CBSE Class 11

NCERT Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure.

Question 1.
Explain the formation of chemical bond.
Answer:
Inert gases or noble gas elements are present in group 18 also called zero group. This means that their valency is zero or their atoms can exist independently of their own. Let us have a close look at their electronic configurations given in the table

With the exception of first member helium which has only two electrons in its valence shell, the rest of the elements have eight electrons.

In 1916, G. N. Lewis and Kossel stated that the stability of noble gas elements is due to the presence of eight electrons in their valence shells (except helium) or due to the presence of complete octet. This is known as the octet rule. According to the rule, the atoms of different elements take part in chemical combination in order to complete their octet (to have eight electrons in the outermost or valence shell) or duplet (to have two valence electrons)in some cases such as H, Li, Re etc. Thus, the atoms take part in chemical combination or bond formation in order to complete their octet and to have the electronic configuration of nearest noble gas atoms.

Question 2.
Write the Lewis dot symbols of the following elements :
Be, Na, B, O, N, Br.
Answer:
Please remember that only the valence electrons are shown in the Lewis dot symbols of the atoms of the elements.

Question 3.
Write the Lewis dot symbols of the following atoms and ions :
S and S^2- ; P and P^3- ; Na and Na⁺; Al and Al³⁺ ; H and H^–
Answer:

Question 4.
Draw the Lewis structures of the following molecules and ions :
PH₃, H₂S, SiCl₄, BeF₂, AlI₃, C0₃^2-, HCOOH
Answer:

Question 5.
Define octet rule. Write its significance and limitations.
Answer:
the atoms of different elements take part in chemical combination in order to complete their octet (to have eight electrons in the outermost or valence shell) or duplet (to have two valence electrons)in some cases such as H, Li, Re etc. Thus, the atoms take part in chemical combination or bond formation in order to complete their octet and to have the electronic configuration of nearest noble gas atoms.

Significance : The octet rule i. e., the tendency of the atom of an element to acquire eight electrons in its valence shell is the basis for the formation of chemical bonds. This applies to both ionic and covalent bonds. For more details
Significance of Lewis Symbols
The Lewis symbols of the atoms have a lot of information to convey.
(a) They represent the number of valence or outermost electrons.
(b) No of valence electrons help in calculating the valency or valence of the element which is also called group valeme
i. e.„ the valence of the group to which a particular element belongs. The group valence is equal to either the number of dots in the Lewis symbol of the element or “Eight – no. of dots.” For example,

Limitations: The main limitation of the octet rule is that in certain molecules, the central atom may have either incomplete or expanded octet. But they are still quite stable. For more details,
Limitations of Octet Rule or Octet Theory
We have discussed the implication of octet rule which states that the atoms of same or different elements take part in chemica I combination to complete their octet either by transference or by sharing of electrons.
The octet rule can be applied to large number of covalent molecules, particularly those of organic nature. However, there are certain exceptions to the octet rule. These are illustrated as follows :
(a) Incomplete octet of the central atom. There are many stable covalent molecules in which the central atom has less tha t eight electrons after sharing i.e. it has an incomplete octet. For example,

(b) Old electron molecules. There are certain molecules in which even single or odd electrons may be present on th; atoms. For example,

(c)Expanded octet of the central atom. There are many examples in which the central atom may have even more that eight electrons after sharing i.e. it may have expanded octet. For example,

(d) Shapes of the molecules. The octet rule or octet theory gives no idea about the, shapes of the molecules which may be tri-atomic, tetra-atomic or polyatomic in nature.

Question 6.
Write the favourable conditions for the formation of ionic bond.
Answer:
The favourable conditions for the formation of ionic bond are :
(i) Low ionization enthalpy for the element to form positive ion or cation.
(ii) High electron gain enthalpy for the element to form negative ion or anion.
(iii) High magnitude of lattice energy or lattice enthalpy.
(i). Ionization enthalpy. In the formation of positive ion or cation, one of the atoms is to lose electrons and for this ionization enthalpy is needed. As already stated, it is the amount of energy required to remove the most loosely bound electron from an isolated gaseous atom. Thus, lesser the ionization enthalpy required, easier will be the formation of cation. For example,
A(g) → A⁺(g) + e^– ; ∆_iH₁ = + ve
The alkali metals and alkaline earth metals present in the s-block normally form cations since they have comparatively low ionization enthalpies.

(ii). Electron gain enthalpy. The electrons released in the formation of cation are to be accepted by the other atom . taking part in the ionic bond formation. The electron accepting tendency of an atom depends upon the electron gain
enthalpy. It may be defined as the energy released when an isolated gaseous atom takes up an electron to form anion. Greater the negative electron gain enthalpy, easier will be formation of anion. For example,
B(g) + e^– → B^–(g) ; ∆_egH The halogens present in group 17 have the maximum tendency to form anions as they have very high negative electron gain enthalpy.
The members of the oxygen family (group 16) such as oxygen also form anions but not so easily because energy is needed t( form divalent anion (^2- is comparatively high.
The details have been discussed in the unit 3 on Chemical families and periodic properties.

(iii). Lattice energy or enthalpy. The ionic compounds exist as crystalline solids and the arrangement is known as crysta lattice. Since the ions are charged species, energy known as lattice energy or enthalpy is released in the attraction o ‘ the ions. It may be defined as :
the energy released when one mole of crystalline solid is formed by the combination of oppositely charged ions.
It is denoted as U.
A⁺ (g) + B^– (g) → A⁺ B^– (s) ; U = – ve (Released)
Thus, greater the magnitude of lattice energy, more will be the stability of the ionic bond or ionic compound. The lattice energy depends upon the following factors :

(a) Size of the ions. The size of the ions influences the lattice energy. Smaller the size, lesser will be the internuclea distance and, thus, greater will be the lattice energy. For example, lattice energy of NaCl (769 8 kJ mol1) is mon than that of KC1 (698-7 kJ mol-1) because the radius of Na+ (95 pm) ion is smaller as compared to K+ (133 pm) ion.

(b) Charge on the ions. Greater the magnitude of charge on the ions, higher will be inter ionic attraction and thus greater will be the value of lattice energy.

Thus, we conclude that if the magnitude of lattice energy and negative electron gain enthalpy is greater than that of the ionization enthalpy required, a stable chemical bond will be formed. In case, it is less then the bond will not be formed.The Lattice Energy can be determined with the help of Born-Haber cycle. For the details, consult section 6.18 (Chapter 6 – Thermodynamics).

