Download 1 Chemical Bonding and Molecular Structure

1
Chemical Bonding and Molecular Structure
1.
What are molecular ?
Ans : A group of atoms which combine and exist as single species having characteristics
properties are known as “Molecules”.
2.
What is a chemical bond ?
Ans : The attractive force which keeps different components (atoms/ions) combined
together in a molecule is called chemical bond.
3.
Which theories have been put forward to explain chemical bonding ?
Ans : To understand the chemical behaviour of atoms several principles have been put
forward like Kossel-lewis approach, VSEPR principle, valance bond theory.,
Molecular orbital theory etc.
4.
Who proposed the 1st successful theory to explain the function of chemical bond
?
Ans : The 1st successful attempt of explaining the formation of chemical bond in terms of
electrons was given by Kossel and Lewis in 1916.
5.
Explain the main points of “Lewis” hypothesis ?
Ans : Lewis proposed an atom has positively charged kernel (nucleus and electrons of
inner orbit), while 8 electrons can be accommodated in outer shell.
These 8 electrons are assumed to occupy the 8 corners (ends) of a cube surrounding
the kernel.
Thus valence e– of Na-atom occupies only one corner (end), while inert gases
(except He) occupy 8 corners of the cube.
Lewis noted that, when atoms are combined by chemical bond, they attain stable
structure of octet of eight electrons. e.g. The atom of Na element losses one electron
and attain stable octet structure while atom of chlorine receives one electron (Cl–)
and attain stable octet structure. Thus, Na loses one e to form Na+ while Cl atom
gains one e to form Cl¯, thereby achieving stable configurations.
Certain molecules like O2, Cl2, F2 etc attain octet configurations by sharing of es .
6.
What are Lewis’s symbol ? Give their significance ?
Ans : The electrons of only outermost shall (valence es ) take part in chemical bonding.
Lewis introduced simple notations to represent these valence electrons, these
notations are called “Lewis symbols”
In this approach, the valence es are shown by dots surrounding the symbol of an
element.
The number of this valence electrons is useful in calculation of valency of group.
Some examples are given as under …
•
••
•
•
•
Li
Be • B • • C • • N •
•
••
•
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7.
Ans :
Explain formation of NaCl or electrovalent bond according to Kossel’s approach
?
Halogen elements being highly electronegative and alkali metals being highly
electropositive, are different from inert gases.
Halogens receive one e and become negatively charged while alkali metals lose
one e and become positively charged.
These positively and negatively charged ions attain stable configurations like inert
elements (except He) having general configuration of ns2np6.
The positive and negative ions are stablized by electrostatic attraction.
In the formation of ionic compound, the atoms having less ionization enthalpy
combine with atoms having more electronegativity and the bond formed by this
combination is called ionic bond or electrovalent bond.
The capacity of formation of ionic bond of an element is called electrons lost OR
gained by element.
The formation of NaCl can be given as under :
(a)
Na
→
Na+
+
e–
1
[Ne] 3s
[Ne]
(b)
Cl
+
[Ne] 3s2 3p5
e–
→
Cl–
[Ne] 3s23p6 or [Ar]
(c)
Na+
+
Cl–
→ NaCl
Thus, the bond formed by electrostatic attraction between positive ion and negative
ion is called electrovalent bond OR ionic bond.
8.
What is “electronic theory of chemical bonding” given by Kossel and Lewis ?
Ans : According to this theory atoms combine either by transfer of their valence electrons
or by sharing of valence electrons to attain octet configurations in their valence
shells.
9.
Ans :
(i)
(ii)
2
State the factors on which formation of ionic bond depends.
A Formation of ionic bond depends on following factors :
Ease of formation of positive and negative ions from neutral atoms.
The formation of positive and negative ions depends on values of ionisation
enthalpy and electron gain enthalpy of respective atoms.
M(g) → M(g)n+ + ne— (Ionisation )
X(g) + e–
→ X–(g) (Electron gain)
Ionisation reaction is always endothermic while electron gain reaction is generally
exothermic.
Thus elements having low ionisation enthalpy and high electronegativity form
ionic bonds easily.
Arrangement of positive and negative ions in the lattice structure of crystalline
compound
The positive and negative ions get arranged in 3-dimensional arrangement having
a strong Coulombic force of attraction between them. The structure formed as a
result of this is called crystal lattice structure or crystal configuration.
The stability of a crystal is decided by crystal lattice enthalpy. The energy required
to remove (separate) one mole of gaseous ions to infinite distance from one mole of
solid crystal is called as “crystal Lattice enthalpy”.z
The lattice enthalpy of NaCl is 788 kJmol–1.
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10.
What is a covalent bond ? Explain its formation ?
Ans : When two or more atoms of any elements attain octet structure by sharing of their
valence es , the bond formed is called “Covalent bond”.
If each atom shares only one electron the bond formed is called “Single Covalent
bond”. e. g.
H
•×
××
× Cl × , H × C • H
H×
H,
H
•
•
×
• ×
××
ו
H
If each atom shares two electrons the bond formed is called “double covalent
bond”. e. g.
H
••
••
••
••
• O • • O • ; • O • × C ו O •
×
•
•
•
•
•
•
×
•
•
• H
×
×
×C
×
;
× × C×
×
•
H•
•
H
If each atoms shares there electrons, the bond formed is called “triple covalent
bond”.
• N •• ••
•
•
• • N •
,
× ×
H • ×C × × C × • H
× ×
11.
Explain the types of covalent bonds ?
Ans : There are 3 types of covalent bonds based on type of sharing and type atoms of
combining elements :
(a) Polar Covalent-Bond :
Ease of formation of positive and negative ions from neutral atoms
The atoms of different elements having different electronegativity of positivity take
part in bond formation. Such molecules are polar. e. g.
HCl
(b)
Non-polar Covalent – Bond :
The atoms of different elements having different electronegativity of positivity take
part in bond formation. Such molecules are polar. e. g.
This type of covalent is formed by atoms belonging to the same element and hence
having same electronegativity or electropositivity
Such compounds are non-polar.
H •• H
(c)
••
••
• Cl • •
•
Cl
•
• •
•
••
••
Co-ordinate Covalent- Bond :
In formation of this type of bond, the electron-pair is donated by any one atom for
sharing. This bond is shown by an arrow (→). Where the direction of arrow is from
atom donating the electron pair towards the atom receiving it.
H
H
ו •×
•
H × B ×× N •× H
ו •×
H
3
××
× Cl ×
H •
×
××
H
H
H
H
B
N
H
H
H
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12.
State limitations of “Octet theory” or Limitations of “Electronic theory of
chemical bonding” by Kossel- Lewis.
Ans : Shapes of molecules cannot be known.
