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JEE • CHEMISTRY • CHEMICAL BONDING & MOLECULAR STRUCTURE

Chemical bonding & molecular structure practice · Mixed difficulty

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151 Medium
151 of 494
Analyzing +4 | -1
Which amongst the following incorrect statement?
View answer and explanation
Correct option: A
O2+O_2^ + ion is diamagnetic
Option A is correct because O2+O_2^+ has 15 electrons and is paramagnetic, not diamagnetic.
152 Medium
152 of 494
Understanding +4 | -1
Assertion: HF has more dipole moment than HCl.

Reason: For above, higher the electronegativity difference, higher the dipole moment.
View answer and explanation
Correct option: A
Both Assertion (A) and Reason (R) are true and
Reason (R) is the correct explanation of Assertion
(A)
Option A is correct because both the assertion and the reason are true, and the reason correctly explains why HF has a higher dipole moment than HCl.
153 Medium
153 of 494
Analyzing +4 | -1
The correct set from the following in which both pairs are in correct order of melting point is
View answer and explanation
Correct option: A
LiF  >  LiCl,   MgO  >  NaCl LiF\,\, > \,\,LiCl,\,\,\,MgO\,\, > \,\,NaCl\,
Option A (LiF>LiCl,MgO>NaClLiF > LiCl, MgO > NaCl) is correct because melting point is directly proportional to lattice energy, which depends on charge and size of ions.
154 Medium
154 of 494
Analyzing +4 | -1
Which pair of diatomic species do not have same bond order?
View answer and explanation
Correct option: A
B2−,C2{B}_2^{-}, {C}_2
Option A (B2−,C2{B}_2^{-}, {C}_2) is correct because B2−{B}_2^{-} has a bond order of 1.5, while C2{C}_2 has a bond order of 2.
155 Medium
155 of 494
Remembering +4 | -1
The element which exists in both hard and soft form is:
View answer and explanation
Correct option: C
C\mathrm{C}
Option C (C\mathrm{C}) is correct because carbon exists as both diamond (hard) and graphite (soft).
156 Easy
156 of 494
Analyzing +4 | -1
Which among the following has only two π\pi-bonds and no σ\sigma-bond?
View answer and explanation
Correct option: A
C2\mathrm{C}_2
Option D is correct because the cyclic hydrocarbon with equal C-C bond lengths between 1.34 and 1.54 Å is benzene, which has a C-C-C bond angle of 120° due to sp2 hybridization.
157 Easy
157 of 494
Understanding +4 | -1
The correct order of repulsion is
View answer and explanation
Correct option: B
L.P−L.P>L.P−B.P>B.P−B.PL.P - L.P > L.P - B.P > B.P - B.P
Option B (L.P−L.P>L.P−B.P>B.P−B.PL.P - L.P > L.P - B.P > B.P - B.P) is correct because lone pair-lone pair repulsions are the strongest, followed by lone pair-bond pair, and then bond pair-bond pair repulsions.
158 Easy
158 of 494
Analyzing +4 | -1
Which of the following orders are correct?