Question 7.
Predict the shapes of the following molecules using the V.S.E.P.R. model.
BeCl₂, SiCl₄, AsF₅, H₂S, PH₃,
Answer:
The shapes of the covalent molecules can be predicted with the help of V.S.E.P.R. theory.

Predicting the Shapes of Covalent Molecules or Ions
The shape of a covalent molecule or ion can be predicting theoretically by calculating the total number of electron pains (shared pairs as well as lone pairs) around the central atom in the molecule or ion. This can be done by the following relation.
= 1/2[No. of valence electrons on the central atom + No. of atoms linked to the central atom by covalent bonds]
(i) For positive ion (or cation), subtract the number of electrons equal to the positive charge from the valence electrors on the central atom.
(ii) For negative ion (or anion), add the number of equal to the negative charge to the valence electrons on the central atom.
The number of bond pairs or shared pairs = No. of atoms linked to the central atom by single bonds.
The number of lone pairs = Total number of electron pairs-No. of bond pairs or shared pairs.

Question 8.
Although geometries of NH₃ and H₂O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
Answer:
In \(\ddot { N }\)₃ and in H₂\(\ddddot { O } \) molecules the central atom is surrounded by four electron pairs. The geometries of these molecules are expected to be tetrahedral.
They however, get distorted on account of the presence of lone electron pairs. On comparison, the bond angle in NH₃ (107°) is more than in H₂O (104.5°).
This is on account of greater magnitude of force of lone pair : lone repulsion in H₂O molecule as compared to lone pair : shared pair repulsion in NH₃ molecule.
Ammonia molecule (NH₃)
In ammonia, the atomic number (Z) of nitrogen is 7 and its electronic configuration is 2, 5. Out of the five electrons present in the valence shell of nitrogen atom, three form shared pairs (bond pairs) with the electrons of three hydrogen atoms.
The remaining two electrons constitute one lone pair.

Thus, the nitrogen atom is surrounded by three shared pairs and one lone pair of electrons. According to VSEPR theory, the geometry of the molecule is irregular.
In order to have minimum force of repulsion in all the four electron pairs around the nitrogen atom, the shape is expected to be tetrahedral but the presence of one lone electron pair distorts the shape of the NH3 molecule.

Actually, Ip – bp repulsion is greater than the bp – bp repulsion. As a result, the two N-H bonds on both sides of the lone pair are stretched a little inwards. Thus, the bond angle in the molecule is 107° and is slightly less than the bond angle of a regular tetrahedron. The actual shape of the NH₃molecule is pyramidal .

Question 9.
How do we express bond strength in terms of bond order ?
Answer:
Bond strength is directly proportional to bond order, i.e., greater the bond order, more is the bond strength. For example,

Question 10.
Define bond length.
Answer:
Bond Length. Bond length may be defined as :
the average equilibrium distance between the centres of the two bonded atoms.
The bond length of different covalent bonds are determined by X-ray diffraction methods. For a covalent bond, it is the sum of covalent radii of the bonding atoms. For example, the bond length of C—Cl bond is rc + rC]. The values of the bond lengths are generally expressed in picometre (1 pm — 1(F12 m). The bond length is influenced by the following factors.

(i)Size of the atoms. As stated earlier, bond length is directly linked with the size of the bonding atoms. For example H—Br bond length (141 pm) is more than H—Cl bond (127 pm).
(ii)Multiplicity of bonds. The multiplicity of the bonds between two atoms brings them close to each other. As a result, the bond length decreases. For example, bond length for C—C bond is 154 pm and the value for C = C bond is 134 pm.

(iii)Types of hybridisation. The value of bond length is also influenced by the type of hybridisation. Since s-orbital is smaller in size as compared to the p-orbital, therefore, greater the s-character associated with a particular bond, smaller will be the bond length. For example,
sp³ C—H (111 pm) > sp² C—H bond (110 pm) > sp C—H bond (108 pm).

Question 11.
Explain the important aspects of resonance with respect to the \({ CO }_{ 3 }^{ 2- }\) ion
Answer:
Resonance
Covalent molecules are normally expressed as Lewis structures. But in some cases, a single Lewis structure fails to explain all the characteristics of the molecule. For example, the Lewis structure of ozone is represented as follows :

The two bonds in the molecule are expected to have different values. But the experimental studies have shown that both the bonds have the same lengths (128 pm). Moreover, the bond length is intermediate between the single and double bonds.

The problems can be solved if we consider that ozone is capable of existing in two forms in which the positions of the double bond and single bond are interchangeable.
Such structures for the molecule are known as resonating or contributing structures separated by (sign for resonance). None of these structures can independently explain all the properties of ozone.
It is believed that the actual structure is inbetween these two contributing structures and is known as resonance hybrid as

Thus, O to O bond length in the molecule is the same for both the bonds i.e. 128 pm. Resonance may be defined as :
the phenomenon as a result of which a molecule can be expressed in different forms none of which can explain all the properties of the molecule. The actual structure of the molecule is called resonance hybrid.

Question 12.
H₃PO₃ can be represented by the structures 1 and 2 as shown below. Can these structures be taken as the canonical forms of the resonance hybrid representing H₃PO₃ ? If not cite reasons for the same.
Answer:

The basic requirement for the resonance is that the canonical structures must differ in the arrangement of the electron pairs and not of the atoms. The structures (1) and (2) are not the canonical structures because in structure (1) phosphorus atom is linked with H (P—H bond) while in the other (2), it is linked with OH group (P—Oil bond).

Question 13.
Write the resonating structures for SO₃, No₂ and \({ NO }_{ 3 }^{ – }\)
Answer:


Note : Please note that in many cases, the co-ordinate bond is shown simply as covalent bond with positive (+) charge on donor atom and negative (-) charge on acceptor atom.

Question 14.
Using Lewis dot symbols, show electron transfer between the following atoms to form cations and anions :
(a) Na and Cl (b) K and S (c) Ca and O (d) Al and N.
Answer:

Question 15.
Although both CO₂ and H₂o are triatomic molecules, the shape of H₂o molecule is bent while that of Co₂ is linear. Explain this on the basis of dipole moment.
Answer:
Carbon dioxide is a linear molecule in which the two C = 0 bonds are oriented in the opposite directions at an angle of 180°. The dipole moment of C = O bond is 2.3D but due to linear structure of Co₂, the bond dipoles of two C = O bonds cancel with each other.

Therefore, the resultant dipole moment of the molecule is zero. On the other hand, H20 is a polar molecule having dipole moment 1.84 D. Actually H₂o molecule has a bent structure in which the O—H bonds are oriented at an angle of 104.5° and do not cancel the bond dipole moments of each other.

