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.
) p 3 s 3 p 2 s 2 s 1 ( K ) s 4 p 3 s 3 p 2 s 2 s 1 ( K 2 2 6 2 6 electron 1 loses 1 6 2 6 2 2 gains 2 2 6 2 5 2 2 6 2 6
1 electron (Ar configuration) Cl (1s 2s 2p 3s 3p )Cl (1s 2s 2p 3s 3p )
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 atoms 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.
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 Na e Na (g) I ) g ( S n Sublimatio ) s ( (g) E e of Addition ) g ( * D 2 / 1 on Dissociati ) g ( 2 Cl Cl Cl 2 1 A -) S ( U formation Crystal ) g ( ) g ( Cl NaCl Na
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 H = S + I + 2 1
D – Ea – U
The most important of these energy terms are I, Ea and U, since these are considerably greater than the remaining terms 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 Mg+ + e– Mg+
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:
Al e Al E1 E 1 = 577kJ 2 E 2 AlAl e E2 = 1816kJ 3 E 2 3 Al Al e E3 = 2745kJ
It is on this account 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.
Characteristics of Electrovalent Compounds
Melting and Boiling Point: Due to the powerful electrostatic force between the ions in a crystal of an electrovalent compound considerable energy is 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 melts 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.
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.
Cl Cl H
H
Hydrogen molecule Chlorine molecule (Both formed by equal sharing of electrons between the atoms, i.e., by non-polar bonds)
H 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.
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 .
Magnitude of dipole moment ||(chargeqin esu)(Distanrcein )
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
1 Cm = 2.998 10 D 10 10 998 . 2 29 18 11 or 3.336 10 Cm 10 998 , 2 1 D 1 29 30
Dipole moment is a vector quantity and is often indicated by an arrow parallel to the line joining the point of charge and pointing towards the negative end e.g., H F.
% Ionic character of a covalent bond = 100
character ionic 100% assuming moment dipole l Theoretica moment dipole al Experiment
Illustration 1: 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 || = 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 = 100 10 165 . 4 10 336 . 3 29 29 = 80.1% 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 F C H H H H C Cl Cl Cl Cl C Cl H H H d 75 . 1 D 0 0D 1.86D Hydrogen fluoride Methane 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 the difference is greater than 1.9, the bond is correspondingly more ionic”. For example, when the electro negativity difference is 0.8,
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.,
compound) (ionic compound) (Covalent 4(b.p. 33K) andNaCl(b.p. 1713K) SiCl
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 passage 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 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.
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.
character covalent increasing CaI CaBr CaCl CaF2 2 2 2
Similarly, in case of trihalides of aluminium, the covalent character increases with increase in size of halide anion i.e.
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.
BRAIN TEASER 1:
SnCl2 is solid whereas SnCl4 is liquid, why?
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).
Electro-negative atom H X or X H
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.
H X H X H X H X H X
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 N H H H N H H H N H H H N H H H N H H H ii) Water O H H O H H O H H O H H
iii) Acetic acid
O C H3 O H O O H CH3
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-Nitrophenol H -O O N O
o-Hydroxy benzoic acid
H
-O O
O
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.
H2O NH3 HF CH4 SnH4 GeH4 SnH4 SbH3 AsH3 PH3 HCl HBr HI H2Se H2Se H2S 100 0 -100 VIA VIIIA VA IVA
Molecular weight increasing
M el ti ng po int s (°C ) inc re a si ng H2O NH3 CH4 SnH4 GeH4 SnH4 SbH3 HCl HBr HI H2Se H2Se H2S 100 0 -100 VIA VIIIA VA IVA
Molecular weight increasing
Bo il ing po int s (° C) inc re a si ng (a) (b) PH3 HF -200 -200
ii) Ice has less density than water. The explanation of this fact is as follows: In the crystal structure of ice, the O-atom is surrounded by four H-atoms. Two H-atoms are linked to O-atom by covalent bonds as shown (by normal covalent bond) and the remaining two atoms 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.