Certain molecules are stable although their octet is not completed. e. g.
••
••
BeCl2 •• Cl •× Be •× Cl ••
• • although they
• • have more than 8 es around central
Certain molecules are stable
atom (expanded octets)e. g. PCl5 , SF6 etc.
In certain cases, the stability decreases with increase in size of central atom
eventhough the octet is completed.e. g.
NH3 > PH3 > AsH3 > SbH3 > BiH3
— Stability decreases →
13.
State the important steps for Lewis dot representations.
Ans : The main steps are as follows :
The total no. of electrons needed for drawing such a structure are obtained by
taking the sum of valence electrons of all the atoms involved in formation of
molecule.
For anions, the no. of es equal to the negative charge are added and for cations the
no. of es equalto the positive charge are deducted from total sumof es .
e. g. in CO 23 −
, 2es are added while in NH +4
, 1 e is deducted.
The least electronegative atom becomes the central atom.
The other atoms are arranged surrounding the central atom and the electrons are
placed around each atom so that their octets are completed.
The electron-pair which is shared between atoms and hence forms the covalent
bond is called “bonding- pair”, while the electron pair which is not involved in
bonding is called “non-bonding” or “lone-pair”.
14.
Give Lewis electronic structures of H2, O2, O3, NF3, CO32– and HNO3.
Ans : Refer table – 1.1 Page 3~4 of Text Book
15.
Define the terms (i) Bond-length (ii) Bond-angle (iii) Bond-energy (iv) Bondorder.
Ans : (i) Bond – Length : The equilibrium distance between the nuclei of two bonded
atoms in a molecule. It is measured by X-ray diffraction method or by
spectroscopic method. It is expressed in picomaters (1 pm = 10–12 M)
(ii) Bond- angle : The angle between the orbitals possessing bonding electronpairs around the central atom of a molecule/ion. It is expressed in degrees and
determined by spectroscopic methods. Bond-angles can give some information
about the shapes of the molecules.
e. g. The bond angle in H2O is 104° 30’ indicating angular shape.
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(iii) Bond-energy/enthalpy :
The energy, needed for breaking one mole of bonds in one mole gaseous
substance is called bond-energy or bond- enthalpy. It is expressed in KJ/Mole.
Higher value of bond-enthalpy indicates higher stability.
e. g. H2(g) → H(g) + H(g)
B. E. = 435.8 KJ/Mol.
(iv) Bond – Order
The no. of bonds present between atoms in a molecule is called its bond order.
e. g. in H2 (H — H) — bond order is one
in O2 ( O = O) – bond order is two
in N2 (N ≡ N) – bond order is three
Higher the bond-order, less is the bond- length and greater is the stability of the
molecule.
16.
Explain the Born-Haber cycle taking example of formation of NaCl.
Ans : The formation of ionic bond and the stability of ionic crystal depends on lattice
enthalpy.
More the value of lattice enthalpy of the product more will be the stability and
easier will be the formation of ionic bond.
The value of lattice energy will be more only when the attraction force between two
atoms is more. Born –Haber first of all gave value of total enthalpy or lattice
enthalpy by studying the enthalpies of different steps involved in the formation of
ionic compounds by ionic bonds.
Born-Haber Cycle can be explained by taking example of different enthalpy
changes involved in the formation of NaCl crystalline compound from its
components in standard state . Generally the simple equation for the formation of
NaCl crystal can be written as follows :
1
Na(s) + Cl 2 (g ) → NaCl (s) ∆ 1H = − 411 kJ mol -1
2
But different steps of formation of NaCl crystal are as follows :
(i)
First of all, sodium element/atom which is in solid form at room
temperature is sublimed and the value of sublimation enthalpy (∆sH) is 108
kJmol–1.]
Sublimation enthalpy
∆ s H = + 108 kJmol –1
Na(s) 
→ Na(g )
(ii)
The value of ionisation enthalpy required to remove electron from one
mole Na (g) atom to infinite distance is 496 kJ mol–1.
Ionization enthalpy
Na(g ) 
→ Na + (g ) + e −
(iii)
(iv)
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∆ i H = + 496 kJmol –1
Gaseous chlorine element exists as diatomic molecule of room temperature.
In one mole solid NaCl crystal structure formation only one chlorine atom is
required. Hence decomposition of chlorine molecule becomes essential. The
enthalpy required for this is called decomposition enthalpy.
1
decomposition
∆ D H = +121 kJmol −1
Cl 2(g ) 
→ Cl(g )
2
2
The gaseous atom of chlorine obtained this way accepts one electron and
changes to chloride ion. The enthalpy released for this is called electron gain
enthalpy (∆egH)
Electron gain enthalpy
∆ eg H = – 349 kJmol −1
Cl(g ) + e − 
→ Cl—
(g )
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(v)
One mole gaseous positive ion (Na+) and one mole gaseous negative ion
(Cl–) combine with each other and form one mole of solid ionic crystal. The
enthalpy released during this is called lattice enthalpy which is expressed as
‘U’
Lattice enthalpy
Na +(g ) + Cl (g ) − →
NaCl (s) ∆ U H = – 787 kJmol −1
By the above method, the crystal structure of any ionic solid can be known
by the study of each step of the different steps according to Born-Haber
cycle, and the value of enthalpy of formation – lattice enthalpy can be
obtained. The above different steps can be shown as below :
All the above different steps can be written in the form of equation as
follows :
∆fH = ∆sH + ∆ D H + ∆iH + ∆egH + ∆UH
2
The enthalpy of formation of crystal of one mole NaCl can be obtained by
adding up the different enthalpies as below :
∆fH = 108 + 121 + 496 + (–349) + (–787)
=
– 411 kJ mol–1
Thus, the value of enthalpy of formation for one mole NaCl obtained is
negative which indicates its stability.
17.
Explain the phenomenon of resonance?
Ans : The experimentally observed properties of certain molecules cannot be explained
by a single Lewis Structure. e. g. Ozone molecule can be explained by following
two structures :
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In both the structures, O – O single bond and O = O double bond are present.
O – O single bond length is 148 pm and O = O double bond length is 121 pm; but
the bond length between any oxygen-oxygen in ozone molecule is same, and its
value is 128 pm.
Thus, the bond length between oxygen-oxygen in ozone molecule is in middle of
single bond and double bond, which cannot be explained by any one of the
structures I and II as shown in given fig.
Hence, the idea of resonance was presented to explain the real structures of
molecules like ozone, CO2 etc.
Thus, according to the concept of resonance, in such cases a number of structures
with same energy, same position of nucleus as well as bonding electron pairs and
non bonding electron pairs are taken as the resonance structures of the hybrid
structure which represents the molecule accurately. According to given fig,
structures I and II, represent the resonance structures of structure III which is the
real structure of ozone. Such structures are called hybrid resonance structures.
