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JEE • CHEMISTRY • CHEMICAL BONDING & MOLECULAR STRUCTURE

Chemical bonding & molecular structure practice · Mixed difficulty

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451 Medium
451 of 494
Understanding +4 | -1
Which of the following boils at higher temperature?
View answer and explanation
Correct option: D
KClKCl
Option D (KClKCl) is correct because it is an ionic compound, and ionic compounds generally have higher boiling points than covalent compounds due to strong electrostatic forces.
452 Easy
452 of 494
Understanding +4 | -1
Which one of the following is the correct set of hybridisation and shape with respect to molecule?
View answer and explanation
Correct option: C
BCl3 Asp2,\mathrm{BCl}_3 \mathrm{~Asp}^2, triangular planar
Option C (BCl3 Asp2,\mathrm{BCl}_3 \mathrm{~Asp}^2, triangular planar) is correct because Boron in BCl3 has sp2 hybridization and a trigonal planar shape.
453 Hard
453 of 494
Application +4 | -1
The bond between atoms of two elements of atomic number 37 and 53 is:
View answer and explanation
Correct option: B
Ionic
Option B (Ionic) is correct because elements with atomic numbers 37 (Rubidium) and 53 (Iodine) form an ionic bond due to their large electronegativity difference.
454 Easy
454 of 494
Analyzing +4 | -1
In which of the following molecules, the type of hybridisation changes when
View answer and explanation
Correct option: B
AlH3Al{H_3} combines with H−{H^ - }
Option B (AlH3Al{H_3} combines with H−{H^ - }) is correct because the hybridization of Al changes from sp2 in AlH3 to sp3 in AlH4-.
455 Medium
455 of 494
Analyzing +4 | -1
Select the correct statements regarding shapes of PCl5, BrF5 and IF7PC{l_5},\,Br{F_5}\,{\text{and}}\,I{F_7}
View answer and explanation
Correct option: C
One of the given compound is square pyramidal
Option C is correct because BrF5Br{F_5} has a square pyramidal shape, while PCl5PC{l_5} is trigonal bipyramidal and IF7I{F_7} is pentagonal bipyramidal.
456 Easy
456 of 494
Remembering +4 | -1
Who introduced the term covalent bond ?
View answer and explanation
Correct option: B
Langmuir
Option B (Langmuir) is correct because Langmuir introduced the term covalent bond.
457 Easy
457 of 494
Analyzing +4 | -1
Assertion : The bond between two identical nonmetal atoms has a pair of electrons with identical spin.

Reason : Electrons are transferred fully from one atom to another.
View answer and explanation
Correct option: D
If the assertion and reason both are false.
Option D (If the assertion and reason both are false) is correct because the bond between two identical nonmetal atoms has electrons with opposite spins, and electrons are equally shared, not transferred.
458 Medium
458 of 494
Understanding +4 | -1
Which of the following statements is true for an ion having sp3{sp}^3 hybridisation?
View answer and explanation
Correct option: D
H{H}-atoms are situated at the corners of tetrahedron
Option D is correct. An ion with sp3 hybridization has a tetrahedral geometry, with the H-atoms (or other atoms) situated at the corners of a tetrahedron.
459 Medium
459 of 494
Remembering +4 | -1
The O-H bond distance in water molecule is:
View answer and explanation
Correct option: C
0.96 A00.96\,{A^0}
Option C (0.96 A00.96\,{A^0}) is correct because the O-H bond distance in a water molecule is an experimentally determined value close to 0.96 Angstroms.
460 Medium
460 of 494
Analyzing +4 | -1
The ONO{ONO} angle is maximum in :
View answer and explanation
Correct option: D
NO2+{NO}_2^{+}
The correct answer is NO2+{NO}_2^{+} because it has a linear shape due to spsp hybridization, resulting in a maximum ONO angle of 180 degrees.
461 Medium
461 of 494
Understanding +4 | -1
Consider following statements.
(a) He2\mathrm{He}_2 molecule does not exist.
(b) Energy of σ2pz\sigma 2 p_z molecular orbital is always higher than that of π2px\pi 2 p_x and π2py\pi 2 p_y molecular orbital.
(c) The formation of molecular orbitals may be described by linear combination of atomic orbitals.
The correct statement(s) is/are
View answer and explanation
Correct option: B
(a) and (c) only
Option B ((a) and (c) only) is correct because He2 molecule does not exist due to zero bond order, and molecular orbitals are formed by linear combination of atomic orbitals (LCAO).
462 Medium
462 of 494
Analyzing +4 | -1
Common feature between N2+\mathrm{N}_2^{+}and O2+\mathrm{O}_2^{+}is that;

(I) both are isoelectronic.

