Q: Which of the following gives the correct increasing order of the number of lone pairs on the central atom?

IF₇ < IF₅ < ClF₃ < XeF₂. Central-atom lone pairs increase as IF7 (0) < IF5 (1) < ClF3 (2) < XeF2 (3), which is option 0. The stored ordering misplaces XeF2 and IF5.

Q: Among the molecules CH₄, NH₃, and H₂O, which statement is incorrect?

The H–O–H angle in H₂O is greater than the H–C–H angle in CH₄. Bond angles: CH4 109.5, NH3 107, H2O 104.5. So 'H-O-H angle > H-C-H angle' (option 1) is the incorrect statement. The stored choice (all angles > 90 degrees) is actually true.

Question 1

Although the electronegativity gap between N and F is larger than that between N and H, ammonia has a dipole moment of 1.5 D while nitrogen trifluoride has only 0.2 D. The reason is that

Accepted Answer

in NH₃, the atomic dipole and bond dipole point in the same direction, whereas in NF₃ they point in opposite directions. In ammonia (NH₃), the dipoles from the N-H bonds add up in the same direction, resulting in a significant overall dipole moment. In contrast, in nitrogen trifluoride (NF₃), the dipoles from the N-F bonds are directed opposite to the dipole from the nitrogen atom, which cancels out much of the overall dipole moment, leading to a much smaller value.

Question 2

When N₂ is converted into N₂⁺, the dissociation energy of the N–N bond ______, and when O₂ is converted into O₂⁺, the dissociation energy of the O–O bond ______.

Accepted Answer

decreases, increases. N2 -> N2+ removes a bonding electron, dropping bond order 3 -> 2.5, so the bond energy DECREASES. O2 -> O2+ removes an antibonding (pi*) electron, raising bond order 2 -> 2.5, so it INCREASES. Hence 'decreases, increases'.

Question 3

From the ions listed below, which pair has geometries that can be accounted for by the same type of orbital hybridization? NO₂⁻, NO₃⁻, NH₂⁻, NH₄⁺, SCN⁻

Accepted Answer

NO₂⁻ and NO₃⁻. Both NO₂⁻ and NO₃⁻ exhibit geometries that can be explained by sp² hybridization. In NO₂⁻, the central nitrogen atom forms one double bond and one single bond, leading to a bent shape, while in NO₃⁻, the nitrogen forms three equivalent bonds with oxygen, resulting in a trigonal planar geometry, both of which are consistent with sp² hybridization.

Question 4

Atoms A and B have electronegativities of 1.20 and 4.0, respectively. What is the percentage ionic character of the A–B bond?

Accepted Answer

72.24%. The percentage ionic character of a bond can be estimated using the difference in electronegativities between the two atoms. In this case, the significant difference of 2.8 between atoms A and B indicates a high ionic character, which is calculated to be approximately 72.24%.

Question 5

In the phosphate ion, PO₄³⁻, what is the formal charge on an oxygen atom that is singly bonded to phosphorus in a P–O bond?

Accepted Answer

−1. The formal charge on an oxygen atom in a singly bonded state can be calculated using the formula: formal charge = valence electrons - (non-bonding electrons + 0.5 * bonding electrons). For the singly bonded oxygen in PO₄³⁻, it has 6 valence electrons, 6 non-bonding electrons, and 1 bonding electron from the P–O bond, resulting in a formal charge of -1.

Question 6

Among the following species, which one has no unpaired electrons?

Accepted Answer

O₂²⁻. O₂²⁻ has all its electrons paired, resulting in a stable electronic configuration with no unpaired electrons, unlike the other species listed which have unpaired electrons due to their molecular orbital arrangements.

Question 7

Which pair of ions has the same total number of electrons and the same molecular geometry?

Accepted Answer

ClO₃⁻ and SO₃²⁻. ClO3- and SO3- (each 42 electrons) are both trigonal pyramidal (AX3E), so they share electron count and geometry. The stored pair CO3^2- (32 e, planar) and SO3^2- (42 e, pyramidal) matches on neither.

Question 8

Which of the following gives the correct increasing order of the number of lone pairs on the central atom?

