1

### 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
2

### 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   3

### 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
4

### JEE Main 2018 (Online) 15th April Morning Slot

In graphite and diamond, the percentage of p-characters of the hybrid orbitals in hybridisation are respectively :
A
33 and 25
B
33 and 75
C
50 and 75
D
67 and 75

## Explanation

Hybridization of carbon in graphite is sp2.

So,   %   of p - in graphite = ${2 \over 3} \times 100$ = 67%

Hybridization of carbon in diamond is sp3 .

So,   %   of p - in diamond = ${3 \over 4} \times 100$ = 75%

NEET