Q: The option(s) in which at least three molecules follow Octet Rule is(are)

Option A: $ CO _2, C _2 H _4, NO $ and HCl 1. $ CO _2$ : Carbon dioxide follows the octet rule. Carbon forms double bonds with two oxygen atoms, allowing it to have 8 electrons in its valence shell. 2. $ C _2 H _4$ : Ethylene follows the octet rule. Each carbon forms a double bond with the other carbon and single bonds with two hydrogen atoms, resulting in 8 electrons in the valence shell of the carbons. 3. NO: Nitric oxide does not follow the octet rule. It has an odd number of electrons,

Q: Thermal decomposition of AgNO 3 produces two paramagnetic gases. The total number of electrons present in the antibonding molecular orbitals of the gas that has the higher number of unpaired ele

2 AgNO 3 s ⟶ s 2 Ag s + 2 NO 2 g + O 2 g Both the NO 2 and O 2 gases are paramagnetic. NO 2 g has 1 unpaired electron and O 2 g has 2 unpaired electrons. According to MOT, electronic configuration of O 2 is σ 1 s 2 σ * 1 s 2 σ 2 s 2 σ * 2 s 2 σ 2 p 2 2 π 2 p x 2 = π 2 λ y 2 π * 2 p x 1 = π * 2 ρ y 1 Total number of electrons present in antibonding molecular orbitals = 6

Question 1

The correct order of dipole moments for the given species is

Accepted Answer

The dipole moment of a molecule depends on its geometry and the vector sum of its individual bond dipoles. BF _3 has a symmetrical trigonal planar geometry. The three B-F bond dipoles cancel each other out completely, resulting in a net dipole moment of zero ( = 0). NH _4^+ has a symmetrical tetrahedral geometry. The four N-H bond dipoles cancel each other out, resulting in a net dipole moment of zero ( = 0). Thus, ( BF _3) = ( NH _4^+) = 0. Both NH _3 and NF _3 have a trigonal pyramidal geometr

Question 2

Consider the following species: SOCl _2, XeOF _4, ClF _3, ClF _5, XeF _5^+, SO _3^ 2- , XeF _3^+, SF _4 List-I contains different molecular shapes and List-II contains total number of species with the

Accepted Answer

Let us determine the hybridization and shape of each given species by finding the number of bond pairs (bp) and lone pairs (lp) on the central atom: SOCl _2: Central atom S has 6 valence electrons. It forms 3 bonds (one with O, two with Cl) and has 1 lone pair. Steric number = 4 (sp^3). Shape is Trigonal Pyramidal. XeOF _4: Central atom Xe has 8 valence electrons. It forms 5 bonds (one with O, four with F) and has 1 lone pair. Steric number = 6 (sp^3d^2). Shape is Square Pyramidal. ClF _3: Centr

Question 3

The correct order of ONO bond angle in the given species is

Accepted Answer

For NO _2^+, the central nitrogen atom is sp hybridized with no lone pairs, resulting in a linear geometry and a bond angle of 180^ . For NO _2, the central nitrogen atom is sp^2 hybridized with one unpaired electron. The repulsion from a single electron is less than that from a bond pair, so the bond angle opens up to be greater than 120^ (approximately 134^ ). For NO _3^-, the central nitrogen atom is sp^2 hybridized with no lone pairs, resulting in a trigonal planar geometry and a bond angle

Question 4

Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT statement(s) is(are)

