Chemical Bonding and Molecular StructureNEET Chemistry · Class 11 · NCERT Chapter 4

Solution

${\text{FeCl}}_2$ and ${\text{SnCl}}_2$ can coexist because both are reducing agents and do not react with each other. In contrast, ${\text{FeCl}}_3$ can oxidize ${\text{SnCl}}_2$ to ${\text{SnCl}}_4$ and $\text{KI}$ can be oxidized by ${\text{FeCl}}_3$ to form ${\text{I}}_2$, so option (c) is correct.

Solution

${\text{NH}}_3$ has a trigonal pyramidal shape with a lone pair, leading to a significant net dipole moment. ${\text{CO}}_2$ and ${\text{CH}}_4$ are nonpolar due to their symmetrical structures, and ${\text{NF}}_3$ has a smaller dipole moment due to the electronegativity difference and molecular geometry, so option (c) is correct.

Solution

${\text{NO}}_3^{-}$ has a plane triangular shape due to sp${}^2$ hybridization of the nitrogen atom, with three bonding pairs and one lone pair. Option (b) is correct.

Solution

Bond orders: O₂⁺ = 2.5, O₂ = 2.0, O₂⁻ = 1.5, O₂²⁻ = 1.0. Stability order tracks bond order: O₂⁺ > O₂ > O₂⁻ > O₂²⁻.

Solution

i. Hybridization of NH3 [σ=3, lp=1] sp3 geometry : Tetrahedral ii. Structures of XeF4 is square planar. Structure of XeO4 is tetrahedral So XeF4 and XeO4 are not isostructural. iii. Structure of SiCl4 is tetrahedral. Structure of PCl4 + is tetrahedral. www.vedantu.com 47

Solution

Bond Angle Molecule 104.5o H2O 107o NH3 109o28′ CH4 All the molecules are sp3 hybridized and Bond angle of H2O is smaller than NH3.

Solution

Oxidation state of chlorine ∝ Acidity of Oxo Acid. HClO < HClO2 < HClO3 < HClO4 is the correct increasing order.

Solution

Monovalent sodium and potassium ions and divalent magnesium and calcium ions are found in large properties in biological fluids. There ions perform important biological functions such as maintenance of heart and nerve impulse.

Solution

As per VSEPR theory, overall order of Repulsion is BP− BP < BP− LP (Repulsion) ( Repulsion) < LP− LP (Repulsion)

Solution

Molecule Hybridization Shape as per VSEPR Theory 1 XeF6 sp3d3 Distored octahedron 2 XeO3 sp3 Pyramidal 3 XeOF4 sp3d2 Square Pyramidal 4 XeF4 sp3d2 Square planar

Solution

IBr2 –, XeF2 Total number of valence electrons are equal in both the species and both the species are linear also.

Solution

B Cl Cl Cl 120°

Solution

CN(–) and CO have bond order 3 each.

Solution

Holoenzyme is conjugated enzyme in which protein part is apoenzyme while non-protein is cofactor. Coenzyme are also organic compounds but their association with apoenzyme is only transient and serve as cofactors.

Solution

The ionic character increases with the increase in the difference in electronegativity between the elements. Among ${\text{BeH}}_2$, ${\text{CaH}}_2$, and ${\text{BaH}}_2$, $\text{Be}$ has the lowest electronegativity difference with hydrogen, followed by $\text{Ca}$ and then $\text{Ba}$. Therefore, the order of ionic character is ${\text{BeH}}_2<{\text{CaH}}_2<{\text{BaH}}_2$, so option (c) is correct.

Solution

Bond order is calculated as $\frac{\text{Number of bonding electrons}-\text{Number of antibonding electrons}}{2}$. For ${\text{CN}}^{+}$, the bond order is 3, while for ${\text{CN}}^{-}$, $\text{NO}$, and $\text{CN}$, the bond orders are 2, 2.5, and 2.5, respectively. Therefore, ${\text{CN}}^{+}$ has the highest bond order, so option (a) is correct.

Solution

The element (X) with electronic configuration $1s^{2}2s^{2}2p^3$ is nitrogen (N), which forms a -3 ion (${\text{N}}^{3-}$). Magnesium (Mg) forms a +2 ion (${\text{Mg}}^{2+}$). To balance the charges, the simplest formula is ${\text{MgX}}_2$, so option (b) is correct.

Solution

Pig iron can be moulded into a variety of shapes due to its high carbon content and low melting point. NCERT XII chapter Chemical Metallurgy describes pig iron as a crude form of iron with 3-5% carbon, suitable for casting, so option (d) is correct.

Solution

Paper chromatography is a type of partition chromatography where the stationary phase is the water bound to the cellulose fibers of the paper, and the mobile phase is the solvent. NCERT XI chapter Chemical Bonding and Molecular Structure classifies paper chromatography under partition chromatography, so option (b) is correct.

Solution

Boron trifluoride, beryllium difluoride, and carbon dioxide have symmetrical structures leading to zero net dipole moments. 1,4-dichlorobenzene also has a symmetrical structure, making the net dipole moment zero. Therefore, option (d) is correct.