The molecular dipole moment of H₂o (p = 1.84 D) is the resultant of the individual values of the bond dipole moment of two O—H bonds.

Question 16.
Write the significance and applications of dipole moment.
Answer:
Applications of Dipole Moment
Some useful applications of dipole moment are given :
(i) In predicting the nature of the molecules. Molecules with specific dipole moments are polar in nature while those with zero value are non-polar. Thus, BeF₂ (μ= 0D) is non-polar while H₂0 (μ = 1.84 D) is polar.

(ii) In comparing the relative polarities of the molecules. The relative polarities of the molecules can be compared by their dipole moment values. Greater the value, more is polarity.

(iii) In determining the shapes of molecules. The dipole moment values are quite helpful in determining the general shapes of the molecules having three or more atoms. In case such molecules have dipole moments, then their shapes will not be symmetrical.

They will be either bent or angular, (e.g. H₂S with// = 0.90 D) or unsymmetrical (e.g. NH3 with// = 1.47 D). However, for molecules with zero dipole moment values, the shapes will be either linear (BeF₂, Co₂ etc.) or symmetrical (CH₄, CCl₄ etc.)

(iv) In calculating the percentage ionic character. The dipole moment values help in calculating the percentage ionic characters of polar bonds. It is the ratio of the observed dipole moment or experimentally determined dipole moment to the dipole moment for the complete transfer of electrons.

For example, the observed dipole moment of HCl molecule is 1.03 D. For complete transfer of electrons (100% ionic character) the charge on H⁺ and Cl^– ions would be equal to one unit (4.8 × 10^-10 esu) each. The bond length of H—Cl bond is 1.275 × 10^-8 cm. Therefore,
Dipole moment for complete electron transfer (μ_ionic) = q × d
= (4.8 × 10^-10esu) × (1.275 × 10^-8 cm)
= 6.12 × 10^-18 esu-cm = 6.12D
Observed dipole moment (μ_observed) = 1.03 D

In a similar manner, the percentage ionic character for other polar bonds can also be calculated.

(v) In distinguishing between cis and trans geometrical isomers. In organic compounds, dipole moment can halp in making a distinction between the cis and trans geometrical isomers of a substance. In general, cis isomer (less symmetrical) has more dipole moment value as compared to the trans isomer. For example, the dipole moment cis-but-2-ene is higher than that of trans-but-2-ene.

(vi) In identifying the position of the substituents in the disubstituted aromatic rings. The relative positions of the substituents such as methyl (- CH₃) groups present at ortho, meta and para positions can be identified from their dipole moments. The dipole moment of para-isomer is zero. Out of the ortho and meta isomers, the dipole moment of the ortho iosmer is higher.

Question 17.
Define electronegativity. How does it differ from electron gain enthalpy ?
Answer:
Difference between electron gain enthalpy and electronegativity.
The main points of distinction are given in a tabular form.

Question 18.
Explain with the help of suitable example polar covalent bond.
Answer:
We have studied that according to Lewis concept, a covalent bond represents a shared pair of electrons. If the atoms joined by covalent bond are the same then the shared electron pair is equidistant to both of them. In other words, it lies exactly in th: centre of the bonding atoms. As a result, no poles are developed and the bond is called non-polar covalent bond. Th: corresponding molecules are known as non-polar molecules. For example,

On the other hand, in case different atoms are linked to each other by covalent bond, then the shared electron pair will not lie in the centre because the bonding atoms differ in electronegativities.

For example, in hydrogen chloride molecule, chlorine with greater electronegativity (3.0) will have greater control over the electron pair as compared to hydrogen which is less electronegativ : (2.1). Since the electron pair is displaced more towards chlorine atom, the latter will acquire a partial negative charge (δ-).

At the same time, the hydrogen atom will have a partial positive charge (δ + ) with the magnitude of charge same as on the chlorine atom. Such a covalent bond is known as polar covalent bond or simply polar bond and is represented as follows :

The molecule with such a bond is called polar covalent molecule. It may be noted that greater the difference in the electronegativity of the bonding atoms, more will be the polarity of the bond. For example,

Question 19.
Arrange the bonds in order of increasing ionic character in the molecules :
LiF, K₂o, N₂, SO₂ and ClF₃
Answer:
N₂ < SO₂ < ClF₃v < K₂o < LiF.

Question 20.
The skeletal structure of CH3COOH as shown below is correct but some of the bonds are wrongly shown. Write the correct Lewis structure of acetic acid.

Answer:
Only the skeletal arrangement of the atoms in the above structure is correct. But it is not according to the Lewis concept as well as tetravalent nature of carbon. The correct structure of acetic acid is

Question 21.
Apart from tetrahedral geometry for methane (CH₄), another possible geometry is square planar with four ‘FT atoms at the comers of the square with the ‘C’ atom at its centre. Explain why CH₄ is not square planar.
Answer:
The tetrahedral and square planar structures of CH₄ are shown as follows :

According to VSEPR theory, the shared electron pairs around the central atom in a covalent molecule are so arranged in space that the force of repulsion in them is the minimum. Now, in a square planar geometry, the bond angle is 90° while it is 109°-28′ in tetrahedral geometry. This clearly shows that the electron repulsions are less in the tetrahedral geometry as compared to the square planar geometry. Thus, methane cannot be represented by square planar structure.

Question 22.
Explain why BeH₂ molecule has zero dipole moment although the Be—H bonds are polar.
Answer:
BeH₂ is a linear molecule (H—Be—H) with bond angle equal to 180°. Although the Be—H bonds are polar on account of electronegativity difference between Be (1.5) and FI (2.1) atoms, the bond polarities cancel with each other. The molecule has resultant dipole moment of zero.

Question 23.
Both NH₃ and NF₃ have identical shapes and same state of hybridisation. Both N—H and N—F bonds have almost the same electronegativity difference. But still, the two molecules have different dipole moment values. How will you account for it ?
Answer:
Both NH₃ and NF₃ have pyramidal geometries and the electronegativity difference of bonding atoms in both NH₃ (3.0 -2.1= 0.9) and NF₃ (4.0 – 3 0 = 1.0) molecules are also nearly the same. But the dipole moment of NH₃ (1.46D) is more than that of NF₃ (0.24D).

This is explained on the basis of the difference in the directions of the dipole moments. In NH₃, the dipole moments of the three N—H bonds are in the same direction as that of the lone electron pair.

But in NF3, the dipole moments of the three N—F bonds are in the direction opposite to that of the lone pair. Therefore, the resultant dipole moment in NH₃ is more than in NF₃.