Such similar example like benzene : carbonate ion (CO32–) also can explain the
resonance.
Resonance give stability to the molecule because the energy of resonance hybrid
structure is less than those of structures shown separately. Thus, decrease in energy
is responsible for the stability of such molecules.
18.
Give hybridization, shapes and examples of molecules in which the central atom
has no non-bonding electron pairs in their lewis electronic structures.
Ans : Refer table 1.3 Page 7 of text book
19.
Give hybridistaion, shapes and examples of molecules in which the central atom
has no non-bonding pairs in their lewis structures ?
Ans : Refer table – 1.3 Page 7 of Text Book.
20.
Give hybridisation, shapes and examples of molecules in which the central atom
has non-bonding and bonding electron pairs around the central atom ?
Ans : Refer table – 1.4 and 1.5 Page 8 and 9 of Text Book.
21.
Explain the main points of VSEPR theory.
Ans : Lewis Approach was incomplete to explain shapes of molecules. Only the
hypotheses about the structures of molecules containing covalent bond can be
made.
First of all, in 1940. Sidwick and Powell presented the theory of repulsion existing
between electron pairs present in valence shells of atoms.
In 1957, Nyholm and Gillespie presented this theory in the developed form. The
main hypotheses of this theory are as follows :
• The shape of the molecule depends upon the number of electron pairs (bonding
and non-bonding electron pairs) in valence shells around the central atom.
• The electron pairs present in valence shell being negatively charged, repel each
other.
• These electron pairs possess the tendency of obtaining such an arrangement in
the space that the repulsion between them is minimum and as a result there is
maximum distance between them.
• If there are two or more than two resonance structures in the molecule,
VESPER principle can be applied to any of the structures.
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• The magnitude of the repulsion produced between electron pair is as follows :
Lp-Lp > Lp-Bp > Bp-Bp
Thus, VSPER principle is helpful in determining geometrical shapes produced due
to presence of electron pair in the molecule. Especially, the geometric structures of
compounds of elements of p-block can be explained by this principle.
Why is the bond-angle in SO2 less than 120° although it has sp2 hybridisation
around the central atom ?
Ans : In SO2, the central sulphur atom has 2 bonding pairs and 1 non-bonding pair
around it.
The higher repulsion between the bonding and non-bonding pairs squeezes the
bond angle and makes it a little less i.e. 119.5°.
22.
23.
Explain the bonding, hybridization and structure of methane molecule ?
Ans : The electronic structure of central carbon atom in methane molecule in ground
state is … 6C : 1s2 2s2 2px 1 2py1 2pz0
As four hydrogen atoms are combined with central carbon atom, four half filled
orbitals will be required. For this, the electron structure of carbon in excited state
*
2
1 2p1 2p1 2p1
will be…
6C : 1s 2s
x
y
z
Thus, carbon atom in its excited state utilizes sp3 hybrid orbitals, combines with
half filled 1s1 orbitals of hydrogen atom and forms four covalent bonds.
Thus, in methane molecule, there are four bonding electron pairs (Bp) between
carbon and hydrogen atoms. According to Sidgwick-Powell rule, there is minimum
repulsion between these four bonding electron pairs. As a result, tetrahedral
structure corresponding to sp3 hybridisation is observed in which the bond angle as
in regular tetrahedral structure that is 109° 28′ is observed.
24.
Explain the structure, bonding and hybridisation of ammonia molecule ?
Ans : In the molecule of NH3, nitrogen atom has five valence electrons, out of which
three electrons combine with electrons of 1s orbital of three hydrogen atoms by
sharing and form three covalent bonds.
Even after that nitrogen atoms has two electrons (or one electrons pair) which
remains uncombined even after formation of bonds. Thus nitrogen has one nonbonding and three bonding electron pairs around it and hence it has sp3 hybridised
orbitals around it.
Then non bonding electron pair of nitrogen atom repels the three bonding electron
pairs around nitrogen according to Sidgwick-Powell rule. As a result, these electron
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pairs are pushed inside. As a result the bond angle between them decreases and
instead of regular tetrahedral bond angle of 109° 28′ it is obtained as 107°.
25.
Explain the structure, bonding and hybridisation of water molecule ?
Ans : In H2O molecule, there are total six valence electrons around central oxygen atom,
out of which, two valence electrons form two covalent bond with 1s orbital
electrons of two hydrogen atoms but remaining four electrons (or two electron
pairs) become non bonding electron pairs.
Thus there are 2 bonding pairs and 2 non-bonding pairs of electrons around the
central oxygen in water resulting in sp3 hybridisation.
According to Sidgwick-Powell rule there is maximum repulsion force between
these two non bonding electron pairs and as a result they remain away from each
other. When it happens so, they go near the bonding electron pairs due to repulsion
is produced between them. Because of this, bonding electron pairs are pushed
inside, in more proportion and there is noticeable decrease in the bond angle. In
H2O molecule – even though there is sp3 hybridisation but the bond angle is
reduced to 104° 30′ from a bond angle 109° 28’.
26.
Explain the polarity in covalent bonds ?
Ans : Any chemical bond does not possess completely ionic or completely covalent
nature.
In the molecules like H2, N2, O2, Cl2 etc. the covalent bond is formed by sharing of
electrons between its two atoms; but when the experimental values of bond lengths
and bond enthalpies are studied, it is found that there is some contribution of ionic
bond.
Heteronuclear diatomic molecules like hydrogen fluoride (HF), the electron pair of
covalent bond remains dragged more towards atom of fluorine because the electronegativity of fluorine (in comparison to hydrogen atom) is more. As a result the
partial negative charge (–δ) is produced on atom of fluorine and partial positive
charge (+δ) is produced on atom of hydrogen. Its real structural formula can be
written as …
Dipole Moment is found in the molecule due to this property of polarity.
For e.g. In a molecule of water, one oxygen atom and two hydrogen atoms combine
to form two covalent bonds. The electro negativity of oxygen being more relative to
that of atom of hydrogen, the electron pairs remain dragged towards atom of
oxygen. This produces partial positive charge (+δ) on hydrogen atom and partial
negative charge (–δ) on oxygen atom.
In addition due to the two non-bonding electron pairs on atom of oxygen element,
property of polarity is as shown below :
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The value of dipole moment of ammonia (NH3) is not zero, so it is not linear but the
structure of the molecule is pyramidal.
27.
Why is dipole moment of NH3 more than that of NF3 ?
Ans : In both these molecules there is one non-bonding electron pair on nitrogen.
