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Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - JEE MCQ


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30 Questions MCQ Test - Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure

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Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 1

In which of the following species is the underlined "carbon" having sp3-hybridisation?

   [AIEEE-2002]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 1





From this figure, we can clearly see that in CH3CH2OH, we have no pi bond. While other compounds have pi bond. For carbon, only sp3 hybridization has no pi bond. So CH3CH2OH has sp3 hybridized carbon atom.

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 2

Which of the following is paramagnetic? 

[NEET 2013]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 2

Paramagnetic species are those who have an unpaired electron. For CO, we have 14 electrons. For O2- we have 17 electrons. For CN- we have 14 electrons and for NO+ we have 14 electrons. So, we can see that only O2- have an unpaired electron.

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Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 3

 A square planar complex is formed by hybridisation of which atomic orbital? 

[AIEEE-2002]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 3

For square planar shape, we need all the orbitals lyng in a single plane. In option b s, px, py and dx2-y2 all lie in a single plane. This will give us a square plane shape.

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 4

The reason for double helical structure of DNA is operation of: 

[AIEEE-2003]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 4

Due to H-bonding, DNA structures itself in a double helix. The two polynucleotide chains or strands of DNA are linked up by hydrogen bonding between the nitrogenous base molecules of their nucleotide monomers.

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 5

Which one of the following pairs of molecules will have permanent dipole moments for both members ?

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 5

In the given set, NO2 and O3 are non-linear. O3 and NO2 both are bent shaped molecules. So, they will have permanent dipole moments.  CO2 is a linear molecule with sp hybridisation. So it won't have any dipole moment. SiF4 has tetrahedral shape but all bonds are identical and so that cancel out each other resulting in net dipole moment - 0

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 6

The pair of species having identical shapes for molecules of both species is:

[AIEEE-2003]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 6

Both XeF2 & CO2 have linear structures. Both are actually linear Tri-atomic molecules.

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 7

The correct order of bond angles (smallest first) in H2S, NH3, BF3 and SiH4 is:

[AIEEE-2004]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 7

For H2S, bond angle is around 90°. SiH4 is sp3 hybridized with no lone pair, so the bond angle is 109.28’. In NH3, due to the presence of lone pair on central N atom, bond angle decreases from 109°28’. BF3 is sp2 hybridized, so its bond angle is 120.So the order is;-
   BF3 > SiH4 > NH3 > H2S
  120   109     107     92.5
 

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 8

 The bond order in NO is 2.5 while that in NO+ is 3. Which of the following statements is true for these two species?     

[AIEEE-2004]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 8

Bond length is inversely proportional to bond -order bond order in NO+  = 3, NO = 2.5Thus bond length in NO> NO+.Bond length is inversely proportional to bond -order bond order in NO+  = 3, NO = 2.5Thus bond length in NO> NO+

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 9

The states of hybridization of boron and oxygen atoms in boric acid (H3BO3) are respectively:

[AIEEE-2004]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 9

According to me the correct option is b and not say because in boric acid Boron is sp2 hybridised due to an filled orbital while oxygen is sp3 hybridised.

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 10

Which one of the following has the regular tetrahedral structure? 

[AIEEE-2004]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 10


From the above figure, we can see that BF4- has regular tetrahedral structure.

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 11

 The maximum number of 90° angles between bond pair-bond pair of electrons is observed in:

[AIEEE-2004]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 11


Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 12

 Beryllium and aluminium exhibit many properties which are similar. But, the two elements differ in:

[AIEEE-2004]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 12

Be (Z = 4) has maximum covalency of 4 while Al (Z = 13) has maximum covalency of 6.

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 13

 Lattice energy of an ionic compound depends upon: 

[AIEEE-2005]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 13

Lattice energy of a compound depends on both charge and size with charge the first priority.

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 14

The molecular shapes of SF4, CF4 and XeF4 are:

 [AIEEE-2005]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 14

Hybridisation of all compounds
Hybridisation = valence electrons +sigma bonds/2
for SF4 = 6+4/2 =5 which means sp3d hybridisation , Now to get lone pairs subtract bonds from this number which is 5-4 =1 that's why 1 lone pair on SF4
for CF4 = 4+4/2 = 4, sp3 hybridisation
lone pairs = 4-4 =0
for XeF4 = 8+4/2 = 6, sp3d2 hybridisation
lone pairs = 6-4 =2
SF4 has sea-saw shape where one lone pair is in triangular plane.
CF4 is tetrahedral in shape because of sp3 hybridisation
XeF4 has square planar structure and 2 lone pairs are up & down of square plane.
 

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 15

Of the following sets which one does not contain isoelectronic species? 

[AIEEE-2005]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 15

In option a, b and d all  species have the same no of electrons.(There are 50 electrons in option a, 14 electrons in option b and 32 electrons in option d)However, SO3-2 has 42 electrons while CO3-2 and NO3- have 32 electrons. SO option c doesn’t contain isoelectronic species.

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 16

 The number and type of bond between two carbon atom in calcium carbide are:

[AIEEE-2005]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 16


There is a triple bond , between two carbons

Among three bonds one one is sigma, bond and the remaining two are π bond.

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 17

Which of the following molecules/ions does not contain unpaired electron ?

[AIEEE-2006]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 17

O2 has two unpaired electron in the doubly degenerate 1πg* orbital. In O22-, the two electrons are accommodated in the 1ng* orbital meaning the orbitals are full and hence all the electrons are paired. So no unpaired electrons are present.