Question 24.
What is meant by hybridisation of atomic orbitals ? Describe the shapes of sp, sp² and sp³ hybrid orbits.
Answer:
Hybridisation is of different types depending upon the nature and number of orbitals taking part in hybridisation. Let us discuss the hybridisation involving 5, p and d orbitals.
Hybridisation involving s and p-orbitals
The hybridisation involving 5 and p orbitals is of three types depending upon the orbitals that are involved. These are sp3, sp2 and sp types :

Question 25.
What is meant by the hybridisation of atomic orbitals ? What is the change in hybridisation (if any) of A1 atom in the reaction ?
AlCl₃ + Cl^– → AlC\({ l }_{ 4 }^{ – }\).
Answer:
Hybridisation is of different types depending upon the nature and number of orbitals taking part in hybridisation. Let us discuss the hybridisation involving 5, p and d orbitals.
In AlCl₃, the central Al atom is sp² hybridised while in the ion [AlCl₄]-, it has sp³ hybridisation.

Question 26.
Is there any change in the hybridisation of B and N atoms as a result of the following reaction ?
BF₃ + :NH₃ → F₃B.NH₃
Answer:
In BF₃, B atom is sp² hybridised. It forms an addition compound with \(\ddot { N }\)₃ (Where N atom is sp³ hybridised). In doing so, it takes up an electron pair from NH₃. This indicates that B atom undergoes a change in hybridisation from sp² to sp³. At the same time there is no change in hybridisation of N atom.

Question 27.
Draw diagrams showing the formation of a double and a triple bond between the carbon atoms in the (C₂H₄ and C₂H₄molecules.
Answer:
Formation of ethene (CH₂ = CH₂). In order to illustrate sp² hybridisation, let us discuss the orbital structure of ethene (H₂C = CH₂) molecule. Out of three hybridised orbitals belonging to each carbon atom, one is involved in the axial overlap with the similar orbital of the other carbon resulting in a sigma bond.

The remaining two hybridised orbitals of the carbon atoms form sigma bonds with the lx orbital of the hydrogen atoms. The unhybridised orbitals (2p^z) of both the carbon atoms participate in the sidewise overlap leading to a pi (π) bond as shown in the Fig. 4.27.

Question 28.
What is the total number of sigma and pi bonds in the following molecules ?
(a) C₂H₂ (b) C₂H₄
Answer:

Question 29.
Considering X-axis as the internuclear axis, which out of the following atomic orbitals will form a sigma bond ?
(a) 1s and 1s
(b) 1s and 2p_x
(c) 2p_y and 2p_y
(d) 1s and 2s.
Answer:
The sigma bond is formed by axial overlap along internuclear axis and is present in the following cases.
(a)1s and 1s
(b) 1s and 2p_x
(d) 1s and 2s.
2p_y and 2p_y atomic orbitals are involved in the sidewise overlap leading to the formation of π-bond.

Question 30.
Which hybrid orbitals are used by the carbon atoms in the following molecules ?
(a) H₃C—CH₃
(b) H₃C—CH=CH₂
(C) CH₃—CHO
(d) CH₃COOH
Answer:

Question 31.
What do you understand by bond pairs and lone pairs of electrons ? Illustrate by giving one example of each type.
Answer:
The electron pair involved in sharing between two bonding atoms is called shared pair or bond pair. At the same time, the electron pair which is not involved in sharing is called lone pair of electrons. For example, in BeH₂ molecule, the central Be atom is surrounded by two bond pairs. In the molecule of NH₃, the central B atom has three shared pairs and one lone pair around it.

Question 32.
Distinguish between a sigma and a pi bond.
Answer:
Types of Orbital Overlap (Sigma and Pi Bonds)
We have studied that the covalent bonds are formed by the overlap of the atomic orbitals. Depending upon the type of the overlapping, the covalent bonds are of two types known as sigma (\(\sigma \) ) and pi (π) bonds. These are described as below :
Sigma (\(\sigma \) ) bond. A sigma bond is formed when the two half filled atomic orbitals belonging to the bonding atoms overlap along their internuclear axes i.e. the line joining the centres of the nuclei of the two atoms.

Thus, the bond formed as a result of the axial overlap is known as \(\sigma \) bond. Since the axial overlap is quite large, the energy released during the overlap is also large. As a result, the sigma bond is a stable bond.
In most of the covalent molecules, s and p orbitals are involved in the bond formation. The axial overlap involving these orbitals is of three types :
(a) s-s overlap. This involves the axial overlap of the half filled v-atomic orbitals of the combining atoms. The cr-bond is
called s-s bond.
(b) s-p overlap. This involves the overlap of the s-atomic orbital belonging to one atom and p-atomic orbital to the other atom along their axes. The bond is known as s-p bond.
(c) p-p overlap. In this case, the atomic orbitals taking part in the axial overlap are both half filled p-orbitals. The \(\sigma \) -bond formed is called p-p bond.

Question 33.
Explain the formation of H? molecule on the basis of valence bond theory.
Answer:
Formation of hydrogen (H₂) molecule
In the light of the above discussion, let us consider the combination between atoms of hydrogen HA and HB. If eA and eB be their respective electrons, then attractive and repulsive forces may be shown in the Fig. 4.9.

Although the number of new attractive and repulsive forces is the same, but the magnitude of the attractive forces is more. Thus, when two hydrogen atoms approach each other the overall potential energy of the system decreases. Therefore, a stable molecule of hydrogen gets formed.

The potential energy changes which take place in the formation of a hydrogen molecule may also be shown graphically as follows: When the two hydrogen atoms are far separated (at infinite distance) the potential energy is zero. As they start coming closer to each other from infinite distance, energy is lost correspondingly.

Ultimately, a point of minimum energy is attained when the attractive and repulsive (or maximum stability) forces balance each other. At this stage, the two hydrogen atoms are bonded together to form a molecule of hydrogen.

M The distance between the centres of their nuclei is called bond length. In c case of hydrogen molecule, H—H bond length is 74 pm. It may be kept — in mind that the two hydrogen atoms can not be brought closer than 74 pm.

If it happens, then the repulsive forces will become more than the 3 attractive forces. As a result, the potential energy of the system will increase as shown by the dotted line.

The total decrease in the potential energy when one mole bonds of a particular type are formed between the atoms in the gaseous state is called bond energy. For example, the bond energy of H—H(g) is 433 kJ mol^-1.

It may be noted that the same energy is needed to break the molecules into the atoms and is known as bond dissociation energy. Thus, the bond dissociation energy of hydrogen is also 433 kJ mol^-1.

