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JEE • CHEMISTRY • CHEMICAL BONDING & MOLECULAR STRUCTURE

Chemical bonding & molecular structure practice · Mixed difficulty

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 421-450 of 494 questions

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421 Easy
421 of 494
Analyzing +4 | -1
Which of the following has the highest dipole moment
View answer and explanation
Correct option: C
H2O{H_2}O
Option C (H2O{H_2}O) is correct because water has a bent structure and a significant difference in electronegativity between oxygen and hydrogen, resulting in a high dipole moment due to the two lone pairs of electrons on oxygen.
422 Medium
422 of 494
Understanding +4 | -1
Which one of the following has not triangular pyramidal shape?
View answer and explanation
Correct option: D
BCl3\mathrm{BC}{\mathrm{l}_3}
Option D (BCl3\mathrm{BC}{\mathrm{l}_3}) is correct because it has a trigonal planar shape due to sp² hybridization, while the others have a triangular pyramidal shape due to sp³ hybridization with a lone pair.
423 Medium
423 of 494
Remembering +4 | -1
The bond length (pm)({pm}) of F2,H2,Cl2{F}_2, {H}_2, {Cl}_2 and I2{I}_2, respectively is
View answer and explanation
Correct option: A
144,74,199,267144,74,199,267
Option A (144,74,199,267144, 74, 199, 267) is correct because it represents the increasing trend of bond lengths down the halogen group due to increasing atomic size.
424 Easy
424 of 494
Analyzing +4 | -1
In which of the following bond angle is maximum
View answer and explanation
Correct option: C
PCl5PC{l_5}
Option C (PCl5PC{l_5}) is correct because it has a bond angle of 120° due to sp3d hybridization, which is the largest among the given options.
425 Medium
425 of 494
Understanding +4 | -1
Glycerol has strong intermolecular bonding therefore it is
View answer and explanation
Correct option: D
Viscous
Option D (Viscous) is correct because glycerol's strong intermolecular hydrogen bonding due to the presence of three -OH groups leads to high viscosity.
426 Medium
426 of 494
Analyzing +4 | -1
Among the interhalide species: IF2⊖,IF3,IF4⊖\mathrm{IF}_2^{\ominus}, \mathrm{IF}_3, \mathrm{IF}_4^{\ominus} and IF7\mathrm{IF}_7
View answer and explanation
Correct option: C
The anionic species are both isoelectronic and isostructural to XeF2\mathrm{XeF}_2 and XeF4\mathrm{XeF}_4
Option C is correct because the anionic species IF2- and IF4- are isoelectronic and isostructural with XeF2 and XeF4, respectively.
427 Medium
427 of 494
Understanding +4 | -1
Which of the following compounds is a good conductor of electricity in solution state?
View answer and explanation
Correct option: D
ionic compounds
Option D (ionic compounds) is correct because ionic compounds dissociate into ions in solution, and these ions are charge carriers, making the solution conductive.
428 Medium
428 of 494
Analyzing +4 | -1
Which of the following order is correct :-
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Correct option: D
All of these
Option D (All of these) is correct because all the given boiling point/melting point orders are accurate due to factors like hydrogen bonding and increasing molecular weight.
429 Easy
429 of 494
Analyzing +4 | -1
Which of the following pairs of species have identical shapes
View answer and explanation
Correct option: C
XeF4Xe{F_4} and ICl4−ICl_4^ -
The correct answer is C (XeF4Xe{F_4} and ICl4−ICl_4^ -) because both species have a square planar shape due to the same number of bond pairs and lone pairs.
430 Medium
430 of 494
Analyzing +4 | -1
Dipole moment is highest for:
View answer and explanation
Correct option: C
CHF3\mathrm{CH}{\mathrm{F}_3}
Option C (CHF3\mathrm{CH}{\mathrm{F}_3}) is correct because the C-F bond is significantly more polar than C-Cl, leading to a higher dipole moment for the molecule.
431 Medium
431 of 494
Analyzing +4 | -1
According to the molecular orbital theory which of the following statement is incorrect?
[LUMO = lowest unoccupied molecular orbital]
View answer and explanation
Correct option: D
C2{C}_2 is paramagnetic but C22−{C}_2^{2-} is diamagnetic.
Option D (C2{C}_2 is paramagnetic but C22−{C}_2^{2-} is diamagnetic.) is incorrect because C2{C}_2 is diamagnetic, not paramagnetic.
432 Medium
432 of 494
Understanding +4 | -1
The important condition(s) for Lewis dot structure of covalent compounds is/are
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because all the listed statements are fundamental conditions for forming valid Lewis dot structures.
433 Medium
433 of 494
Analyzing +4 | -1
Which of the following is paramagnetic in nature?
View answer and explanation
Correct option: C
B2{B_2}
Option C (B2{B_2}) is correct because its molecular orbital configuration results in unpaired electrons, making it paramagnetic.
434 Easy
434 of 494
Analyzing +4 | -1
Given below are two statements: one is labeled as Assertion (A) and the other is labeled as Reason (R).
Assertion (A) Dipole –dipole interactions are the only non-covalent interactions, resulting in hydrogen bond formation.
Reason (R) Fluorine is the most electronegative element and hydrogen bonds in HFHF are symmetrical.
In the light of the above statements, choose the most appropriate answer from the options given below.
View answer and explanation
Correct option: A
A is false but R is true.
Option A is correct because the assertion is false as dipole-dipole interactions are not the ONLY non-covalent interactions, and the reason is true.
435 Medium
435 of 494
Analyzing +4 | -1
Select pair of compounds in which both have different hybridization but have same molecular geometry
View answer and explanation
Correct option: B
ICl2−, BeCl2ICl_2^ - ,\,BeC{l_2}
Option B (ICl2−, BeCl2ICl_2^ - ,\,BeCl_2) is correct because ICl2−ICl_2^- has sp3dsp^3d hybridization and linear geometry, while BeCl2BeCl_2 has spsp hybridization and linear geometry.
436 Medium
436 of 494
Understanding +4 | -1
Statement I : According to wave mechanics, the atomic orbitals can be expressed by wave functions which represent the magnitude of the electron waves. These are obtained from the solution of Schrodinger wave equation
Statement II : The magnitude of H-bonding depends on the physical state of the compound. It is maximum in the gaseous state and minimum in the solid state
View answer and explanation
Correct option: B
Both Statement I and II are incorrect
Option B (Both Statement I and II are incorrect) is correct because wave functions represent the probability of finding an electron, and H-bonding is strongest in the solid state.
437 Medium
437 of 494
Understanding +4 | -1
In compound XX, all the bond angles are exactly 109∘28′109^\circ 28\prime , XX is:
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Correct option: C
Carbon tetrachloride
Option C (Carbon tetrachloride) is correct because it has a tetrahedral structure with all bond angles exactly 109∘28′109^\circ 28\prime due to its symmetrical arrangement.
438 Medium
438 of 494
Analyzing +4 | -1
For CO\mathrm{CO} molecule, correct statement is
View answer and explanation
Correct option: C
Antibonding 2s2 s orbital will donate electron for the formation of CO⊕\mathrm{CO}^{\oplus}
The correct answer is C because CO donates electrons from its bonding 2s orbital, not the antibonding.
439 Easy
439 of 494
Analyzing +4 | -1
Asser tion : Fluorine molecule has bond order one.