Accepted Answer

IF₇ < IF₅ < ClF₃ < XeF₂. Central-atom lone pairs increase as IF7 (0) < IF5 (1) < ClF3 (2) < XeF2 (3), which is option 0. The stored ordering misplaces XeF2 and IF5.

Question 9

Among the molecules CH₄, NH₃, and H₂O, which statement is incorrect?

Accepted Answer

The H–O–H angle in H₂O is greater than the H–C–H angle in CH₄. Bond angles: CH4 109.5, NH3 107, H2O 104.5. So 'H-O-H angle > H-C-H angle' (option 1) is the incorrect statement. The stored choice (all angles > 90 degrees) is actually true.

Question 10

Choose the sequence that correctly represents the decreasing order of repulsion between electron pairs:

Accepted Answer

lone pair - lone pair > lone pair - bond pair > bond pair - bond pair. The correct option reflects the fact that lone pairs of electrons occupy more space and repel each other more strongly than bond pairs, leading to the order of repulsion being highest for lone pair-lone pair interactions, followed by lone pair-bond pair interactions, and lowest for bond pair-bond pair interactions.

Question 11

Which pair of species has an interatomic bond angle of 109° 28′?

Accepted Answer

NH₄⁺, BF₄⁻. The interatomic bond angle of 109° 28′ is characteristic of tetrahedral molecular geometry, which is exhibited by both NH₄⁺ and BF₄⁻ due to their four bonding pairs of electrons and no lone pairs.

Question 12

For the molecules XeF₂, XeF₄, and XeF₆, what are the respective numbers of lone pairs present on the xenon atom?

Accepted Answer

3, 2, 1. The correct option indicates that xenon in XeF₂ has three lone pairs, in XeF₄ it has two lone pairs, and in XeF₆ it has one lone pair. This is consistent with the molecular geometry and valence shell electron pair repulsion theory, which accounts for the bonding and lone pairs around the xenon atom.

Question 13

The hybridisation of the underlined atom is altered in which of the following transformations?

Accepted Answer

AlH₃ is converted into AlH₄⁻. AlH3 (sp2) -> AlH4- (sp3) changes the Al hybridisation, but H2O -> H3O+ stays sp3 and NH3 -> NH4+ stays sp3. So only the first transformation alters hybridisation, not 'all of the above'.

Question 14

The fall in bond angle from NH₃ (106°) to SbH₃ (101°) down group 15 of the periodic table is mainly because of

Accepted Answer

a decrease in electronegativity. The bond angle decreases from NH₃ to SbH₃ primarily due to the decrease in electronegativity of the central atom, which leads to a weaker bond character and less effective lone pair repulsion, resulting in a smaller bond angle.

Question 15

Which of the following pairs of species has identical bond order?

Accepted Answer

CN⁻ and NO⁺. CN⁻ and NO⁺ both have a bond order of 3, as calculated from their molecular orbital diagrams, indicating that they share the same stability and strength of the bond.

Question 16

When N₂ and O₂ each lose one electron to form N₂⁺ and O₂⁺, respectively, which statement is incorrect?

Accepted Answer

N₂⁺ turns diamagnetic. N₂⁺ is diamagnetic because it has all of its electrons paired, resulting in no net magnetic moment. This occurs after the removal of an electron from the bonding molecular orbital, which does not affect the pairing of the remaining electrons.

Question 17

In BF₃, the B–F bond dissociation energy is 646 kJ mol⁻¹, while in CF₄ the C–F bond dissociation energy is 515 kJ mol⁻¹. What best explains why the B–F bond in BF₃ has the higher bond dissociation energy?

Accepted Answer

There is appreciable pπ–pπ overlap between B and F in BF₃, whereas such overlap is not possible between C and F in CF₄.. The presence of significant pπ–pπ overlap between boron and fluorine in BF₃ enhances the bond strength, resulting in a higher bond dissociation energy compared to CF₄, where such overlap is not feasible due to the larger size of carbon.

Question 18

From the salts LiCl, RbCl, BeCl₂ and MgCl₂, which pair represents the compound with the highest ionic character and the one with the lowest ionic character, respectively?