The polarity of all the three N-H bonds is towards the nitrogen in NH3 as nitrogen
is more electronegative than hydrogen and hence the net dipole moment increases.
In NF3, the polarity of nitrogen is less than fluorine and hence the polarity of N-F
bond is towards fluorine atom. As all the N-F bonds are in different directions the
net dipole moment is reduced.
This can be explained by the figure given below.
28.
The net dipole-moment of BF3 and BeF2 are zero- explain ?
Ans : BeF2 is a linear molecule and the dipoles of two polar Be-F bonds remain in exactly
opposite directions.
As dipole moment is a vector quantity the two dipoles of Be-F cancel each-other
and hence the net dipole-moment is zero making the molecule non-polar.
BF3 is a planar and triangular molecule containing three polar B-F bonds with their
dipoles oriented towards fluorine atoms.
The three bonds are oriented at an angle of 120° and the resultant dipole-moment
of any two bonds is equal and opposite to the third and thus the net moment is
zero making the molecule non-polar.
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29.
Ans :
Explain valence-bond theory giving example of the formation of hydrogen
molecule ?
In 1927, Heitler and London first of all proposed the valance bond theory. Its
detailed study and development were carried out by scientists Linus Pauling and
Slater.
According to this theory, when half filled atomic orbitals come near to each other,
they get overlapped and as a result a covalent bond is formed.
Thus, one half filled orbital of one atom overlaps with half filled orbital of other
atom and forms bond. Thus, the combining atoms share their valence electrons and
form covalent bond.
Sometimes, out of the two combining atoms, the orbital of one atom may be
completely filled and the other orbital of the other atom may be vacant, even then
they both overlap and form a special type of covalent bond (which is called coordinate covalent bond).
Consider two atoms, A and B of hydrogen element which possess nuclei NA and
NB. The electrons in the two atoms are shown as eA and eB.
When these two atoms are at very far distance, there is no inter attractive force
between them. When they come nearer to each other, the attractive and repulsive
forces are produced between them.
The attractive forces are produced because of following factors :
(i) The attractive forces produced between nuclei of the atoms and their own
electrons or NA – eA and NB– eB.
(ii) The attractive force produced by nucleus of one atom and electron of other
atom. NA– eBand NB – eA.
In the same way, repulsive forces are produced because of the following factors :
(i) Repulsive force produced between electrons of both the atoms i.e.,eA – eB.
(ii) Repulsive force produced between nuclei of both the atoms shown as
NA – NB.
Attractive forces try to take both the atoms nearer to each other while repulsive
forces try to push them apart.
It is proved from the experiments that the magnitude of attractive forces is more
than that of repulsive forces.
As a result both the atoms go near to each other and their potential energy
decreases. Both the atoms go near to each other to such an extent that attractive
forces are balanced by repulsive forces and the system attains minimum energy.
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At this stage, both hydrogen atoms combine with each other and remain at certain
definite distance and forms stable hydrogen molecule. This distance or bond length
is 74 p.m.
30.
Give main assumptions of valence-bond theory ?
Ans : According to valence bond theory, the half filled atomic orbitals having same
symmetry and same energy overlap with each other and form covalent bond.
As a result there is pairing of electrons in valence orbitals. The strength of covalent
bond is directly proportional to the magnitude of overlapping of orbitals. The main
assumptions of valence bond theory can be given as follows.
(i)
Generally there must not be much difference in energies of overlapping
atomic orbitals.
The overlapping orbitals must be half filled and the spin of the electrons in
(ii)
them must be in opposite direction to each other.
(iii) There must be overlapping of atomic orbitals to proper extent, so that
chemical bond formation can take place.
31.
Explain the types of overlapping taking place between atomic orbitals according
to valence-bond theory ?
Ans : When two atoms come near to each other, there is overlapping of atomic orbitals.
This overlapping can be positive, negative or zero which depends upon the
properties of the overlapping atomic orbitals.
The different overlapping arrangements of s and p orbitals are shown as under.
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The assumptions of overlapping of atomic orbitals are applicable similarly to
homonuclear, heteronuclear, diatomic and polyatomic molecules.
The chemical bonding in molecules having poly nuclear atoms like CH4, NH3 and
H2O can be explained with the help of valence bond theory. Their shapes and bond
angles can also be known.
The shapes of the molecules… CH4, NH3 and H2O are tetrahedral, pyramidal and
bent respectively. These geometrical shapes can be explained with reference to
overlapping of atomic orbitals.
32.
“Simple orbital overlap cannot explain the directional characteristics of bonds”explain this statement taking example of CH4 molecule ?
Ans : The electronic structure of central atom carbon in CH4 molecule, in its ground state
is [He] 2s2 2px 12py1 2pz0 ; and in excited state it is [He] 2s12px12py12pz1.
Thus, four half filled orbitals of carbon atom overlap with 1s orbitals of four
hydrogen atom, which are also half filled and form four C – H bonds.
Three p-orbitals of carbon atom are at 90° to one another and so three C – H bonds
are found at 90° to one another.
But 2s orbital of carbon and 1s orbital of hydrogen atom are symmetrically
spherical, hence, they overlap with each other in any direction. Hence, the direction
of fourth C – H bond is not determined clearly.
But this assumption is not consistent with tetrahedral molecular structure of CH4 in
which bond angle is 109° 28’. Hence, it can be said that the overlapping of atomic
orbitals does not possess vector properties.
The same disagreement is also observed in case of molecules like NH3 and H2O.
Explain the nature of covalent bonds based on the type of overlapping of orbitals ?
33.
Ans : The covalent bond formed by overlapping of atomic orbitals can be divided in to
two types.
(i)
σ-Bond :
This type of σ - covalent bond is obtained by end to end overlapping of the ends of
two atomic orbital having inter-nuclear axis. It is also called axial overlapping. This
type of overlapping is obtained by overlapping of atomic orbitals as shown below :
(a) s-s Overlapping : In this type, overlapping of two half filled s-orbitals having
same axis occurs.
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(b) s-p overlapping : In this type, there is overlapping between half filled s-orbital
of one atom and half filled p-orbital of other atom.
(c) Overlapping of p-p orbitals : In this type, there is overlapping of half filled porbitals of two approaching atoms.
(ii)
π – bond :
When the axes of atomic orbitals that are overlapping remains parallel to each other
and perpendicular to the inter-nuclear axis, the type of covalent bond formed is
called π−bond.
The strength of any type of σ or π bond is proportional to the magnitude of
overlapping of atomic orbitals.
In σ bond formation, overlapping of atomic orbitals is more and as a result σ - bond
is strong.
In π - bond formation the magnitude of overlapping is less because of overlapping
of atomic orbitals is sidewise and as a result π - bond is weaker relative to σ - bond.