H H H -H H H H H H H Å 76 . 2 1.80Å .96Å 0.96Å 0 molecule ter wa
Open cage-like tetrahedral crystal structure of ice. Circles indicate oxygen atoms. Bonds represented by solid line are normal covalent bonds while those represented by dotted lines are hydrogen bonds.
BRAIN TEASER 2:
Although HF forms stronger hydrogen bond than H2O, Hv of H2O is greater than that of HF why?
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.
A B x x xx x x A B x x xx x x + 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 N H H + H+ H N H 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 xxOxx C O x x x x 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 non-polar (organic) solvents.
iii) Conductivity: Like covalent compounds, these are also bad conductors of electricity. The solutions or fused mass do not allow the passage of electricity.
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 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
Orbitals on Shape Number of Number of central atom bond pairs lone pairs Bond angle
BeCl2 2 Linear 2 0 180° BF3 3 Plane triangle 3 0 120° CH4 4 Tetrahedral 4 0 109°28 NH3 4 Pyramidal 3 1 107°48 NF3 4 Pyramidal 3 1 102°30 H2O 4 Bent (V-shape) 2 2 104°27 F2O 4 Bent (V-shape) 2 2 102°
PCl5 5 Trigonal bipyramid 5 0 120° & 90°
SF4 5 Trigonal bipyramid 4 1 101°36 & 86°33
ClF3 5 T-shape 3 2 87°40
XeF2 5 Linear 2 3 180°
SF6 6 Octahedral 6 0 90°
BrF5 6 Square pyramidal 5 1 84°30
XeF4 6 Square planar 4 2 90°
Some examples using the VSEPR Theory
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 PCl
5 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.
Note: Thus PCl5 is highly reactive, and in the solid state it splits into [PCl4]+ and [PCl 6]
– ions, which have tetrahedral and octahedral structures respectively.
Cl Cl Cl Cl Cl P Structure of PCl5 molecule
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 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.
Chlorine trifluoride molecule
Cl F F F Cl F F F Cl F F F I II III
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.
Sulphur hexafluoride SF6: The electronic structure of S is 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 in the outer shell: hence the structure is octahedral. There are no lone pairs, so the structure is completely regular with bond angles of 90°. S F F F F F F
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.
4. In CH4,the C atom in its ground state has the electronic configuration 1s2, 2s2, 1 x p
2 , 2p1y and only has two unpaired electrons, and so can form only two bonds. If the C atom is excited, then the 2s electrons may be unpaired, giving 1s2, 2s1, 1
x p
2 ,2p1y, 1 x p
2 . There are now four unpaired electrons which overlap with singly occupied s orbitals on four H atoms.
1s 2s 2p
2px 2py 2pz
Electronic structure of carbon atom - groun state
2s
2p
Carbon atom - excited state
Carbon atom having gained four electrons from H atoms in CH4
molecule. sp3
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 ( and 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 ( bond): The bond formed by the overlapping of two half filled atomic orbitals along their axis is known as sigma bond. bond is a strong bond because overlapping in it takes place to large extent. The hybrid orbitals always from bond.
a) s – s overlapping Molecular axis b) s – p overlapping c) p – p overlapping head on overlap pz pz p-p overlap M.O.
2. Pi bond ( 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, bond formed is a weak bond. bond overlapping takes place only at the sides of two lobes. A bond is formed when a bond already exists between the combining atoms.
M.O. p-p overlapping
p p
Example:
In A – B molecule the bond formed is bond.
A In
B, molecule thereareoneandonebonds A In bonds two and one are there molecule , B
Thus, all the single bonds are bonds. Double bond has one and one bond. Triple bond has one and two 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’
Promotion of an electron 2s 2s 2p sp3 E n er g y Hybridisation
Excited State Hybridised State
Ground State 2p 109.5° C H H H H H H H H
Shape and formation of methane molecule
sp2 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 bonds with H atoms. The unhybridized 2p, orbitals of the two carbon atoms undergo a side-wise overlap forming a bond. Thus the carbon to carbon double bond in ethene is made of one bond and one bond. Since the energy of a bond is less than that of a bond, the two bonds constituting the ethene molecule are not identical in strength. The molecule is a planar one.