(II) both have same bond order.

(III) both are paramagnetic.

(IV) both have unpaired, electron in same orbital.
View answer and explanation
Correct option: C
II and III
Option C is correct because both N2+ and O2+ are paramagnetic and have the same bond order of 2.5.
463 Medium
463 of 494
Analyzing +4 | -1
Assertion : Bond order of O2\mathrm{O}_2 and BN\mathrm{BN} is same.

Reason : O2\mathrm{O}_2 and BN\mathrm{BN} are isoelectron
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C is correct. The assertion is true because both O2\mathrm{O}_2 and BN\mathrm{BN} have a bond order of 2.5. However, the reason is false because O2\mathrm{O}_2 and BN\mathrm{BN} are not isoelectronic.
464 Medium
464 of 494
Remembering +4 | -1
How much energy is required to dissociate one mole of H2\mathrm{H}_2 molecule?
View answer and explanation
Correct option: A
435.8 kJ mol−1435.8 \mathrm{~kJ} \mathrm{~mol}^{-1}
Option A (435.8 kJ mol−1435.8 \mathrm{~kJ} \mathrm{~mol}^{-1}) is correct because the bond dissociation energy for one mole of H2 is 435.8 kJ/mol.
465 Medium
465 of 494
Understanding +4 | -1
The correct order of decreasing polarisability of ion is:
View answer and explanation
Correct option: C
I−,Br−,Cl−,F−{\mathrm{I}^ - },\mathrm{B}{\mathrm{r}^ - },\mathrm{C}{\mathrm{l}^ - },{\mathrm{F}^ - }
Option C (I−,Br−,Cl−,F−{\mathrm{I}^ - },\mathrm{B}{\mathrm{r}^ - },\mathrm{C}{\mathrm{l}^ - },{\mathrm{F}^ - }) is correct because polarizability increases with increasing size of the ion.

