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JEE • CHEMISTRY • CHEMICAL BONDING & MOLECULAR STRUCTURE

Chemical bonding & molecular structure practice · Medium

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211 Medium
211 of 342
Analyzing +4 | -1
In which of the following ionisation processes, the bond order has increased and the magnetic behaviour has changed
View answer and explanation
Correct option: B
NO→NO+NO \to N{O^ + }
Option B (NO→NO+NO \to N{O^ + }) is correct because the bond order increases from 2.5 to 3.0, and the magnetic behavior changes from paramagnetic to diamagnetic upon ionization.
212 Medium
212 of 342
Analyzing +4 | -1
Which among the following is/are paramagnetic in nature?
View answer and explanation
Correct option: D
Both (1) and (3)
Option D is correct because both B2B_2 and N2+N_2^+ have unpaired electrons in their molecular orbitals, making them paramagnetic.
213 Medium
213 of 342
Analyzing +4 | -1
Which amongst the following is incorrect statement?
View answer and explanation
Correct option: C
O2+{O}_2^{+}ion is diamagnetic.
Option C is correct. O2+ is paramagnetic, not diamagnetic, because it has an unpaired electron in its molecular orbital configuration.
214 Medium
214 of 342
Understanding +4 | -1
Which of the following halogens has the highest bond energy
View answer and explanation
Correct option: B
Cl2Cl_2
The correct answer is B (Cl2Cl_2) because chlorine has the highest bond energy among the given halogens due to its smaller size and strong interatomic attraction.
215 Medium
215 of 342
Application +4 | -1
Two elements XX and YY have following electronic configuration
X=1s2 2s22p6  3s23p6  4s2X = 1{s^2}\,2{s^2}2{p^6}\,\,3{s^2}3{p^6}\,\,4{s^2}
Y=1s2 2s22p6  3s23p5 Y = 1{s^2}\,2{s^2}2{p^6}\,\,3{s^2}3{p^5}\,
The expected compound formed by combination of XX and YY will be expressed as
View answer and explanation
Correct option: A
XY2X{Y_2}
Option A (XY2X{Y_2}) is correct because X forms a +2 ion and Y forms a -1 ion, leading to the compound XY2.
216 Medium
216 of 342
Application +4 | -1
The bond order of CO\mathrm{CO} molecule on the basis of molecular orbital theory is:
View answer and explanation
Correct option: C
33
Option C (33) is correct because the bond order is calculated from the number of bonding and antibonding electrons in the molecular orbital configuration.
217 Medium
217 of 342
Analyzing +4 | -1
Some of the properties of the two species, NO3−\mathrm{NO}_3^ - and H3O+{\mathrm{H}_3}{\mathrm{O}^ + } are described below. Which one of them is correct?
View answer and explanation
Correct option: A
Dissimilar in hybridization for the central atom with different structure
Option A (x > y) is correct because the enthalpy of combustion of CH4 is more negative than that of CH3OH, indicating that more heat is released during the combustion of CH4.
218 Medium
218 of 342
Understanding +4 | -1
Which bond in a molecule of ethyl magnesium bromide is ionic in nature?
View answer and explanation
Correct option: C
Mg−Br{Mg}-{Br} bond
Option C (Mg−Br{Mg}-{Br} bond) is correct because magnesium is a metal and bromine is a nonmetal, leading to a significant electronegativity difference and an ionic bond.
219 Medium
219 of 342
Analyzing +4 | -1
Decreasing order of bond angle is
View answer and explanation
Correct option: A
BeCl2>NO2>SO2BeC{l_2} > N{O_2} > S{O_2}
Option A is correct because the bond angle decreases with increasing lone pair repulsion, and BeCl2 has no lone pairs, NO2 has one, and SO2 has one.
220 Medium
220 of 342
Analyzing +4 | -1
Bond order of N2+N_2^{+}and N2−N_2^{-}are same. Which relation is correct for N2+N_2^{+}and N2−?N_2^{-}?
View answer and explanation
Correct option: B
Bond energy of N2+>N_2^{+}> Bond energy of N2−N_2^{-}
Option B is correct because the bond energy of N2+N_2^+ is greater than the bond energy of N2−N_2^- due to the electron being removed from a bonding orbital in N2+N_2^+ and added to an antibonding orbital in N2−N_2^-.
221 Medium
221 of 342
Understanding +4 | -1
Which does not contain sigma bond:
View answer and explanation
Correct option: C
B2B_2
Option C (B2B_2) is correct because the last molecular orbitals are π2px\pi 2 p x and π2py\pi 2 p y, which form only π\pi-bonds, and thus it does not contain a sigma bond.
222 Medium
222 of 342
Analyzing +4 | -1
Assertion : ClF3{ClF}_3 has T\mathrm{T}-shape structure.

