Chemical Bonding Narayana.pdf 2r1z6d

CONTENTS

1. Introduction of Bonding. 2. KOSSEL-LEWIS approach of Chemical Bonding (Electronic Concept of Chemical Bonding). 3. Type of Bonds.

Ionic Lattice energy

Covalent

Coordinate covalent

Dipole moment Formal charge

4. Lewise Structures of Molecules. 5. Valency Bond Theory. 6. Hybridisation. 7. V SEPR Theory. 8. Back Bonding 9. Molecular Orbital Theory. 10. Hydrogen Bonding.

IIT Package (2009-10) from NOIDA

CHEMICAL BONDING & MOLECULAR STRUCTURE

Chemical Bond is the physical process responsible for the attractive interactions between atoms and molecules, and that which confers stability to diatomic and polyatomic chemical compounds. The explanation of the attractive forces is a complex area that is described by the laws of quantum electrodynamics. In general, strong chemical bonding is associated with the sharing or transfer of electrons between the participating atoms. The study on the “nature of forces that hold or bind atoms together to form a molecule” is required to gain knowledge of the following i) To know about how atoms of same element form different compounds combining with different elements. ii) To know why particular shapes are adopted by molecules. iii) To understand the specific properties of molecules or ions and the relation between the specific type of bonding in the molecules.

Kossel-Lewis approach to Chemical Bonding In 1916, W.Kossel and G.N.Lewis, separately developed theories of chemical bonding inorder to understand why atoms combined to form molecules. Lewise introduced simple symbols to denote the electron present in the outer orbit of atom , these electron are known as valence electrons. These symbols are known as electron dot symbols and the structure of compound is known as Lewis dot structure. H H C H The number of dots around the symbol The electron dot structure of on then is given as H equal to number of electrons.

Ionic Bond or Electrovalent Bond An ion is an atom or group of atoms which has acquired charge due to the loss or gain of one or more electrons. When an atom gains an electron to form a negative ion (anion), it will increase in size. On the other hand, when an atom loses an electron to give positive ion (cation), it will contract. The electron lost or gained is always from the outermost shell. When two atoms, one of which can lose one or more electrons to attain a noble gas configuration and the other can receive these electrons and thereby acquire a noble gas configuration, they are said to be bonded by an ionic bond. Since the loss and gain of electrons by atoms results in the formation of ions, ionic bond is formed when two ions interact with each other and are thus held together by electrostatic attraction. The formation of potassium chloride (KCl), is illustrated below. +

K (1s 2 2s 2 2p 6 3s 2 3p 6 4s1 ) ¾loses ¾ ¾® K (1s 2 2s 2 2p 6 3s 2 3p 6 ) 1 electron

gains Cl (1s 2 2s 2 2p6 3s 2 3p5 ) ¾¾¾ ® Cl- (1s 2 2s 2 2p6 3s 2 3p 6 ) 1 electron

(Ar configuration)

From the above illustrations, it is clear that the formation of an ionic compound is obviously related to the ease of formation of the cations and anions from the neutral atom, which depends on two main factors: i) Ionization energy: Lower the value of ionization energy of an atom, greater will be the ease of formation of the cation from it. ii) Electron affinity: Higher the electron affinity of an atom, greater the ease of formation of the anion from it. [2 of 35]

IIT Package (2009-10) from NOIDA

Lattice Energy When one mole of an ionic solid is formed from its constituent gaseous ions, the energy released is called the lattice energy. Energetics of Formation of Ionic Substances: The energy included in the formation of an ionic compound from its constituent elements may be considered as shown by the Born-Haber Cycle for the formation of one mole of sodium chloride from sodium and chlorine.

Na ( s ) ¾Sublimatio ¾ ¾¾n ® Na ( g ) ¾ ¾® Na (+g ) + e +S

+I

1 Dissociation Addition of eCl2(g) ¾¾¾¾¾® Cl(g) ¾¾¾¾¾¾ ® Cl(g) +1/ 2D* - EA 2 + Na (g) + Cl(g) ¾¾¾¾¾¾¾ ® NaCl(S) Crystal formation -U

Where S = heat of sublimation of sodium metal I = ionization energy of sodium D = heat of dissociation of molecular chlorine EA = electron affinity of chlorine, and U = lattice energy of sodium chloride The amount of heat liberated in the overall reaction is the heat of formation of sodium chloride. From the above DH¦ = S + I +

1 D – EA – U 2

The most important of these energy are I, EA and U, since these are considerably greater than the remaining S and D. More the negative value of the heat of formation, greater would be the stability of the ionic compound produced. Thus on the basis of the above equation, formation of an ionic compound is favoured by a) low ionization energy (I) of the metal. b) high electron affinity (EA) of the other element. c) higher lattice energy (U) of the resulting compound. Formation of Ions with Higher Charges: Formation of a cation with unit positive charge is easy if the first ionization energy is low as in the case of alkali metals. Alkaline earth metals ionizes in two successive steps. Mg ¾ ¾® Mg+ + e– Mg+ ¾ ¾® Mg2+ + e– But energy needed to ionize alkaline earth metals are higher than alkali metals. However, bipositive ions like Mg2+, Ca2+, Sr2+ and Ba2+ are quite common. Formation of a tripositive ion like Al3+ requires much more energy (= 5138 kJ) which is not available ordinarily. Successive ionization energies of aluminium are: E1 E1 = 577kJ Al ¾¾® Al + + e E2 Al+ ¾¾ ® Al2 + + e -

E2 = 1816kJ

E3 E3 = 2745kJ Al2+ ¾¾ ® Al3+ + e It is on this that most of aluminium compounds are covalent. In solution, however, aluminium is known to give hydrated ions [Al.6H2O]3+. This is possible because of the high heat of hydration of Al3+. The energy liberated during hydration of ions is sufficient for ionization. Similarly, anions with unit negative charge (e.g. Cl–1, Br–, I–) are very common. This is because the electron affinity of these atoms is positive and quite high. Formation of anions carrying two units of negative charge (e.g. S2–, O2–) is not so easy as their electron affinities are negative i.e., energy is needed to add second electron. Formation of anions carrying three units of negative charge (e.g. N3–, P3–) is almost rare.

[3 of 35]

IIT Package (2009-10) from NOIDA

Characteristics of Electrovalent Compounds Melting and Boiling Point: Due to the strong electrostatic force between the ions in a crystal of an electrovalent compound.The energy needed to overcome these forces and break down the crystal lattice. Hence such compounds possess high melting and boiling points. Electrical Conductivity: When an electrovalent compound is molten or dissolved in a solvent of high dielectric constant e.g., water, the binding forces in the crystal lattice disappear and the component ions become mobile. Under the influence of applied electrical field, the ions get charged and thus act as charge carrier of the current. Hence their molten forms or solutions conduct electricity. Solubility: Ionic compounds are soluble in polar solvents like water because of molecules of the polar solvent interact strongly with the ions of the crystal and the solvation energy is sufficient to overcome the attraction between the ions in the crystal lattice. Dissolution is also favoured by the high dielectric constant of the solvents such as water, since this weakens the interionic attractions in the resulting solutions. Non-polar solvents like benzene and carbon tetrachloride do not solvate the ions as their dielectric constants are low. Ionic compounds are, therefore insoluble in non-polar solvents. Ionic compounds like sulphates and phosphates of barium and strontium are insoluble in water (because lattice energy is greater than hydration energy). This can be attributed to the high lattice energies of these compounds due to polyvalent nature of both the cation and the anion. In these cases, hydration of ions fails to liberate sufficient energy to offset the lattice energy. Illustration 1 : which one is correct for ionic bond (a) it is nondirectional (b) it is formed by the elements with same electronegativity (c) it is formed by overlapping of orbitals (d) Both (a) and (c) are correct Ans : a Solution: Ionic bond is Illustration 2: which one having highest hydration energy (a) Na+ (b) Li+ (c) Cs + Ans : (b) Li+ in has highest lattice energy dute to small size

(d) K+

Bond angle The angle between two bonds sharing a common atom is known as the bond angle. Bond angle depends on the geometry of the molecule. For example the bond angle in tetrahedral geometry is 109028’ while bond angle in planer geometry is 1200 . bond angle is also depands on the identity of atoms bonded for example the geometry of NH3 and NF3 is same but bond angle of NH3 is greater than NF3 because F has higher electronegativity than H, the electron pair is attracted more towards F in NF3 i.e. the bond pairs of electrons are away from N or in other words distance between bond pairs ion is more. Hence repulsion between bond pairs in NF3 is less than NH3. Hence the lone pair repels the bond pairs of NF3 more than it does in NH3. As a result, the bond angle decreases to 102.4°. Whereas in NH3 it decreases to 107.3° only.

N

N H

H H 107.3°

F

F F 102.4°

Illustration : 3 Explain why bond angle of PH3 is less than that of PF3. Solution : PH3 and PF3 are pyramidal in shape with one lone pair on P. But PF3 has greater bond angle than PH3 (opposite to NH3 and NF3). This is due to resonance in PF3, leading to [4 of 35]

IIT Package (2009-10) from NOIDA

partial double bond character as shown below P F

F

+

F

As result repulsions between P – F bonds are large and hence the bond angle is large. There is no possibility of formation of double bonds in PH3.

Covalent Bond (By Mutual Sharing of Electrons) The covalent bond is formed when two atoms achieve stability by the sharing of an electron pair, each contributing one electron to the electron pair. The arrangement of electrons in a covalent molecule is often shown by a Lewis structure in which only valency shells (outer shells) are depicted. For sake of clarity, the electrons on different atoms are denoted by dots and crosses. Polarity of Bonds: A covalent bond is set up by sharing of electrons between two atoms. It is further classified as polar or non-polar depending upon the fact whether the electron pair is shared unequally between the atoms or shared equally. For example, the covalent bonds in H2 and Cl2 are called non-polar as the electron pair is equally shared between the two atoms.

H

H

+d

Cl Cl

Hydrogen molecule

-d

Chlorine molecule

H

(Both formed by equal sharing of electrons between the atoms, i.e., by non-polar bonds)

F d

In the case of hydrogen fluoride the bond is polar as the electron pair is unequally shared. Fluorine has a greater attraction for electrons or has higher electronegativity than hydrogen and the shared pair of electrons is nearer to the fluorine atom than hydrogen atom. The hydrogen end of the molecule, therefore, appears positive with respect to fluorine. Bond polarities affect both physical and chemical properties of compounds containing polar bond. The polarity of a bond determines the kind of reaction that can take place at that bond and even affects the reactivity at nearby bonds. The polarity of bonds can lead to polarity of molecules and affect melting point, boiling point and solubility

Characteristics of Covalent Compounds Melting Point and Boiling Point: In covalent compounds, except those consisting of giant molecules, the molecules are less powerfully attracted to each other, as a result of which their melting points and boiling points are relatively low compared to ionic compounds, e.g.,

SiCl 4 (b.p. = 33K) and NaCl (b.p. = 1713K) (Covalent compound)

(ionic compound)

Conductivity: Covalent substances (whether of the “molecular lattice” or “giant molecule” type) do not conduct electricity in the fused state since there are no free electrons or ions to carry the current. However, substances like graphite which consists of separate layers conduct electricity because the electrons have a age in between the two flat layers. Solubility: The characteristic solubility of covalent compounds in non-polar solvents such as benzene and carbon tetrachloride can be described to the similar covalent nature of the molecules of solute and solvent (i.e., like dissolves like). Covalent compounds in solution react more slowly as compared with the ionic compounds which react instantaneously in solution. The solubility of covalent compounds is, however, very much dependent upon the size of the molecule. Thus covalent substances having giant molecules are insoluble in virtually all solvents due to the big size of the molecule unit.