Question 34.
Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
Answer:
The molecular orbitals are formed by the linear combination of the wave functions of the participating atomic orbitals. They may combine either by addition or by subtraction. Let yA and yB represent the wave functions (or amplitude) of the two combining atomic orbitals A and B taking part in chemical combination.
(a) Combination by addition. When the two electron waves are in phase i.e. they have the same sign*, they will add up to give a new wave function expressed as φ (or \(\phi \)) = φ_A +φ_B.

(b) Combination by subtraction. When the two electron waves are out of phase i.e. they have opposite sign of the wave functions, then they will combine by subtraction and the resulting wave function φ* (or <()*) = φ_A – φ_B as shown below :

We know that the probable electron density is given by φ² and not by φ. On squaring we get :
φ² = φ_A² + φ_B² + 2φ_Aφ_B
φ*² = φ_A² + φ_B² – 2φ_Aφ_A
Here yA2 and yB2 represent the probable electron density in the two atomic orbitals whereas φ2 and φ*² indicate the electron density in the two molecular orbitals. These are called Bonding Molecular orbital (φ²) and Antibond ing Molecular orbital (φ*sup>2).

• Bonding Molecular Orbital (φ2) is formed by the linear combination of wave functions of the combining atomic orbitals [L.C.A.O) by addition.

• Antibonding Molecular Orbital (φ*sup>2) is formed by the linear combination of wave functions of the combining atomic orbitals (L.C. A. O) by subtraction. Let us illustrate both these orbitals by the combination of the atomic orbitals of hydrogen atoms

The formation of bonding molecular orbital when two atomic orbitals of hydrogen atoms (Is orbital) combine by the addition of their wave functions is shown below :

Similarly, the antibonding molecular orbital arising from the subtraction of the wave functions of the two participating atomic orbitals is shown as follows :

Thus we conclude that whenever two atomic orbitals combine, two molecular orbitals are formed. One of them is bonding molecular orbital (φ²M 0) while the other is antibonding molecular orbital (φ*²M.O)-

Question .35.
Use molecular orbital theory to explain why Be₂ molecule does not exist.
Answer:
The atomic no. (Z) of Be is 4. This means that 8 electrons are to be filled in the M.O of Be₂. The configuration is :

As the bond order comes out to be zero, the molecule of Be2 does not exist.

Question 36.
Compare the relative stability of the following species and indicate their magnetic properties :
O₂, \({ O }_{ 2 }^{ + }\),\({ O }_{ 2 }^{ – }\) (superoxide), \({ O }_{ 2 }^{ 2- }\) (peroxide)
Answer:
Bond orders of the different species are ;
O₂(2.0), \({ O }_{ 2 }^{ + }\)(2.5), \({ O }_{ 2 }^{ – }\)(l.5), \({ O }_{ 2 }^{ 2- }\)(1.0)
Relative stability : \({ O }_{ 2 }^{ + }\) > O₂ > \({ O }_{ 2 }^{ – }\) > \({ O }_{ 2 }^{ 2- }\)
For magnetic properties; consult section 4.32.

Question 37.
Write the significance of plus and minus signs in representing the orbitals.
Answer:
Plus and minus signs have been given to identify the nature of the electron waves. Plus (+ve) sign denotes crest while negative (-ve) sign denotes trough.

Question 38.
Describe the hybridisation in case of PCl₅. Why are the axial bonds longer as compared to equatorial bonds ?
Answer:
The hybridisation in the molecule of PCl₅ is same as in PF₅ because the central atom is the same in both the molecules and the two halogens chlorine (Cl) and fluorine (F) also belong to the same group (group 17). They are expected to have same hybridisation as well as same geometry. The P atom is sp²d hybridised and the molecule has trigonal bipyramidal geometry. The details of hybridisation of PF₅

Question 39.
Define hydrogen bond. Is it weaker or stronger than the van der Waals forces ?
Answer:
Hydrogen bond represents strong dipole dipole interactions between the H and the highly electronegative atoms (F, O and N) present in different bonds belonging to either different molecules or same molecule van der Waals forces are also dipole forces but are comparatively weak. For the details of hydrogen bond, consult section 4.34. For the details of van der Waals, forces, consult unit-5 on States of Matter.

Question 40.
What is meant by the term bond order ? Calculate the bond order of: N₂, O₂, \({ O }_{ 2 }^{ – }\) and \({ O }_{ 2 }^{ 2- }\); .
Answer:
Significance of Bond Order. The bond order conveys very important information about the molecules and ions. These are briefly discussed.
(a) Stability of molecules or ions. The stability of the molecules and their ions can be predicted in terms of bond order. If bond ‘ order is positive, (N_b > N_a), the molecule or ion will be stable. Incase, it is zero (N_b = N_a) or negative (N_b< N_a), the molecule or ion will be unstable. A fractional value of the bond order suggests that the molecule is unstable. However, it still exists.

(b) Bond dissociation enthalpy. Bond dissociation enthalpy is directly proportional to the bond order. Thus, more the value of bond order, greater will be the bond dissociation enthalpy and vice versa. This is further supported by the available data.

(c) Bond Length. Bond order is inversely proportional to bond length. This means that greater the bond order, smaller will be the bond length.

(d) Number of bonds. The value of bond order also predicts the number of covalent bonds in the molecule. Bond order 1, 2, 3 signifies the presence of single, double and triple bond respectively in a particular molecule, However, in some cases, the bond order is fractional (e.g. 0-5) and it is quite difficult to explain its existence.

We hope the NCERT Solutions for Class 11 Chemistry at Work Chapter 4 Chemical Bonding and Molecular Structure, help you. If you have any query regarding NCERT Solutions for Class 11 Chemistry at Work Chapter 4 Chemical Bonding and Molecular Structure, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties.

Question 1.
What is the basic theme of organisation in periodic table ?
Answer:
The basic theme of organisation in periodic table is to group the elements with same characteristics (physical and chemical) together so that it may become rather easy to follow these characteristics. Since these mainly depend upon the valence shell electronic configurations, the elements placed in a group have therefore same valence shell configuration of their atoms. In fact, the valence shell electronic configuration of the elements placed in a group gets repeated after definite gaps of atomic numbers (8, 8, 18, 18, 32).

Question 2.
Which important property did Mendeleev use to classify the elements in periodic table and did he stick to that ?
Answer:
The property used by Mendeleev is known as Mendeleev’s Periodic Law. According to the law,

Physical and chemical properties of the elements are a periodic function of their mass numbers.

Mendeleev did stick to it and classified elements into groups and periods. This classification was perhaps the first systematic way in order to study the characteristics of the elements. No doubt, later on many defects were pointed out and scientists challenged the basis of the classification.