Accepted Answer

RbCl and BeCl₂. By Fajans' rules the large low-charge Rb+ gives RbCl the highest ionic character, while the small highly-polarising Be2+ makes BeCl2 the most covalent (lowest ionic character). So the pair is RbCl and BeCl2, not LiCl and RbCl.

Question 19

Bond order is often used as a measure of the stability of a molecule. Although H₂, Li₂ and B₂ each have the same bond order, they do not possess equal stability. Which of the following gives their correct order of stability?

Accepted Answer

H₂ > B₂ > Li₂. The stability of a molecule is influenced not only by bond order but also by factors such as bond length and the presence of antibonding orbitals. H₂ is the most stable due to its strong single bond and lack of antibonding electrons, while B₂ has a lower stability than H₂ because it has one antibonding electron, and Li₂ is the least stable due to its longer bond length and presence of two antibonding electrons.

Question 20

Which of the following species shows paramagnetism?

Accepted Answer

NO. NO is paramagnetic because it has an unpaired electron in its molecular orbital configuration, which allows it to be attracted to a magnetic field. In contrast, the other species listed either have all paired electrons or do not exhibit this property.

Question 21

Which pair of compounds among LiCl, RbCl, BeCl2, and MgCl2 shows the highest and the lowest ionic character, respectively?

Accepted Answer

RbCl and BeCl2. RbCl exhibits high ionic character due to the large size and low charge density of Rb+, which leads to weaker lattice energy, while BeCl2 has a covalent character because of the small size and high charge density of Be2+, resulting in significant polarization of the chloride ions.

Question 22

Graphite is a soft, slippery solid and has an unusually high melting point. What accounts for this unusual behavior?

Accepted Answer

Its carbon atoms are arranged in extended sheets of strongly bonded rings, with weak forces between the sheets. Graphite is soft/slippery yet high-melting because carbon atoms form strongly bonded hexagonal sheets with only weak van der Waals forces between layers. It is crystalline, so the stored statement is wrong.

Question 23

Given that the ionic radius of Li+ is 60 pm and that of F- is 136 pm, what crystal structure does LiF adopt and what is its coordination number?

Accepted Answer

Same as NaCl, coordination number 6. Li+/F- radius ratio = 60/136 = 0.44, which lies in the 0.414-0.732 octahedral range, so LiF takes the NaCl (rock-salt) structure with coordination number 6, not the antifluorite type.

Question 24

Which of the following is present as a covalent crystal in the solid state?

Accepted Answer

Silicon. Silicon is a covalent (network) solid like diamond, whereas iodine, sulphur and phosphorus form molecular crystals. So the covalent crystal is silicon, not iodine.

Question 25

Given the ionic radii Na+ = 0.095 nm, Cl− = 0.181 nm, Zn2+ = 0.074 nm, S2− = 0.184 nm, Ti4+ = 0.068 nm, O2− = 0.140 nm, and Cs+ = 0.169 nm, identify the correct statement(s) using the radius-ratio criterion.

Accepted Answer

All of the above.. All the statements are correct based on the radius-ratio criterion, which helps determine the coordination number and arrangement of ions in a crystal lattice. Na+ fits well in octahedral voids due to its size relative to Cl−, Zn2+ is small enough to occupy tetrahedral voids, and Cs+ can be arranged in a simple cubic lattice with Cl− due to its larger ionic radius.

Question 26

The fall in bond angle from NH3 (106°) to SbH3 (101°) as we move down group 15 is mainly because of

Accepted Answer

a decrease in electronegativity. Going N -> P -> As -> Sb, the central atom's electronegativity falls, so the bonding electron pairs lie farther from the central atom and closer to the H atoms. This reduces bond pair-bond pair repulsion near the centre, allowing the lone pair to push the bonds closer together, so the H-X-H angle decreases.

Question 27

What is the hybridization of iodine in ICl7?

Accepted Answer

sp3d3. Iodine in ICl7 undergoes hybridization to accommodate seven bonding pairs, which requires the use of one s, three p, and three d orbitals, resulting in an sp3d3 hybridization.

Question 28

What is the molecular geometry of XeOF4?