I. B.E. : C−C<C=C<C≡C{C}-{C}<{C}={C}<{C} \equiv {C}

II. B.E. : H−F>H−Cl>H−Br>H−{H}-{F}>{H}-{Cl}>{H}-{Br}>\mathrm{H}- I

III. B.E. :sp−sp>sp2−sp2>sp3−sp3: s p-s p>s p^2-s p^2>s p^3-s p^3

IV. B.E.:C≡C>C≡N>N≡NB . E .: {C} \equiv {C}>{C} \equiv {N}>{N} \equiv {N}
View answer and explanation
Correct option: D
I, II and III only
Option D (I, II and III only) is correct because the bond energy order for C≡C, C≡N, and N≡N is C≡C < C≡N < N≡N.
159 Medium
159 of 494
Analyzing +4 | -1
Which of the following species contains only π\pi-bond(s)?
View answer and explanation
Correct option: D
C2\mathrm{C}_2
The correct answer is D (C2\mathrm{C}_2) because the molecular orbital configuration of C2\mathrm{C}_2 indicates the presence of only π\pi bonds.
160 Hard
160 of 494
Application +4 | -1
The dipole moment of HBrHBr is 1.6×10−301.6 \times {10^{ - 30}} cm and inter &ndash; atomic spacing is 1  Ao1\,\,\mathop \mathrm{A}\limits^\mathrm{o}. The %\% ionic character of HBrHBr is
View answer and explanation
Correct option: B
10
Option B (10) is correct because the percentage ionic character is calculated by dividing the actual dipole moment by the theoretical dipole moment and multiplying by 100.
161 Medium
161 of 494
Analyzing +4 | -1
The stability of the species NO2, NO2+   and  NO2−N{O_2},\,NO_2^ + \,\,\,{\text{and}}\,{\text{ }}NO_2^ - is in the order
View answer and explanation
Correct option: A
NO2+ > NO2>NO2−NO_2^ + {\text{ > }}N{O_2} > NO_2^ -
Option A (NO2+>NO2>NO2−NO_2^+ > NO_2 > NO_2^-) is correct because the stability is related to the bond angle, with NO2+NO_2^+ having the largest bond angle and thus being the most stable.
162 Easy
162 of 494
Understanding +4 | -1
Which one has a pyramidal structure
View answer and explanation
Correct option: B
NH3{NH}_3
Option B is correct because ammonia (NH3{NH}_3) has a pyramidal structure due to the presence of a lone pair of electrons on the nitrogen atom.
163 Medium
163 of 494
Analyzing +4 | -1
Correct order of bond angle is:
View answer and explanation
Correct option: A
OF2<H2O<NH3<Cl2O\mathrm{OF}_2 < \mathrm{H}_2 \mathrm{O} < \mathrm{NH}_3 < \mathrm{Cl}_2 \mathrm{O}
Option A is correct because the bond angles increase in the order OF2 < H2O < NH3 < Cl2O due to the electronegativity of the surrounding atoms and the presence of lone pairs.
164 Medium
164 of 494
Analyzing +4 | -1
The electronic configuration of 44 elements K,L,M  and  NK,L,M\,\,\mathrm{and}\,\,N are,
K=1s2,2s22p1     L=1s2,2s22p6M=1s2,2s22p4    N=1s2,2s22p3\begin{array}{l} K = 1{s^2},2{s^2}2{p^1}\,\,\,\,\,L = 1{s^2},2{s^2}2{p^6}\\ M = 1{s^2},2{s^2}2{p^4}\,\,\,\,N = 1{s^2},2{s^2}2{p^3} \end{array}
The element that would form a diatomic molecule with double bond is:
View answer and explanation
Correct option: C
MM
Option M (MM) is correct because the electronic configuration 1s2,2s22p41{s^2},2{s^2}2{p^4} indicates that the element needs two more electrons to complete its octet, thus forming a double bond in a diatomic molecule.
165 Medium
165 of 494
Analyzing +4 | -1
If Na+\mathrm{N}{\mathrm{a}^ + } ion is larger than Mg2+\mathrm{M}{\mathrm{g}^{2 + }} ion and S2−{\mathrm{S}^{2 - }} is larger than Cl−\mathrm{C}{\mathrm{l}^ - } ion, which of the following will be stable soluble in water?
View answer and explanation
Correct option: D
Magnesium sulphide
Option D (Magnesium sulphide) is correct. MgS has higher lattice energy due to higher charges, making it less soluble compared to the other options.

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166 Medium
166 of 494
Understanding +4 | -1
Which atomic orbital is always involved in sigma bonding only?
View answer and explanation
Correct option: A
ss
Option A is correct because s atomic orbitals are spherically symmetrical and can only participate in sigma (σ\sigma) bonding due to their non-directional nature.
167 Medium
167 of 494
Understanding +4 | -1
Statement-I: H−H\mathrm{H}-\mathrm{H} bond is weaker than D\mathrm{D} - D\mathrm{D} bond