Question 3.
What is the basic difference in approach between Mendeleev’s periodic law and Modern periodic law ?
Answer:
The atomic mass of the elements is the basis of periodicity according to Mendeleev’s periodic law. On the other hand, the atomic number of the elements constitutes the same according to Modern periodic law. For more details,

• Modern Periodic Law

Dmitri Mendeleev, a Russian Chemist, was the first to make a very significant contribution in the formation of the periodic table. He studied the chemical properties of the large number of the elements and in 1869 and came out with a statement that the chemical properties of the elements are a periodic function of their atomic masses. When Mendeleev came to know about the work of Lother Meyer, he modified his statement and gave a law called Mendeleev-Lother Meyer Periodic Law or simply Mendeleev’s Periodic Law.
Modern Periodic Law
It states that In the year 1913, an English physicist, Henry Moseley, a young physicist from England, studied the frequencies of the X-rays which were emitted when certain metals were bombarded with high speed electrons. He found that in all the cases, the square root of the frequency (Vv) was directly proportional to the atomic number of the atom of the metal. Upon investigation, he further stated that there was no co-relation between the frequency and the atomic mass.

Actually, when Moseley plotted a graph between and atomic number of the different metals, a straight line was obtained. But it was not the case when a graph was plotted between (4v) and atomic mass of the metals. These studies led Moseley to believe that atomic number and not the atomic mass is the fundamental property of an element. Therefore, atomic numbers must form the basis of the classification of the elements in the periodic table. In the light of the above observations. Moseley gave the modern periodic law which states that

Question 4.
On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
Answer:
The sixth period corresponds to sixth shell. The sub-shells present in this shell are 6s, 4f, 5p and 6d. The maximum number of electrons which can be present in these sub-shell are 2 + 14 + 6 + 10 = 32. Since the number of elements in a period correspond to the number of electrons in the shells, therefore, sixth period should have a maximum of 32 elements.

Question 5.
In terms of period and group where will you locate the element with Z = 114 ?
Answer:
We know that the gaps of atomic numbers in a group are 8, 8, 18, 18, 32. The element which proceeds the element with Z = 114 in the same group must have atomic number equal to 114 – 32 = 82. This represents lead (Pb) which is present in 6th period and group 14 of the p-block. This means that the element with Z = 114 must also belong to the group 14 of p-block and it must be a member of 7th period.

Question 6.
Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
Answer:
The element is chlorine (Cl) with atomic number (Z) = 17.

Question 7.
Which element do you think would have been named by :
(i) Lawrence Berkeley laboratory
(ii) Seaborg’s group ?
Answer:
(i) Lawrencium (Lr) with atomic number (Z) = 103
(ii) Seaborgium (Sg) with atomic number (Z) = 106.

Question 8.
Why do elements in a group have similar physical and chemical properties ?
Answer:
The elements in a group have the same valence shell electronic configurations of their atoms. However, they differ in the atomic sizes which tend to increase. Therefore, the elements in a group have similar chemical properties while their physical characteristics show a little variation.

Question 9.
What do atomic radius and ionic radius really mean to you ?
Answer:
For the definition and details of atomic radius,

Variation of Atomic Radius in the Periodic Table

We shall discuss the variation or change of the atomic radius along the period from left to the right and also down the group.
Variation in a period. Along a period, the atomic radii of the elements generally decrease from left to the right.
The atomic radii (metallic radii for Li, Be and B while covalent radii for the remaining elements) of the elements present in the second period are given :

The trend in atomic radii have also been shown in the figure 3.5. From the values, it is clear that the alkali metal (Li) placed on the extreme left has the maximum atomic radius while the halogen (F) on the extreme right has the minimum value.

Explanation. In moving from left to the right in a period, the nuclear charge gradually increases by one unit and at the same time one electron is also being added in the electron shell. Due to increased nuclear charge from left to the right, the electrons are also getting attracted more and more towards the nucleus. Consequently, the atomic size is expected to decrease as shown in case of the elements of second period.

Ionic Radii
Ionic compounds are crystalline solids and the ionic radii are related to the ions present in them. The ions formed by the loss of one or more electrons from the neutral atom is known as cation (positive ion) while the one formed by the gain of electrons by the neutral atom is called anion (negative ion). Ionic radius may be defined as :

the effective distance from the centre of the nucleus of the ion upto which it exerts its influence on the electron cloud.

Since the electron cloud may extend itself to a very large distance from the nucleus it may not be possible to determine the ionic radius experimentally. However, some theoretical methods have been given to calculate the radius of the ion and the values as given by Pauling are most acceptable.

But it is quite easy to determine the inter-nuclear distance between two oppositely charged ions (say Na⁺ and Cl^– ions) in the crystal lattice by X-ray studies. If the absolute radius of one of them is known, that of the other can be obtained by subtracting the same from the inter-nuclear distance.

Similarly for the details of ionic radius,  In a group, the atomic radii increase downwards.

For example, in sodium chloride crystals, the inter-nuclear distance is 276 pm and the radius of the Na⁺ ion is 95 pm (Pauling method). Therefore, the radius of CL^– ion is 276 – 95 = 181 pm.

It may be noted that the value of the ionic radius is linked with the atomic radius and it varies accordingly. Thus, the ionic radius always increases down the group and decreases along the period provided the ions involved have the same magnitude of charge.

This is on account of the increase in the number of electron shells and also due to increase in the magnitude of shielding or screening effect. In a period, the trend is the reverse.

The atomic radii decrease from left to the right because the electrons are filled in the same shell and no new shell is formed. The ionic radii of the elements also follow the same trend.

The atomic radii of the elements influence the other periodic properties such as ionisation enthalpy, electron gain enthalpy and electronegativity.

Question 10.
How do atomic radii vary in a period and in a group ? How do you explain the variation ?
Answer:
Variation in a group. The atomic radii of the elements in every group of the periodic table increase as we move downwards. Covalent Radii of the alkali metals present in group 1 are given. Since all of them are metallic in nature, their metallic radii have also been given for reference.

The trend in the atomic radii are also shown in the figure 3.6. From the values, it is quite clear that the atomic radius of lithium (Li) is the minimum while that of cesium (Cs) is the maximum. The value of the last element francium (Fr) is not known since being as radioactive in nature, it is unstable.

Explanation. On moving down a group, there is an increase in the principal quantum number and thus, increase in the number of electron shells. Therefore, the atomic size is expected to increase. But at the same time, there is an increase in the atomic number or nuclear charge also. As a result, the atomic size must decrease. However, the effect of increase in the electron shells is more pronounced than the effect of increase in nuclear charge. Consequently, the atomic size or atomic radius increases down a group. This is well supported by the values given in Table 3.5 for the alkali metals.