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 18

Among the following mixtures, dipole-dipole as the major interaction, is present in:

[AIEEE-2006]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 18

Acetonitrile and Acetone both are polar molecules hence dipole dipole interaction exist between them between KCL and water Ion dipole interaction is found and in Benzene ethanol and benzene carbon tetrachloride dispersion force is present.

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 19

 A metal, M foms chlorides in its +2 and +4 oxidation states. Which of the following statements about these chlorides is correct? 

[AIEEE-2006]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 19

Because according to Fajan's rule, compounds of metal in lower oxidation state are more ionic than those in a higher state.

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 20

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

[AIEEE-2006]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 20

Out of the four, SF4 will not have equal bonds, because of it structure .Trigonal bipyramidal. Sulfur in SF4 is in the formal +4 oxidation state. sulfur has total of six valence electrons, two form a lone pair. The fluorine-sulfur-fluorine bond angles are 120° in the trigonal plane and 90° to the two fluorine atoms above and below the plane.

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 21

The decreasing value of bond angles from NH3 (106)° to SbH3 (101)° down group-15 of the periodic table is due to: 

[AIEEE-2006]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 21

With the increase in electronegativity, bond pairs move farther and farther away from the central atom. As a result bp-bp repulsion decreases and the angle decreases accordingly.
 

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 22

In which of the following ionizion processes, the bond order has increased and the magnetic behaviour has changed

 [AIEEE-2007]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 22

N2: bond order 3, paramagnetic
N2-: bond order 2.5, paramagnetic
C2: bond order 2, diamagnetic
C2+: bond order 1.5, paramagnetic
NO: bond order 2.5, paramagnetic
NO+: bond order 3, paramagnetic
O2: bond order 2, paramagnetic
O2+: bond order 2.5, paramagnetic
Therefore, option a is correct

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 23

Which of the following hydrogen bonds is the strongest  

[AIEEE-2007]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 23

Since F is most electronegative,so H bon by its hydride will be most strong.

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 24

The charge/size ratio of a cation determines its polarzing power. Which one of the following sequences represents the increasing order of the polarizing power of the cationic species, K+, Ca+2 Mg+2, Be+2    

[AIEEE-2007]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 24

Since, Be belongs to the 2nd group, so its size is the least. Therefore its polarisability power will be most. Mg is in 3rd period and its polarisability power will be less than Be. FO rCA and K, both belong to the same period but Ca has more charge. SO its polarisability will be more than k .

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 25

Give reason carbon oxygen bond lengths in formic acid are 1.23 Å & 1.36Å and both the carbon oxygen bonds in sodium formate have the same value i.e. 1.27 Å                [JEE' 88]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 25

Carbon-oxygen bond lengths in formic acid are 1.23 Å and 1.36 Å, but both the carbon-oxygen bonds in sodium formate have the same value, 1.27 Å. In formic acid the carbon-oxygen bond lengths are different due to hydrogen bonding.

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 26

In allene (C3H4), the type(s) of hybridisation of the carbon atoms is (are) 

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 26

The structure of Allene is H2C=C=CH2. We can see that there is a sp hybridised carbon atom in the middle and 2 sp2 hybridised carbon atom on extreme sides. So the correct answer is sp and sp2

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 27

Mark True or False: BeCl2 can be easily hydrolyed               [JEE'99]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 27

BeCl2 is hydrolysed due to high polarizing power and presence of vacant p-orbitals in Be-atom Chromic acid. Therefore CrO3 is acid anhydride of H2Cr2O4.

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 28

o-hydroxybenzaldehyde is a liquid at room temperature, while p-hydroxybenzaldehyde is a high melting solid due to           [JEE' 99]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 28

o-hydroxybenzadehyde is a liquid at room temperature while p-hydroxybenzaldehyde is high melting solid because there is intramolecular H-bonding in o-hydroxybenzaldehyde (weak) while intermolecular H-bonding in p-isomer (strong).

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 29

N2, O2, F2, Cl2 in increasing order of bond dissociation energy.                   [JEE' 88]

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 29

The correct answer is F2<Cl2<O2<N2
Bond dissociation energy of F2 is less than Cl2
F2 < Cl2
Bond dissociation energies of O2 (O=O) and N2(N=N) are more than Cl2 or F2 which contains only single bonds. Out of N2 (with a triple bond) and O2 (with a double bond), bond dissociation energy of N2 is more than O2. Therefore correct order is : F2<Cl2<O2<N2

Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 30

 Arrange in increasing order of dipole moment. [JEE'96]

(I) Toluene, (II) m-dichcorobenzene (III) O-dichlorobenzene (IV) P-dichlorobenzene

Detailed Solution for Test: JEE Previous Year Questions- Chemical Bonding & Molecular Structure - Question 30

The increasing order of dipole moment is IV<I<II<III.

o−, m−, p− derivates has α=60o, 120o and 180o and thus, resultant vector has zero dipole moment in p-derivative because of symmetrical structure. Also dipole moment of m-dichlorobenzene is more than toluene. o− and m−dichlorobenzene have higher dipole moments than toluene due to high electronegativity of chlorine than CH3​ group. Further, to the o-dichlorobenzene has higher dipole moment due to lower bond angle than the m-isomer.

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