Statement-II: H−H\mathrm{H}-\mathrm{H} bond is longer than D\mathrm{D} - D\mathrm{D} bond
View answer and explanation
Correct option: C
I is true & II is false
Option C is correct because the H-H bond is indeed weaker than the D-D bond due to the higher vibrational frequency of H, but the bond lengths are approximately equal.
168 Hard
168 of 494
Application +4 | -1
A diatomic molecule has a dipole moment of 1.2 D. If its bond length is equal to 10−10 m10^{-10} \mathrm{~m} then the fraction of an electronic charge on each atom will be
View answer and explanation
Correct option: D
25%25 \%
Option D (25%25 \%) is correct because it is calculated by dividing the dipole moment by the product of bond length and electronic charge, then multiplying by 100.
169 Easy
169 of 494
Understanding +4 | -1
The unequal sharing of bonded pair of electrons between two atoms in a molecule gives rise to:
View answer and explanation
Correct option: B
Polar covalent bond
The correct answer is B (Polar covalent bond) because unequal sharing of electrons leads to a dipole moment and a polar covalent bond.
170 Medium
170 of 494
Understanding +4 | -1
The bond length the species O2,O2+{O}_2, {O}_2^{+}and O2−{O}_2^{-}are in the order of
View answer and explanation
Correct option: A
O2+>O2>O2−{O}_2^{+}>{O}_2>{O}_2^{-}
Option D (O2−>O2>O2+{O}_2^{-}>{O}_2>{O}_2^{+}) is correct because bond length decreases as bond order increases, and the bond orders are O2+{O}_2^{+} (2.5) > O2{O}_2 (2) > O2−{O}_2^{-} (1.5).
171 Easy
171 of 494
Remembering +4 | -1
Which is present in peroxides?
View answer and explanation
Correct option: C
O22−\mathrm{O}_2^{2 - }
Option C (O22−\mathrm{O}_2^{2 - }) is correct because the peroxide ion is represented as O22-.
172 Easy
172 of 494
Application +4 | -1
Using MO theory predict which of the following species has the shortest bond length
View answer and explanation
Correct option: A
O2+{O_2}^ +
Option A (O2+{O_2}^ +) is correct because it has the highest bond order, leading to the shortest bond length.
173 Easy
173 of 494
Remembering +4 | -1
The dd-orbital used in sp3d{sp}^3 d hybridisation forming square pyramidal geometry is:
View answer and explanation
Correct option: B
dx2−y2d_{x^2-y^2}
Option B (dx2−y2d_{x^2-y^2}) is correct because in sp3d{sp}^3 d hybridization forming square pyramidal geometry, the dx2−y2d_{x^2-y^2} orbital is involved.
174 Medium
174 of 494
Analyzing +4 | -1
There will be no change in hybridization when:
View answer and explanation
Correct option: D
SO2\mathrm{SO}_2 is oxidized to SO3\mathrm{SO}_3
Option D is correct because in the oxidation of SO2 to SO3, the hybridization of sulfur remains sp2.
175 Medium
175 of 494
Analyzing +4 | -1
Assertion: SO4−2\mathrm{SO}_4^{-2} is square planar in shape.

Reason : SO4−2\mathrm{SO}_4{ }^{-2} has sp3d\mathrm{sp}^3 \mathrm{d} hybridisation.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D is correct because SO4-2 is tetrahedral, not square planar, and it has sp3 hybridization, not sp3d.
176 Medium
176 of 494
Analyzing +4 | -1
Pick up the incorrect statement.
View answer and explanation
Correct option: D
O2,H2{O}_2, {H}_2 show bond dipole due to polarisation.
Option D is correct because O2 and H2 are nonpolar molecules and do not exhibit bond dipoles due to polarization since they are composed of the same element.
177 Medium
177 of 494
Understanding +4 | -1
In which of the following, the bond length between hybridised carbon atom and other carbon atom is minimum?
View answer and explanation
Correct option: A
Propyne
Option A (Propyne) is correct because propyne contains a carbon-carbon triple bond, which has the shortest bond length compared to single or double bonds.
178 Easy
178 of 494
Application +4 | -1
The order of increasing lattice energy of the following salts is
View answer and explanation
Correct option: B
NaBr<NaCl<BaO<CaONaBr < NaCl < BaO < CaO
Option D (NaBr<NaCl<CaO<BaONaBr < NaCl < CaO < BaO) is incorrect. Lattice energy depends on charge and size; higher charge and smaller size lead to greater lattice energy. The correct order considers both factors.
179 Medium
179 of 494
Analyzing +4 | -1
Which of the following two are isostructural?
View answer and explanation
Correct option: A
XeF2,IF2−\mathrm{Xe}{\mathrm{F}_2},\mathrm{IF}_2^ -
Option A (XeF2,IF2−\mathrm{Xe}{\mathrm{F}_2},\mathrm{IF}_2^ -) is correct because both molecules have a linear structure.
180 Medium
180 of 494
Analyzing +4 | -1
Which of the following species is not expected to exist
View answer and explanation
Correct option: C
Be2\mathrm{Be}_2
Option C is correct because Be2 has a bond order of 0, indicating that it is not expected to exist.
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