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

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.

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 O^₂- have an unpaired electron.

For square planar shape, we need all the orbitals lyng in a single plane. In option b s, p_x, p_y and d_x2-y2 all lie in a single plane. This will give us a square plane shape.

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.

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

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

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
 

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⁺

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.

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

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

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

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.
 

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.

There is a triple bond , between two carbons

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

O₂ has two unpaired electron in the doubly degenerate 1πg* orbital. In O₂^2-, 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.

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.

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

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.

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.
 

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

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

Since, Be belongs to the 2nd group, so its size is the least. Therefore its polarisability power will be most. Mg is in 3^rd 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 .

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.

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

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.

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).

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

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.