Reason : The number of electrons in the anti-bonding molecular orbitals is two less than that in bonding molecular orbitals.
View answer and explanation
Correct option: A
If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.
Option A is correct because both the assertion and the reason are correct, and the reason correctly explains why the fluorine molecule has a bond order of one.
440 Medium
440 of 494
Understanding +4 | -1
In which of the following state of compound the magnitude of H{H}-bonding will be maximum and in which case it will be minimum ?
View answer and explanation
Correct option: C
Maximum == Solid, Minimum == Gas
Option C is correct because hydrogen bonding is strongest in the solid state due to the close proximity of molecules and weakest in the gaseous state due to large intermolecular distances.
441 Medium
441 of 494
Remembering +4 | -1
Among the following, which is water insoluble?
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Correct option: D
Magnesium fluoride
Option D (Magnesium fluoride) is correct because fluorides of alkaline earth metals like magnesium are generally insoluble in water.
442 Easy
442 of 494
Understanding +4 | -1
Which of the following species assumes planar structure
View answer and explanation
Correct option: B
CH3+CH_3^ +
Option B (CH3+CH_3^ +) is correct because CH3+CH_3^+ has a planar structure with the carbon atom being sp2sp^2 hybridized.
443 Medium
443 of 494
Analyzing +4 | -1
Assertion(A) All F−S−F{F}-{S}-{F} angles in SF4{SF}_4 are greater than 90∘90^{\circ} but less than 180∘180^{\circ}.