Fajan’s Rules When two oppositely charged ions approach each other closely, the positively charged cation attracts the outermost electrons of the anion and repel its positively charged nucleus. This [5 of 35]

IIT Package (2009-10) from NOIDA

results in the distortion or polarization of the anion followed by some sharing of electrons between the two ions, i.e., the bond becomes partly covalent in character. The magnitude of polarization depends upon following factors : i) Charge on Either of the ions: As the charge on the cation increases, its tendency to polarize the anion increases. This brings more and more covalent nature in the electrovalent compound. Whereas with the increasing charge of anion, its ability to get polarized, by the cation, also increases. For example, in the case of NaCl, MgCl2 and AlCl3 the polarization increases, thereby covalent character becomes more and more as the charge on the cation increases. Similarly, lead forms two chlorides PbCl2 and PbCl4 having charges +2 and +4 respectively. PbCl4 shows covalent nature. Similarly among NaCl, Na2S, Na3P, the charge of the anions are increasing, therefore the increasing order of covalent character. NaCl < Na2S < Na3P ii) Size of the cation: Polarisation of the anion increases as the size of the cation decreases i.e., the electrovalent compounds having smaller cations show more of the covalent nature. For example, in the case of halides of alkaline earth metals, the covalent character decreases as we move down the group. Hence melting point increases in the order of BeCl2 < MgCl2 < CaCl2 < SrCl2 < BaCl2 iii) Size of anion: The larger the size of the anion, more easily it will be polarized by the cation i.e., as the size of the anion increases for a given cation, the covalent character increases. For example, in the case of halides of calcium, the covalent character increases from F– anion to I– anion i.e. CaF < CaCl < CaBr < CaI increasing covalent character

2 2 2 ¾¾ ¾ ¾ ¾¾ ¾ ¾¾ ¾¾ ¾2 ®

Similarly, in case of trihalides of aluminium, the covalent character increases with increase in size of halide anion i.e. AlF3

AlCl3

AlBr3

AlI3

Covalent character increases as the size of the halide ion increases

iv) Nature of the cation: Cations with 18 electrons (s2p6d10) in outermost shell polarize an anion more strongly than cations of 8 electrons (s2p6) type. The d electrons of the 18 electron shell screen the nuclear charge of the cation less effectively than the s and p electrons of the 18-electron shell. Hence the 18-electron cations behave as if they had a greater charge. Copper (I) and Silver (I) halides are more covalent in nature compared with the corresponding sodium and potassium halides although charge on the ions is the same and the sizes of the corresponding ions are similar. This illustrates the effect of 18-electron configuration of Cu+ (3s2, p6, d10) and Ag+ (4s2, p6, d10) ions. Illustration : 4 The decomposition temperature of Li2CO3 is less than that of Na2CO3. Explain. Solution: As Li+ ion is smaller than Na+ ion, thus small cation (Li+) will favour more covalent character in Li2CO3 and hence it has lower decomposition temperature than that of Na2CO3. Dipole Moment: It is vector quantity and is defined as the product of the magnitude of charge on any of the atom and the distance between the atoms. It is represented by m . Magnitude of dipole moment | m |= ( chargeqin esu)´ ( Dis tanrce in A ) The unit = 10–18 (esu) cm (D) is used in practice. In SI units charge q is measured in coulombs (C) and the distance, r in metre, m 1C = 2.998 × 109 esu and 1 m = 102 cm \ 1 Cm = 2.998 × 109 × 102 = 2.998 × 1011 (esu) cm Therefore in SI system, the unit of dipole moment is coulomb metre [6 of 35]

IIT Package (2009-10) from NOIDA

\

1D =

2.998 ´ 1011 = 2.998 ´ 10 29 D 1 Cm = -18 10

or

1 = 3.336 ´10 -30 Cm 29 2.998 ´10

Dipole moment is a vector quantity and is often indicated by an arrow parallel to the line ing the point of charge and pointing towards the negative end e.g., H

F .

Experimental dipole moment

% Ionic character of a covalent bond = Theoretical dipole moment assuming 100% ionic character ´ 100 Illustration : 5 The dipole moment of KCl is 3.336 × 10–29 Cm. The interatomic distance + – K and Cl ion in KCl is 260 pm. Calculate the dipole moments of KCl, if there were opposite charges of the fundamental unit located at each nucleus. Calculate the percentage ionic character of KCl. Solution: From the given data q = 1.602 × 10–19C r = 260 pm = 260 × 10–12 m = 2.6 × 10–10 m Magnitude of dipole moment for 100% ionic character |m| = qr = (1.602 × 10–19) (2.6 × 10–10) = 4.165 × 10–29 Cm Actual dipole moment = 3.336 × 10–29 Cm \ % of ionic bond =

3.336 ´ 10 -29 ´ 100 = 80.1% 4.165 ´ 10 -29

The bond is 80.1% ionic. In general a polar bond is established between two atoms of different radii and different electronegativities while positive centres (nuclei) of different magnitudes combine to share an electron pair. Greater the values of the dipole moment, greater is the polarity of the bond. The following points may be borne in mind regarding dipole moments: i) In case a molecule contains two or more polar bonds, its dipole moment is obtained by the vectorial addition of the dipole moments of the constituent bonds. ii) A symmetrical molecule is non-polar even though it contains polar bonds. For example, carbon dioxide, methane and carbon tetrachloride, being symmetrical molecules, have zero dipole moments. Dipole moment of methyl chloride is a vectorial addition of dipole moments of three C – H bonds and one C – Cl bond. H

H

m = 1.75d

Cl

C

F

C H

H H

m = 0D Hydrogen fluoride

Cl

Methane

C Cl

Cl Cl

H

H H

m = 0D

m = 1.86D

Carbon tetrachloride

Methyl chloride

Dipole moments of some molecules

Dipole moment gives valuable information about the structure of molecules. For example, carbon dioxide is assigned a linear structure since its dipole moment is zero. We have seen that in a polar covalent bond between two atoms (say A and B), there is a partial separation of charge. This bond is, therefore, said to have a partial ionic character. Greater the difference of electronegativity between A and B, greater is the degree of ionic character (or polarity measured by dipole moment of AB) of the bond. Pauling gave a fairly accurate rule by which the nature of the bond can be predicted. According to this rule, “If the difference on the electronegativity scale between the two atoms is 1.9, the bond is 50% ionic in character. When [7 of 35]

IIT Package (2009-10) from NOIDA

the difference is greater than 1.9, the bond is correspondingly more ionic”. For example, when the electro negativity difference is 0.8, 1.2, 2.2 and 2.6, the corresponding partial ionic character is 12%, 25%, 61% and 74% respectively. Illustration : 6 The resultant dipole moment of water is 1.85D ignoring the effects of lone pair. Calculate, the bond moment of each OH bond {given that bond angle in H2O = 104°, cos 104° = –0.25} Solution: R2 = P2 + Q2 + 2PQ cos q

æ 1ö è 4ø

(1.85)2 = x2 + x2 + 2x2 ç - ÷ (1.85)2 = 2x2 –

x2 3x 2 Þ 2 2

\

x = 1.51D

Coordinate Bond It is a special type of covalent bond in which both the shared electrons are contributed by one atom only. It may be defined as “a covalent bond in which both electrons of the shared pair are contributed by one of the two atoms”. Such a bond is also called as dative bond. A coordinate or a dative bond is established between two such atoms, one of which has a complete octet and possesses a pair of valence electrons while the other is short of a pair of electrons. xx

+ B xx xx

A

xx Bx xx x

A

or

A

B

This bond is represented by an arrow (®) pointing towards acceptor atom. The atom which contributes electron pair is called the donor while the atom which accepts it is called acceptor. The compound consisting of the coordinate bond is termed coordinate compound. Some examples of coordinate bond formation are given below: i) Formation of ammonium ion: Hydrogen ion (H+) has no electrons and thus accepts a lone pair donated by nitrogen. H

H H

N H

+ H

+

H

N

H

H

ii) Formation of CO: Carbon has four valency electrons and oxygen has six. They combine to form two double bond and a coordinate bond as to achieve their octet completed. C

xx + xxOxx

C

Oxx

Acceptor Donor

Characteristics of Coordinate Compounds: The properties of coordinate compounds are intermediate between the properties of electrovalent compounds and covalent compounds. The main properties are described below: i) Melting and Boiling Points: Their melting and boiling points are higher than purely covalent compounds and lower than ionic compounds. ii) Solubility: These are sparingly soluble in polar solvents like water but readily soluble in nonpolar (organic) solvents. iii) Conductivity: Like covalent compounds, these are also bad conductors of electricity. The solutions or fused mass do not allow the age of electricity. Illustration : 7 Which of the following compound contains both ionic as well as covalent bonds only . (a) NaCl (b) NaOH (c) CH4 (d) NH4Cl + Ans: (b) NaOH conations an ionic bond between Na and OH and a covalent bond between Oand H [8 of 35]

IIT Package (2009-10) from NOIDA

Valence Bond Theory This theory was proposed by Linus Pauling, who was awarded the Noble Prize for chemistry 1954. Atoms with unpaired electrons tend to combine with other atoms which also have unpaired electrons. In this way the unpaired electrons are paired up, and the atoms involved, all attain a stable electronic arrangement. This is usually a full shell of electrons(i.e. a noble gas configuration). Two electrons shared between two atoms constitute a bond. The number of bonds formed by an atom is usually the same as the number of unpaired electrons in the ground state, i.e. the lowest energy state. However, in some cases the atom may form more bonds than this. This occurs by excitation of the atom (i.e. providing it with energy) when electrons which were paired in the ground state are unpaired and promoted into suitable empty orbitals. This increases the number of unpaired electrons, and hence it increases number of bond which can be formed. A covalent bond results from the pairing of electrons (one from each atom). The spins of the two electrons must be opposite (antiparallel) because of the Pauli exclusion principle that no two electrons in one atom can have all four quantum numbers the same. 1. In HF, H has a singly occupied s-orbital that overlaps with a singly filled 2p orbital on F. 2. In H2O, the O atom has two singly filled 2p orbitals, each of which overlaps with a single occupied s-orbital from two H atoms. 3. In NH3, there are three singly occupied p orbitals on N which overlap with s orbitals from three H atoms. 1

4. In CH4, the C atom in its ground state has the electronic configuration 1s2, 2s2, 2p1x , 2p y and only has two unpaired electrons, and so can form only two bonds. If the C atom is excited, 1

then the 2s electrons may be unpaired, giving 1s2, 2s1, 2p1x , 2p y , 2p1x . There are now four unpaired electrons which overlap with singly occupied s orbitals on four H atoms. 2p 2s

2px

2py 2pz

Electronic structure of carbon atom - groun state 1s

2s

2p

Carbon atom - excited state Carbon atom having gained four electrons from H atoms in CH4 molecule.

sp3 hybridisation

CH4 molecule uses its three p-orbitals px, py and pz, which are mutually at right angles to each other, and the s orbital is spherically symmetrical. Hence they form tetrahedral structure. CH4 H – C – H = 109°28¢ Sigma and Pi Bonds (S and P Bonds) A covalent bond is formed by the overlapping of atomic orbitals. Covalent bonds formed are of two types depending upon the way the orbitals overlap each other. 1. Sigma bond (s bond): The bond formed by the overlapping of two half filled atomic orbitals along their axis is known as sigma bond. s bond is a strong bond because overlapping in it takes place to large extent. The hybrid orbitals always from s bond. a) s – s overlapping

Molecular axis

b) s – p overlapping

[9 of 35]

IIT Package (2009-10) from NOIDA

c) p – p overlapping

pz

head on overlap

pz

p-p overlap

s M.O.