Promotion of an electron 2s 2s 2p sp2 E n er g y Hybridisation Excited State Ground State 2p Pure p-orbital sp2 hybrid orbitals
C H C H H H pz p z sp2 sp2 sp2 sp 2 sp2 H H H H sp2
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.
Types of atomic orbitals used Hybridisation Orientation of Examples hybrid orbitals
1. one s + one p-orbital sp Linear BeF2,BeCl2,C2H2,HgCl2
2. one s + two p-orbitals sp2 Trigonal planar 2
3 3 4 2 3,C H ,NO ,CO BF
3. one s + three p-orbitals sp3 Tetrahedral
4 2 4 4 4 4 4,CCl ,SiF ,NH ,SO ,ClO CH
4. one s + three p + d sp3d Trigonal bipyramidal PF5,PCl5
5. one s + three p + two d 3 2
d
sp Octahedral SF5,[CrF6]3,IF5 6. one s + three p+three d sp3d3 Pentagonal Bipyramidal IF
7
7. One d + one s + two p dsp2 Square planar
Only in complexes like 2 4 2 4) ,[PtCl ] ) CN ( Ni [ 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 Polyatomic Cation ] G V [ 2 1 X [V G a] 2 1 X [V G c] 2 1 X
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
Calculate the value of X and decide the hybrid state of central atom as follows : X 2 3 4 5 6 7 Hybrid state sp sp2 sp3 sp3d sp3d2 sp3d3 5 PF COCl2 NH4 ClO4 ] 5 5 [ X12 X12[24] X12[451] X12[071] = 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
3 NO IF5 CO2 XeF4 ] 1 5 0 [ X12 X12[57] X12[04] X12[48] = 3 = 6 = 2 = 6 2 sp sp3d2 sp sp3d2 6 PCl PH3 SF3 SF4 ] 1 5 6 [ X12 X12[35] X12[361] X12[46] = 6 = 4 = 4 = 5
Hybrid state Hybrid state Hybrid state Hybrid state
2 3d
sp sp3 sp3 sp3d
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 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
1s, *1s, 2s, *2s, 2px, [2py = 2pz], [*2py = *2pz], *2px For N2 and lower molecules
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.) =
2 ABMO in electrons of No. -BMO in electrons of No.
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 2 Arrange the species O2, O2–,O 2
2– and O 2
+ in the decreasing order of bond order and
stability and also indicate their magnetic properties. Solution: The molecular orbital configuration of O2, O2–,O 2 2– and O 2 2+ are as follows: O2 = 1s2, *1s2, 2s2, *2s2, 2p x 2, 2p y 2, 2p z 2, *2p y 1 = *2p z 1 Bond order = 2 2 6 -10
, No. of unpaired electrons = 2
paramagnetic O2– = 1s2, *1s2, 2s2, *2s2, 2p x 2, 2p y 2, 2p z 2, *2p y 2 = *2p z 1 Bond order = 2.5 2 5 -10
, No. of unpaired electrons = 1
paramagnetic O22– = 1s2, *1s2, 2s2, *2s2, 2p x 2, 2p y 2, 2p z 2, *2p y 2 = *2p z 2 Bond order = 1 2 8 -10
, No. of unpaired electrons = 0 diamagnetic O2+ = 1s2, *1s2, 2s2, *2s2, 2p x 2, 2p y 2, 2p z 2, *2p y 1 = *2p z 0 Bond order = 2.5 2 5 -10
, No. of unpaired electrons = 1 paramagnetic
Now as the bond order decreases in the order O2+ O 2 O2
– O 2
2–
So, same will be the stability order of the above species because stability is directionally proportional to bond order.
PRAYAS - I
Q.1 Which of the following has zero dipole moment ?
(A) NH3 (B) H2O (C) BCl3 (D) SO2
Q.2 Which of the following has maximum bond energy ?
(A) H – O – H (B) H – S – H (C) H – Te – H (D) H – Se – H Q.3 Which of the following statements is false ?