34.
What is “molecular orbital theory” and discuss its important points ?
Ans : The molecular orbital theory was first of all presented by Mulliken and Hund in
1932.
According to them, the description of molecular orbital is similar to that of atomic
orbital. As electrons of an atom are arranged in atomic orbitals (such as s, p, d….
etc) in the same way the electrons of the molecule are arranged in molecular
orbitals.
The probability distribution of electrons around the nucleus of the atom can be
shown by atomic orbitals. Similarly, the probability distribution of electrons
around a group of nuclei in a molecule can be shown by molecular orbitals.
The filling of electrons in molecular orbitals is carried out obeying Auf-bau
principle, Pauli’s Exclusion principle and Hund’s rule of maximum multiplicity.
Important points of molecular orbital theory are as under…
(i) As the electrons of the atom are arranged in different atomic orbitals, similarly
electrons of the molecule are arranged in different molecular orbitals.
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e-CHEMTEST / CH-1 / Sem-II / Ph : 30004433
(ii)
The atomic orbitals having similar energy and proper symmetry combine with
each other in some definite way and give molecular orbitals.
(iii) The electrons in the atomic orbitals are under the effect of only one nucleus,
while the electrons in the molecular orbitals are under the effect of two or
more nuclei. Thus atomic orbitals are mono centric, while molecular orbitals
are poly centric.
(iv) The number of molecular orbitals formed is equal to the number of combining
atomic orbitals. Half the number of these molecular orbitals are called
Bonding Molecular Orbitals (BMO) and remaining half of these molecular
orbitals are called Anti-bonding Molecular Orbitals (ABMO).
(v) The energy of the bonding molecular orbitals is less in comparison to the antibonding molecular orbitals.
35.
Ans :
Explain the formation of molecular orbitals by linear combination of atomic
orbitals (LCAO)?
According to quantum mechanics, the formation of molecular orbitals between any
two atoms is done by linear combination of acceptable wave functions or Linear
combination of atomic Orbitals.
When two hydrogen atoms combine and H2 molecule is formed, acceptable wave
functions of each hydrogen atom Ψ1s(1) and Ψ1s(2) undergo linear combination in
two ways and hence two acceptable wave functions ΨMO and Ψ∗MO are formed.
This can be shown as under…
ΨMO = Ψ1s(1) + Ψ1s(2)
Ψ∗MO = Ψ1s(1) – Ψ1s(2)
Here, (1) and (2) are the numbers used for two atoms of hydrogen. Two molecular
wave functions formed here describe two molecular orbitals.
The molecular orbital indicated by ΨMO is called bonding molecular orbital and by
Ψ∗ MO is called antibonding molecular orbital.
When the molecular orbital is formed by complementary overlapping of two 1s
orbital, it is spread over the nuclei of both the atoms. The total energy of this
molecular orbital is less than the total energy of both the atomic orbitals. It is called
bonding molecular orbital.
When the molecular orbital is formed by opposing overlapping of two 1s orbitals
its total energy is more than the total energy of both the atomic orbitals, it is called
antibonding molecular orbital.
36.
State the conditions for combination of atomic orbitals as per MO-theory.
Ans : When the molecular orbitals are formed by linear combination of atomic orbitals,
some of the requirements should be satisfied. These requirements are also called
the conditions for the linear combination of atomic orbitals. They are as follows…
(i) The atomic orbitals of the combining atoms must possess similar energies.
(ii) The combining atoms must be as near as possible so that the overlapping can
be maximum on the axis of atomic orbitals.
(iii) The symmetry of the atomic orbitals of combining atoms must be same.
37.
Based on the symmetry of molecular orbitals explain their types and also discuss
their nodal planes ?
Ans : Consider the linear combination of different atomic orbitals to form different
molecular orbitals as shown in the figure below:
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e-CHEMTEST / CH-1 / Sem-II / Ph : 30004433
16
In the above figure the horizontal line is the bond axis while (+) and (–) signs are
the signs of wave function; they do not indicate electric charge.
If the rotation of axis is carried out by 1800 angle, the positions of (+) and (–) signs
in (a) and (b) do not change. Hence, these molecular orbitals possess cylindrical
symmetry. These orbitals are expressed by σ - symbol. In (a) the probability of
electron in the region between two nuclei is maximum and so this type of
molecular orbital is called bonding molecular orbital (σ−orbital) .
In (b) the probability of electron in the region between two nuclei is less and so this
type of molecular orbital resist the entry of electrons. Hence, it is called antibonding molecular orbitals (σ *-orbital).
The molecular orbitals shown by (c) and (d) are of π- type because, the positions of
the signs of wave functions change by rotation of bond axis by an angle of 180°.
In (c) the probability of electron in the region between two nuclei is more and so it
is π - type bonding molecular orbital (π).
In (d) the probability of electron in the region between two nuclei is decreasing and
so it is an anti-bonding molecular orbital (π∗). In this type of molecular orbital there
is a nodal plane between the two nuclei. The probability of finding electron in the
nodal plane is zero.
Generally σ and σ * - type molecular orbitals are formed by combination of s — s ,
s — p or pz – pz atomic orbitals.
By linear combination of Px — Px OR Py — Py atomic orbitals, π and π∗ molecular
orbitals are formed.
e-CHEMTEST / CH-1 / Sem-II / Ph : 30004433
38.
Give the order of energies of different molecular orbitals in different Homonuclear
diatomic molecules containing atoms having atomic nos. from 1 to 10 ?
Ans : The types of molecular orbitals formed when diatomic molecules are formed by
bringing two atoms of elements from hydrogen to neon near to each other along
with their relative energies can be shown as under
The increasing order of molecular orbitals for molecules from H2 to N2 is shown as
under:
σ 1s < σ∗1s < σ 2s < σ∗ 2s < (π 2px = π 2py) < σ 2pz < (π∗ 2px =π∗ 2py) < σ∗2pz
The order changes for molecular orbitals for molecules from O2 to Ne2 , which is
given as under:
σ 1s < σ∗1s < σ 2s < σ∗ 2s< σ 2pz < (π 2px = π 2py) < (π∗ 2px =π∗ 2py) < σ∗2pz
39.
Ans :
(i)
(ii)
(iii)
What is molecular electronic configuration and what information can be
obtained from it?
The distribution of electrons in different molecular orbitals is called molecular
electronic configuration. Some important information is obtained from electronic
configuration of molecule.
Stability of Molecule : Suppose Nb is the number of electrons in bonding
molecular orbital and Na is the number of electrons in anti bonding molecular
orbitals, then :
(i)
For Nb > Na , molecule is more stable.
For Nb < Na molecule becomes unstable.