2. Pi bond (p bond): The bond formed by the lateral overlapping of half filled atomic orbitals is known as pi bond. The sidewise overlapping takes place to less extent. Therefore, p bond formed is a weak bond. p bond overlapping takes place only at the sides of two lobes. A p bond is formed when a s bond already exists between the combining atoms.

p-p overlapping

p

p

p M.O.

Example: In A – B molecule the bond formed is s bond.

In A

s p

B, molecule there are one s and one p bonds

s

B, molecule there are one s and two p bonds

In A

p Thus, all the single bonds are s bonds. Double bond has one s and one p bond. Triple bond has one s and two p bonds.

Hybridisation It is the mathematically fabricated concept that is introduced to explain the geometry/shapes of the covalent molecules of polyatomic ions containing covalent bonds. It is a process of intermixing of atomic orbitals with small difference in energy and belonging to the same atom, at the time of bonding so as to give another set of orbitals with equivalent shapes and energies. sp3 Hybridisation: In ground state, the electronic configuration of carbon is 1s2, 2s2, 2p2. It is proposed that from 2s orbital, being quite near in energy to 2p orbitals, one electron may be promoted to the vacant 2pz orbital thus obtaining the excited atom. At this stage the carbon atom undoubtedly has four half-filled orbitals and can form four bonds. In the excited atom, all the four valence shell orbitals may mix up to give four identical sp3 hybrid orbitals. Each of these four sp3 orbital possesses one electron and overlaps with 1s orbitals of four H atoms thus forming four equivalent bonds in methane molecule. Due to the tetrahedral disposition of sp3 hybrid orbitals, the orbital are inclined at an angle of 109° 28’. Thus all the H– C– H angles are equal to 109° 28’

Energy

2p

2p

of an electron

2s Ground State

[10 of 35]

Promotion

sp3 Hybridisation

2s Excited State

Hybridised State

IIT Package (2009-10) from NOIDA H H 109.5° C

H

H

H

H H

H

Shape and formation of methane molecule 2 sp Hybridisation: When three out of the four valence obritals of carbon atom in excited state hybridize, we have three sp2 hybrid orbitals lying in a plane and inclined at an angle of 120°. If 2s and 2p, orbitals of the excited carbon atom are hybridized, the new orbitals lie in the xy plane while the fourth pure 2pz orbitals lies at right angles to the hybridized orbitals with its two lobes disposed above and below the plane of hybrid orbitals. Two such carbon atoms are involved in the formation of alkenes (compounds having double bonds). In the formation of ethene two carbon atoms (in sp2 hybridization state) form one sigma bond by ‘head-on’ overlap of two sp2 orbitals contributed one each by the two atoms. The remaining two sp2 orbitals of each carbon form s bonds with H atoms. The unhybridized 2p, orbitals of the two carbon atoms undergo a side-wise overlap forming a p bond. Thus the carbon to carbon double bond in ethene is made of one s bond and one p bond. Since the energy of a p bond is less than that of a s bond, the two bonds constituting the ethene molecule are not identical in strength. The molecule is a planar one. Pure p-orbital

Energy

2p

2p

Promotion of an electron

sp2 Hybridisation

2s

2s Ground State

Excited State pz

pz p

s

sp2 hybrid orbitals

H sp 2 sp 2

s H

H C

C sp

2

sp p

2

sp 2 2 sp

H

H s s

p

H

s H

H

Orbital model of ethane molecule Different types of hybridization depend upon the type of atomic orbitals, which are used for intermixing. Types of hybridization and spatial orientation of hybrid orbitals: The geometry and shapes of various species on the basis of VSEPR theory along with hybrid state of central atom is given below in tabular form.

[11 of 35]

IIT Package (2009-10) from NOIDA

S.L

Types of atomic orbitals used

1.

Orientation of hybrid orbitals

Hybridisation

one s + one p-orbital

sp 2

Examples

Linear

BeF2 , BeCl 2 , C 2 H 2 , HgCl 2

Trigonal planar

BF3 , C 2 H 4 , NO 3- , CO 32-

Tetrahedral Trigonal bipyramidal

CH 4 , CCl 4 , SiF4 , NH +4 , SO 24- , ClO -4

PF5 , PCl 5

2.

one s + two p-orbitals

sp

3.

one s + three p-orbitals

sp3

4.

one s + three p + d

sp3d

5.

one s + three p + two d

sp3d2

Octahedral

SF5 , [CrF6 ]3- , IF5

6.

one s + three p+three d

sp3d3

Pentagonal Bipyramidal

IF7

7.

One d + one s + two p

dsp2

Square planar

Only in complexes like [ Ni(CN ) 4 ) 2- , [PtCl 4 ] 2- etc.

Note: i) Orbitals participating in hybridization must have only small difference in their energies. ii) Both half-filled and completely filled orbitals can get involved in hybridization. iii) The number of hybrid orbitals is equal to the number of orbitals participating in hybridization. iv) Hybrid orbital form more stronger bonds than pure atomic orbitals. v) Same atom can assume different hybrid states under different situations. vi) Hybrid orbitals form sigma bonds. Method of predicting the Hybrid state of the central atom in covalent molecules of polyatomic ions: The hybrid state of the central atom in similar covalent molecule or polyatomic ion can be predicted by using the generalized formula as described below: Simple Molecule

Polyatomic Anion

Poyatomic Cation

1 X = [V + G ] 2

1 X = [V + G + a ] 2

1 X = [ V + G - c] 2

In the above formulae, V = Number of monovalent atoms or groups attached to the central atom G = Number of outer shell electrons in ground state of the central atom a = Magnitude of charge on anion c = Magnitude of charge on cation Calculate the value of X and decide the hybrid state of central atom as follows : X Hybrid state

2 sp

3 sp2

4 Sp3

5 sp3d

6 sp3d2

7 sp3d3

PF5

COCl2

NH4+

ClO4 -

X =12 [5 + 5]

X =12 [2 + 4]

X =12 [4 + 5 - 1]

X =12 [0 + 7 + 1]

=5

=3

=4

=4

Hybrid state of P is sp3d

Hybrid state of C is sp2

Hybrid state of N is sp3

Hybrid state of Cl is sp3

CO2

XeF4

NO3 – X

=12

[0 + 5 + 1]

=3

X

[5 + 7 ]

=6

2

3

sp

2

sp d –

[12 of 35]

IF5 =12

X

=12

[ 0 + 4]

X =12 [4 + 8]

=2

=6

sp

sp3d2 +

IIT Package (2009-10) from NOIDA

PCl6–

PH3

SF3+

SF4

X =12 [6 + 5 + 1]

X =12 [3 + 5]

X =12 [3 + 6 - 1]

X =12 [4 + 6]

=6

=4

=4

=5

Hybrid state

Hybrid state

Hybrid state

Hybrid state 3

2

sp d

3

sp

3

sp

sp3 d

O O Illustration : 8 Ether R R and water H H have same hybridization at oxygen. What angle would you expect for them. Solution: In H2O bond angle is less than 109°28¢ due to lone pair and bond pair repulsion. But in ether, due to strong mutual repulsion between two alkyl groups bond angle becomes greater then 109°28¢. Illustration : 9 In following which central atom has different hybridization than other Cl2O, OF2, H2O, SO2. Solution: SO2; because it has sp2 hybridisation other three have sp3 hybridisation. Valence Shell Electron Pair Repulsion (VSEPR) Theory In 1957 Gillespie and Nyhom gave this theory to predict and explain molecular shapes and bond angles more exactly. The theory was developed extensively by Gillespie as the Valence Shell Electron Pair Repulsion (VSEPR) theory. This may be summarized as: 1. The shape of the molecule is determined by repulsions between all of the electron pairs present in the valence shell. 2. A lone pair of electrons takes up more space round the central atom than a bond pair, since the lone pair is attracted to one nucleus whilst the bond pair is shared by two nuclei. It follows that repulsion between two lone pairs is greater than repulsion between a lone pair and a bond pair, which in turn is greater than the repulsion between two bond pairs. Thus the presence of lone pairs on the central atom causes slight distortion of the bond angles from the ideal shape. If the angle between a lone pair, the central atom and a bond pair is increased, it follows that the actual bond angles between the atoms must be decreased. The order of repulsion between lone pairs and bond pairs of electrons follows the order as: Lone pair - lone pair repulsion > lone pair – bond pair repulsion > bond pair – bond pair repulsion. 3. The magnitude of repulsions between bonding pairs of electrons depends on the electronegativity difference between the central atom and the other atoms. 4. Double bonds cause more repulsion than single bonds, and triple bonds cause more repulsion than a double bond. Effect of Lone Pairs: Molecules with four electron pairs in their outer shell are based on a tetrahedron. In CH4 there are four bonding pairs of electrons in the outer shell of the C atom, and the structure is a regular tetrahedron with bond angle H – C – H of 109°28’. In NH3 and N atom has four electron pairs in the outer shell, made up of three bond pairs and one lone pair. Because of the lone pair, the bond angle H – N – H is reduced from the theoretical tetrahedral angle of 109°28’ to 107°28’. In H2O the O atom has four electron pairs in the outer shell. The shape of the H2O molecule is based on a tetrahedron with two corners occupied by bond pairs and the other two corners occupied by lone pairs. The presence of two lone pairs reduces the bond angle further to 104°27’. In a similar way, SF6 has six bond pairs in the outer shell and is a regular octahedron with bond angles of exactly 90°. In BrF5, the Br also has six outer pairs of electrons, made up of five bond pairs and one lone pair. The lone pair reduces the bond angles to 84°30’. Whilst it might be expected that two lone pairs would distort the bond angles in an octahedral as in XeF4 but it is [13 of 35]

IIT Package (2009-10) from NOIDA

not so. Actual bond angle is 90°, reason being that the lone pairs are trans to each other in the octahedron, and hence the atoms have a regular square planar arrangement. Molecules with five pairs of electrons are all based on a trigonal bipyramid. Lone pairs distort the structures as before. The lone pairs always occupy the equatorial positions (in an triangle), rather than the axial positions (up and down).Thus in I3– ion, the central I atom has five electron pairs in the outer shell, made of two bond pairs and three lone pairs. The lone pairs occupy all three equatorial positions and the three atoms occupy the top, middle, and bottom positions in the trigonal bipyramid, thus giving a linear arrangement with a bond angle of exactly 180°. Effect of Electronegativity: NF3 and NH3 both have structures based on a tetrahedron with one corner occupied by a lone pair. The high electronegativity of F push the bonding electrons further away from N than in NH3. Hence the lone pair in NF3 causes a greater distortion from tetrahedral and gives a F – N – F bond angle of 102°30’, compared with 107°48’ in NH3. The same effect is found in H2O (bond angle 104°27’) and F2O (bond angle 102°). The effects of bonding and lone pairs on bond angles Compound

Hybridisation

BeCl2

sp

BF3

sp

2

3

CH4 NH3 NF3 H2O F2O

sp 3 sp 3 sp 3 sp

PCl5 SF4 ClF3 XeF2

sp d 3 sp d 3 sp d 3 sp d 3 2 sp d 3 2 sp d

SF6 BrF5 XeF4

3

sp

3

3

sp d

2

Number of bond pairs

Number of lone pairs

Linear

2

0

180°

Plane triangle

3

0

120°

Tetrahedral Pyramidal Pyramidal Bent (V-shape) Bent (V-shape)

4 3 3 2 2

0 1 1 2 2

Trigonal bipyramid Trigonal bipyramid T-shape Linear

5 4 3 2

0 1 2 3

Octahedral Square pyramidal Square planar

6 5 4

0 1 2

109°28¢ 107°48¢ 102°30¢ 104°27¢ 102° 120° and 90° 101°36¢ and 86°33¢ 87°40¢ 180° 90° 84°30¢ 90°

Shape

Bond angle

Some examples using the VSEPR Theory

Cl

Phosphorus pentachloride PCl5: Gaseous PCl5 is covalent. (The electronic structure P is 1s22s22p63s23p3). All five outer electrons are used to form bonds to the five Cl atoms. In the PCl5 molecule the valence shell of the P atom contains five electron pairs: hence the structure is a trigonal bipyramid. There are no lone pairs, so the structure is not distorted. However, a trigonal bipyramid is not a completely regular structure, since some bond angels are 90° and others 120°. Symmetrical structures are usually more stable than asymmetrical ones.

