Skip to main content
Back to CHEMICAL BONDING & MOLECULAR STRUCTURE

Your Progress

0 / 342 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • CHEMISTRY • CHEMICAL BONDING & MOLECULAR STRUCTURE

Chemical bonding & molecular structure practice · Medium

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 301-330 of 342 questions

Preparing practice and saved progress…

301 Medium
301 of 342
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.
302 Medium
302 of 342
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.
303 Medium
303 of 342
Remembering +4 | -1
Among the following, which is water insoluble?
View answer and explanation
Correct option: D
Magnesium fluoride
Option D (Magnesium fluoride) is correct because fluorides of alkaline earth metals like magnesium are generally insoluble in water.
304 Medium
304 of 342
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.
305 Medium
305 of 342
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.
306 Medium
306 of 342
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.
307 Medium
307 of 342
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.
308 Medium
308 of 342
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.
309 Medium
309 of 342
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.
310 Medium
310 of 342
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.
311 Medium
311 of 342
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.
312 Medium
312 of 342
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.
313 Medium
313 of 342
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.
314 Medium
314 of 342
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.
315 Medium
315 of 342
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.
316 Medium
316 of 342
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.
317 Medium
317 of 342
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).
318 Medium
318 of 342
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.
319 Medium
319 of 342
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.
320 Medium
320 of 342
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.
321 Medium
321 of 342
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.
322 Medium
322 of 342
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.
323 Medium
323 of 342
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.
324 Medium
324 of 342
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.
325 Medium
325 of 342
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.
326 Medium
326 of 342
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).
327 Medium
327 of 342
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.
328 Medium
328 of 342
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.
329 Medium
329 of 342
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.
330 Medium
330 of 342
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.
Revision notes and key takeaways

Review CHEMICAL BONDING & MOLECULAR STRUCTURE in your textbook or class notes, then use this set to check your understanding.

Showing 301-330 of 342 questions

Your practice

0 / 342 answered · 342 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Answer feedback and difficulty

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.