Reason : It has two lone pair arrange at 180∘180^{\circ} (Angle).
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
The correct answer is C (If Assertion is True but the Reason is False) because ClF3{ClF}_3 does have a T-shape structure, but the two lone pairs are arranged in the equatorial positions to minimize repulsion, not at 180 degrees.
223 Medium
223 of 342
Analyzing +4 | -1
Which of the following is diamagnetic
View answer and explanation
Correct option: D
None
Option D (None) is correct because oxygen molecule, boron molecule, and N2+ ion all have unpaired electrons, making them paramagnetic, not diamagnetic.
224 Medium
224 of 342
Remembering +4 | -1
The decreasing order of bond dissociation energies of C−C,C−H\mathrm{C}-\mathrm{C}, \mathrm{C}-\mathrm{H} and H−H\mathrm{H}-\mathrm{H} bonds in
View answer and explanation
Correct option: A
H−H>−C−H>−C−C−\mathrm{H}-\mathrm{H}>-\mathrm{C}-\mathrm{H}>-\mathrm{C}-\mathrm{C}-
Option A (H−H>−C−H>−C−C−\mathrm{H}-\mathrm{H}>-\mathrm{C}-\mathrm{H}>-\mathrm{C}-\mathrm{C}-) is correct because the bond dissociation energy is related to bond strength, and smaller atoms generally form stronger bonds.
225 Medium
225 of 342
Understanding +4 | -1
Assertion (A): The order of thermal stability of

Li2CO3<Na2CO3<K2CO3<Rb2CO3<Cs2CO3\mathrm{Li}_2 \mathrm{CO}_3< \mathrm{Na}_2 \mathrm{CO}_3< \mathrm{K}_2 \mathrm{CO}_3< \mathrm{Rb}_2 \mathrm{CO}_3< \mathrm{Cs}_2 \mathrm{CO}_3

Reason (R): As we go

Li2CO3→Na2CO3→ K2CO3→Rb2CO3→Cs2CO3\mathrm{Li}_2 \mathrm{CO}_3 \rightarrow \mathrm{Na}_2 \mathrm{CO}_3 \rightarrow \mathrm{~K}_2 \mathrm{CO}_3 \rightarrow \mathrm{Rb}_2 \mathrm{CO}_3 \rightarrow \mathrm{Cs}_2 \mathrm{CO}_3

ionic character of carbonates increases.
View answer and explanation
Correct option: A
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Option A is correct because both the assertion and the reason are true, and the reason correctly explains the assertion regarding the thermal stability of alkali metal carbonates.
226 Medium
226 of 342
Analyzing +4 | -1
Given below are two statements:

Assertion (A): Double bond in carbon molecule (C2)\left(C_2\right) is consist of two pi-bonds.

Reason (R): Due to presence of four electrons in two pi-molecular orbitals.

