1
MCQ (Single Correct Answer)

JEE Main 2017 (Online) 8th April Morning Slot

A metal ‘M’ reacts with nitrogen gas to afford ‘M3N’. ‘M3N’ on heating at high temperature gives back ‘M’ and on reaction with water produces a gas ‘B’. Gas ‘B’ reacts with aqueous solution of CuSO4 to form a deep blue compound. ‘M’ and ‘B’ respectively are :
A
Li and NH3
B
Ba and N2
C
Na and NH3
D
Al and N2

Explanation

2
MCQ (Single Correct Answer)

JEE Main 2017 (Online) 8th April Morning Slot

The number of S = O and S − OH bonds present in peroxodisulphuric acid and pyrosulphuric acid respectively are :
A
(2 and 2) and (2 and 2)
B
(2 and 4) and (2 and 4)
C
(4 and 2) and (2 and 4)
D
(4 and 2) and (4 and 2)

Explanation

Peroxidisulphuric Acid (H2 S2 O8) :

Number of S = O, bonds = 4

Number of S $$-$$ OH, bond = 2

Pyrosulphuric Acid (H2 S2 O7) :

Number of S = O, bonds = 4

Number of S $$-$$ OH, bonds = 2
3
MCQ (Single Correct Answer)

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.
4
MCQ (Single Correct Answer)

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

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