Question 11.
What do you understand by isoelectronic species ? Name a species that will be isoelectronic with each of the following atoms or ions.
(i) F^– (ii) Ar (in) Mg²⁺ (iv) Rb⁺.
Answer:
Isoelectronic species are those species (atoms/ions) which have same number of electrons (iso-same ; electronic- electrons). The isoelectronic species are listed :
(i) Na⁺
(ii) K⁺
(iii) Na⁺
(iv) Sr²⁺.

Question 12.
Consider the following species :
N^3-, O^2-, F^–, Na⁺, Mg²⁺, Al³⁺ .
(a) What is common in them ?
(b) Arrange them in order of increasing radii.
Answer:
(a) All of them are isoelectronic in nature and have 10 electrons each.

(b) In isoelectronic species, greater the nuclear charge, lesser will be the atomic or ionic radius. Based upon that, the increasing order of atomic radii are :
Al³⁺ < Mg²⁺ < Na⁺ < F^– < O^2- < N^3-.

Question 13.
Explain why are cations smaller and anions larger in radii than their parent atoms.
Answer:
Cations (positive ions) are formed by the loss of electrons from the parent atoms. Therefore, they have smaller radii than the parent elements. On the other hand, anions are formed when the parent atoms take up one or more electrons. They have therefore, bigger radii than the parent atoms. For illustration,

Question 14.
What is the significance of the terms— ‘isolated gaseous atom’ and ‘ground state’ while defining ionization enthalpy and electron gain enthalpy ?
Answer:
(a) Significance of term ‘isolated gaseous atom’ : The three states of matter as we know differ in the interparticle spaces. These are the maximum in the solid state while least in the gaseous state.

The atoms in the gaseous state are far separated in the sense that they do not have any mutual attractive and repulsive interactions. These are therefore, regarded as isolated atoms.

When an atom in the gaseous state is isolated, its electron releasing tendency and electron accepting tendency are both absolute in nature.

It means that their values of ionisation enthalpy and of electron gain enthalpy are not influenced by the presence of the other atoms. It is not possible to express these when the atoms are in the liquid or solid state due to the presence of inter atomic forces.

(b) Significance of ground state : The ground state means that in a particular atom, the electrons associated are in the lowest energy state i.e.. they neither lose nor gain electrons. This represents the normal energy state of an atom. Both ionisation enthalpy and electron gain enthalpy are generally expressed with respect to the ground state of an atom only.

Question 15.
The energy of an electron in the ground state of the hydrogen atom is – 2.18 × 10^-18 J. Calculate the ionisation enthalpy of atomic hydrogen in terms of J mol^-1.
Answer:
The ionisation enthalpy is for 1 mole atoms. Therefore, ground state energy of one mole atoms may be expressed as:
E_{(groun state)} = (-2.18 × 10^-18J) × (6.022 × lO²³ mol^-1) = – 1.312 × 10⁶Jmol^-1
Ionisation enthalpy = E_x – E_{groun state}
= 0 – (-1.312 × 10⁶ J mol^-1) = 1-312 × 10⁶ J mol^-1

Question 16.
Among the second period elements, the actual ionization enthalpies are in the order : Li<B<Be<C<O<N<F< Ne.
Explain why
(i) Be has higher ∆_iH₁ than B ?
(ii) O has lower ∆_i8H₁ than N and F ?
Answer:
(i)We know that the cations or positive ions are formed when the neutral atoms (generally metal atoms) lose electrons. They can do so only in the gaseous state since they are isolated and interatomic forces are the minimum. Now energy is needed to overcome the force of attraction between the nucleus and electrons so that the latter may be released. This is known as ionization enthalpy. It may be defined as :
The minimum amount of energy which is needed to remove the most loosely bound electron from a neutral isolated gaseous atom in its ground state to form a cation also in the gaseous state.
Atom (g) → Positive ion (g) + e^– ; IE (∆_iH₁)
(Cation)
Na (g) → Na⁺ (g) + e^– ; IE (∆_iH₁)
The energy required can also be expressed in the form of ionization potential which is the minimum amount of potential needed to remove the most loosly held electron from an isolated neutral gaseous atom in its ground state.

(ii) ∆_iH₁ values of the there elements present in second period are given :
N(1402 kJ mol^-1) ; O(1314 kJ mol^-1) ; F(1681 kJ mol^-1).
• ∆_iH₁ of O is expected to more than that of N but is actually lesser because the electronic configuration of N is more symmetrical as well as stable in comparison to O. For electronic configuration.
• ∆_iH₁ of O is less than that of F because the ionization enthalpy in general increases along a period.

Question 17.
∆_iH₁ value of Mg is more as compared to that of Na while its ∆_iH₂ value is less. Explain.
Answer:
The electronic configuration of the two elements and their first and second ionization enthalpies are.,given :

∆_iH₁ value of Mg is more than that of Na due to greater symmetry and smaller size. But ∆_iH₂ value of Na is higher because Na⁺ ion has the configuration of noble gas element neon while Mg⁺ ion does not have a symmetrical configuration.

Question 18.
What are the various factors due to which the ionisation enthalpy of the main group elements tends to decrease down the group ?
Answer:
Variation down a group. The ionization enthalpies of the elements decrease on moving from top to the bottom in any group. The trend is quite evident from the data given in the table 3.10. The ∆_iH₁ values of the members belonging to the alkali metals of group 1 have been given separately in the fig 3.10.

The decrease in ionization enthalpies down any group is because of the following factors :
(i) There is an increase in the number of the main energy shells (n) in moving from one element to the other.
(ii) There is also an increase in the magnitude of the screening effect due to the gradual increase in the number of inner electrons.
Although the nuclear charge qtlso increases down the group which is likely to result in increased ionization enthalpy but its effect is less pronounced than the two factors listed above. The net result is the decrease in ionization enthalpies in a group in the periodic table.

Question 19.
The first ionisation enthalpy values (kJ mol^-1) of group 13 elements are :

How would you account for their deviation from the general trend ?
Answer:
The decrease in ∆_iH₁ value from B to Al is quite expected because of the bigger size of Al atom. But the element Ga has ten 3d electrons present in the 3d sub-shell which do not screen as much as is done by s and p electrons. Therefore, there is an unexpected increase in the magnitude of effective nuclear charge resulting in increased 3 Classification of Elements and Periodicity in Properties value. The same explanation can be offered in moving from In to T1. The latter has fourteen 4f electrons with very poor shielding effect. This also results in unexpected increase in the effective nuclear charge resulting in inflated ∆_iH₁ value.