In the light of the above statements, choose the most appropriate answer from the options given below:
View answer and explanation
Correct option: D
Both (A) and (R) are correct and (R) is the correct explanation of (A)
The correct answer is D (Both (A) and (R) are correct and (R) is the correct explanation of (A)) because the carbon molecule (C2C_2) indeed has a double bond consisting of two pi-bonds, which is due to the presence of four electrons in two pi-molecular orbitals.
227 Medium
227 of 342
Understanding +4 | -1
In which of the following hydrogen bond is present?
View answer and explanation
Correct option: B
Ice
Option B (Ice) is correct because hydrogen bonding is significant in ice due to the presence of hydrogen atoms bonded to highly electronegative oxygen atoms.
228 Medium
228 of 342
Remembering +4 | -1
Dipole-dipole interaction energy between stationary polar molecules is proportional to
View answer and explanation
Correct option: C
1r3\frac{1}{{{r^3}}}
Option C (1r3\frac{1}{{{r^3}}}) is correct because the dipole-dipole interaction energy between stationary polar molecules is inversely proportional to the cube of the distance between them.
229 Medium
229 of 342
Analyzing +4 | -1
N2{\mathrm{N}_2} and O2{\mathrm{O}_2} are converted into monocations, N2+\mathrm{N}_2^ + and O2+\mathrm{O}_2^ + respectively. Which of the following is wrong?
View answer and explanation
Correct option: D
N2+\mathrm{N}_2^ + become diamagnetic
Option D (N2+\mathrm{N}_2^ + become diamagnetic) is wrong because the removal of an electron from N2N_2 (which is diamagnetic) results in N2+N_2^+ which is paramagnetic due to the presence of an unpaired electron.
230 Medium
230 of 342
Remembering +4 | -1
Which has the highest bond energy?
View answer and explanation
Correct option: B
Triple bond
Option B (Triple bond) is correct because triple bonds have the highest bond energy due to the presence of three bonds between the atoms.
231 Medium
231 of 342
Analyzing +4 | -1
In which of the following, unpaired electrons are present? KO2,AlO2−,BaO2,NO2+\mathrm{K}{\mathrm{O}_2},\mathrm{AlO}_2^ - ,\mathrm{Ba}{\mathrm{O}_2},\mathrm{NO}_2^ +
View answer and explanation
Correct option: C
only KO2\mathrm{K}{\mathrm{O}_2}
The correct answer is C (only KO2\mathrm{K}{\mathrm{O}_2}) because KO2\mathrm{K}{\mathrm{O}_2} contains the superoxide ion which has one unpaired electron.
232 Medium
232 of 342
Analyzing +4 | -1
Among the following, the molecule of high dipole moment is
View answer and explanation
Correct option: C
H2O{H}_2 {O}
Option C (H2O{H}_2 {O}) is correct because it has the highest dipole moment among the given molecules due to its bent geometry and electronegativity difference between oxygen and hydrogen.
233 Medium
233 of 342
Analyzing +4 | -1
Of which of the following hybridisation is not the same as in the other three
View answer and explanation
Correct option: D
ClF3Cl{F_3}
Option D is correct because ClF3ClF_3 has sp3dsp^3d hybridization, while BrF5BrF_5, ICl4−ICl_4^-, and XeF4XeF_4 all have sp3d2sp^3d^2 hybridization.
234 Medium
234 of 342
Understanding +4 | -1
Which statement is wrong?
View answer and explanation
Correct option: D
dsp3ds{p^3}-hybrid orbitals are directed towards the corners of a regular tetrahedron
The correct answer is D (dsp3ds{p^3}-hybrid orbitals are directed towards the corners of a regular tetrahedron) because dsp3ds{p^3} hybridization leads to a square planar or other geometries, not a tetrahedral geometry. Tetrahedral geometry is associated with sp3sp^3 hybridization.
235 Medium
235 of 342
Understanding +4 | -1
The number of unpaired electrons in O2{\mathrm{O}_2} molecule is:
View answer and explanation
Correct option: C
22
Option C (2) is correct because the molecular orbital configuration of O2 has two unpaired electrons in the antibonding π* orbitals, making it paramagnetic.
236 Medium
236 of 342
Remembering +4 | -1
Which postulation provide the basis for the modern concepts regarding ion formation by electron transfer and the formation of ionic crystalline compounds?
View answer and explanation
Correct option: A
Kossel's postulations
Option A (Kossel's postulations) is correct because Kossel's theory explains ion formation through electron transfer and the formation of ionic compounds.
237 Medium
237 of 342
Analyzing +4 | -1
If Na+\mathrm{Na}^{+}ion is larger than Mg2+\mathrm{Mg}^{2+} ions and S2−\mathrm{S}^{2-} ion is larger than Cl−\mathrm{Cl}^{-}ion, which of the following will be stable salt in water?
View answer and explanation
Correct option: C
Magnesium sulphide
Option C (Magnesium sulphide) is correct because MgS has the highest lattice energy due to higher charges and smaller sizes, making it the most stable salt in water (least soluble).
238 Medium
238 of 342
Analyzing +4 | -1
N(SiH3)3\mathrm{N}\left(\mathrm{SiH}_3\right)_3 has -
View answer and explanation
Correct option: B
sp2\mathrm{sp}^2 hybridisation, planar shape
Option B is correct because N(SiH3)3\mathrm{N(SiH_3)_3} has sp2\mathrm{sp^2} hybridization and a planar shape due to back bonding from N to Si.
239 Medium
239 of 342
Understanding +4 | -1
Which of the following statement are correct:
(a) O2\mathrm{O}_2 is paramagnetic and O3\mathrm{O}_3 is diamagnetic
(b) O2\mathrm{O}_2 and B2\mathrm{B}_2 both are paramagnetic
(c) B2\mathrm{B}_2 and C2\mathrm{C}_2 both are paramagnetic
(d) O2\mathrm{O}_2 is paramagnetic but B2\mathrm{B}_2 is diamagnetic.
View answer and explanation
Correct option: A
a and b
Option A is correct because both statements (a) and (b) accurately describe the magnetic properties of oxygen and boron molecules based on their electronic configurations.
240 Medium
240 of 342
Analyzing +4 | -1
Which one of the following is not the resonance structure of CO2{CO}_2 ?
View answer and explanation
Correct option: D
O≡C=O{O} \equiv {C}={O}
The correct answer is D (O≡C=O{O} \equiv {C}={O}) because it does not represent a valid resonance structure of CO2 due to incorrect valency.
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