Chemistry Test 107

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Chemistry Test 107

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Weekly Quiz Competition

Question 1
4 / -1

Select the incorrect statement on the basis of following conversion

A
X may be MnO₂

B
Y can be K₂MnO₄

C
Z can be KMnO₄

D
Z is coloured.due to d-d transition
Question 2
4 / -1

Among the following which is strongest oxidizing agent ?

A
Cl₂

B
F₂

C
Br₂

D
I₂

Solution

due to highest SOP value.
Question 3
4 / -1

Back bonding is negligible in :-

A
(SiH₃)₃N

B
H₃Si–N=C=O

C
O(SiH₃)₂

D
P(SiH₃)₃

Solution

3^rd pd – 3^rd pd
Question 4
4 / -1

Which of the following is the best order of Lewis acid strength of BF₃, BCl₃ and BBr₃ ?

A
BF₃ > BCl₃ > BBr₃

B
BF₃ = BCl₃ = BBr₃

C
BF₃ < BCl₃ < BBr₃

D
BBr₃ > BF₃ > BCl₃

Solution

Due to back bonding
Question 5
4 / -1

In which of following back bonding is not possible :-

A
BF₃

B
B₃N₃H₆

C
C₆H₆

D

Solution

Due to absence LP & vacant orbital.
Question 6
4 / -1

Which of the following reaction is not possible?

A
(CH₃)₂O + BF₃ → [(CH₃)₂O → BF₃]

B
NH₃ + BF₃ → [NH₃ → BF₃]

C
(CH₃)₂O + BF₄^–→ [(CH₃)₂O → BF₄^–]

D
N(CH₃)₃+B(CH₃)₃→ [N(CH₃)₃→ B(CH₃)₃]

Solution

BF₄^– → obey octet rule so it can't act as a Lewis acid.
Question 7
4 / -1

Which of the following oxidation states are most stable for lead and tin respectively :

A
+ 4, + 2

B
+ 2, + 4

C
+ 4, + 4

D
+ 2, + 2

Solution

Due to inert pair effect

Question 8

4 / -1

Match the column (for B₂H₆) :-

Column-I		Column-II
(a)	Maximum atom in same plane	(P)	0
(b)	3C-4 e^– bridge bonds	(Q)	2
(c)	Maximum replaceable H-atoms	(R)	6
(d)	To complete the octet of 1'B' atom e^– required are	(S)	4

a-S, b-P, c-Q, d-R

a-R, b-P, c-S, d-Q

a-R, b-S, c-P, d-Q

a-P, b-Q, c-R, d-S

Solution

Maximum no. of atoms in same plane = 6
No. of 3C - 4e^– bonds = 0
Maximum no. of replacable H = 4

Question 9
4 / -1

The ratio of number of 3C-2e^– bond and 2C-2e^– bond in B₂H₆ is :-

A
4 : 2

B
2 : 4

C
1 : 1

D
4 : 1
Question 10
4 / -1

Which is correct for diborane :-

A
Diborane has 12 valence electron

B
Diborane is dimer of BH₃

C
Diborane is a lewis acid

D
All of these

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Re-Attempt Weekly Quiz Competition

Chemistry Test 107

Result

Chemistry Test 107

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Question 1

X may be MnO2

Y can be K2MnO4

Z can be KMnO4

Z is coloured.due to d-d transition

Question 2

Cl2

F2

Br2

I2

Solution

Question 3

(SiH3)3N

H3Si–N=C=O

O(SiH3)2

P(SiH3)3

Solution

Question 4

BF3 > BCl3 > BBr3

BF3 = BCl3 = BBr3

BF3 < BCl3 < BBr3

BBr3 > BF3 > BCl3

Solution

Question 5

BF3

B3N3H6

C6H6

Solution

Question 6

(CH3)2O + BF3 → [(CH3)2O → BF3]

NH3 + BF3 → [NH3 → BF3]

(CH3)2O + BF4– → [(CH3)2O → BF4– ]

N(CH3)3+B(CH3)3→ [N(CH3)3→ B(CH3)3]

Solution

Question 7

+ 4, + 2

+ 2, + 4

+ 4, + 4

+ 2, + 2

Solution

Question 8

a-S, b-P, c-Q, d-R

a-R, b-P, c-S, d-Q

a-R, b-S, c-P, d-Q

a-P, b-Q, c-R, d-S

Solution

Question 9

4 : 2

2 : 4

1 : 1

4 : 1

Question 10

Diborane has 12 valence electron

Diborane is dimer of BH3

Diborane is a lewis acid

All of these

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X may be MnO₂

Y can be K₂MnO₄

Z can be KMnO₄

Cl₂

F₂

Br₂

I₂

(SiH₃)₃N

H₃Si–N=C=O

O(SiH₃)₂

P(SiH₃)₃

BF₃ > BCl₃ > BBr₃

BF₃ = BCl₃ = BBr₃

BF₃ < BCl₃ < BBr₃

BBr₃ > BF₃ > BCl₃

BF₃

B₃N₃H₆

C₆H₆

(CH₃)₂O + BF₃ → [(CH₃)₂O → BF₃]

NH₃ + BF₃ → [NH₃ → BF₃]

(CH₃)₂O + BF₄^–→ [(CH₃)₂O → BF₄^–]

N(CH₃)₃+B(CH₃)₃→ [N(CH₃)₃→ B(CH₃)₃]

Diborane is dimer of BH₃