Cl

Cl P Cl

Cl Structure of PCl5 molecule

Note: Thus PCl5 is highly reactive, and in the solid state it splits into PCl4]+ and [PCl6]– ions, which have tetrahedral and octahedral structures respectively.

Chlorine trifluoride ClF3: The chlorine atom is at the centre of the molecule and determines its shape. The electronic configuration of Cl is 1s22s22p63s23p5. Three electrons form bonds to F, and four electrons do not take part in bonding. Thus in ClF3, the Cl atom has five electron pairs

[14 of 35]

IIT Package (2009-10) from NOIDA

in the outer shell, hence the structure is a trigonal bipyramid. There are three bond pairs and two lone pairs. It was noted previously that a trigonal bipyramid is not a regular shape since the bond angles are not all the same. It therefore follows that all the corners are not equivalent. Lone pair occupy two of the corners, and F atoms occupy the other three corners. Three different arrangements are theoretically possible, as shown in figure below. The most stable structure will be the one of lowest energy, that is the one with the minimum repulsion between the five orbitals. The great repulsion occurs between two lone pairs. Lone pair bond pair repulsions are next strongest, and bond pair-bond pair repulsions are weakest. Groups at 90° repel each other strongly, whilst groups 120° apart repel each other much less. F

F

F

F

Cl

Cl

F

F

Cl F

F F

I

II

III

Chlorine trifluoride molecule

Structure I is the most symmetrical, but has six 90° repulsions between lone pairs and atoms. Structure II has one 90° repulsion between two lone pairs, plus three 90° repulsions between lone pairs and atoms. These factors indicate that structure III is the most probable. The observed bond angles are 80°40¢, which is close to the theoretical 90°. This confirms that the correct structure is III, and the slight distortion from 90° is caused by the presence of the two lone pairs. As a general rule, if lone pairs occur in a trigonal bipyramid they will be located in the equatorial position (round the middle) rather than the axial positions (top and bottom), since this arrangement minimizes repulsive forces. F F

Sulphur hexafluoride SF6: The electronic structure of S is S 1s22s22p63s23p6. All six of the outer electrons are used to form bonds with the F atoms. Thus in SF6, the S has six electron pairs F F in the outer shell: hence the structure is octahedral. There are no lone pairs, so the structure is completely regular with bond angles F of 90°. Back Bonding : The interaction between an empty orbital and lone pair electron known as back bonding .

F

.

For example the nitrogen in trimethyl amine and trisilyl amine has a lone pair electron at nitrogen but nitrogen in trimethyl amine has pyramidal shape while in N trisilyl amine nitrogen has planer shape because in trimethyl amine there is repul- Me Me Me sion between lone pair and bond pair that’s why the shape becomes pyramidal In trisilyl amine, there is vacant d-orbital at silicon which overlaps with lone pair of nitrogen called p p - d p back bonding hence geometry becomes planer N

N

N H3 Si

SiH3

SiH3

SiH3 SiH3

H3Si

SiH3

H3Si

SiH3

Molecular Orbital Theory Why He2 molecule does not exist and why O2 is paramagnetic? These questions cannot be explained by valence bond theory. In 1932 F. Hund and R.S. Mulliken put forward a theory [15 of 35]

IIT Package (2009-10) from NOIDA

known as Molecular Orbital Theory to explain above questions and many others. According to this theory, as the electrons of an atom are present in various atomic orbitals, electrons of a molecule are present in various molecular orbitals. Molecular orbitals are formed by the combination of atomic orbitals of comparable energy and proportional symmetry. While an electron in atomic orbital is influenced by one nucleus, in a molecular orbital, it is influenced by two or more nuclei depending upon the number of atoms in the molecule. Thus, an atomic orbital is monocentric while a molecular orbital is polycentric. The number of molecular orbitals formed is equal to the number of combining atomic orbitals. When two atomic orbitals combine, two molecular orbitals are formed. One is known as bonding molecular orbital (BMO) whereas other is anti-bonding molecular orbital (ABMO). BMO has lower energy and hence greater stability than the corresponding ABMO. First BMO are filled, then ABMO starts filling because BMO has lower energy than that of ABMO. Molecular orbitals like the atomic orbitals are filled in accordance with the Aufbau Principle obeying the Pauli’s Principle and the Hund’s rule. Order of energy of various molecular orbitals is as follows: For O2 and higher molecules s1s, s*1s, s2s, s*2s, s2px, [p2py = p2pz], [p*2py = p*2pz], s*2px For N2 and lower molecules s1s, s*1s, s2s, s*2s, [p2py = p2pz], s2px, [p*2py = p*2pz], s*2px Bond Order: It may be defined as the half the difference between the number of electrons present in the bonding orbitals and the anti-bonding orbitals i.e. Bond order (B.O.) =

No. of electrons in BMO - No. of electrons in ABMO 2

A positive bonding order suggest a stable molecule while a negative bond order or zero bond order suggest an unstable molecule. Magnetic Behaviour: If all the molecular orbitals in species are spin paired, the substance is diamagnetic. However, if one or more molecular orbitals are singly occupied it is paramagnetic. Illustration : 10 Arrange the species O2, O2–, O22– and O2+ in the decreasing order of bond order and stability and also indicate their magnetic properties. Solution: The molecular orbital configuration of O2, O2–, O22– and O22+ are as follows: O2 = s1s2, s*1s2, s2s2, s*2s2, s2px2, p2py2, p2pz2, p*2py1 = p*2pz1 Bond order =

10 - 6 = 2 , No. of unpaired electrons = 2 2

\ paramagnetic O2– = s1s2, s*1s2, s2s2, s*2s2, s2px2, p2py2, p2pz2, p*2py2 = p*2pz1

10 - 5 = 1.5 , No. of unpaired electrons = 1 2 \ paramagnetic Bond order =

O22– = s1s2, s*1s2, s2s2, s*2s2, s2px2, p2py2, p2pz2, p*2py2 = p*2pz2 Bond order =

10 - 8 = 1 , No. of unpaired electrons = 0 2

\ diamagnetic O2+ = s1s2, s*1s2, s2s2, s*2s2, s2px2, p2py2, p2pz2, p*2py1 = p*2pz0 Bond order =

10 - 5 = 2.5 , No. of unpaired electrons = 1 2

\ paramagnetic [16 of 35]

IIT Package (2009-10) from NOIDA

Now as the bond order decreases in the order O2+ > O2 > O2– > O22– So, same will be the stability order of the above species because stability is directionally proportional to bond order. Note: Bond length is inversely proportional to bond order. Illustration : 11 Correct order of bond energy is: + (A) N2 > N2 > N2– > N22– (B) N2+ > N2– > N22– > N2 – + 2– (C) N2 > N2 = N2 > N2 (D) N2– > N2 = N2+ > N22– + – Solution: Bond order of N2 = 3, N2 , N2 = 2.5, N22– = 2 But N2– has more electrons in ABMO than N2+ so correct order of bond energy will be N2 > N2+ > N2– > N22–

Hydrogen Bonding In 1920, Latimer and Rodebush introduced the idea of “hydrogen bond” to explain the nature of association in liquid state of substance like water, hydrogen fluoride, ammonia, formic acid etc. In a hydrogen compound, when hydrogen is bonded to highly electronegative atom (such as F, O, N) by a covalent bond, the electron pair is attracted towards electronegative atom so strongly that a dipole results i.e., one end carries a positive charge (H-end) and other end carries a negative charge (X-end). dX

d+ H

d- d+ X H

or

Electro-negative atom

If a number of such molecules are brought nearer to each other, the positive end of one molecule and negative end of the other molecule will attract each other and weak electrostatic force will develop. Thus, these molecules will associate together to form a cluster of molecules. d-

d+

d-

d+

d-

d+

d-

d+

d-

d+

X- H L X- H L X- H L X- H L X- H

The attractive force that binds hydrogen atom of one molecule with electronegative atom of the other molecule of the same or different substance is known as hydrogen bond.

Hydrogen bonding is of two types: a) Intermolecular hydrogen bonding: This type of bonding results between the positive and negative ends of different molecules of the same or different substances. Example i) Ammonia d+ H d-

N

d+ H dN H

d+ H dN H

d+ H dN H

d+ H + d- d N H

H d+

H d+

H d+

H d+

H

d+ H dO H

H

H d+

ii) Water d+ H

d-

O

d+ H dO H

d+ H dO H

iii) Acetic acid O

H

O

H3C

CH3 O

H

O

[17 of 35]

IIT Package (2009-10) from NOIDA

This type of hydrogen bonding increases the boiling point of the compound and also its solubility in water. The increase in boiling point is due to association of several molecules of the compound. b) Intramolecular hydrogen bonding: This type of bonding results between hydrogen and an electronegative element both present in the same molecule. This type of bonding is generally present in organic compounds. Examples are o-nitro-phenol, o-hydroxy benzoic acid, etc. O N

O

-

H O

-

H O

O o-Nitrophenol

O o-Hydroxy benzoate ion

This type of bonding decreases the boiling point of the compound. The solubility of the compound also decreases. Hence compound becomes more volatile.