Reason(R) The lone pair-bond repulsion is weaker than bond pairbond repulsion.
View answer and explanation
Correct option: C
If (A) is correct but (R) is incorrect.
The correct answer is C (If (A) is correct but (R) is incorrect) because the assertion about F-S-F angles in SF4 is true, but the reason regarding lone pair-bond pair repulsion is stated incorrectly.
444 Medium
444 of 494
Understanding +4 | -1
Consider the following statements
I. The canonical forms have no real existence.
II. The molecule does not exist for a certain fraction of time in one canonical form and for other fractions of time in other canonical forms.
III. There is no such equilibrium between the canonical forms as we have between tautomeric forms (keto and enol) in tautomerism.
IV. The molecule such as has a single structure which is the resonance hybrid of the canonical forms and which cannot such as be depicted by a single Lewis structure.
Which of the above statements are correct?
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because all the statements accurately describe the nature of canonical forms and resonance hybrids.
445 Medium
445 of 494
Understanding +4 | -1
Which of the following ion has maximum polarsing power.
View answer and explanation
Correct option: B
Al+3A{l^{ + 3}}
Option B (Al+3Al^{+3}) is correct because, according to Fajan's rule, a higher positive charge and smaller size of the cation lead to greater polarizing power.
446 Medium
446 of 494
Analyzing +4 | -1
Which of the following has the minimum bond length?
View answer and explanation
Correct option: B
O2+\mathrm{O}_2^ +
The correct answer is O2+\mathrm{O}_2^ + because it has the highest bond order (2.5) among the given options, leading to the shortest bond length.
447 Medium
447 of 494
Understanding +4 | -1
Which of the following gives correct arrangement of compounds involved based on their bond strength
View answer and explanation
Correct option: A
HF>HCl>HBr>HI{HF}>{HCl}>{HBr}>{HI}
Option A (HF>HCl>HBr>HI{HF}>{HCl}>{HBr}>{HI}) is correct because bond strength decreases as the size of the halogen increases, with HF having the strongest bond due to fluorine's small size and high electronegativity.
448 Medium
448 of 494
Analyzing +4 | -1
The d-orbital which is not involved in hybridization of central atom in ICl4−{ICl}_4^{-}:
View answer and explanation
Correct option: D
dxy,dxz,dyzd_{x y}, d_{x z}, d_{y z}
Option D (dxy,dxz,dyzd_{xy}, d_{xz}, d_{yz}) is correct because the hybridization of the central atom in ICl4−\mathrm{ICl}_4^{-} is sp3d2sp^3d^2, which involves dz2d_{z^2} and dx2−y2d_{x^2-y^2} orbitals, but not dxy,dxz,dyzd_{xy}, d_{xz}, d_{yz} orbitals.
449 Medium
449 of 494
Analyzing +4 | -1
Assertion : B2{B}_2 molecule is diamagnetic.

Reason : The highest occupied molecular orbital is of σ\sigma type.
View answer and explanation
Correct option: D
If both the Assertion and Reason are incorrect.
Option D (If both the Assertion and Reason are incorrect) is correct because the B2B_2 molecule is paramagnetic due to unpaired electrons in the π\pi molecular orbitals.
450 Medium
450 of 494
Analyzing +4 | -1
Which one of the following molecules has zero dipole moment?
View answer and explanation
Correct option: A
NO2+{NO}_2^{+}
Option A (NO2+{NO}_2^{+}) is correct because the individual bond dipoles cancel each other out due to the linear structure, resulting in a zero net dipole moment.
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