(ii)
Bond Order and Stability :
The stability of molecule is directly proportional to the value of bond order.
The difference between total number of electrons in bonding molecular orbitals
(Nb) and the number of electrons in anti-bonding molecular orbitals (Na) when
divided by two, the value obtained is called the bond order.
1
Bond order = [N b − N a ]
2
If the value of bond order is positive (Nb > Na) the molecule becomes stable but
negative or zero value of bond order (Nb < Na OR Nb = Na) it indicates instability
of molecule.
Magnetic properties:
If the molecular electronic configuration of any molecule consists of unpaired
electrons then such a molecule is said to be paramagnetic while if all the electrons
are paired, then the molecule is said to be diamagnetic.
40.
Draw the molecular orbital diagrams of homonuclear diatomic molecules of
atoms from hydrogen to neon and discuss the bond-orders and magnetic
properties?
Ans : Refer Page nos. 19, 20 and 21 in Text Book.
41.
Draw the molecular orbital diagrams for NO and CO molecules and discuss
their stability, bond-order and magnetic properties?
Ans : CO ( carbon monoxide) molecule :
CO is a hetero nuclear diatomic molecule.
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e-CHEMTEST / CH-1 / Sem-II / Ph : 30004433
There is difference in energy levels of atomic orbitals of carbon and oxygen atoms.
As compared to carbon atom, the energy level of respective orbitals of oxygen atom
is at lower level i.e. possess less energy. As a result it is more stable.
Total fourteen electrons (six electrons of carbon atom and eight electrons of oxygen
atom) are there in carbon monoxide molecule.
These fourteen electrons are arranged in molecular orbitals obtained by
combination of atomic orbitals of carbon and oxygen.
It can be seen from the energy diagram of carbon monoxide that all the electrons in
it are paired and so it is diamagnetic. The energy diagram of CO molecule is shown
under.
The arrangement of fourteen electrons present in carbon monoxide molecule, can
be written in molecular orbitals as follows :
Ele. configuration of CO
Bond Order
Magnetic Property
18
:
: (σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (π2px)2=(π2py)2 (σ2pz)2
1
1
Nb − Na ] =
[10 − 4] = 3
[
2
2
: No unpaired electrons. ∴ CO is diamagnetic.
e-CHEMTEST / CH-1 / Sem-II / Ph : 30004433
NO-Molecule :
NO is a hetero nuclear diatomic molecule.
Atoms of two different elements nitrogen and oxygen take part in its formation.
There is difference in the energy of oxygen and nitrogen atom. The energy of
oxygen is less than that of corresponding nitrogen atom.
There are total 15 electrons (seven electrons of nitrogen atom and eight electrons of
oxygen atom). The arrangement of these 15 electrons can be made in molecular
orbitals as shown in Fig.
As there is one unpaired electron in NO molecule, it is paramagnetic and as the
bond order is a fraction, and hence NO molecule is unstable.
The electrons of 1s orbitals (inner shells) do not take part in formation of bond and
so they are called non-bonding electrons.
Ele. configuration of NO
Bond Order
Magnetic Property
19
: (σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (σ2pz)2 (π2px)2=(π2py)2 (π*2px)1
1
1
N b − Na ] =
[10 − 5] = 2.5
[
2
2
: One unpaired electrons. ∴ NO is paramagnetic.
:
e-CHEMTEST / CH-1 / Sem-II / Ph : 30004433
42.
What are Van-der Waal forces of attraction ?
Ans : Scientist van der Waals mentioned the existence of weak attractive forces between
molecules of any compound on the basis of the study of their deviation from ideal
gas behaviour.
The existence of such weak attractive forces cannot be explained by any other
chemical attractive forces and so they are in general known as van-der Waal’s
attractive forces.
There is presence of van der Waal’s force of attraction in components of ionic
substances and also in components of covalent substances.
It’s magnitude is very less (about 42 kJ mole–1) as compared to other attractive
forces, hence it is mostly covered by other attractive forces.
This attractive force prevails up to a very less distance of about 4.4 Å .
The electrons on the surface of one molecule experience attraction by the nucleus of
other molecule. Hence, van der Waals attraction force is produced.
As the strength of this attractive force is different between molecules of different
substances, there is a difference in melting points and boiling points observed in
them.
43.
State the factors on which the Vander- Waal’s forces of attraction depend.
Ans : These intermolecular forces depend upon the following factors :
Shapes of molecules
(i)
Number of electrons in molecules
(ii)
(iii) Contact surface of molecules
(iv) Average intermolecular distance
From the study of these four factors, it can be understood that, at normal
temperature, from amongst the elements of the same group nitrogen is in gaseous
form while phosphorus is in solid form.
44.
What is “Hydrogen Bond”, how is it formed and give its consequences ?
Ans : Nitrogen, oxygen and fluorine are strong electronegative elements. When such
elements combine with hydrogen atom through covalent bond, the electrons
combined by sharing in covalent bond remains dragged towards more
electronegative element.
Thus, partial positively charged hydrogen atom of one hydride develops a strong
attraction with more electronegative element of the other hydride which is called
hydrogen- bond (H-bond).
Hydrogen bond is weaker than covalent bond. Thus, the attraction force produced
between electronegative elements having non-bonding electron pair with positively
charged hydrogen atom is called hydrogen bond.
Hydrogen bond is shown by dotted (……………….) line. e. g. In molecule of HF,
hydrogen atom of one molecule and fluorine atom of other molecule form H-bond
as shown below :
Here, hydrogen bond works as a bridge between two atoms.
The hydrogen bonds in NH3, H2O and HF molecule can be shown as under.
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NH3, H2O and HF in their liquid states possess strong hydrogen bond. Hence their
boiling points and melting points are very high as compared to those of hydrides
of other elements of the same group viz. the higher melting points and boiling
points of hydrides of first elements of group 15, 16 and 17 (NH3, H2O, HF) support
presence of hydrogen bond in them.
The energy of hydrogen bond is about 40 kJ mol–1, which is higher than the van der
Waals’ forces.
The density of water is maximum at 277 K.
In the temperature range of 273 K to 277 K the density of water being more than
that of ice, hence ice floats on water.
Ice is a solid crystal of water having hydrogen bonds.
The magnitude of hydrogen bond is maximum in solid state of substance and is
minimum in gaseous state of a substance.
The effect of hydrogen bond is also observed on the structures and the properties of
the compounds.
45.
Give importance of H-Bond.
Ans : The importance of H-bond can be given as under:
(i)
As hydrogen bond is formed in water, its evaporation is slow at normal
temperature. Hence, water on the surface of the earth is retained in large
proportion.
The storage of water in animal and vegetation cell is due to hydrogen bond.