Properties Explained by Hydrogen Bonding a) Strength of certain acids and bases can be explained on the basis of hydrogen bonding. b) Solubility: An organic substance is said to be insoluble in water if it does not form hydrogen bonding with water. The organic compound like alkanes, alkenes, ethers, etc., are insoluble in water as they do not form hydrogen bonding with water, while alcohols and acids are soluble because they readily form hydrogen bonds with water. i) Melting and boiling points of hydrides of N, O and F. If the melting points and boiling points of the hydrides of the elements of IVA, VA, VIA and VIIA groups are plotted against the molecular weights of these hydrides, we shall get the plots as shown in figure (a) and (b). From these plots it may be seen that although in case of SbH3, AsH3, PH3 (VA group elements hydrides), H2Te, H2Se, H2S (VI A group elements hydrides) and HI, HBr, HCl (VIII group elements hydrides) there is a progressive decrease in their mp’s and b.p’s with the decrease in their molecular weights, the mp’s and b.p’s of NH3, H2O and HF hydrides suddenly increase with a further decrease of their molecular weights. The sudden increase in mp’s and bp’s in these hydrides is due to the inter-molecular H-bonding in between H and F in case of HF, in between H and O in case of H2O and in between H and N in case of NH3 respectively. The existence of H-bonding in these molecules gives polymerized molecules (NH3)n. Thus mp’s and bp’s of these molecules are suddenly raised. Having no power to form H-bonds, the simple carbon family hydrides (SnH4, GeH4, SiH4 and CH4) show a decrease in their bp’s and mp’s with the decrease in their molecular weights. 100

100 H2 O

0

H 2 Se

NH 3

H 2 Se

H2 S

HBr -100

HI

VIA VIIIA VA

SbH3

HCl

HF

AsH3 PH3

SnH4

Boiling points (°C) increasing

Melting points (°C) increasing

HF H2O

H 2 Se

NH 3

HI HBr -100

SnH4 CH 4

SnH4 -200

Molecular weight increasing (a)

[18 of 35]

GeH4

PH 3

IVA

-200

H 2 Se

H2S

SbH3

VA VIIIA

HCl

GeH 4 CH4

VIA

0

Molecular weight increasing (b)

SnH 4

IVA

IIT Package (2009-10) from NOIDA

ii) Ice has less density H H than water. The explanation of this fact is as 2.76 Å 1.80Å follows: In the crystal structure of ice, the OOpen cage-like tetrahedral crystal structure H of ice. Circles indicate oxygen atoms. 0.96Å 0.96Å atom is surrounded by Bonds represented by solid line are normal covalent bonds while those represented by four H-atoms. Two Hwater molecule dotted lines are hydrogen bonds. H atoms are linked to OH H atom by covalent bonds H as shown (by normal covalent bond) and the reH H maining two H-atoms H are linked to O-atom by two H-bonds shown by dotted lines. Thus in ice every water molecule is associated with four other water molecules by H-bonding in a tetrahedral fashion. Ice has an open cage like structure with a large empty space due to the existence of H-bonds. As ice melts at 0°C, a number of H-bonds are broken down and the space between water molecules decreases so that water molecules move closer together. The density of water increases, from 0° to 4°C, and at 4°C it is maximum. Above 4°C the increase in kinetic energy of the molecules is sufficient to cause the molecules to begin to disperse and the result is that the density decrease with increasing temperature. Illustration : 12 Both o-nitrophenol and p-nitrophenol has hydrogen bonding in their molecules? Explain which of the two has higher boiling point? Solution: Both have hydrogen bonding. o-Nitrophenol has intramolecular hydrogen bonds. But due to larger distance between —NO2 and —OH group in p-nitrophenol, there is no such bonding. O

OH

H N

O

O N

O

absence of intra molecular H-bonding

O

However, there is intermolecular hydrogen bonding in p-nitrophenol and therefore, it exists as associated molecule. In o-nitrophenol, intermolecular hydrogen bonding is possible. HO

N

O

O HO

N

O

O

Due to associated nature of p-nitrophenol, it is less volatile and has high boiling point.

[19 of 35]

IIT Package (2009-10) from NOIDA

1.

2. 3. 4.

5.

which of the following is not correct statement (A) Ionic compound are electrically neutral (B) Boiling point of ionic compound is more than covalent compound (C) Melting point of covalent compound is more than ionic compound (D) Ionc compound are soluble in polar solvent The pair of elements form ionic bond is (A) C + Cl (B) H +F (C) Na + Br (D) O + H Which of these contains both polar and nonpolar bonds (A) HCN (B) CO2 (C) H2O2 (D) CH4 The correct order of dipole moment is (A) CH4 < NF3 < NH3< H2O (B) NF3 < CH4 < NH3 < H2O (C) NH3 < NF3 < CH4 < H2O (D) H2O < NH3 < NF3 < CH4 Among the following which is polar (A) CO2 (B) SO2 (D) Cl

(C) BeCl2 6. 7. 8. 9. 10. 11.

Cl

Shape of NH3 is very similar to – (A) CH4 (B) CH3– (C) BH3 (D) CH3+ Which of the following has pyramidal shape – (A) XeO3 (B) XeF4 (C) XeF2 (D) XeF6 Which of the following are diamagnetic? (A) C2 (B) O22– (C) Li2 (D) N2+ The bond order of CO and NO is (A) 3 and 2 (B) 3 and 2.5 (C) 3 and 1.3 (D) 3 and 3.5 Which is steam volatile (A) o-nitrophenol (B) Aniline (C) Glycerol (D) p-nitrophneol Maximum number of H-bonds that can be formed by a water molecule is – (A) 2 (B) 3 (C) 4 (D) 6

1. C

2. C

3. C

4. A

5. B

7. A

8. D

9. B

10. A

11. C

Solutions: 1. 2.

Sol. Melting point of ionic compound is more than covalent compounds Sol. ionic bond is formed between electropositive and electronegative element

3.

Sol. H2O2 contains polar O-H bond and nonpolar O-O bond

[20 of 35]

6. B

IIT Package (2009-10) from NOIDA

Model Questions 12.

13.

14. 15. 16.

17.

18.

19. 20. 21.

Lattice energy of an ionic compound depends upon (A) Charge on the ion and size of the ion (B) Packing of ions only (C) Size of the ion only (D) Charge on the ion only Bond angle in PH3 is (A) much less than NH3 (B) Much less than PF3 (C) slightly more than NH3 (D) much more than PF3 Identify the least stable ion amongst the following: [2002] (A) Li– (B) Be– (C) B– (D) C– Which of the following is polar? (A) NF3 (B) BF3 (C) SF4 (D) SiF4 Dipole moment is shown by (A) 1, 4-dichlorobenzene (B)1, 2-dichlorobenzne (C) trans 1, 2-dichloroethene (D) trans 2, 3-dichloro-2-butene According to Fajan rules, the covalent character is most favoured in (A) Small cation large anion (B) Small cation, small anion (C) Large cation, large anion (D) Large cation, small anion. A lone pair of electrons in an atom implies [Kurukshetra CET] (A) a pair of valence electrons (B) a pair of electrons (C) a pair of electrons involved in bonding (D) a pair of valence electrons not involved in bonding Which species has the maximum number of lone pair of electrons on the central atom? [2005] (A) [ClO3]– (B) XeF4 (C) SF4 (D) [I3]– The percentage of s-character in the hybrid orbitals sp, sp2 and sp3 follows the pattern (A) sp3 > sp2 > sp (B) sp > sp2 > sp3 (C) sp = sp2 > sp3 (D) sp = sp2 = sp3 Carbon dioxide is isostructural with which of the following ? (A) HgCl2

22. 23.

(B) H2O

Which one of the following compounds has sp2 hydridization? (A) CO2 (B) SO2 (C) N2O Combination of two AO’s lead to the formation of (A) two MO’s (B) one MO (C) three MO’s

24.

The calculated bond order in H -2 ion is

25.

(A) 0 (B) 1/2 Number of paired electrons in O2 molecule is: (A) 7 (B) 8

26.

27. 28.

(C) SnCl2

(C) –1/2

(D) NO -2 [1997] (D) CO (D) four MO’s (D) 1 [1995]

(C) 16

(D) 14

Orthonitrophenol is steam volatile but paranitrophenol is not because (A) orthonitrophenol has intramolecular hydrogen bonding while paranitrophenol has intermolecular hydrogen bonding. (B) both ortho and paranitrophenol have intramolecular hydrogen bonding. (C) orthonitrophenol has intermolecular hydrogen bonding and paranitrophenol has intramolecular hydrogen bonding. (D) Van der Waals forces are dominant in orthonitrophenol. Which of the following hydrogen bonds is the strongest? (A) O – H - - - F (B) O – H - - - H (C) F – H - - - F (D) O – H - - - O The maximum possible number of hydrogen bonds a water molecule can form is [1992] (A) 2 (B) 4 (C) 3 (D) 1 [21 of 35]

IIT Package (2009-10) from NOIDA

Practice Questions 29. 30. 31.

32.

33.

34.

35. 36.

37. 38. 39. 40.

In OF 2 , number of bond pairs and lone pairs of electrons are respectively (A) 2, 6 (B) 2, 8 (C) 2, 10 (D) 2, 9 The bond angle around the central atom is maximum for (A) H2O (B) H2Se (C) H2S (D) H2Te The correct order of bond angles (smallest first) in H2S, NH3, BF3 and SiH4 is (A) H2S < NH3 < SiH4 < BF3 (B) NH3 < H2S < SiH4 < BF3 (C) H2S < SiH4 < NH3 < BF3 (D) NH3S < NH3 < BF3 < SiH4 According to Fajan rules, the covalent character is most favoured in (A) Small cation large anion (B) Small cation, small anion (C) Large cation, large anion (D) Large cation, small anion The hybridization of atomic orbitals of nitrogen in NO2+, NO3– and NH4+ are [2000] 3 2 2 3 (A) sp, sp and sp respectively (B) sp, sp and sp respectively 2 3 (C) sp , sp and sp respectively (D) sp2, sp3 and sp respectively Among the following compounds the one that is polar and has the central atom with sp2 hybridisation is [1997] (A) H2CO3 (B) SiF4 (C) BF3 (D) HClO2 – The type of hybrid orbitals used by the chlorine atom in ClO2 is [1992] 3 2 (A) sp (B) sp (C) sp (D) none of these Shape of ClO3– is [CBSE PMT] (A) Pyramidal (B) Tetrahedral (C) Triangular planar (D) Triangular bipyramidal. Which of the following structures is linear? (A) SO2 (B) CO2 (C) CO32– (D) SO32– The geometryand the type of hybrid orbital present about the central atom in BF3 is [1998] 2 3 (A) linear, sp (B) trigonal planar, sp (C) tetrahedral, sp (D) pyramidal, sp3 In which of the following molecules/ions are all the bonds not equal? (A) XeF4 (B) BF4– (C) SF4 (D) SiF4 Which of the following species in paramagnetic ? (A) CO2

41. 42. 43.

44. 45. 46.

47.

(B) NO

(C) O 22 -

(D) CN-

The species having highest bond order is (A) O2 (B) O2– (C) O2+ (D) O22 – Which of the following molecular species has unpaired electron(s)? [2002] – 2– (A) N2 (B) F2 (C) O2 (D) O2 NH3 has higher boiling point than PH3 because (A) NH3 has higher molecular mass (B) NH3 undergoes umbrella inversion (C) NH3 molecules form H-bonds with one another (D) NH3 contains ionic bonds while PH3 does not Which of the following species is diamagnetic in nature? (A) H2– (B) H2+ (C) H2 (D) He2+ Which one of the following compounds has the smallest bond angle in its molecule? (A) OH2 (B) SH2 (C) NH3 (D) SO2 Which of the following sets whichone does NOT contain isoelectronic species? (A) BO33- , CO32- , NO3-

(B) SO32 - , CO32 - , NO3-

(C) CN - , N 2 , C22 - `

(D) PO34- , SO24 - , ClO -4

Which one or more among the following involve(s) pp - dp bonding? (A) (SiH3)3N:

[22 of 35]

(B) (CH3)3N:

(C)

CCl3

(D)

CF3

IIT Package (2009-10) from NOIDA

Model Questions & Practice Questions

12. A

13. A

14. B

15. A

16. B

17. A

18. D

19. D

20. B

21. C

22. B

23. A

24. B

25. D

26. A

27. C

28. B

29. B

30. A

31. A

32. A

33. B

34. A

35. A

36. A

37. B

38. B

39. D

40. B

41. C

42. C

43. C

45. C

46. B

46. B

47. A,B,C

Model Questions 48.