(A) there will be 25% s character in hybrid orbitals if the angle between two hybrid orbital is 105° (B) in HClO4. HClO3 and HClO2 . number of valance electrons will be 32.26 and 20. respectively (C) bond length is inversely proportional to bond order
(D) dipole moment of CH2Cl2 is zero Q.4 The bond present in CuSO4.5H2O will be
(A) ionic bond and hydrogen bond (B) covalent bond
(C) coordinate bond (D) all of the above
Q.5 Ionic compound is (A) H2O (B) HCl (C) Csl (D) NH3 Q.6 If A (s) A+ (g) + e ... 600 KJ mol–1/2 2 1 B2 (g) + e B– (g) ... 260 KJ mol—1 A (s) + 2 1 B2 (g) AB (s) ... – 550 KJ mol–1 then lattice energy of AB (s) will be
(A) – 160 (B) – 890 (C) – 420 (D) + 360
Q.7 Which of the following ions gives cyanide ion in aqueous solution ?
(A) potassium ferrocyanide (B) potassium ferricyanide (C) potassium cyanide (D) potassium cobaltcyanide Q.8 Nature of bond between two nonmetal atoms will be
(A) Van der waals (B) covalent (C) ionic (D) all of the above
Q.9 If the formula of a compound is X2 Y5, then what will be the numbers of electrons present in the valance shells of X and Y, respectively ?
(A) 5 and 6 (B) 6 and 3 (C) 2 and 3 (D) 5 and 2
Q.10 The compound with two lone pairs and two bond electrons is
(A) SCl2 (B) PH3 (C) NH3 (D) HF
Q.11 Which of the following has maximum melting point ?
(A) SiC (B) Al4Cl3 (C) CO2 (D) Cl2
Q.12 Covalent compound is
(A) K2O (B) Cl2O (C) CaO (D) MgO
Q.13 A molecule MX3 has zero dipole moment. The orbitals used by M in the formation of bond will
be-(A) pure p (B) sp2 (C) sp (D) sp3d
Q.14 Which of the following ions has an atom in a state of sp2 hybridisation ?
(A) BeF3– (B) NF
3 (C) OF2 (D) H3O
Q.15 Which of the following statements is false for BF3 ?
(A) it is a Lewis acid (B) it fiorms an addition produced with ammonia (C) it is a planar molecule (D) it has ionic character
Q.16 Maximum bond length is shown in
(A) CO2 (B) CH4 (C) NH3 (D) H2O
Q.17 Which of the following does not have H - bond ?
(A) water (B) phenol (C) liquefied HCl (D) Liquefied NH3 Q.18 Coordinate bond is formed in
(A) NaCl (B) NH4Cl (C) Cl2 (D) BCl3
Q.19 Both the compounds of which pair have linear geometry ?
(A) SO2, SnCl2 (B) CO2, BeH2 (C) SO2., CO2 (D) SO2, H2O Q.20 XPO4 is a compound of a metal X, then the formula of the chloride of X will be
(A) XCl3 (B) XCl (C) X2Cl3 (D) X2Cl2
Q.21 Which of the following has pyramidal geometry ?
(A) PCl3 (B) SO3 (C) CO3–2 (D) NO
3 – Q.22 Select the correct statement.
(A) (CF3)3N is a base, whereas (CH3)3N is not
(B) PbCl2 melts at 606° C. whereas PbCl4 melts at 114°C
(C) NaCl is soluble in petrol (D) Dipole moment of BF3 is high Q.23 False statement is
(A) SnCl2 is white and SnI2 is coloured (B) Boiling point of H2O is higher than H2S (C) Water is capable of extinguishing fire of petrol (D) Glucose forms H-bonds with water
Q.24 Lewis structure of carbon suboxide (C3O2) in ground state is
(A) O : C : : : C : C ::: O : (B) : O :: C : C : C :: O : (C) : :: C :: C :: C :: : (D) : O :: C :: C :: C :: O : Q.25 Which of the following hypothesis justifies that the bond angle of H2S is 92° ?
