Chemistry Chapter 10 Entry Test MCQs

a. CHCl3
b. NF3
c. PCl5
d. BF3

c. PCl5

a. sp3, sp2
b. sp, sp
c. sp, sp2
d. sp2, sp

d. sp2, sp

a. π-bonds and σ-bonds involve in hybridization
b. It explains trivalent nature of carbon
c. Hybridization explains paramagnetic behaviour of oxygen
d. All are incorrect

d. All are incorrect

a. pz-pz
b. s-s
c. s-p
d. none

d. none

a. H2S
b. NH3
c. H2Se
d. H2O

c. H2Se

a. Square planar structure
b. Triangular structure.
c. Octahedral structure.
d. Tetragonal structure

c. Octahedral structure.

a. 104.5o, 1.84 D
b. 102.5o, 1.56 D
c. 109.5o 1.84 D
d. 107.5o, 1.56 D

a. 104.5o, 1.84 D

a. 2 sigma and 1 pi bonds.
b. 1 sigma and 1 pi bonds.
c. 2 sigma and 2 pi bonds.
d. 1 sigma and 2 pi bonds.

a. 2 sigma and 1 pi bonds.

a. sp2
b. sp3
c. sp3d2
d. sp

b. sp3

a. NaCl < LiCl < BeCl2
b. LiCl < NaCl < BeCl2
c. BeCl2 < NaCl < LiCl
d. BeCl2 < LiCl < NaCl

a. NaCl < LiCl < BeCl2

a. CH3+
b. NH3
c. NH4+
d. H2O

c. NH4+

a. H2S
b. SO2
c. NH3
d. H2O

a. H2S

a. sp hybridization and tetrahedral geometry
b. sp2 hybridization and trigonal planar
c. sp3 hybridization and tetrahedral geometry
d. sp2 hybridization and trigonal planar

c. sp3 hybridization and tetrahedral geometry

a. NH3
b. BF3
c. H2O
d. BeCl2

d. BeCl2

a. Increases upto 1200
b. Increases upto 109.50
c. Increases upto 1800
d. Decreases upto 109.50

b. Increases upto 109.50

a. Similar size of C and Cl
b. Regular tetrahedral structure
c. Its planner structure
d. Similar electron affinities of C and Cl

b. Regular tetrahedral structure

a. BF3 involves sp3 hybridization and NF3 involves sp2 hybridization
b. BF3 involves sp2 hybridization and NF3 involves sp3 hybridization
c. Both BF3 and NF3 involve sp3 hybridization
d. Both BF3 and NF3 involve sp2 hybridization

b. BF3 involves sp2 hybridization and NF3 involves sp3 hybridization

hybridization states

a. sp2, sp2, sp2
b. sp2, sp2, sp
c. sp3, sp2, sp
d. sp3, sp2, sp2

b. sp2, sp2, sp

a. BF3 , NF3 , AlCl3
b. SF4 , CH4 , SeF4
c. BF3 , BCl3 , BBr3
d. NF3 , BCl3 , NH3

c. BF3 , BCl3 , BBr3

a. H2S < NH3 < SiH4 < BF3
b. H2S < SiH4 < NH3 < BF3
c. H2S < NH3 < BF3 < SiH4
d. NH3 < H2S < SiH4 < BF3

a. H2S < NH3 < SiH4 < BF3

a. PH3 > AsH3 > SbH3 > NH3
b. SbH3 > AsH3 > PH3 > NH3
c. NH3 > PH3 > AsH3 > SbH3
d. SbH3 > AsH3 > NH3 > PH3

c. NH3 > PH3 > AsH3 > SbH3

a. NH3, CH3+, H3O+
b. NO2, SO2, CO2
c. CH3+, BF3, NH3
d. NO3-1, CH3+, CO3-2

d. NO3-1, CH3+, CO3-2

a. sp, sp2 and sp3 respectively
b. sp, sp3 and sp2 respectively
c. sp2, sp3 and sp respectively
d. sp2, sp and sp3 respectively

a. sp, sp2 and sp3 respectively

a. The carbon atom in CO2 is sp2 hybridized
b. The nitrogen atom in NH3 is sp3 hybridized
c. sp hybrid orbitals lie at 180o to each other
d. sp2 hybrid orbitals are coplanar, and at 120o to each other.

a. The carbon atom in CO2 is sp2 hybridized

a. Both are polar
b. Both have sp-hybridized carbon
c. Both are acidic
d. Both contain four lone pairs

b. Both have sp-hybridized carbon

Similar Posts

Leave a Reply

Your email address will not be published. Required fields are marked *