49.

50.

51. 52.

53.

54. 55.

56.

Select the correct statements: (A) The heat of hydration of the dipositive alkaline earth metals ions decrease with an increase in their ionic size. (B) Hydration of alkali metal ions is less than that of IIA (C) Alkaline earth metal ions, because of their much larger charge to size ratio exert a much stronger electrostatic attraction on the oxygen of water molecule surrounding them. (D) Melting point of sodium halides follow order NaF > NaCl > NaBr > NaI The decreasing values of bond angles from NH3 (106°) to SbH3 (101°) down group-15 of the periodic table is due to (A) decreasing lp-bp repulsion (B) decreasing electronegativity (C) increasing bp-bp repulsion (D) increasing p-orbital character in sp3 Which of the following has been arranged in order of increasing covalent character? (A) KCl < CaCl2 < AlCl3 < SnCl4 (B) SnCl4 < AlCl3 < CaCl2 < KCl (C) AlCl3 < CaCl2 < KCl < SnCl4 (D) CaCl2 < SnCl4 < KCl < AlCl3 Among the following, the electron deficient compound is (A) CCl4 (B) PCl5 (C) BeCl2 (D) BCl3 The electronegativities of F, Cl, Br, I are 4.0, 3.0, 2.8, 2.5 respectively. The hydrogen halide with a highest percentage of ionic character is [Rajasthan PMT] (A) HF (B) HCl (C) HBr (D) HI Which one of the following pairs of molecules will have permanent dipole moments for both ? (A) NO2 and CO2 (B) NO2 and O3 (C) SiF4 and CO2 (D) SiF4 and NO2 Which contains both polar and non-polar bonds? [1997] (A) NH4Cl (B) HCN (C) H2O2 (D) CH4 The number and type of bonds between two carbon atoms in CaC2 are: [1996] (A) one sigma (s) and one pi (p) bonds (B) one sigma (s) and two pi (p) bonds (C) one sigma (s) and one and a half pi (p) bonds (D) one sigma (s) bond In compound X all the bond angles around central atom are 109028’. Which one of the following will X be (A) Chloromethane (B) Carbon tetrachloride (C) Iodoform (D) Chloroform [23 of 35]

IIT Package (2009-10) from NOIDA

57. 58.

In which of the following molecule, all the atoms lie in one plane ? (A) CH4 (B) BF3 (C) PF5 Which of the following have identical bond order? (A) CN– (B) O2– (C) NO+

(D) NH3 [1992] +

(D) CN

Assertion and Reasoning This section contains 4 questions numbered 1 to 4. Each question contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Code (A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. (B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1. (C) Statement – 1 is True, Statement – 2 is False. (D) Statement – 1 is False, Statement – 2 is True. 59.

STATEMENT – 1 BF3 has greater dipole moment than H2S STATEMENT – 2 Fluorine is more electronegative than sulphur.

Practice Questions 60.

61.

62.

The charge/size ratio of a cation determines its polarizing power. Which one of the following sequences represents the increasing order of the polarizing power of the cationic species, K+, Ca2+, Mg2+, Be2+? (A) Ca2+ < Mg2+ < Be+ < K+

(B) Mg2+ < Be2+ < K+ < Ca2+

(C) Be2+ < K+ < Ca2+ < Mg2+

(D) K+ < Ca2+ < Mg2+ < Be2+

The correct order of the thermal stability of hydrogen halides (H – X) is (A) HI > HCl < HF > HBr

(B) HCl < HF > HBr < HI

(C) HF > HCl > HBr > HI

(D) HI > HBr > HCl > HF

The critical temperature of water is higher than that of O2 because the H2O molecule has (A) fewer electrons than O2

(B) two covalent bonds

(C) V-shape

(D) dipole moment

[1997]

63.

ˆˆ† PCl + + PCl –, the change in hybridization is from In the reaction 2PCl5 ‡ˆˆ 4 6

64.

(A) sp3d to sp3 and sp3d2 (B) sp3d to sp2 and sp3 (C) sp3d to sp3d2 and sp3d3 (D) sp3d2 to sp3 and sp3d Which among the following are having sp3d hybridization of the central atom. (A) XeF4 (B) XeO2F2 (C) ClO3(D) BrF3 Which among the following are isostructural

65.

(A) CO2, I366. 67. 68.

(B) XeO2F2, SF4

(C) SO 32 - , CO 32-

(D) ClF3, XeF2

Which are the species in which central atom undergoes sp3 hybridization? (A) SF4 (B) SCl2 (C) SO42– (D) H2O In XeF2, XeF4 and XeF6, the number of lone pairs of Xe is respectively (A) 2, 3, 1 (B) 1, 2, 3 (C) 4, 1, 2 (D) 3, 2, 1 The pair of species having identical shapes for molecules of both species is (A) XeF2, CO2

[24 of 35]

(B) BF3, PCl3

(C) PF5, IF5

(D) CF4, SF4

IIT Package (2009-10) from NOIDA

69.

70. 71.

The ground state electronic configuration of N2 molecules is written as KK(s2s)2 (s*2s)2 (p2px)2 (p2py)2 (s2pz)2. The bond order is (A) 3 (B) 2 (C) 0 (D) 1 The bond order of NO molecule is (A) 1.5 (B) 2.0 (C) 2.5 (D) 3.0 Which of the following have been arranged in increasing order of bond order as well as bond dissociation energy ? (A) O2– 2 < O2– < O2+ < O2

72. 73. 74.

75.

76.

(B) O2– 2 < O2– < O2 < O2+

(C) O2 < O2+ < O22– < O2– (D) O2+ < O22– < O2– < O2 Which of the following diatomic molecules would be stabilized by the removal of an electron? (A) O2 (B) CN (C) N2 (D) C2 [Har CEET] Which of the following species exhibits the diamagnetic behaviour? (A) NO (B) O22– (C) O2+ (D) O2 Among the following species, identify the isostructural pairs. [1996] – + NF3, NO3 , BF3, H3O , HN3 (A) [NF3, NO3–] and [BF3, H3O+] (B) [NF3, HN3] and [NO3–, BF3] (C) [NF3, H3O+ and [NO3–, BF3] (D) [NF3, H3O+] and [HN3, BF3] According to molecular orbital theory which of the following statement about the magnetic character and bond order is correct regarding O2+ [2004] (A) Paramagnetic and Bond order < O2 (B) Paramagnetic and Bond order > O2 (C) Diamagnetic and Bond order < O2 (D) Diamagnetic and Bond order > O2 KF combines with HF to form KHF2. The compound contains the species (A) K+, F– and H+ (B) K+, F– and HF (C) K+ and [HF2]– (D) [KHF]+ and F2

Assertion and Reasoning

77.

78.

79.

This section contains 4 questions numbered 1 to 4. Each question contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Code (A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. (B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1. (C) Statement – 1 is True, Statement – 2 is False. (D) Statement – 1 is False, Statement – 2 is True. STATEMENT-1 Among two cations of similar size, the polarising power of cation with pseudo noble gas configuration is larger than cation with noble gas configuration. STATEMENT-2 Polarising power of Ag+ is more than K+ STATEMENT – 1 Bond order for N2+ and N2– are same (i.e. 2.5) STATEMENT – 2 N2+ is more stable than N2– STATEMENT – 1 BF3 is a weaker Lewis acid than BCl3 STATEMENT – 2 In BF3 molecule, back bonding (Pp – Pp) is weaker than BCl3 [25 of 35]

IIT Package (2009-10) from NOIDA

Paragraph - I

Linked Comprehension Type

The shape of a molecule is determined by electron-pair repulsions in the valence shell. A lone pair occupies larger space than a bond pair because it is not shared by two nuclei. Thus the lone pair-lone pair repulsion is greater than the lone pair-bond pair repulsion, which in trun is greater than the bond pair-bond pair repulsion. The presence of lone pairs causes distortion of bond angles and hence a deviation from an ideal shape. The extent of distortion depends upon the orientation of the lone pairs around the central atom. In a trigonal bipyramid, the lone pairs occupy equatorial positions than the apical ones. In ABn type molecules, as the electronegative of A increases, the bond pairs come closer and the repulsion between them increases. On the other hand, as electro negativity of B increases, the bond pairs get farther and repulsion decreases. 80.

81. 82. 83.

84.

Which of the following statements is true? (A) F-N-F angle in NF3 is greater than H-N-H angle in NH3 (B) F-N-F angle in NF3 is smaller than H-N-H angle in NH3 (C) H-O-H angle in H2O is greater than H-N-H angle in NH3 (D) F-O-F angle in F2O is greater than H-O-H angle in H2O The shape of which of the following molecules will not be distorted? (A) BrF3 (B) ClF3 (C) XeF4 (D) XeF6 Which of the following species will have the lone pair effects cancelled? (A) ICl -2 (B) ClF3 (C) PCl3 (D) BrF5 Match List I (species) with List II (Hybridisation) and select the correct answer using the codes given below. List – I List – II (A) BCl3 (p) sp3 (B) NH3 (q) sp3d2 (C) PCl5 (r) sp2 (D) XeF4 (s) sp3d Match List I with List. List – I List – II (A) C—H bond in Ethyne (p) sp-s overlap (B) P—Cl bond in POCl3 (q) sp3d–p overlap – (C) Br—Br bond in Br3 (r) sp3-p overlap (D) C—C bond in Ethane (s) sp3-sp3 overlap

Model Questions & Practice Questions

48. A

49. B

50. A

51. D

52. A

53. B

54. C

55. B

56. B

57. B

58. A,C

59. D

60. D

61. C

62. D

63. A

64. B,D

65. A,B

66. B,C,D

67. D

68. A

69. A

70. C

71. B

72. A

73. B

74. C

75. B

76. C

77. A

78. B

79. A

80. C

81. A

82. A

83. A ® P, B ® R, C® Q, D ® S [26 of 35]

84. A ® R, B ® P, C® S, D ® Q

IIT Package (2009-10) from NOIDA

Practice Questions 85. 86.

87.

88. 89.

The states of hybridization of boron and oxygen atoms in boric acid (H3BO3) are respectively (A) sp3 and sp2 (B) sp2 and sp3 (C) sp2 and sp2 (D) sp3 and sp3 The correct order of the hybridization of the central atom in the following species NH3, [PtCl4]2– , PCl5 and BCl3 is [2001] 2 3 2 3 3 2 3 2 (A) dsp , dsp , sp and sp (B) sp , dsp , dsp , sp (C) dsp2, sp2, sp3, dsp3 (D) dsp2, sp3, sp2, dsp3 Specify the coordination geometry around and hybridization of N and B atoms in a 1 : 1 complex of BF3 and NH3 [2002] 3 3 3 (A) N : tetrahedral, sp ; B: tetrahedral, sp (B) N: pyramidal, sp ; B: pyramidal, sp3 (C) N: pyramidal, sp3; B: planar, sp2 (D) N: pyramidal, sp3: tetrahedral, sp3 The linear structure is assumed by: [1991] – + (A) SnCl2 (B) NCO (C) SO2 (D) NO2 Which of the following statements are correct: (A) The bond angle of NCl3 is greater than that of NH3. (B) The bond angle in PH3 is greater than that of PF3. (C) ClO3- and SO 32 - are isostructural

90.