Accepted Answer

square pyramidal. The molecular geometry of XeOF4 is square pyramidal because it has a central xenon atom surrounded by four oxygen atoms and one lone pair, resulting in a shape that features a square base with one atom above the plane.

Question 29

Arrange the following species in the order of increasing bond angle:

ClO2, Cl2O, ClO2−

Accepted Answer

ClO2− < Cl2O < ClO2. Bond angles: ClO2- ~110.5 deg, Cl2O ~111 deg, ClO2 ~117.4 deg. ClO2 is widest because it is an odd-electron radical (single non-bonding electron exerts less repulsion than a lone pair), giving the order ClO2- < Cl2O < ClO2.

Question 30

Which of the following species exhibits a trigonal bipyramidal arrangement?

Accepted Answer

XeO3F2. In XeO3F2 the Xe atom forms three Xe=O and two Xe-F bonds with no lone pair, giving five electron domains and a trigonal bipyramidal shape. [XeF8]2- is square antiprismatic, not TBP.

Question 31

The geometries of some interhalogen compounds are listed below. Which pairing is incorrectly matched?

Accepted Answer

BrF5: trigonal bipyramid. BrF5 has 5 bond pairs and 1 lone pair (sp3d2), giving a square pyramidal shape, not trigonal bipyramidal. IF7 (pentagonal bipyramid), BrF3 (T-shaped), and ICl3 (planar dimer) are correctly described.

Question 32

This item has two statements. From the four options that follow, select the one that most accurately describes them.

Statement 1: The ion [HeH]+ is more stable than H2+.
Statement 2: The two species mentioned above have the same bond order.

Accepted Answer

Statement 1 is correct, while Statement 2 is incorrect.. [HeH]+ is indeed more stable than H2+ due to its favorable bonding interactions, while the bond orders of the two species differ, making Statement 2 incorrect.

Question 33

Which of the following are isostructural pairs?
A. SO4²− and CrO4²−
B. SiCl4 and TiCl4
C. NH3 and NO3−
D. BCl3 and BrCl3

Accepted Answer

A and B only. Isostructural pairs are compounds that have the same structural arrangement of atoms. SO4²− and CrO4²− both have tetrahedral geometries, while SiCl4 and TiCl4 also share a similar tetrahedral structure, making both pairs isostructural.

Question 34

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

Accepted Answer

SO3²−, CO3²−, NO3−. The set SO3²−, CO3²−, NO3− does not contain isoelectronic species because the total number of electrons in each ion is different: SO3²− has 26 electrons, CO3²− has 24 electrons, and NO3− has 24 electrons, thus they do not share the same electron configuration.

Question 35

Which of the following molecules has the least bond angle?

Accepted Answer

H2S. H2S has a bond angle of approximately 92 degrees due to its larger sulfur atom and the presence of two lone pairs, which exert less repulsion compared to the lone pairs in H2O and NH3, resulting in a smaller bond angle.

Question 36

Which pair of species has molecules with the same molecular geometry for both members?

Accepted Answer

XeF2 and CO2. Both XeF2 and CO2 have a linear molecular geometry due to their symmetrical arrangement of atoms around the central atom, resulting in bond angles of 180 degrees.

Question 37

Arrange the following molecules in increasing order of their bond angles: H2S, NH3, BF3 and SiH4.

Accepted Answer

H2S < NH3 < SiH4 < BF3. The bond angles increase with the presence of lone pairs and the hybridization of the central atom. H2S has the smallest angle due to its bent shape and lone pairs, followed by NH3 with a larger angle due to one lone pair, SiH4 with a tetrahedral shape and no lone pairs, and finally BF3 with a trigonal planar shape, which has the largest bond angle.

Question 38

Nitric oxide (NO) has a bond order of 2.5, whereas the nitrosonium ion (NO+) has a bond order of 3. Which statement correctly compares their bond lengths?

Accepted Answer

The bond length of NO is greater than the bond length of NO+. A higher bond order indicates a stronger bond, which typically results in a shorter bond length. Since NO+ has a bond order of 3 compared to NO's bond order of 2.5, NO+ has a stronger bond and thus a shorter bond length.

Question 39

Which of the following species has a regular tetrahedral geometry?