Accepted Answer

(i) Ne _2 ( 1 ~s ^2 ) ( ^* 1 ~s ^2 ) ( 2 ~s ^2 ) ( ^* 2 ~s ^2 ) ( 2 p _ z ^2 ) ( 2 p _ x ^2= 2 p _ y ^2 ) ( ^* 2 p _ x ^2= ^* 2 p _ y ^2 ) ( ^* 2 p _ z ^2 ) B.O. = 6-6 2 =0 (ii) F _2 ( 1 ~s ^2 ) ( ^* 1 ~s ^2 ) ( 2 ~s ^2 ) ( ^* 2 ~s ^2 ) ( 2 p _ z ^2 ) ( 2 p _ x ^2= 2 p _ y ^2 ) ( ^* 2 p _ x ^2= ^* 2 p _ y ^2 ) (iii) O _2^ ( 1 s^2 ) ( ^* 1 s^2 ) ( 2 s^2 ) ( ^* 2 s^2 ) ( 2 p_z^2 ) ( 2 p_x^2= 2 p_y^2 ) ( ^* 2 p_x^1= ^* 2 p_y ) B.O. = 6-1 2 =2.5 O _2 ( 1 s^2 ) ( ^* 1 s^2 ) ( 2 s^2 ) ( ^* 2 s^2 ) ( 2

Question 5

The correct statements (s) about intermolecular forces is(are)

Accepted Answer

(i) Ion - Ion Interaction energy 1 r Ion - dipole Interaction energy 1 r^2 Ion - dipole Interaction energy approaches zero more rapidly as ' r ' increases (ii) Rotating Polar molecules Interaction energy 1 r ^6 . (iii) Dipole - induced dipole forces are independent of temperature. (iv) Non-polar species show London dispersion forces.

Question 6

The option(s) in which at least three molecules follow Octet Rule is(are)

Accepted Answer

Option A: $ CO _2, C _2 H _4, NO $ and HCl 1. $ CO _2$ : Carbon dioxide follows the octet rule. Carbon forms double bonds with two oxygen atoms, allowing it to have 8 electrons in its valence shell. 2. $ C _2 H _4$ : Ethylene follows the octet rule. Each carbon forms a double bond with the other carbon and single bonds with two hydrogen atoms, resulting in 8 electrons in the valence shell of the carbons. 3. NO: Nitric oxide does not follow the octet rule. It has an odd number of electrons,

Question 7

Based on VSEPR model, match the xenon compounds given in List-I with the corresponding geometries and the number of lone pairs on xenon given in List-II and choose the correct option.

Accepted Answer

XeF _2 2 sigma bonds and 3 lone pairs on Xe , number of hybrid orbitals =5, sp ^3 ~d hybridisation geometry will be trigonal bipyramidal. P - 5 XeF _4 4 sigma bonds and 2 lone pairs on Xe , number of hybrid orbitals =6, sp ^3 ~d ^2 hybridisation , geometry will be octahedral. Q - 3 XeO _3 3 sigma bonds and 1 lone pairs on Xe , number of hybrid orbitals =4, sp ^3 hybridisation , geometry will be tetrahedral. R - 2 XeO _3 ~F _2 5 sigma bonds and 0 lone pairs on Xe , number of hybrid orbitals =5, s

Question 8

The correct molecular orbital diagram for F 2  molecule in the ground state is

Accepted Answer

The electronic configuration of fluorine atom is  1 s 2 2 s 2 2 p 5 . The fluorine molecule has 18  electrons. The s - p mixing occurs when the s  and p  orbitals have similar energies. The energy difference between 2 s &#160; and &#160; 2 p orbitals in O , &#160; F , &#160; and &#160; Ne is greater than that in Li , &#160; Be , &#160; B , &#160; C , &#160; and &#160; N . Hence, the molecular orbital electronic configuration of  F 2  is  &#963; 1 s 2 &#963; * 1

Question 9

Among I 3 + , SiO 4 4 - , SO 2 Cl 2 , XeF 2 , SF 4 , ClF 3 , Ni CO 4 , XeO 2 F 2 , PtCl 4 2 - , XeF 4 , and SOCl 2 , the total number of species having sp 3  hybridised central atom is

Accepted Answer

I 3 + ion has sp 3 hybridisation and bent shape containing two bond pairs and two lone pairs. Structure of SiO 4 4 &#8722; is tetrahedral structure and Si  is sp 3 hybridised. The molecular geometry of SOCl 2 is trigonal pyramidal and its electron geometry is tetrahedral. In this molecule sulphur undergo  sp 3  hbridisation, XeF 2 molecule is linear molecule but it is sp 3 d hybridised. In  SF 4 , sulphur undergo  sp 3 d  hybridisation and molecule ha

