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NEET Chemistry MCQ
Molecular Bonding Mcq Neet Chemistry
Quiz 7
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Q.1
A hybrid orbital form s and p orbital can contribute to
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a) σ bond only
0%
b) π bond only
0%
c) either σ or π bond
0%
d) cannot be predicted
Explanation
s orbitals can not undergo side ways overlapping hence only σ bonding Answer: (a)
Q.2
Which orbital is used by oxygen atom to form a sigma bond with other oxygen atom in O2 molecule
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a)pure p-orbital
0%
b) sp2 hybrid orbital
0%
c)sp3 hybrid orbital
0%
d)sp-hybrid orbital
Explanation
Since two Oxygen are connected by double thus hybridization of O is sp2Answer: (b)
Q.3
the hybrid state of b in BF4- is
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a) sp2
0%
b) sp
0%
c)sp3
0%
d)no specific
Explanation
In BF4- number of valence electrons=3 +4×7=+1=32 Now 32÷8=4(Q) + 0 (R) since Q+ R=4+0=4 sp3 hybridization Answer: (c)
Q.4
C2H2 is iso structural with
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a) H2O2
0%
b) NO2
0%
c)SnCl2
0%
d)CO2
Explanation
C2H2 and CO2 both are linear Answer:(d)
Q.5
The weakest bond among the following is
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a) ionic
0%
b) covalent
0%
c) metallic
0%
d) hydrogen
Explanation
Answer: (d)
Q.6
Which of the following structure is most expected for molecule XeOF4
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a)Tetrahedral
0%
b) Square pyramidal
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c)Square planar
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d)Octahedral
Explanation
Total number of valence electrons in XeOF4 are=8+ 6 +4 ×7=42 Now 42 ÷8=5(Q1) + 2(R1) and 2 ÷2=1 Sum of all quotients and final reminder=5+1+0=6 State of hybridization of X in XeOF4 is sp3d2 Since there are six orbitals and five atoms , therefore one orbital will be occupied by loan pair of electrons. Thus molecule have square pyamidal geometry Answer: (b)
Q.7
Molecular shapes of SF4, CF4 and XeF4 are
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a) same with 2,0 and 1 lone pair of electrons respectively
0%
b) same with 1,1 and 1 lone pair of electrons respectively
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c)different with 1,0 and 2 lone pairs of electrons respectively
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d)different with 1,1 and 1 lone pairs of electrons respectively
Explanation
SF4 (sp3d , trigonal bipyramidal with one equatorial position lone pair)CF4 (sp3, tetra hedral, no loan pair)XeF4 (sp3d2, square planar, two loan pairs)Answer: (c)
Q.8
The hybrid state of S in SO3 is similar to that of
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a) C in C2H2
0%
b) C in C2H4
0%
c)C in CH4
0%
d)C in CO2
Explanation
Number of valence electrons in SO3 are =6 + 3 × 6=24 Now 24 ÷ 8=3(Q) SO3 is sp2 hybridized While C in C2H4 sp2 hybridized Answer:(b)
Q.9
Two hybrid orbitals have a bond angle of 120°,. The percentage of s character in the hybrid orbital is nearly
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a) 25%
0%
b) 33%
0%
c) 50%
0%
d) 66%
Explanation
sp2 Hybrid orbitals are inclined at an angle of 120° and contain 33% s-character Answer: (b)
Q.10
The d orbital involved in sp3d hybridization is
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a)dxy
0%
b) dz2
0%
c)dx2 - y2
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d)dzx
Explanation
Answer: (b)
Q.11
In a change from PCl3 → PCl5, the hybrid state of P changes from
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a) sp2 to sp3
0%
b) sp3 to sp2
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c)sp3 to sp3d
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d)sp3 to dsp2
Explanation
Total number of valence electrons in PCl3 =5+ 3 ×7=26 NOw 26÷8=3(q1 + 2 (R1) and R1 ÷ 2=1(Q2 + 0 (R2) By adding all quotients and final remainder=3+1+0=4 ∴ Hybridization in of P in PCl3 is sp3 in PCl5 Total number of valence electrons=5 +5×7=40 Now 40÷8=5 (Q) + 5(R) Thus hybridization is sp3d option 'c' is correctAnswer: (c)
Q.12
In OF2, Oxygen has hybridization of
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a) sp
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b) sp2
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c)sp3
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d)none of these
Explanation
Total number of valence electrons in OF2=6 + 7×2=20Now 20 ÷8=2(Q1 + 4(R1)Now 4 ÷ 2=2(Q2 + 0 (R2Q1 + Q2=2+2=4 Thus Hybridization is sp3 Answer:(c)
Q.13
Which one is appreciably soluble in water
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a) CS2
0%
b) C2H5OH
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c) CCl4
0%
d) CHCl3
Explanation
C2H5OH forms H bonds with water hence it is soluble in water Answer: (b)
Q.14
The geometrical arrangement and shape of I3- are respectively
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a)Trigonal bipyramidal geometry, linear shape
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b) Hexagonal geometry, T -sphape
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c)Triangular planar geometry, triangular shape
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d)Tetrahedral geometry, pyramidal shape
Explanation
Total valence electrons in I3- are=7 ×3 + 1=22Now 22 ÷ 8=2(Q1 + 6(R1)And R1 ÷ 2=6 ÷ 2=3(Q2 + 0 (R2)Sum of all quotients and final reminder=2+3+0=5 ∴ Hybridization of I is sp3d Thus I3- has trigonal bipyramidal geometry with linear shapeAnswer: (a)
Q.15
In which of the following molecules/ions, the central atom does not involved a d-orbital in the hybridization process
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a) I3-
0%
b) SF6
0%
c)[Cu(NH3)4]2+
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d)MnO4-
Explanation
Number of valance electrons in MnO4-are=7+ 4 × 6 + 1=32 Now 32 ÷ 8=4(Q)+ 0(R)sp3 hybridization , thus d orbital is not involvedAnswer: (d)
Q.16
Pair of molecules having identical geometry is
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a) BF3, NH3
0%
b) BF3, AlF3
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c)BeF2, H2O
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d)BCl3, PCl3
Explanation
BF3 and AlF3 both are planar triangular Answer:(b)
Q.17
Which of the following will be planar trigonal?