Accepted Answer

BF4−. BF4− has a regular tetrahedral geometry because it has four bonding pairs of electrons and no lone pairs, allowing the electron pairs to be spaced evenly around the central boron atom.

Question 40

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

Accepted Answer

sp3d2 hybridization. In sp3d2 octahedral geometry each bond makes 90 degrees with four neighbours, giving 12 distinct 90-degree bond-pair/bond-pair angles, more than dsp3 (6) or sp3/sp2 (none).

Question 41

Lattice energy of an ionic compound depends upon

Accepted Answer

Charge on the ion and size of the ion. By the Born-Lande/Coulombic relation, lattice energy is proportional to the product of ionic charges and inversely proportional to the inter-ionic distance. Hence it depends on both the charge on the ions and the size of the ions.

Question 42

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

Accepted Answer

O2²−. O2²− has all its electrons paired due to the addition of two electrons, which fill the molecular orbitals completely, resulting in no unpaired electrons.

Question 43

The charge/size ratio of a cation determines its polarizing power. Which one of the following sequences represents the increasing order of the polarizing power of the cationic species, K+, Ca2+, Mg2+, Be2+?

Accepted Answer

K+ < Ca2+ < Mg2+ < Be2+. Polarizing power rises with charge and falls with size. Order: K+ (1+, large) < Ca2+ < Mg2+ < Be2+ (2+, smallest). So increasing order is K+ < Ca2+ < Mg2+ < Be2+.

Question 44

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

Accepted Answer

NO → NO+. In the ionization of NO to NO+, the removal of an electron from a bonding orbital increases the bond order, resulting in a stronger bond. Additionally, this process changes the magnetic behavior from paramagnetic to diamagnetic, as the unpaired electron is removed.

Question 45

Which one of the following pairs of species have the same bond order?

Accepted Answer

CN− and NO+. CN- has 14 electrons (bond order 3) and NO+ also has 14 electrons (bond order 3), so they share the same bond order. CN+ (12 e-) has bond order 2 and O2- has bond order 2.5.

Question 46

The bond dissociation energy of B−F in BF3 is 646 kJ mol−1, whereas that of C−F in CF4 is 515 kJ mol−1. The correct reason for higher B−F bond dissociation energy as compared to that of C−F bond is

Accepted Answer

significant pπ−pπ interaction between B and F in BF3 whereas there is no such interaction between C and F in CF4. Boron has a vacant 2p orbital, so fluorine lone pairs donate into it giving significant pi(p-p) back-bonding in BF3, which strengthens (shortens) the B-F bond. Carbon in CF4 has no vacant p orbital for such interaction, so C-F is weaker. Hence the correct reason is significant p-pi-p-pi interaction in BF3.

Question 47

Using MO theory, predict which of the following species has the shortest bond length?

Accepted Answer

O2²+. Bond orders: O2+ = 2.5, O2- = 1.5, O2(2-) = 1.0, O2(2+) = 3.0. The highest bond order, O2(2+) with BO = 3, has the shortest bond length.

Question 48

In which of the following pairs, the two species are not isostructural ?

Accepted Answer

PF5 and BrF5. PF5 and BrF5 are not isostructural because PF5 has a trigonal bipyramidal geometry, while BrF5 has a square pyramidal geometry due to the presence of a lone pair on the bromine atom, leading to different spatial arrangements.

Question 49

In which of the following pairs of molecules/ions, the species are not likely to exist ?

Accepted Answer

H2+, He2. H2+ is a stable molecular ion formed from hydrogen, while He2− is not likely to exist because helium is a noble gas with a complete electron shell, making it unlikely to form a stable anion.

Question 50

Stability of the species Li2, Li2− and Li2+ increases in the order of:

Accepted Answer

Li2− < Li2+ < Li2. The stability of these lithium species is influenced by their electron configurations; Li2 has a stable bond with a complete electron configuration, while Li2− has an extra electron that leads to increased repulsion, making it less stable, and Li2+ has one less electron, resulting in a stronger bond and thus higher stability than Li2−.

JEE Main Chemistry: Chemical Bonding and Molecular Structure questions with solutions

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