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[email protected] P. JOY
IIT-JEE
CLASS TEST - 3 (INORGANIC)
CHEMISTRY
Dear student following is an Easy level [ ] test paper. A score of 23 marks in 10 minutes would be a satisfactory performance: Q.No. 1 to 11 (+3, –1). (M.M. 33) Single option correct Q.1
When BF3 + NH3 ® BF3NH3, how does the hybridization of the boron atom change, if at all ? (A) sp3 → sp2 (B) sp2 → sp3d (C) No change (D) sp2 → sp3
Q.7
The percentage of p-character in the orbitals forming S-S bonds in S8 is (A) 25 (B) 33 (C) 50 (D) 75
Q.2
Choose the pair in which carbon of each molecule shows more than one hybridisation state : (A) HC ≡ CH, CH2 = CH2 (B) CH2 = CH – CH = CH2, CH3 – CH3 (C) CH3 – CH = CH2, CH2 = C = CH2 (D) (CH3)2 CH – CH3, CH2 = CH2.
Q.8
What is the geometry of the short lived CH22+ ion.
Q.3
The formal charge on the central oxygen atom in O3 molecule is : (A) 0 (B) +1 (C) –1 (D) –2
Q.4
Which statement is incorrect for OSF4 : (A) S atom has sp3d hybridisation (B) S atom has steric number = 5 (C) O atom is at one of the two axial positions having S=O bond. (D) O atom is at one of the equatorial position having S=O bond.
H O
H H
N
Q.6
(B) 2
Statement-1 : ICl3 has T-shaped. Statement-2 : Cl is more electronegative than I. (A) Statement-1 is true, Statement-2 is True; Statemen-2 is a correct explanation for statment-1. (B) Statement-1 is true, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statment-2 is False
H H | H N
(C) 1
(D) Statement-1 is False, Statement-2 is True
(D) 3 Q.11
Choose the incorrect statment
(D) Trigonal planar
Q.10
Caffeine
(A) 4
(C) T-shaped
Which of the following corresponds to hybridisation having 'd' orbitals(1) AX6 (2) AX5E (3) AX4E2 (4) AX3E2 Note : E-denotes lone pair X-denotes bonded pair (A) Only 1 (B) Only 1 & 2 (C) Only 2, 3 & 4 (D) All of these
N N H | H H
O
(B) Tetrahedral
Q.9
How many π bond(s) is/ are present between carbon atoms in the compound given below
Q.5
(A) Linear
+
(A) Both H 2 O and H3 O have same hybridisation
Choose the correct statement (A) Linear shape is assumed by only those molecule which has steric number two (B) Both CO2 & BeCl2 have linear shape
+
(B) Both NH 3 and NH4 have same hybridisation (C) Both BF 3 and BF 4 – have same hybridisation (D) All of the above
(C) Both XeF2 and XeF6 have same no. of lone pairs of electron around Xe. (D) All of the above
&
&
CHEMISTRY
IIT JEE
(CLASS TEST - 3)
(INORGANIC)
ANSWER KEY
Name : .......................................................................................................... A
B
C
D
A
B
C
M.M. 33
Roll No. : ..................................
D
A
1.
5.
9.
2.
6.
10.
3.
7.
11.
4.
8.
B
C
D
P. JOY
IIT-JEE
CLASS TEST - 3 (INORGANIC)
CHEMISTRY
Que.
1
2
3
4
5
6
7
8
9
10
11
Ans.
D
C
B
C
C
C
D
A
D
B
B
Sol.1 (D) Sol.7 (D) In S8 molecule, S has sp3 hybridisation state. So percentage of p-character in S–S bond orbitals is 25 & 75 respectively.
3
2
sp BF3 + NH3
sp BF3
H3N:
Sol.2 (C) 2
sp
sp sp
sp
(A) H – C ≡ C – H , CH2 = CH2 2
2
2
3
2
2
2
3
3
2
3
2
sp sp sp sp CH – CH3 , CH2 = CH2
In options (A), (B), (D) in both compounds, C shows only one type of hybridisation state while in option (C), C has more than one hybridisation state in both compounds. Sol.3 (B) :O ..
.. O
Sol.9 (D) S.No. 1. 2. 3. 4.
Formula AX6 AX5E AX4E2 AX3E2
F
F F
Cl
3
(sp d hybridisation with steric no.-5)
H O N
H H | H N
..
Cl
(A) Molecules having steric number - 5 can also have linear geometry. ..
— Eg. :XeF .. 2, I3 have linear geometry..
(B) According to VSEPR theory σ σ O=C=O π π
N N H | H H
O
Cl
Sol.11 (B)
Sol.5 (C) Caffeine has only one C = C π-bond. H H
I
Steric number = σ b.p. + l.p. = 3 + 2 = 5 Molecular geometry = Trigonal bipyramidal Electronic geometry = T-shaped.
Trigonal bipyramidal geometry.
Steric number = σ b.p. + l.p. = 2 + 0 = 2 (sp hybridisation) Molecular geometry = electronic geometry = Linear geometry.
Caffeine
σ σ Cl — Be — Cl
Sol.6 (C)
(B) H
(C) F
.. N H
B
3
sp H
,
H
F
H
H
2
sp ,
H
N H H B H
= 2 + 0 = 2 (sp hybridisation) Molecular geometry = electronic geometry
3
= Linear geometry. CO2 & BeCl2 have linear shape so this is correct statement.
+
H
3
sp H ,
F
O
sp
3
..
H
:O:
Steric number = σ b.p. + l.p.
+
H
(A)
Hybridization sp3d2/d2sp3 sp3d2/d2sp3 sp3d2/d2sp3 sp3d/dsp3
..
S
N = s + l.p. N=6+0=6 N=5+1=6 N=4+2=6 N=3+2=5
Sol.10 (B) Applying VSEPR theory for ICl3 molecule
.. O .. :
Sol.4 (C) O
.. S:
Sol.8 (A) CH22+ ion has linear geometry. [H – C – H]2+
Formal charge on central oxygen atom = 6 – (3 + 2) = 6 – 5 = 1.
F
.. S ..
.. S .. S8 molecule
2
2 sp sp sp sp sp sp (C) CH3 – CH = CH2 , CH2 = C = CH2
.. S ..
.. S ..
3 3 sp sp sp sp sp sp (B) CH2 = CH – CH = CH2 , CH3 – CH3
sp CH3 (D) sp3 CH3
.. S ..
.. : S .. S ..
2
sp
H
..
(C) In XeF 2 , Xe has 3 lone pairs around it and ..
—
in XeF6 , Xe has only I lone pair around it. 3
H
sp
