NEETprep Bank NEET Chemistry Chemical Bonding and Molecular Structure Questions Solved


Which of the following would have a permanent dipole moment ?

(a) BF3

(b) SiF4

(c) SF4

(d) XeF4

Concept Videos :-

#15 | Dipole Moment

Concept Questions :-

(c) 

SF4 have μ >0 It has permanent dipole moment

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In the case of alkali metals, the covalent character decreases in the order

(a) MCl > MI > MBr > MF

(b) MF > MCl >  MBr > M

(c)MF > MCl > MI > MBr

(d) MI > MBr > MCl > MF

Concept Videos :-

#15 | Dipole Moment

Concept Questions :-

(d) According to Fajan's rule,

Covalent character 1size of cationsize of anion

In the given options, cation is same but anions are different. Among halogens the order of size is

                      F < Cl < Br < I

Order of covalent character is

                   MI > MBr > MCl > MF

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Which of the following two are isostructural? 

(a)XeF2, and IF_2

(b)NH3, and BF3

(c)CO32-, and SO32-

(d)PCl5, and ICl5

Concept Videos :-

#1 | Introduction

Concept Questions :-

Types of Bonding

(a) Isostructural Compounds having same structure and same hybridisation are known as isostructural species. e.g. XeF2 and IF_2   are sp3d hybridised and both have linear shape.

F-I-F    F-Xe-F

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Which one of the following species does not exist under normal conditions ?

(a) Be2+

(b) Be2

(c) B2

(d) Li2

 

Concept Videos :-

#17 | MOT| Linear Combination of Atomic Orbitals
#18 | MOT| Formation of Bonding & Anti|Bonding Orbitals
#19 | MOT| Molecular Orbital Diagram

Concept Questions :-

(b) Molecules with zero bond order, do not exist.

According to molecular orbital theory,

(a) Be2+ (4+4-1=7)                              = σ1s2, σ*1s2 , σ2s2,  σ*2s1       Bond order (BO) = 4-32 = 0.5(b) Be2(4+4=8)= σ1s2, σ*1s2 , σ2s2,  σ*2s2                              BO = 4-42=0(c) B2(5+5=10) =  σ1s2, σ*1s2 , σ2s2,  σ*2s2, π2px1 π2py1(d) Li2 (3+3=6) =  σ1s2, σ*1s2 , σ2s2                             BO = 4-22=1Thus, Be2 does not exist under normal conditions.

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Which one of the following is not paramagnetic ?

(a) NO

(b) N2+

(c) CO

(d) O2-

Concept Videos :-

#17 | MOT| Linear Combination of Atomic Orbitals
#18 | MOT| Formation of Bonding & Anti|Bonding Orbitals
#19 | MOT| Molecular Orbital Diagram

Concept Questions :-

(c) Paramagnetic character is shown by those atoms or molecules which have unpaired electrons.

In the given compounds CO is not paramagnetic since, it does not have unpaired electrons. The configuration of CO molecule is 

CO(14) = σ1s2, σ*1s2, σ2s2, σ*2s2, σ2px2, π2py2 π2pz2 

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In a regular octahedral molecule, MX6 the number of X-M-X bonds at 180°is

(a) 3

(b) 2

(c) 6

(d) 4

Concept Videos :-

#7 | Hybridization in case of Carbon| Intro
#8 | Hybridization in case of Carbon| sp3
#9 | Hybridization in case of Carbon| sp2
#10 | Hybridization in case of Carbon| sp
#11 | Hybridization in case of other Molecules

Concept Questions :-

Hybridisation

(a) In octahedral structure MX6, the six hybrid orbitals (sp3d2) are directed towards the corners of a regular octahedral with an angle of 90°. According to following structure of MX6, the number of X-M-X bonds at 180° must be three. 

                              

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The angular shape of ozone molecule (O3) consists of 

(a) 1 sigma and 2 pi-bonds

(b) 2 sigma and 2 pi-bonds

(c) 1 sigma and 1 pi-bonds

(d) 2 sigma and 1 pi-bonds

Concept Videos :-

#4 | Covalency

Concept Questions :-

Covalent Bonding

(d) In case of single bond, there is only one s-bond in case of double bond, there is one and one p-bonds. Thus, angular shape of ozone (O3) contains 2and 1 p-bonds as

                        

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Which of the following is not isostructural with SiCl4

(a) SCl4     

(b) SO42-

(c) PO43-

(d) NH4+

Concept Videos :-

#7 | Hybridization in case of Carbon| Intro
#8 | Hybridization in case of Carbon| sp3
#9 | Hybridization in case of Carbon| sp2
#10 | Hybridization in case of Carbon| sp
#11 | Hybridization in case of other Molecules

Concept Questions :-

Hybridisation

(a) SCl4 (Seesaw structure)  is not isostructural with  Sicl4 because shows tetrahedral structure due to involvement of repulsion between lone pair and bond pair of electrons.

SO42- shows tetrahedral structure due to sp3 hybridisation.

PO43- shows tetrahedral structure due to sp3 hybridisation.

NH4shows tetrahedral structure due to sp3 hybridisation

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In which of the following molecules are all the bonds not equal ?

[2006]

(a) ClF3

(b) BF3

(c) AlF3

(d) NF3

Concept Videos :-

#7 | Hybridization in case of Carbon| Intro
#8 | Hybridization in case of Carbon| sp3
#9 | Hybridization in case of Carbon| sp2
#10 | Hybridization in case of Carbon| sp
#11 | Hybridization in case of other Molecules

Concept Questions :-

Hybridisation

(a) ClF3 all bonds are not equal due to trigonal-bipyramidal (sp3d hybridisation) geometry of ClF3 molecule.

BF3 and AlF3 show trigonal symmetric structure due to sp2 hybridisation.

NF3 shows pyramidal geometry due to sp3 hybridisation.

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The correct order of increasing bond angles in the following species is

(a)Cl2O < ClO2ClO2-

(b) ClO2 < Cl2O < ClO2-

(c) Cl2O < ClO2- < ClO2  

(d) ClO2- < ClO2 < Cl2

Concept Videos :-

#4 | Covalency

Concept Questions :-

Covalent Bonding

(c) According to VESPR theory repulsion order

  lp - lp > lp - bp > bp - bp

As the number of lone pairs of electrons increases, bond angle decreases due to repulsion between lp-lp, Moreover, as the electronegativity of central atom decreases, bond angle decreases.

Hence, the order of bond angle is

(Cl is less electronegative as compared to O).

 

clo2- =111,cl2O ,110, Clo2 =117

 

explanation is given, its can be explained on the multiple concept like valence electron,hybridisation,electro negativity, but fact is ,its a experiemental science , so we need to develop theory to explain the observation, this is basis of evolution of VBT to CFT.

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