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a) PCl3
0%
b) NH3
0%
c)ClF3
0%
d)AlCl3
Explanation
Hybridization state in PCl3 is sp3Hybridization state in NH3 is sp3 Hybridization state in ClF3 is sp3d While AlCl3 hybridization is sp2 Thus option 'd' is correct We can calculate the hybridization state using quotients and remainder systemAnswer: (d)
Q.18
Compound in which central atom assumes sp3d hybridization is
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a) SO3
0%
b) PCl5
0%
c)SO2
0%
d)PCl3
Explanation
Total number of valence electron in PCl5=5+5×7=40 Now 40÷8=5(Q) + 0(R) ∴ Hybridization of P in PCl5 is sp3d Answer:(b)
Q.19
Hydrogen fluoride is a liquid unlike other halides because
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a) HF molecules associate due to hydrogen bonding
0%
b) F2 is highly reactive
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c) HF is the weakest acid of all halogens
0%
d) Fluorine atom is the smallest of all halogens
Explanation
Answer: (a)
Q.20
Which of the following will be least polar
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a)N - H
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b) C - H
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c)O - H
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d)H - F
Explanation
electro negativity between H and C is minimumAnswer: (b)
Q.21
hich of the following compound has bond angle as nearly as 90°
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a) NH3
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b) H2S
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c)H2O
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d)CH4
Explanation
In H2S the central atom is SP3 hybridized but due to low electronegative of S and bond pair - bond pair repulsions, the bond angel is near to 90°Answer: (b)
Q.22
Which of the following molecules has largest bond angle
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a) H2O
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b) NH3
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c)CH4
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d)CO2
Explanation
CO2 has sp hybridization and hence has the largest bond angle of 180° Answer:(d)
Q.23
If the central atom in a certain molecule has two lone pairs and three bond pairs, the shape of the molecule could be
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a) T shape
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b) trigonal planar
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c) Trigonal bipyramidal
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d) distorted tetrahedral
Explanation
Answer: (a)
Q.24
Which of the following bonds is strongest
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a)F - F
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b) I - I
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c)Cl - Cl
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d)Br - Br
Explanation
Bond strength decreases with increase in the size of halogenbut due to interelectronic repulsion in F due to smaller size bond strength is less than chlorineAnswer: (c)
Q.25
Which of the following molecules will have polar bonds but zero dipole moment
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a) O2
0%
b) CHCl3
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c)CF4
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d)none of these
Explanation
O2 is non polarCHCl3 is polar but dipole moment is non zero as dipole do not cancel each otherCF4 have polar bond C-Cl but have regular tetra hedral geometry thus cancels dipole momentAnswer: (c)
Q.26
The pair of molecules having identical geometry is
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a)BCl3 , PCl3
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b) BF3, NF3
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c)CCl4 , CH4
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d)CHCl3, CH3Cl
Explanation
BCl3 : planar but PCl3 : pyramidalBF3 : : planar but NF3 : pyramidalCHCl3 and CH3Cl have same sp3 hybridization but differs due to three polar bonding in CHCl3 CCl4 and CH4 have four bond pairs each and hence tetrahedral geometry Answer:(c)
Q.27
Which of the following has one loan pair of electrons on central atom
0%
a) H2
0%
b) CH4
0%
c) NH4-
0%
d) NCl3
Explanation
Answer: (d)
Q.28
From among the following triatomic species, the least angle around the central atom is in
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a)O3
0%
b) I3-
0%
c)NO2-
0%
d)H2S
Explanation
O3 have sp2 hybridization hence bond angle is 120≥I3- have sp3d It has linear shape with three equatorial positions occupied by lone pairs . As such bond angle is 180° I NO2- have sp2 hybridization hence bond angle around 120≥ In H2S the central atom is SP3 hybridized but due to low electronegative of S and bond pair - bond pair repulsions, the bond angle is near to 901° Thus H2S have least bond angle from given species Answer: (d)
Q.29
Which of the following compound has dipole moment approximately equal to that of chlorobenzene
0%
a) o-Dichloro benzene
0%
b) m-Dichlorobenzene
0%
c)p-dichlorobenzne
0%
d)p-Chloronitrobenzene
Explanation
Resultant will be obviously nearly equal to that of chlorobenzneAnswer: (b)
Q.30
The bond angle in H2S is
0%
a) > NH3
0%
b) same as BeCl2
0%
c)> H2Se < H2
0%
d)same as in CH4
Explanation
in a group bond angles decreases as the electronegativity of the central atom decreases H2O > H2S > H2Sc Answer:(c)
0 h : 0 m : 1 s
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