WBJEE 2015 Solved Chemistry Question Paper – Part 4

Find WBJEE 2015 Solved Chemistry Question Paper – Part 4 in this article. This paper consists of 5 questions (#56 to #60) from WBJEE 2015 Physics paper. Detailed solution of these questions has been provided so that students can match their solutions.

Importance of Previous Years’ Paper:

Previous years’ question papers help aspirants in understanding exam pattern, question format, important topics and assessing preparation. It has also been seen that sometimes questions are repeated in WBJEE Exam. So, this paper will certainly boost your confidence.

About WBJEE Exam

WBJEE is a common entrance examinations held at state level for admission to the Undergraduate Level Engineering and Medical courses in the State of West Bengal. The Chemistry section of WBJEE 2015 engineering entrance exam consists of 40 questions.

56. In aqueous alkaline solution, two electron reduction of HO2- gives

(A) HO

(B) H2O

(C) O2

(D) O2-

Ans: (A)

Sol.

57. Cold ferrous sulphate solution on absorption of NO develops brown colour due to the formation of

(A) paramagnetic [Fe(H2O)5(NO)]SO4

(B) diamagnetic [Fe(H2O)5(N3)]SO4

(C) paramagnetic [Fe(H2O)5(NO3)](SO4)2

(D) diamagnetic [Fe(H2O)4(SO4)]NO3

Ans : (A)

Sol:

The reaction of cold ferrous sulphate on absorption of NO is shown as below:

FeSO4 + NO + 5H2O → [Fe(H2O)5 (NO)]SO4   

[Fe(H2O)5 (NO)]SO4 ­is a brown ring complex and has three unpaired electron so, it is paramagnetic.

58. Amongst Be, B, Mg and Al the second ionization potential is maximum for

(A) B

(B) Be

(C) Mg

(D) Al

Ans : (A)

Sol:

Beryllium and Magnesium are group two elements and Boron and Alumunium are group 13 elements.

Also, the electronic configuration for Boron is 2s22p1.

The electronic configuration of B+ is 2s2

So, it is difficult to remove 2nd e and hence the second ionization potential is maximum for Boron.

WBJEE 2016 Solved Physics and Chemistry Question Paper

59. In a mixture, two enantiomers are found to be present in 85% and 15% respectively. The enantiomeric excess (ee) is

(A) 85%

(B) 15%

(C) 70%

(D) 60%

Ans : (C)

Sol:

Enantiometric excess reflects the degree to which a sample contains one enantiomer in greater amounts than the other.

So, the enantiomeric excess (ee) is given as: (85 – 15)% = 70%

60. 1,4-dimethylbenzene on heating with anhydrous AlCl3 and HCl produces

(A) 1,2-dimethylbenzene

(B) 1,3-dimethylbenzene

(C) 1,2,3-trimethylbenzene

(D) Ethylbenzene

Ans : (B)

Sol:

On heating 1,4-dimethylbenzene with anhydrous AlCl3 and HCl the following reaction takes place

So, the product formed is 1,3-dimethylbenzene.

Also Get:

WBJEE Sample Papers

WBJEE Previous Years' Question Papers

WBJEE Online Test

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