(A) Lewis structure (B) Valence bond theory (C) Valance bond concept of hybrid orbitals (D) Octet rule
Q.26 Geometry of a molecule in which the central atom has 50% p character, will be
(A) linear (B) tetrahedral (C) trigonal (D) distorted tetrahedral Q.27 Configuration of the cation of which of the following metals is ns2, np6 , ndx type, and the total value of
2 + 6 + x is nine to eighteen ?
(A) alkali metal (B) alkaline earth metal (C) inert metal (D) d block metal Q.28 Pb+4 is less stable than Pb+2, because of
(A) inert pair effect (B) small size
(C) high ionisation potential (D) high electronegatively Q.29 The element showing highest valency is –
(A) Cl (B) I (C) F (D) Br
Q.30 Increasing order of dipole moment in H2O, NH3, NF3 and CCl4 is
(A) CCl4 < NF3 < NH3 < H2O (B) CCl4 > NF3 > NH3 > H2O
Q.31 Species in which peroxide ion is not present is
(A) PbO2 (B) H2O2 (C) SrO2 (D) A and B both
Q.32 If s character decreases in hybrid orbital, then bond angle
(A) decreases (B) increases (C) remains uncertain (D) all are wrong Q.33 Which of the following compounds does not have a coordinate bond ?
(A) SO3 (B) H2SO4 (C) H2SO3 (D) HNO2
Q.34 Maximum covalency of an element is
(A) the number of unpaired s electrons (B) number of unpaired s and p electrons
(C) number of unpaired d electrons (D) number of s and p electrons present in valance shell Q.35 Which of the following molecules consists of only covalent bonds ?
(A) KCl (B) PCl3 (C) NH4Cl (D) BaClO4
Q.36 Number of hybrid orbitals is equivalent to
(A) the atomic orbitals participating in hybridisation (B) electrons participating in the process
(C) number of unpaired electrons in the valence shell (D) vacant orbitals in valence shell
Q.37 When electron cloud of an anion is shared by the cation, then the polar bond so formed exhibits which of the following characters ?
(A) covalent (B) metallic (C) coordinate (D) van der waals
Q.38 Which of the following types of bond in an inorganic compound undergoes dissociation to gives ions ? (A) coordinate (B) covalent (C) electrovalent (D) hydrogen bond
Q.39 What will be the ionic potential , when nature of the oxide of a metal is basic ? (A) > 2.2 and < 3.2 (B) > 2.2 (C) > 3.2 (D) < 2.2 Q.40 Which of the following does not have sp3 hybridisation ?
(A) BF4– (B) H
3O
+ (C) OF
2 (D) BF3
Q.41 Sulphur have maximum covalency –
(A) 3 (B) 4 (C) 5 (D) 6
Q.42 Which of the following has least ionic character ?
(A) Cu2Cl2 (B) KCl (C) CsCl (D) BaCl2
Q.43 Which of the following are correctly matched ? Compound Hybridisation 1. Graphite – sp3 2. NH4+ – sp2 3. XeF2 – sp3d2 4. SF4 – sp3d (A) 4 (B) 3 (C) 2 (D) 1
Q.44 Formula of the phosphate of a metal is MHPO4. What will be the formula of its chloride
(A) MCl (B) MCl2 (C) MCl3 (D) M2Cl2
Q.45 The central atom of which of the following compounds has four bond pairs and two lone pairs
(A) SF4 (B) XeF4 (C) NH+
Q.46 Three soluble salts of three metals A (Atomic weight 7), B (atomic weight 27) and C (atomic weigh 64) on electrolysis liberate 2.1, 2.7 and 9.6 grams of metals, respectively at the electrode. The valencies of the metals are, respectively ,
(A) 1, 3 and 2 (B) 3, 1 and 2 (C) 3, 1 and 3 (D) 2, 2 and 3 Q.47 Incorrect information about Cl2O is
(A) angular structure (B) 110° bond angle (C) four lone pairs (D) two bonds Q.48 Maximum energy is released from which process ?
(A) F + F F2 (B) N + N N2 (C) Cl + Cl Cl2 (D) O + O O2 Q.49 False statements about CO2 is that
(A) its dipole moment is zero (B) its solid form is called dry ice
(C) it is diamagnetic (D) hybridisation state of C and O is same Q.50 What will be the energy of the system on bringing x and y atoms closer upto bond distance ?