(ii)
(iii) Molecules of water forms hydrogen bond with components of the soil in the
land. Hence, the moisture is retained in the land.
(iv) Hydrogen bond is formed in protein molecules present in the muscles of
living organisms through amide (–CONH –) group. Hence, certain functions
of muscles are due to hydrogen bond.
(v)
The effectiveness of medicines increases and becomes faster due to
hydrogen bond.
(vi) Hydrogen bond plays an important role in formation of biochemical
molecules present in cells of living-organisms like nucleic acid, DNA, RNA
etc.
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(vii)
The clothes of synthetic fibers (nylon, Terylene etc) dry faster than cotton
clothes because of hydrogen bond formation by water with cellulose of
cotton clothes.
46.
Explain the types of H-bonds giving examples ?
Ans : Hydrogen bond is mainly of two types :
(i)
Intra molecular Hydrogen Bond
(ii)
Intermolecular Hydrogen Bond
If hydrogen bond is formed between atoms of same molecule, then the type of
hydrogen bond is called intra molecular hydrogen bond e.g. o – Chlorophenol and
Ethanediol
The hydrogen bond formed between two or more different molecules of the same
compound is called intermolecular hydrogen bond e. g. p-Chlorophenol and
Methanol
As the intermolecular hydrogen bond is formed with more than one molecules of
the same compound, the number of such hydrogen bonds formed is more. As a
result, the melting points and boiling points of such compounds are comparatively
higher.
The intra molecular hydrogen bond is formed internally between atoms of the same
molecule. As a result, the number of such hydrogen bonds formed is limited.
Hence, the melting points and boiling points of the compounds having intra
molecular hydrogen bond are relatively lower.
E.g. The boiling point of p-chlorophenol is higher than that of o-chlorophenol
because intra molecular hydrogen bond is present in o- chloro phenol and
intermolecular hydrogen bond is present in p-chloro phenol.
47.
Explain the concept of Metallic-Bond.
Ans : The low ionisation energy of metals indicate that metal atoms have less attraction
for valence electron.
The number of valence electrons in metal atoms are also less (generally 1, 2 or 3).
Hence, a covalent bond is not formed between two atoms of a metal. As a result,
the directional property is not observed in metals.
The valence electrons of atoms of metal are not considered to be specific for any
one nucleus but they are considered to be same for the whole crystal. These
electrons are called delocalized electrons.
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e-CHEMTEST / CH-1 / Sem-II / Ph : 30004433
The delocalized electrons can change their positively charged kernel (positively
charged part of the atom with the nucleus except valence orbit).
This can be described as positively charged spheres in a cloud of delocalized
electrons.
According to Laws of Electrostatics, the atomic kernel being positively charged,
there must be repulsion between them. Even then, the atomic kernels are arranged
very close to each other.
The attraction between atomic kernels and the delocalized electrons keep the
atomic kernels together.
The attraction between positively charged atomic kernels and the delocalized
electrons arranged around it is called the metallic bond.
Due to this metallic bond, the atomic kernels can be accommodated in least
possible space and they maintain a definite parallel distance between them.
48.
Explain the concept of metallic-bond based on electron-sea model ?
Ans : According to this model, the metal crystal is considered as sea of delocalized
valence electrons.
The positively charged kernel is imagined to be floating in it. They are arranged
very close to each other.
Because of the presence of delocalized electrons between such positively charged
kernels, these delocalized electrons possess a very strong attractive force with
positively charged kernels. Such an attraction force is called metallic bond.
49.
What is co-ordinate covalent bond, explain giving a suitable example.
Ans : Co-ordinate covalent bond is a type of covalent bond.
In the formation of a regular covalent bond, the atoms combining through bond
give same number of electrons for sharing. E.g. in H2 molecule both the hydrogen
atoms give one electron each for sharing.
But sometimes, the electron pair required for sharing in the formation of a
molecule/ion, is given by any one of the atoms from the combining atoms and
bond gets formed.
When the electron pair required for sharing is donated by any one of the atoms
belonging to any one element then a special type of covalent bond is formed which
is called co-ordinate covalent bond.
This bonding is expressed by the sign of arrow (→) and the direction of arrow is
from the atom donating the electron pair towards the atom receiving the electron
pair.
Generally, the molecules having one or more non-bonding electron pairs form this
type of bond by donating electron pair.
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e-CHEMTEST / CH-1 / Sem-II / Ph : 30004433
Thus, this type of bond is formed between completely vacant orbitals of any one
atom and the non-bonding electron pair donated for sharing by the other atom
between which the bond is getting formed.
E. g. in NH4+ ion, there are three covalent and one co-ordinate covalent bond.
Extra shots :
Molecular Orbital Theory for Covalent Bond :
Molecular orbital theory (M. O. theory) of chemical bonding is proposed by Mulliken
and Hund.
The probability distribution of an electron around nucleus of an atom is indicated by
atomic orbital. Similarly the probability distribution of an electron around a group of
nucleus in molecule is indicated by molecular orbital.
These orbitals are filled with electrons on the basis of Aufbau principle, Pauli’s
principle and Hund’s rule.
According to the principle of wave mechanics, a molecular orbital is constructed by the
linear combination of atomic orbitals of two atoms between which a bond is formed.
e.g. when H2 molecule is formed by bonding two H-atoms, the acceptable wave
function Ψ1s(1) and Ψ1s(2) combine linearly in two ways as under and therefore two
acceptable molecular wave functions ΨMO and Ψ*MO are formed.
ΨMO = Ψ1s(1) + Ψ1s(2)
Ψ*MO = Ψ1s(1) − Ψ1s(2)
BMO and ABMO :
The molecular orbital having less energy than the energy of two atomic orbitals is
known as bonding molecular orbital (BMO). It is indicated by ΨMO.
The molecular orbital having higher energy than the energy of two atomic orbitals is
known as anti bonding molecular orbital (ABMO). It is indicated by Ψ*MO.
Ψ∗ MO
Ψ1 s(1)
E
A.O.
ABMO
ΨMO
Ψ1 s(2)
A.O.
BMO
M.O.
If electrons are filled in BMO there is an attraction between two hydrogen atoms,
resulting in bond formation.
If electrons are filled in ABMO the atoms resist to form the bond.
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e-CHEMTEST / CH-1 / Sem-II / Ph : 30004433
Molecular Orbitals and Symmetry :
Symbols σ & π are used to indicate bonding molecular orbitals and σ* & π* are used to
indicate anti bonding molecular orbitals.
σ & σ* Molecular orbitals :
In molecular orbitals when rotation of 180° is carried out about bond axis the position
of (+) and (−) signs (Signs of wave functions) do not change, the molecular orbitals
have cylindrical symmetry. These MO are known as σ type molecular orbitals.