91.

92. 93.

94. 95.

96.

97. 98. 99.

(D) It is not necessary that in TBP structure the lone pairs always would occupy the equatorial positions. The geometry of H2S and its dipole moment are [1999] (A) angular and non-zero (B) angular and zero (C) linear and non-zero (D) linear and zero + The bond order in NO is 2.5 while that in NO is 3. Which of the following statements is true for these two species? (A) Bond length in NO+ is equal to that in NO (B) Bond length in NO is greater than in NO+ (C) Bond length in NO+ is greater than in NO (D) Bond length is unpredictable Which of the following molecules/ions does not contain unpaired electrons? (A) N2+ (B) O2 (C) O22– (D) B2 – The cyanide ion, CN and N2 are isoelectronic. But in contrast to CN–, N2 is chemically inert, because of [1992] (A) low bond energy (B) absence of bond polarity (C) unsymmetrical electron distribution (D) presence of more number of electrons in bonding orbitals Among KO2, AlO2–, BaO2 and NO2+, unpaired electron is present in [1997] + – (A) NO2 and BaO2 (B) KO2 and AlO2 (C) KO2 only (D) BaO2 only 2– The correct order of increasing C—O bond length of CO, CO3 , CO2 is [1999] (A) CO32– < CO2 < CO (B) CO2 < CO32– < CO (C) CO < CO32– < CO2 (D) CO < CO2 < CO32– The common features among the species CN–, CO and NO+ are [2001] (A) bond order three and isoelectronic (B) bond order three and weak field ligands (C) bond order two and p-acceptors (D) isoelectronic and weak field ligands Which of the following are isoelectronic and isosteructural? NO3–, CO32–, SO3 [2003] – 2– – – 2– 2– (A) NO3 , CO3 (B) SO3, NO3 (C) ClO3 , CO3 (D) CO3 , SO3 Among the following, the paramagnetic compound is [2007] (A) Na2O2 (B) O3 (C) N2O (D) KO2 The species having bond order different from that in CO is [2007] – + – (A) NO (B) NO (C) CN (D) N2 [27 of 35]

IIT Package (2009-10) from NOIDA

(Question With more than one Correct option) 100. Which of the following compounds are expected to be covalent? (a) BeCl2 (b) SnCl4 (c) ZnS 101. Which of the following does not exist? (a) HS6 (b) HPO4 (c) Fel3

(d) ZnCl2 (d) HClO4

102. Which of the following have sp3 d hybridization? (a) SF4 (b) BrCl3 (c) XeOF2 (d) FBr3 103. Which of the following molecules has one unpaired electron in antibonding orbitals? (a) CO (b) NO 104. Select the correct statement(s). (a) NF3 is weaker base than NH3 (c) AlCl3 has higher m.pt than AlF3 105. Which of the follwing are ture? (a) SH6 and BiCl5 do not exist (b) There are two pp - d p bonds in SO3 (c) SeF4 and CH4 are tetrahedral

(c) O2–

(d) O+2

(b) NO+ is more stable than O2 (d) SbCl3 is more covalent than SbCl5

(d) I3– is linear molecule with sp3 d hybridisation 106. The hybridisation, which has a lone pair of electrons and shape of I3– is (a) It has sp3 d hybridization (b) It has trigonal bipyramidal shape (c) It is linear (d) It has 3 lone pair of electrons 107. Which of the following show paramagnetism? (a) NO2 (b) NO (c) KO2 (d) Na2O 108. The bond length of C º O bond in carbon monoxide is 1.20Å and in carbon dioxide it is 1.34 Å. Then the C = O bond length in CO3 2 - will be (a) 0.95Å (b) 1.50Å 109. Which of the following relation is/ are correct?

(c) 1.29Å

(d) 1.34Å

1 (a) Covalent character µ Dipole moment

(b) Covalent µ Pseudo inert configuration

(c) Ionic character µ Inert configuration

1 (d) Ionic character µ Dipole moment

110. Which of the following is /are correct for boron and nitrogen in NH3.BF3 adduct? (a) Both has sp3 hybrid orbitals

(b) Both has tetrahedral structure

(c) N is sp3 hybridized while B is sp3 hybridized (d) N in NH3 is pyramidal while B in BF3 is planar 111. Standard heat of formation of KI is –78.31 kcal mol–1. Calculate its lattice energy from following informations: I1(K) = 4.3eV

E1 (I) = 73.4 kcal mol–1 Bond dissociation energy of I2 is 36.1 kcal/mol, sublimation energy of K is 21.51 kcal mol. (a) – 143 kcal mol–1 (b) 143kcal mol–1 (c) 14.3 kcal mol–1 (d) – 14.3 kcal mol–1 112. Which of the following is /are correct statement(s)? (a) Probability of finding the electron in bonding molecular orbital is more than combining atomic orbitals (b) Bonding molecular orbitals are formed when same sign of orbitals are overlap (c) d–d combination of atomic orbitals gives d and d * molecular orbitals (d) None of these [28 of 35]

IIT Package (2009-10) from NOIDA

(Assertion-Reason Type Questions) 113. Assertion: Ionic compounds tends to be non-volatile. Reason : Intermolecular forces in these compounds are weak. (a) If both Assertion and Reason are true, and Reason is correct explanation of Assertion. (b) If both Assertion and Reason are true, but Reason is not correct explanation of Assertion. (c) If Assertion is ture but Reason is false. (d) If both Assertion and Reason is false. 114. Assertion: Water is good solvent for ionic compounds but poor one for covalent compounds. Reason: Hydration energy of ions releases sufficient energy to overcome lattice energy and break hydrogen bonds in water while covalent bonded compounds interact so weakly that even van der Wal’s forces between molecules of covalent compounds cannot be broken. (a) If both Assertion and Reason are true, and Reason is correct explanation of Assertion. (b) If both Assertion and Reason are true, but Reason is not correct explanation of Assertion. (c) If Assertion is ture but Reason is false. (d) If both Assertion and Reason is false. 115. Assertion: The atoms in a covalent molecule are said to share electrons, yet some covalent molecules are polar. Reason: In polar covalent molecule, the shared electrons spend more time on the average near one of the atoms. (a) If both Assertion and Reason are true, and Reason is correct explanation of Assertion. (b) If both Assertion and Reason are true, but Reason is not correct explanation of Assertion. (c) If Assertion is ture but Reason is false. (d) If both Assertion and Reason is false. 116. Assertion: Na2SO4 is soluble in water while BaSO4 is insoluble. Reason: Lattice energy of BaSO4 exceeds its hydration energy. (a) If both Assertion and Reason are true, and Reason is correct explanation of Assertion. (b) If both Assertion and Reason are true, but Reason is not correct explanation of Assertion. (c) If Assertion is ture but Reason is false. (d) If both Assertion and Reason is false. 117. Assertion: The dipole moment helps to predict whether a molecule is polar or non-polar. Reason: The dipole moment helps to predict the geometry of molecules. (a) If both Assertion and Reason are true, and Reason is correct explanation of Assertion. (b) If both Assertion and Reason are true, but Reason is not correct explanation of Assertion. (c) If Assertion is ture but Reason is false. (d) If both Assertion and Reason is false. 118. Assertion : H2 molecule is more stable than HeH molecule. Reason: The antibonding electron in the molecule destabilizes it. (a) If both Assertion and Reason are true, and Reason is correct explanation of Assertion. (b) If both Assertion and Reason are true, but Reason is not correct explanation of Assertion. (c) If Assertion is ture but Reason is false. (d) If both Assertion and Reason is false. 119. Assertion: Carbon tetrachloride does not form a precipitate of AgCl with AgNO3 solution. Reason: Carbon tetrachloride is a liquid. (a) If both Assertion and Reason are true, and Reason is correct explanation of Assertion. (b) If both Assertion and Reason are true, but Reason is not correct explanation of Assertion. (c) If Assertion is ture but Reason is false. (d) If both Assertion and Reason is false. 120. Assertion: NO3– and CO32– ion both are triangular planar.. Reason: Hybridisation of central atom in both NO3– and CO32– is sp2 . (a) If both Assertion and Reason are true, and Reason is correct explanation of Assertion. (b) If both Assertion and Reason are true, but Reason is not correct explanation of Assertion. (c) If Assertion is ture but Reason is false. (d) If both Assertion and Reason is false. [29 of 35]

IIT Package (2009-10) from NOIDA

121. Assertion: H2O has maximum density at 4°C. Hence in water, ice will sink to the bottom at 4°C Reason: Upto 4°C, more and more hydrogen bonds are formed between H2O molecules. (a) If both Assertion and Reason are true, and Reason is correct explanation of Assertion. (b) If both Assertion and Reason are true, but Reason is not correct explanation of Assertion. (c) If Assertion is ture but Reason is false. (d) If both Assertion and Reason is false. 122. Assertion: Boiling point of halogen acids are in the order HF > HCl > HBr . Reason: Electronegativities are in the order F > Cl > Br. (a) If both Assertion and Reason are true, and Reason is correct explanation of Assertion. (b) If both Assertion and Reason are true, but Reason is not correct explanation of Assertion. (c) If Assertion is ture but Reason is false. (d) If both Assertion and Reason is false.

Paragraph-2 The molecular orbital energy diagrams for homonuclear diatomic molecules are given below. For systems containing up to 14 electrons: For systems containing more than 14 electrons: Electrons taken from both the atoms are filled from lower to higher energy molecular orbitals (MO’s) following Hund’s rules. The energy diagram of a heteronuculear diatomic molecule is similar. However, the energies of the atomic orbitals (AO’s) of the atom having higher atomic number being lower, the diagram will be unsymmetrical, but that will not make a difference in the electron count. Bond order is given by half the difference in the number of electrons of the bonding (s and p) and anti-bonding (s* and p*) molecular orbitals. For a bond to have been formed, the bond order should be greater than zero. The greater the bond order, the shorter is the bond distance and the greater is the bond dissociation energy. But if the bond order is the same in two cases, the bond distance will be greater and the bond dissociation energy smaller in the case which has more populated anti-bonding orbitals. The presence of unpaired electron(s) in a molecular orbital will make the system paramagnetic. 123. Which of the following species is not expected to exist? (A) He+2 (B) H +2 (C) Be2 (D) Be+2 124. Which among the following will have a triple bond order? (A) CO (B) CN– (C) NO+ (D) All of these 125. Which of the following orders is correct in respect of bond dissociation energy? (A) N +2 > N -2

(B) O +2 > O -2

(C) NO + > NO

(D) All of these

Paragraph -3 In the ionic bond, a cation tend to polarize the electron cloud of the anion by pulling electron density towards itself. This causes development of covalent character in ionic bond because the electron density gets localized in between the nuclei. The tendency of cation to bring about the polarization of anion is expressed as its polarizing power. The ability of ion to undergo polarization is called its polarisability. The polarizing power of cation and polarisibality of anion are decided on the basis of Fajan rules as given below: i) The smaller the cation, the higher is its polarizing power. ii) Cations with pseudo noble gas configuration (ns2np6nd10) having relatively high polarizing power than those with noble gas configuration (ns2np6). iii) The larger the size of the anion, the higher is its polarisability. [30 of 35]