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466 Medium
466 of 494
Analyzing +4 | -1
In which of the given molecules only π\pi bonds are present?
View answer and explanation
Correct option: B
C2\mathrm{C}_2
Option B is correct because the C2 molecule has only pi bonds and no sigma bonds.
467 Medium
467 of 494
Understanding +4 | -1
Decreasing order of C−CC\mathrm{ }-\mathrm{ }C bond length is
(I) C2H4{\mathrm{C}_2}{\mathrm{H}_4} (II) C2H2{\mathrm{C}_2}{\mathrm{H}_2}
(III) C6H6{\mathrm{C}_6}{\mathrm{H}_6} (IV) C2H6{\mathrm{C}_2}{\mathrm{H}_6}
View answer and explanation
Correct option: A
IV>III>I>II\mathrm{IV} > III > I > II
Option D (IV > III > I > II) is correct because single bonds are longer than double bonds, which are longer than triple bonds, and the bond length in benzene is intermediate between single and double bonds.
468 Medium
468 of 494
Analyzing +4 | -1
Which one of the following molecules has the smallest bond angle?
View answer and explanation
Correct option: D
H2Se{\mathrm{H}_2}\mathrm{Se}
Option D (H2Se{\mathrm{H}_2}\mathrm{Se}) is correct because it has the smallest bond angle due to the increasing size of the central atom down the group, leading to decreased bond pair-bond pair repulsion.
469 Medium
469 of 494
Analyzing +4 | -1
The compound with the maximum dipole moment among the following is:
View answer and explanation
Correct option: C
oo-dichlorobenzene
The correct answer is C (oo-dichlorobenzene) because the dipole moments of the C-Cl bonds add vectorially, resulting in a non-zero dipole moment.
470 Medium
470 of 494
Analyzing +4 | -1
Arrange the following species in the correct order of their stability C2,Li,O2+,He2+{C}_2, {Li}, {O}_2^{+}, {He}_2^{+}
View answer and explanation
Correct option: C
He2+<Li2<C2<O2+{He}_2^{+} < {Li}_2 < {C}_2 < {O}_2^{+}
The correct answer is A because the stability of a molecule is directly proportional to its bond order, and the bond orders are He2+ (0.5) < Li2 (1) < C2 (2) < O2+ (2.5).
471 Medium
471 of 494
Analyzing +4 | -1
Which has the lowest bond angle?
View answer and explanation
Correct option: A
NH3\mathrm{N}{\mathrm{H}_3}
Option A (NH3\mathrm{N}{\mathrm{H}_3}) is correct because the presence of a lone pair on the nitrogen atom in ammonia causes greater repulsion, leading to a smaller bond angle compared to the other molecules.
472 Medium
472 of 494
Analyzing +4 | -1
The correct order of decreasing polarity is
View answer and explanation
Correct option: B
HF>H2O>SO2>NH3\mathrm{HF} > {\mathrm{H}_2}\mathrm{O} > S{\mathrm{O}_2} > N{\mathrm{H}_3}
Option B (HF>H2O>SO2>NH3\mathrm{HF} > {\mathrm{H}_2}\mathrm{O} > S{\mathrm{O}_2} > N{\mathrm{H}_3}) is correct because it reflects the decreasing order of polarity based on electronegativity differences and molecular geometry.
473 Medium
473 of 494
Analyzing +4 | -1
Which of the following molecules are expected to exhibit intermolecular H – bonding
(i) Acetic acid (ii) o – nitro phenol
(iii) m – nitro phenol (iv) o –boric acid
Select the correct option
View answer and explanation
Correct option: C
i, iii and iv
Option C is correct because acetic acid, m-nitro phenol, and o-boric acid can form intermolecular hydrogen bonds, while o-nitro phenol forms intramolecular hydrogen bonds.
474 Medium
474 of 494
Understanding +4 | -1
Which statement is wrong?
View answer and explanation
Correct option: D
d2sp2{d^2}s{p^2}-hybrid orbitals are directed towards the corners of a regular tetrahedron
Option D (d2sp2{d^2}s{p^2}-hybrid orbitals are directed towards the corners of a regular tetrahedron) is wrong because d2sp2 hybridization leads to square planar geometry, not tetrahedral.
475 Medium
475 of 494
Understanding +4 | -1
Hydrogen bonding is not present in
View answer and explanation
Correct option: C
Hydrogen sulphide
The correct answer is C (Hydrogen sulphide) because hydrogen bonding requires a hydrogen atom bonded to a highly electronegative atom like F, O, or N, which is not the case in hydrogen sulfide.
476 Medium
476 of 494
Application +4 | -1
The observed dipole-moment of HCl molecule is 1.03 D . If H−Cl\mathrm{H}-\mathrm{Cl} bond distance is 1.275 A∘1.275 \, \mathop {\text{A}}\limits^ \circ and electronic charge is 4.8×10−104.8 \times 10^{-10} esu. What is the percent polarity of HCl?
View answer and explanation
Correct option: C
1.03×100×10−184.8×10−10×1.275×10−8\frac{1.03 \times 100 \times 10^{-18}}{4.8 \times 10^{-10} \times 1.275 \times 10^{-8}}
The correct answer is 1.03×100×10−184.8×10−10×1.275×10−8\frac{1.03 \times 100 \times 10^{-18}}{4.8 \times 10^{-10} \times 1.275 \times 10^{-8}}. Percent polarity is calculated by dividing the observed dipole moment by the calculated dipole moment and multiplying by 100.
477 Medium
477 of 494
Understanding +4 | -1
Shape of O2 F2{O}_2 {~F}_2 is similar to that of:
View answer and explanation
Correct option: B
H2O2{H}_2 {O}_2
Option B (H2O2{H}_2 {O}_2) is correct because both O2 F2{O}_2 {~F}_2 and H2O2{H}_2 {O}_2 have a similar peroxide structure.
478 Easy
478 of 494
Understanding +4 | -1
Hybridization of PP in PCl5PCl_5 is
View answer and explanation
Correct option: A
sp3dsp^3d
Option A (sp3dsp^3d) is correct because in PCl5PCl_5, the phosphorus atom is sp3dsp^3d hybridized, resulting in a trigonal bipyramidal geometry.
479 Medium
479 of 494
Analyzing +4 | -1
In terms of polar character which one of the following orders is correct?
View answer and explanation
Correct option: A
H2S<NH3<H2O<HF{H}_2 {S} < {NH}_3 < {H}_2 {O} < {HF}
Option A is correct because the polar character increases with the electronegativity difference between the bonded atoms, and the order of electronegativity is S < N < O < F.
480 Medium
480 of 494
Analyzing +4 | -1
The pair likely to form the strongest hydrogen bonding:
View answer and explanation
Correct option: B
HCOOHandCH3COOH\mathrm{HCOOH and C}{\mathrm{H}_3}\mathrm{COOH}
The correct answer is B (HCOOH and CH3COOH) because carboxylic acids form strong hydrogen bonds due to dimerization.
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