(A) infinity (B) minimum (C) maximum (D) zero
Q.51 Bond angle of HCN is similar to all of the following except
(A) BeF2 (B) C2H2 (C) OF2 (D) CO2
Q.52 When a molecule breaks, then
(A) it absorbs energy (B) it transmits energy
(C) it neither transmit nor absorbs energy (D) it transmit as well as absorbs energy Q.53 Which of the following statements in true for hybridisation ?
(A) sp3 hybridisation is found in SiF
4. SnCl4. ClO4 –, SO 4 –2, NH 4 + and NH 3
(B) sp hybridisation is found in BeCl2, BeF2 , BeH2, Hg Cl2. ZnCl2, CO2, C2H2 and HCN (C) sp2 hybridisation is found in BeF
3 –, BCl
3, BH3, C2H4 AlCl3, NO3 –1 (D) all of the above are correct.
Q.54 Which of the following statements is true for [Cu(NH3)4]2+ ? (A) tetrahedral configuration and all paired electrons (B) square planar configuration and one unpaired electron (C) square planar configuration and all paired electrons (D) none of the above
Q.55 What is the reason that ionic compounds do not show stereo isomerism
(A) presence of ions (B) nondirectional nature of ionic bond
(C) brittle nature (D) electrostatic forces of attraction between ions Q.56 In which of the following compounds, the coordinate bond is not present ?
(A) CuSO4 (B) H2O (C) BF4– (D) AlCl
4 – Q.57 Bond length of which of the following types of bonds is maximum ?
(A) sp2 – sp2 (B) sp – sp (C) sp3 – sp3 (D) sp3 – sp2 Q.58 Which of the following is correct for CO2 ?
(A) nonpolar with polar bonds (B) polar with nonpolar bonds (C) polar with polar bonds (D) all of the above are wrong Q.59 Which of the following statements is true ?
(A) there is H-bonding in aqueous hydrochloric acid
(B) dipole moment of NF3 is more than the dipole moment of NH3 (C) dipole moment of CCl4 is less than the dipole moment of CHI3 (D) positions of atoms get changed in resonance
Q.60 In a compound, 50% X (valency 1) and 50% Y (valency 1) are present, then the formula of the compound will be
(A) X2Y (B) XY (C) XY2 (D) X2Y3
Q.61 Electronic configuration of an atom is ns2 np2 , which forms 2 and 2 bonds in a hydrocarbon, then hybridisation state and bond angle of the atom will be
(A) sp, 180° (B) sp2, 120° (C) sp3, 109° 28 (D) dsp2, 90° Q.62 Which of the following compound is not known ?
(A) OF2 (B) SF4 (C) SF6 (D) OF6
Q.63 Which of the following bonds should have higher polarity than the remaining three –
(A) C – F (B) C – O (C) C – B (D) C – H
Q.64 Which of the following has maximum melting point ?
(A) ionic crystal (B) covalent crystal (C) metallic crystal (D) molecular crystal Q.65 Inorganic graphite is –
(A) boron nitrate (B) boron nitride (BN)
(C) boron carbonate (D) none of the above
Q.66 The type of bond between the layers in graphite is –
(A) vander waals force (B) covalent
(C) coordinate (D) ionic
Q.67 Anhydrous HCl is –
(A) an acid (B) a base (C) a salt (D) a covalent compound
Q.68 Which of the following has ionic and covalent bonds ?
(A) OF2 (B) KCl (C) AIN (D) KCN
Q.69 Which of the following oxides is most acidic ?
