal decomposition it gives NH3 while other give N2 gas