Question 10

Thermal decomposition of AgNO 3  produces two paramagnetic gases. The total number of electrons present in the antibonding molecular orbitals of the gas that has the higher number of unpaired ele

Accepted Answer

2 AgNO 3 s &#10230; s 2 Ag s + 2 NO 2 g + O 2 g Both the NO 2  and O 2  gases are paramagnetic. NO 2 g has 1 unpaired electron and O 2 g has 2 unpaired electrons. According to MOT, electronic configuration of O 2  is &#963; 1 s 2 &#160; &#963; * 1 s 2 &#160; &#963; 2 s 2 &#160; &#963; * 2 s 2 &#160; &#963; 2 p 2 2 &#160; &#960; 2 p x 2 = &#960; 2 &#955; y 2 &#160; &#960; * 2 p x 1 = &#960; * 2 &#961; y 1 Total number of electrons present in antibonding molecular orbitals = 6

Question 11

For diatomic molecules, the correct statement(s) about the molecular orbitals formed by the overlap of two 2 p z orbitals is(are)

Accepted Answer

When this orbital combine along the internuclear axis then  &#963;  orbital is formed while this orbital approach each other parallel to each other and perpendicular to the internuclear axis  &#960;  orbital is formed.

Question 12

Consider the following compounds in the liquid form: O 2 , HF , H 2 O , NH 3 , H 2 O 2 , CCl 4 , CHCl 3 , C 6 H 6 , C 6 H 5 Cl . When a charged comb is brought near their flowing stream, how many of t

Accepted Answer

Polar compound show deflection by charged comb So,  HF , &#160; H 2 O , &#160; NH 3 , &#160; H 2 O 2 , &#160; CHCl 3 , &#160; C 6 H 5 Cl  shows deflection by charged comb.

Question 13

In an experiment, m grams of a compound X (gas/liquid/solid) taken in a container is loaded in a balance as shown in figure I below. In the presence of a magnetic field, the pan with X is either defle

Accepted Answer

Paramagnetic substance attracted by magnetic field so magnetic balance show downward deflection. While diamagnetic substance show repulsion in magnetic field show magnetic balance show upward deflection.   Compound Magnetic Nature Deflection ( a ) H 2 O Diamagnetic Upward ( b ) K 4 Fe ( CN ) 6 Diamagnetic Upward ( c ) O 2 ( g ) Paramagnetic Downward ( d ) C 6 H 6 Diamagnetic Upward

Question 14

A tin chloride Q undergoes the following reactions (not balanced) Q + C l - &#8594; X Q + M e 3 N &#8594; Y Q + C u C l 2 &#8594; Z + C u C l X is a monoanion having pyramidal geometry. Both Y and Z a

Accepted Answer

Sn has vacant d orbitals in tin(II)chloride which can be used to accommodate electron pairs and form coordinate bond. LP = Lone pairs X is mono anion with sp 3 hybridisation and It has trigonal pyramidal geometry. The above reaction is a Lewis acid-base reaction. The adduct Y formed is a coordination compound, and it is a neutral compound. The central atom in Z is tin, its oxidation state is +4, and it has no lone pair electrons. Z is a neutral compound.

Question 15

Each of the following options contains a set of four molecules. Identify the option(s) where all four molecules possess permanent dipole moment at room temperature.

Accepted Answer

Answer option index: 3

Question 16

Among B 2 H 6 , &#160; B 3 N 3 H 6 , &#160; N 2 O , &#160; N 2 O 4 , &#160; H 2 S 2 O 3 and H 2 S 2 O 8 ,  the number of molecules containing covalent bond between two atoms of the same kind is _

Accepted Answer

Answer option index: 0

Chemical Bonding and Molecular Structure — Previous Year Questions & Solutions