(A) MnO2 (B) Mn2O3 (C) Mn2O7 (D) MnO
Q.70 Which of the following has least bond energy –
(A) C – O (B) C = O (C) C – C (D) C = C
Q.71 Which of the following has minimum bond energy –
(A) O – H (B) Cl – H (C) C – H (D) N – H
Q.72 p – p overlapping is not possible in –
(A) Cl2 (B) O2 (C) N2 (D) H2
Q.73 A bond is formed O2 by overlapping of –
(A) 2py – 2pz (B) 2py – 2s (C) 2pz – 2s (D) 2pz – 2pz Q.74 Highest bond angle will be in –
(A) H2O (B) H2S (C) H2Se (D) H2Te
Q.75 Which of the following ions has tetrahedral geometry –
(A) Na+ (B) NH
4
+ (C) Mg+2 (D) CO
3 –2
Q.76 Which of the followin molecules has hybridisdation on central atom different from that on the remaining three ?
(A) SiH4 (B) CH4 (C) BF4– (D) [Ni(CN
4) –2] Q.77 Which of following has square planar geometry ?
(A) XeF4 (B) SiCl4 (C) NH4+ (D) BF
4 –1
Q.78 Molecule having zero dipole moment is –
(A) CH2Cl2 (B) NF3 (C) BF3 (D) ClO2–1
Q.79 Hybridisation state of carbon changes as follows. sp3 sp2 sp, then the angels of hybrid orbitals will –
(A) remain unchanged (B) go on increasing (C) go on decreasing (D) first inct Q.80 CCl4 is a covalent compound, whereas LiCl is less covalent, because –
(A) C – Cl bond is nonpolar (B) charge on CCl4 is more
(C) Li – Cl bond is polar (D) moment of Li – Cl is not definite Q.81 Shape of xenon hexafluoride is –
(A) tetrahderal (B) distorted pentagonal bipyramidal
(C) square planar (D) octahedral
Q.82 CO2 is isostructural with –
(A) SnCl2 (B) HgCl2 (C) ZnI2 (D) OH2
Q.83 Which of the following statements is correct ?
(A) bond order is a measure of the strength of the bond.
(B) bond order is equal to the number of bonds present in a molecule (C) greater the bond order, more paramagnetic is the molecule (D) none of the above
Q.84 Octahedral shape exists in hybridization –
(A) sp3d (B) sp3d2 (C) sp3d3 (D) none of these Q.85 Relative stabilities of O2, O2– and O
2 +, O
2
2– are in the order – (A) O2 > O2+ > O 2 – > O 2 2– (B) O 2 + > O 2 > O2 – > O 2 2– (C) O2+ > O 2 2– > O 2 – > O 2 (D) O2 2– > O 2 – > O 2 + > O 2 Q.86 Which of the following is not paramagnetic –
(A) S2– (B) NO (C) O
2
– (D) N
2 – Q.87 The dipole moment of a molecule of the type AX4 having a square planar geometry is –
(A) O (B) 4D (C) 2D (D) none of the above
Q.88 The nodle plane in the –bond of ethene is located in – (A) the molecular plane
(B) a plane parallel to the molecular plane
(C) a plane perpendicular to the molecular plane which bisect the carbon-carbon bond at right angle
(D) a plane perpendicular to the molecular plane which contains the carbon-carbons bonds Q.89 The type of bonds present in CuSO4.5H2O are ... only –
(A) electrovalent and covalent (B) electrovalent and co-ordinate (C) electrovalent, covalent and co-ordinate (D) covalent and co-ordinate Q.90 The structure of XeF4 is –
(A) planer (B) tetrahedral (C) square planar (D) pyramidal Q.91 Which of the following is most stable –
Q.92 Shape of molecules is decided by –
(A) sigma bond (B) -bond
(C) both sigma and -bonds (D) neither sigma and -bonds Q.93 The metallic lustre exhibited by sodium is explained by –
(A) diffusion of sodium ions (B) excitation of free proton
(C) oscillations of loose electrons (D) existence of body-centred cubic lattice Q.94 The molecule which has pyramidal shape is –
(A) PCl3 (B) SO3 (C) CO32– (D) NO
3 – Q.95 Sulphuric acid molecule contains –
(A) only covalent bonds (B) covalent and ionic bonds
(C) covalent and co-ordinate bonds (D) covalent, ionic and co-ordinate bonds Q.96 NF3 is –
(A) non-polar compound (B) electrovalent compound
(C) having low value of dipole moment than NH3 (D) having more dipole moment than NH3
Q.97 The maximum possible number of hydrogen bonds in which a water (H2O) molecule can participate –
(A) 4 (B) 3 (C) 2 (D) 6
Q.98 Variable valency is characteristic of –
(A) noble gases (B) alkali metals (C) transition metals (D) non-metallic elements Q.99 PCl5 exists but NCl5 does not because –
(A) nitrogen has no vacant 2d-orbitals (B) NCl5 is unstable (C) nitrogen atom is much smaller than p (D) nitrogen is highly inert Q.100 The type of bond formed between H+ and NH
3 in NH4
+ ion is –
(A) ionic (B) covalent (C) dative (D) hydrogen
PRAYAS - II
Q.1 Weakest bond is
(A) ionic bond (B) covalent bond (C) coordinate bond (D) hydrogen bond Q.2 Which of the following statements is true ?
