IIT-JEE
CHEMISTRY
P. JOY
CLASS TEST - 3 (INORGANIC)
Dear student following is Moderate level [ ] test paper. A score of 15 marks in 10 minutes
would be a satisfactory performance: Q.No. 1 to 11 (+3, –1)
(M.M. 33)
Name : .......................................................................................................... Roll No. : ..................................
CHEMISTRY IIT JEE (CLASS TEST - 3) (INORGANIC) ANSWER KEY M.M. 33
&
&
A B C D
1.
2.
3.
4.
A B C D
5.
6.
7.
8.
A B C D
9.
10.
11.
Please read following SHORT WRITE UP and
ANSWER subsequent questions (1 to 5)
Meperidine is a morphine-like analgesic. At
one stage, it was hoped that it would be free of
many of the undesirable side effects of morphine,
but it is now clear it is not. Meperidine is definitely
addictive-
O
OCH CH
23
CH
3
7
4
5
6
3
2
1
8
9
10
11
12
N
Q.1 How many sp
3
hybridised carbon atoms
are there in meperidine ?
(A) 5 (B) 7
(C) 9 (D) 8
Q.2 What is the hybridisation of the nitrogen
atom ?
(A) sp
2
(B) sp
3
(C) sp (D) None of these
Q.3 How many lone electron pairs are there
in the molecule ?
(A) 5 (B) 7
(C) 9 (D) 8
Q.4 State the bond angle at C-2
(A) 112 (B) 120
(C) 180 (D) 109.5º
Q.5 State the bond angle at C-12.
(A) 112 (B) 120
(C) 180 (D) 109.5º
Single option correct
Q.6 Choose a correct statement for a
chemical change given below
CH
3
CN
®
CH
3
CONH
2
(A) Both product and reactant have same
number of
p
bonds .
(B) Product has more number of
p
bonds
than reactant.
(C) Reactant has more number of
p
bonds
than product.
(D) Cant' be pedicted.
Q.7 How many
s
bonds (sp
2
- s) are present
in the molecule given below
(A) 1 (B) 0 (C) 6 (D) 5
Q.8 In which excited state iodine shows
sp
3
d
3
hybridisation state :
(A) First (B) Second
(C) Third. (D) None
Q.9 The percentage of p-character in the
hybridisation state of carbon in CH
4
,
CH
3
+
and CH
3
–
respectively will be
(A) 75, 50, 50 (B) 75, 66.7, 50
(C) 75, 66.7, 75 (D) 50, 50, 75.
Q.10 Choose the pair in which carbon shows
more than one hybridisation state :
(A) HC º CH, CH
2
= CH
2
(B) CH
2
= CH – CH = CH
2
, CH
3
– CH
3
(C) CH
3
– C = CH
2
, CH
2
= C = CH
2
(D) (CH
3
)
2
CH – CH
3
, CH
2
= CH
2
.
Q.11 The number of shared electrons in x mol-
ecules of CS
2
is :
(A) 2x (B) 4x
(C) 6x (D) 8x.