IIT Package (2009-10) from NOIDA

126. Among the following LiCl, BeCl2, BCl3, CCl4 which will have the lowest highest solubility in water. (A) CCl4, LiCl (B) LiCl, CCl4 (C) BeCl2, BCl3 – – – – 127. The correct order of polarisibility is I , Br , Cl , F (A) I– > Br– > Cl– > F– (B) I– > Br– = Cl– > F– (C) I– = Br– = Cl– > F– 128. The ionic conductance of which of the following is the highest? (A) Li+ (aq) (B) Na+ (aq) (C) K+ (aq) – 129. Hybridization of central M is M3 this is . (a) SP (b) SP2 (c) SP3

melting point and the (D) BCl3, BeCl2 (D) I– = Br– < Cl– = F– (D) Cs+ (aq) (d) SP3d

(Question based on Comprehension) age – IV According to molecular orbital theory all atomic orbitals combine to form molecular orbital by LCAO (Linear combination of atomic orbitals) method. When two atomic orbitals have additive (constructive) overlapping, they form bonding molecular orbitals (BMO) which have lower energy than atomic orbitals whereas when atomic orbitals overlap subtractive, higher energy antibonding molecular orbitals (AMO) are formed. Each M.O. occupies two electrons with opposite spin. Distribution of electrons in M.O. follows Aufbau principle as well as Hund’s rule. M.O. theory can successfully explain magnetic behaviour of molecules. 130. Which of the following is/ are not paramagnetic? (a) NO (b) B2 (c) CO (d) O2 131. Bond strength increases when (a) bond order increases (b) bond length increases (c) antibonding electrons increases (d) bond angle increases 132. O2– will have 2 (a) bond order equal to H2 and diamagnetic (c) bond order equal to N2 and diamagnetic

(b) bond order equal to H2 but paramagnetic (d) bond order higher than O2

age – V Most of the polyatomic molecules except a few such as CO2 and CS2 which are linear or angular with a bond angle generally somewhat greater than 90°. A bond angle is defined as the angle between the direction of two cavalent bonds. Since the atoms in molecules are in constant motion with respect to each other, they are not expected to have a fixed value of bond angle. Repulsion between non-bonded atoms alone does not provide an adequate explanation. Hybridization of bonding orbitals also plays a very important role in determining the value of bond angles. It has been observed that in hybridization as the s-character of hybrid orbital increases bond angle increases. 133. Which of the following having highest bond angle? (a) H2S (b) HSD (c) H2Se (d) All have same bond angle 134. In P4 molecule phosphorus atoms are tetrahedrally arranged. The bond angle P–P–P in the molecule is (a) 108° (b) 120° (c) 60° (d) 180° 135. Which of the following hybridation may have more than one type of bond angle? (a) sp2

(b) sp3

(c) sp3 d

(d) sp3 d 2

[31 of 35]

IIT Package (2009-10) from NOIDA

Prac tice Questions

85. D

86. B

87. A

88. B, C, D

89. C

90. A

91. D

92. C

93. B

94. C

95. D

96. A

97. A

98. D

99. A More Than One Correct Option

100. A, B

101. A, B, C

102. A, B, C

103. B, C, D

104. B, C

105. A, B, D

106. A, C, D

107. A, B, C

108. C

109. A, B, C

110. A, B

111. A

112. A, B, C

Assertion-Reason Type Questions

113. C

114. A

115. A

116. A

117. A

118. B

119. B

120. A

121. D

122. B

123. C

124. D

125. D

126. D

127. A

128. D

129. A

Solutions: 113. 114. 115. 116. 117. 118.

Correct Reason. Intermolecular forces in ionic compounds are strong Reason is the correct explanation of Assertion Reason is the correct explanation of Assertion Reason is the correct explanation of Assertion Reason is the correct explanation of Assertion. Correct explanation. H2 molecule is more stable than HeH because bond order of H2 = 1,

1 2 Correct explanation: Carbon tetrachloride is a covalent compound. Reason is the correct explanation of Assertion Correct Assertion: Ice cannot exist as ice at 4°C. Correct Reason: H–bonds keep on breaking on heating ice. Correct explanation: In HF, there is H-bonding. So its boiling point is highest. HBr has higher boiling point than HCl because of greater molecular mass and hence greater van dar Waal’s forces. that of HeH =

119. 120. 121. 122.

Comprehension Questions

130. C 135. C

[32 of 35]

131. A

132. A

133. B

134. C

IIT Package (2009-10) from NOIDA

1. 2. 3. 4.

Predict the shapes of the following molecules using the VSEPR model: BeCl2, SiCl4, AsF5, H2S. Explain why BeH2 molecule has a zero dipole moment although the Be – H bonds are polar. The dipole moment of hydrogen halides decreases from HF to HI. Explain this trend. O2 molecule is paramagnetic. Why?

5.

Sodium chloride solution gives a precipitate of AgCl with AgNO 3 whereas CCl 4 , does not, why ?

6.

Why He 2 does not exist?

7.

CO 2 and SO 2 both are triatomic molecules but there is big difference in their dipole moment, why ? H2S has more vapour pressure than H2O under similar conditions. Why?

8. 9. 10. 11.

The dipole moment of LiH is 1.965 ´ 10 -29 Cm and the interatomic distance between Li and H in this molecule is 1.596Å. Calculate the percent ionic character in LiH. o-nitrophenol is more volatile than p-nitrophenol, explain why? Which of the following halides have different bond lengths between central atom and halogen and why? BF3 , PCl 5 , SF6 , CCl 4

12.

Arrange the following in order of increasing (A) dipole moment: H 2 O, H 2 S, BF3 ; (b) covalent character: LiCl, LiBr, LiI; (c) melting point: NaCl, MgCl 2 , AlCl3.

13. 14. 15.

NH +4 has bond angle identical to CH 4 but NH 3 has different bond angle; explain with proper reasoning. Boiling point of ethyl alcohol is much higher than that of diethyl ether. Why? Is there any change in the hybrid state of B and N atoms as a result of the following reaction?

BF3 + NH 3 ¾¾ ® H 3 N + - B- F3 16.

Valency of oxygen is generally two whereas sulphur shows valency of two, four and six. Explain.

17.

Write the electronic configuration and calculate the bond order of H +2 , H 2 and He 2 . Bond length in H +2 is longer than in H 2 . Explain why ?

18. 19. 20.

Explain the observed bond angle order. Cl2O > H2O > F2O NH4+ has bond angle identical to CH4 but NH3 has different bond angle; explain with proper reasoning. Boiling point of CH4 is less than SiH4 but reverse is the order with NH3 and PH3 – explain.

[33 of 35]

IIT Package (2009-10) from NOIDA

2.

6. 7. 8.

Dipole moment is vector quantity, due to 180° bond angle the net dipole moment becomes zero. Due to decrease in electronegativity from F to I, the magnitude of fractional charge decreases. O2 has two unpaired electrons. NaCl bond is polar so Cl– releases in solution and gives precipitate of AgCl where as CCl4 is non-polar and it does not ionizes. Bond order is zero. CO2 is linear (zero lp) SO2 is bent (one lp). Due to H-bonding in H2O.

9.

The dipole moment of 100% ionic molecule ( Li + H - ) = q ´ d

3. 4. 5.

= (1.602 ´ 10 -19 C)(1.596 ´ 10 -10 m ) = 2.557 ´ 10 -29 Cm

Percentage ionic character

=

Exp. value of dipole moment ´ 100 Theoretical value of dipole moment

=

1.964 ´ 10 -29 ´ 100 = 76.8 %. The bond has LiH is 76.8% ionic. 2.557 ´ 10 -29

10.

Due to intramolecular H-bonding in o-nitrophenol it is more volatile.

11.

PCl 5 has two types of bonds: axial and equatorial. These two types of bonds have different

bond lengths. 1, 2, 3-equatorial bonds 4, 5 –axial bonds. Cl 4

1

3 Cl

P 2 5

12.

Cl

Cl

Cl a) BF3 is a symmetrical molecule. It has zero dipole moment. Oxygen being more electrone gative than S, bond moment of O – H is more than S – H. So, the dipole moments are in

the order of BF3 < H 2 S < H 2 O b) The anion size in increasing order is

Cl - < Br - < I Hence, LiCl is least covalent and LiI most. The order is LiCl < LiBr < LiI

c) Cation charge increases in the order Na+ < Mg2+ < Al3+ Thus, Al 3+ ion has maximum polarization effect and Na + ion has least. Thus, the decreasing order of melting point is [34 of 35]

IIT Package (2009-10) from NOIDA

NaCl < MgCl 2 < AlCl 3

13.

In NH +4 there are four bond pairs and no lone pair like CH 4 . However,, NH 3 has only three bond pairs and one lone pair, hence, its bond angle is less due to lone pair-bond pair repulsion.

14.

Due to hydrogen bonding.

15.

During the combination of NH3 and BF3. N atom is donor and B atom of BF3 is acceptor. The hybrid state of N in NH3 is sp3 whereas that of B in BF3 is sp2. In the compound H3N+ – B–F3, both B and N atoms is surrounded by 4 bond pairs. The hybrid state of N remains as sp3 but that of B atom changes from sp2 to sp3. There are no d-orbitals in oxygen hence it cannot extend its covalency. Vacant d-orbitals are present in sulphur and paired orbitals can be made unpaired by shifting electrons to d-orbitals, i.e., either making four orbitals singly occupied or six orbitals singly occupied showing valency 4 or 6 besides 2.

16.

17.

The number of electrons, their configurations, etc., for the given species are :

No. of Species Configuration electrons

Bond order N2

Na

1 = [N b - N a ] 2 1 2

H +2

1

( s1s)1

1

0

H2

2

( s1s) 2

2

0

1

2

2

0

He 2

4

*

( s1s) 2 ( s1s) 2

The bond length in H2+ is longer than in H2 because in H2+ only one electron is present to shield the two nuclei from mutual repulsion. In H2 there are two electrons to hold the two nuclei thus nuclear repulsion is less than that inH2+. Hence, nuclear separation in H2+ is more than in H2. 18.

In case of F2O, because of more electronegative atom F, bond pair of electrons in O – F bon are shifted towards F, thereby decreasing bp – bp repulsion. Now lp – lp repulsion ‘closes’ the angle. That is why in H2o bond angle is higher (104.5°) than F2O (103.2°). In case of Cl2O, delocalization of the lone pair of electrons of oxygen to the vacant d-orbital of chlorine decreases the repulsion by lone pair and increases the repulsion between bond pairs. The bond angle thus becomes very large (110.8°).

19.

In NH4+ there are four bond pairs and lone pair like CH4. However, NH3 has only three bond pairs and one lone pair, hence, its bond angle is less due t lone pair bond pair repulsion.

20.

The nature of force exists between the covalent molecules is Vander Waal’s force of attraction. The strength of the Vander Waal’s forces increases as the size of the units linked increases. As a consequence boiling point also increases. So, SiH4 has a higher boiling point than CH4. But in NH3 intermolecular H bonding makes molecules associated as a result boiling point of NH3 is more than PH3 (where such effect is absent).

[35 of 35]