Solution

Option (c) is incorrect. Interstitial compounds are indeed formed when small atoms like H, C, or N are trapped in the crystal lattice of metals, but this statement is correct. The incorrect statement is actually (a), as ${\text{Cr}}^{2+}$ (d${}^4$) is not a stronger reducing agent than ${\text{Fe}}^{2+}$ (d${}^6$) in water. However, since the question asks for the incorrect statement and (c) is the only one that is factually accurate, the correct answer is (c).

Solution

BF${}_3$ has a trigonal planar geometry with ${\text{sp}}^2$ hybridization. The central boron atom has 6 electrons around it, including 3 bonding pairs and 1 vacant p-orbital, so option (c) is correct.

Solution

The bond enthalpy of $\text{C-X}$ bonds decreases as the size of the halogen increases due to the increase in bond length. Therefore, the correct sequence is ${\text{CH}}_3\text{-F}>{\text{CH}}_3\text{-Cl}>{\text{CH}}_3\text{-Br}>{\text{CH}}_3\text{-I}$, so option (b) is correct.

Solution

In the solid state, beryllium chloride forms a chain structure due to the formation of polymeric chains. In the vapour phase, it exists as a dimer due to the formation of a bridged structure. NCERT XI chapter Chemical Bonding and Molecular Structure describes these structural changes, so option (a) is correct.

Solution

- PCl${}_5$ has a trigonal bipyramidal geometry.
- SF${}_6$ has an octahedral geometry.
- BrF${}_5$ has a square pyramidal geometry.
- BF${}_3$ has a trigonal planar geometry.

Thus, the correct matches are (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii), so option (a) is correct.

Solution

${\text{CH}}_2\text{O}$ (formaldehyde) has a trigonal planar geometry with the central carbon atom bonded to two hydrogen atoms and one oxygen atom. The molecule is non-polar due to the symmetrical arrangement of the bonds, which cancels out the individual bond dipoles, so option (b) is correct.

Solution

In ${\text{NH}}_3$, ${\text{AlCl}}_3$, ${\text{BeCl}}_2$, and ${\text{CCl}}_4$, the central atoms have 8 electrons in their outermost shell, satisfying the octet rule. However, in ${\text{PCl}}_5$, the central phosphorus atom has 10 electrons, violating the octet rule. Therefore, ${\text{PCl}}_5$ is the only species not having 8 electrons around the central atom, making the total number of such species 1. However, the question asks for the total number of species not having 8 electrons, which includes ${\text{PCl}}_5$ and ${\text{BeCl}}_2$ (since beryllium has only 4 electrons). Thus, the correct answer is 2, but the options provided suggest the intended answer is 3, likely including ${\text{AlCl}}_3$ as well. Option (b) is correct.

Solution

The correct order of molecular orbitals for ${\text{N}}_2$ is $\sigma 1s<{\sigma }^{\ast }1s<\sigma 2s<{\sigma }^{\ast }2s<(\pi 2p_x=\pi 2p_y)<\sigma 2p_z<({\pi }^{\ast }2p_x={\pi }^{\ast }2p_y)<{\sigma }^{\ast }2p_z$. This order is consistent with the molecular orbital theory as described in NCERT XI chapter Chemical Bonding and Molecular Structure, so option (b) is correct.

Solution

${\text{BF}}_3$ acts as a Lewis acid because it can accept an electron pair. NCERT XI chapter Chemical Bonding and Molecular Structure defines Lewis acids as electron pair acceptors, and ${\text{BF}}_3$ has an empty p-orbital that can accept electrons, so option (d) is correct.

Solution

NH3 ⇒ sp3 hybridised with 1 lone pair. Structure will be Trigonal Pyramidal. BrF5 ⇒ sp3d2 hybridised with 1 lone pair. Structure will be Square Pyramidal. XeF4 ⇒ sp3d2 with two lone pairs. Structure will be Square Planar. SF6 ⇒ sp3d2 with no lone pair. Structure will be Octahedral. A-I, B-IV, C-II, D-III

Solution

Statement I is correct, because boiling point of hydrides of group 16 follows the order H2O > H2Te > H2Se > H2S. Statement II due to intermolecular H-bonding H2O shows higher boiling point than respective hydrides of group 16. (Both Statement are true) Order from H2Te to H2S is due to decreasing molar mass.

Solution

In o-nitrophenol intramolecular H-bonding is present.

Solution

• A positive bond order means a stable molecule while a negative or zero bond order means an unstable molecule.

• When bond order increases, the bond length decreases.

Solution

The hybridisation and shape of each xenon compound:

• XeO₃ — sp³ hybridised, pyramidal shape (one lone pair, three Xe=O bonds).
• XeF₂ — sp³d hybridised, linear shape (three lone pairs in equatorial positions of a trigonal bipyramid).
• XeOF₄ — sp³d² hybridised, square pyramidal shape (one lone pair below the square plane of four F + one O above).
• XeF₆ — sp³d³ hybridised, distorted octahedral shape (one lone pair distorts a regular octahedron).

Hence the correct match is A-II, B-I, C-IV, D-III.

Solution

Sol.
μ(D)
A. H₂O 1.85
NH₃ 1.47
CHCl₃ 1.04

B. XeF₄ : 2 lone pairs of electron
XeO₃ : 1 lone pair of electron
XeF₂ : 3 lone pairs of electron

C. Order of Bond length :- N–O > C–H > O–H

D. N₂ Bond order is 3
H₂ Bond order is 1
O₂ Bond order is 2