(A) and bonds are weak bonds (B) bond is stronger than bond (C) both the bonds are equally strong (D) all the above are wrong Q.3 Which of the following is nondirectional orbital ?
(A) sp (B) sp3 (C) dsp2 (D) s
Q.4 orbital will formed by
(A) overlapping of s-s orbitals (B) overlapping of sp-s orbitals
(C) coaxial overlapping of p-p orbitals (D) collateral overlampping of p-p orbitals Q.5 Which of the following compounds has linear geometry ?
(A) CO2 (B) NO2 (C) SO2 (D) SnCl2
Q.6 Which of the following has an atom having incomplete octet ?
(A) BF3 (B) NH4+ (C) CCl
4 (D) PH3
Q.7 The compound having odd number of electron pairs is
Q.8 Electronic configuration of two elements X and Y are 2, 5 and 2, 7, respectively. If these elements form a covalent compound with each other, then the formula of the compound will be
(A) XY3 (B) X2Y3 (C) XY2 (D) XY
Q.9 What is the oxidation number of carbon in diamond ?
(A) 0 (B) 4 (C) 1 (D) 2
Q.10 The s character in sp3d hybrid orbital will be (A) 5 1 (B) 4 1 (C) 2 1 (D) 4 3
Q.11 Which of the following is present in the lead pencil ?
(A) Pb (B) Charcoal (C) Graphite (D) Fe
Q.12 What is the type of hybridisation and nature of Ni in [Ni(CN)4–2 ion ? (A) dsp2 and diamagnetic (B) sp3d and paramagnetic (C) dsp3 and diamagnetic (D) sp3 and paramagnetic Q.13 Which of the following species as boron in sp3 hybridisation state ?
(A) BF4– (B) BH
3 (C) B2H6 (D) BCl3
Q.14 Hydrolysis of SiCl4 occurs very fast, because
-(A) Si is an electrophile due to vacant orbital in it (B) SiCl4 is ionic
(C) SiCl4 is covalent (D) Si is an element of V A group Q.15 The factor responsible for insolubility of BaSO4 in water is
(A) high dissociation energy (B) high covalent character (C) high hydration energy (D) high lattice energy Q.16 In hybridisation process
(A) the hybrid orbitals formed are nondirectional
(B) the hybrid orbitals of the atom have almost same energy (C) shapes of the hybrid orbitals are different
(D) hybrid orbitals have high attractive forces between them Q.17 The compound soluble in ether is
(A) NaCl (B) RbCl (C) BaCl2 (D) LiCl
Q.18 Which of the following is not a covalent compound ?
(A) NH3 (B) MgCl2 (C) SO2 (D) CCl4
Q.19 Element X is strongly electropositive and Y is strongly electronegative. X and Y are monovalent, then the compound formed from them will be
(A) X+ Y– (B) X - Y+ (C) X - Y (D) X Y
Q.20 Which of the following statements is correct ?
(A) all the covalent compounds are solids at room temperature (B) all the compounds of hydrogen are solids
(C) all the halogen compounds are ionic (D) all the salts are generally ionic in nature
Q.21 The valency of B in BCl3 is 3. This is justified on the basis of –
(A) resonance (B) hybridisation