In σ type molecular orbitals if probability of electrons (i.e. electron density) in the
region between two nuclei is increased (higher), then this kind of molecular orbital is
known as bonding molecular orbital. It is represented by σMO.
If probability of electrons (i.e. electron density) in the region between two nuclei is
decreased (lower), the electrons oppose the bond formation between atoms. Such
orbitals are known as anti bonding molecular orbitals σ*MO.
There is nodal plane between two nuclei in σ*MO. The probability of electrons in nodal
plane is zero.
π & π* Molecular orbitals :
In molecular orbitals, when rotation of 180° is carried out about bond axis, the position
of (+) and (−) signs (signs of wave functions) are changed then the molecular orbitals
are dissymmetric. These dissymmetric molecular orbitals are known as π type of
molecular orbitals.
In π type molecular orbitals, if the probability of the electrons (i.e. electron density)
between two nuclei is increased (higher) then it is known as bonding molecular orbital.
It is represented by πMO.
If probability of electrons (i.e. electron density) between two nuclei is decreased
(lower), the electrons oppose the bond formation between atoms. Such orbitals are
known as anti bonding molecular orbitals π*MO.
There is nodal plane between two nuclei in π*MO. The probability of electrons in nodal
plane is zero.
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e-CHEMTEST / CH-1 / Sem-II / Ph : 30004433
M O Diagram :
Electronic configuration, bond order and magnetic properties of some molecules are as
shown below.
1.
H2 :
H
+
H
→
H2
1
H : 1s
σ∗ 1s
1s
1s
H-atom
H-atom
σ 1s
H 2 -molecule
2.
Electronic configuration
:
Bond Order
:
Magnetic Property
:
He2 :
He
+
2
He : 1s
He
→
H2 : (σ1s)2
1
[N b − N a ] = 1 [2 − 0] = 1
2
2
In H2 all electrons are paired.
∴ H2 is diamagnetic.
Bond order =
He2
σ∗ 1s
1s
1s
He-atom
He-atom
σ 1s
He 2 molecule
Electronic configuration
:
Bond Order
:
Magnetic Property
:
He2 : (σ1s)2 (σ*1s)2
1
1
Bond order = [N b − N a ] = [2 − 2] = 0
2
2
He2 is not possible because bond order is zero.
In He2 all electrons are paired.
∴ He2 is diamagnetic.
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e-CHEMTEST / CH-1 / Sem-II / Ph : 30004433
3.
C2 :
C
+
C
→
C2
C : 1s2 2s2 2px1 2py1
σ∗
π∗
π∗
2px
σ
2p x
2p y
2 pz
2py
2 pz
2p z
2p z
π
π
2px
2p y
2p x
2py
σ∗ 2s
2s
2s
σ 2s
σ∗ 1s
1s
1s
C-atom
σ 1s
C 2 molecule
C-atom
Electronic configuration
:
C2 : (σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (π2px)2=(π2py)2
Bond Order
:
Bond order =
1
[N b − N a ] = 1 [8 − 4] = 2
2
2
C=C
Magnetic Property
27
:
In C2 all electrons are paired.
∴ C2 is diamagnetic.
e-CHEMTEST / CH-1 / Sem-II / Ph : 30004433
4.
N2 :
N
+
N
→
N2
N : 1s2 2s2 2px1 2py1 2pz1
σ∗
π∗
π∗
2px
σ
2p x
2p y
2 pz
2py
2 pz
2p z
2p z
π
π
2px
2p y
2p x
2py
σ∗ 2s
2s
2s
σ 2s
σ∗ 1s
1s
1s
N-atom
σ 1s
N 2 molecule
N-atom
Electronic configuration
:
N2 : (σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (π2px)2=(π2py)2 (σ2pz)2
Bond Order
:
Bond order =
1
[N b − N a ] = 1 [10 − 4] = 3
2
2
N≡N
Magnetic Property
28
:
In N2 all electrons are paired.
∴ N2 is diamagnetic.
e-CHEMTEST / CH-1 / Sem-II / Ph : 30004433
5.
O2 :
O
+
O
→
O2
O : 1s2 2s2 2px2 2py1 2pz1
σ∗ 2pz
π∗
π
2px
2py
π∗
2px
π
2px
2py
2py
2pz
2pz
2py
2px
σ2pz
σ∗ 2s
2s
2s
σ2s
σ∗ 1s
1s
1s
O-atom
σ1s
O2 -molecule
O-atom
Electronic configuration
:
O2 : (σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (σ2pz)2 (π2px)2=(π2py)2
(π*2px)1=(π*2py)1
Bond Order
:
Bond order =
1
[N b − N a ] = 1 [10 − 6] = 2
2
2
O=O
Magnetic Property
29
:
In O2 two electrons are unpaired.
∴ O2 is paramagnetic.
e-CHEMTEST / CH-1 / Sem-II / Ph : 30004433
6.
F2 :
F
+
F
→
F2
F : 1s2 2s2 2px2 2py2 2pz1
σ∗ 2pz
π∗
π
2px
2py
π∗
2px
π
2px
2py
2py
2pz
2pz
2py
2px
σ2pz
σ∗ 2s
2s
2s
σ2s
σ∗ 1s
1s
1s
F-atom
Electronic configuration
Bond Order
σ1s
F2 -molecule
F-atom
:
F2 : (σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (σ2pz)2 (π2px)2=(π2py)2
(π*2px)2=(π*2py)2
:
Bond order =
1
[N b − N a ] = 1 [10 − 8] = 1
2
2
F=F
Magnetic Property
30
:
In F2 all electrons are paired.
∴ F2 is diamagnetic.
e-CHEMTEST / CH-1 / Sem-II / Ph : 30004433
7.
Ne2 :
Ne
+
Ne
→
Ne2
Ne : 1s2 2s2 2px2 2py2 2pz2
σ∗2pz
π∗ 2px
π∗ 2py
π 2px
2px
2py
π 2py
2pz
2pz
2py
2px
σ2pz
σ∗2s
2s
2s
σ2s
σ∗1s
1s
1s
Ne-atom
σ1s
Ne-atom
Ne2-molecule
Electronic configuration
:
Bond Order
:
Magnetic Property
:
Ne2 : (σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (σ2pz)2 (π2px)2=(π2py)2
(π*2px)2=(π*2py)2 (σ*2pz)2
1
[N b − N a ] = 1 [10 − 10] = 0
2
2
Ne2 is not possible because bond order is zero.
Bond order =
In Ne2 all electrons are paired.
∴ Ne2 is diamagnetic.
How far is far, How high is high ?
We will never know until we try.
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