Question 20.
Which of the following pairs of elements would have a more negative electron gain enthalpy ?
(i) O or F
(ii) F or Cl
Answer:
(i) O or F. The negative electron gain enthalpy of F(∆_egH = – 328 kJ mol^-1) is more than that of O(∆_egH = – 141 kJ mol^-1). This is on account of smaller size of the F atom and its greater urge to have a noble gas configuration (only one electron is needed) as compared to oxygen (two electrons are needed).
(ii) F or Cl. The negative electron gain enthalpy of Cl (∆_egH = – 349 kJ mol^-1) is more than that of F(∆_egH = – 328 kJ mol^-1). Actually, the size of F atom (atomic radius = 72 pm) is quite small as compared to Cl atom (atomic radius = 99 pm). This results in greater crowding of electrons in small space around the nucleus in case of F and the attraction for outside electron is less as compared to Cl in which the atomic size is large and the crowding of electrons is less.

Question 21.
Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first ? Justify your answer.
Answer:
For oxygen atom ; O(g) + e^– → 0^–(g) ; (∆_egH) = – 141 kJ mol^-1
O^–(g) e^– → 0^2-(g) ; (∆_egH) = + 780 kJ mol^-1
The first electron gain enthalpy of oxygen is negative because energy is released when a gaseous atom accepts an electron to form monovalent anion. The second electron gain enthalpy is positive because energy is needed to overcome the force of repulsion between monovalent anion and the second incoming electron.

Question 22.
What is basic difference between the terms electron gain enthalpy and electronegativity ?
Answer:
Difference between electron gain enthalpy and electronegativity.
Both electronegativity and electron gain enthalpy are the electron attracting tendencies of the elements but they differ in many respects. The important points of distinction are listed.

Question 23.
How will you react to the statement that the electronegativity of N on Pauling scale is 3-0 in all the nitrogen compounds ?
Answer:
On Pauling scale, the electronegativity of nitrogen, (3-0) indicates that it is sufficiently electronegative. This is quite expected as well due to its very small size and its requirement to have only three electrons to achieve the noble gas configuration. But it is not correct to say that the electronegativity of nitrogen in all the compounds is 3. It depends upon its state of hydridisation in a particular compound. It may be noted that greater the percentage of ^-character, more will be the electronegativity of the element. Thus, the electronegativity of nitrogen increases in moving from sp3 hybridised orbitals to sp hydridised orbitals i.e., as sp³ < sp² < sp.

Question 24.
Describe the theory associated with the radius of an atom as it :
(a) gains an electron.
(b) loses an electron.
Answer:
(a) When an atom gains an electron. It changes into an anion (M + e^– → M^–). In
doing so, there is an increase in the size. Actually, there is no change in the atomic number or nuclear charge but there is an increase in the electrons by one. As a result, electrons experience less attraction towards the nucleus and the anionic radius or size increases. For example, the radius of F atom (72 pm) is less than that of F^– ion (136 pm).

(b) When an atom loses an electron. It changes into a cation (M → M⁺ + e^–).
In doing so, there is a decrease in size. In this case also, the nuclear charge is the same but the electrons have decreased by one. As a result, the electrons experience more attraction towards the nucleus and the cationic radius decreases. For example, radius of Na atom (157 pm) is less than that of Na⁺ ion (95 pm).

Question 25.
What are the major differences between metals and non-metals ?
Answer:
Metals and non-metals as the names suggest are quite different from one another. They differ in physical as well as chemical characteristics.
Distinction based upon physical properties. The main points of distinction are given :

Distinction based upon chemical properties. The main points of distinction are given :

Question 26.
Would you expect the ionization enthalpies of two isotopes of the same element to be same or different ? Justify your answer.
Answer:
The ionization enthalpies are related to the magnitude of nuclear charge as well as the electronic configuration of the elements. Since the isotopes of an element have same nuclear charge as well as electronic configuration, they have the same ionization enthalpies.

Question 27.
Use periodic table to answer the following questions :
(a) Identify the element with five electrons in the outer sub-shell.
(b) Identify the element that would tend to lose two electrons.
(c) Identify the element that would tend to gain two electrons.
Answer:
(a) Element belongs to nitrogen family (group 15) e.g. nitrogen
(b) Eletnent belongs to alkaline earth family (group 2) e.g. magnesium
(c) Element belongs to oxygen family (group 16) e.g. oxygen

Question 28.
The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas among the group 17 elements, it is F > Cl > Br > I. Explain.
Answer:
In group 1 elements (alkali metals) the reactivity of the metals is mainly due to the electron releasing tendency of their atoms, which is related to ionization enthalpy. Since ionization enthalpy decreases down the group, the reactivity of alkali metals increases.
In group 17 elements (halogens-non-metals), the reactivity is linked with electron accepting tendency of the members of the family. It is linked with electronegativity and electron gain enthalpy. Since both of them decrease down the group, the reactivity therefore decreases.
Thus, we conclude that the order of reactivity in two cases is different since in one case it is due to electron releasing tendency and it is because of electron accepting tendency in the other case.

Question 29.
Write the general electronic configuration of s, p, d and /block elements.
Answer:
s-Block elements : ns^1-2 (n varies from 2 to 7)
p-Block elements : ns² np^1-6 (n varies from 2 to 7)
d-Block elements : (n – 1) d^1-10 ns^1-2 (n varies from 4 to 7)
f-Block elements : (n – 2)f^1-14 (n – 1) d^0-1 ns² (n may have value either 6 or 7)

Question 30.
Assign the position of the elements having outer electronic configuration
(i) ns²p⁴ for n = 3
(ii) (n – 1 )d²ns² for n = 4 and
(iii) (n – 2) f⁷ (n – 1 )d¹ ns² for n = 6 in the periodic table.
Answer:
(i) The element is present in third period (n = 3) and belongs to group 16(10 + 2 + 4 = 16).
The element is sulphur : [1s² 2s² 2p⁶ 3s² 3p⁴].
(ii) The element is present in fourth period (.n = 4) and belongs to group 4 (2 + 2 = 4).
The element is titanium : [Ne] 3d² 4s².
(iii) The element is present in sixth period (n = 6) and belongs to group 3.
The element is gadolinium (Z = 64): [Xe] 4f⁷ 5d¹ 6s².

Question 31.
The first (∆_iH₁) and the second (∆_iH₂) ionization enthalpies (in kJ mol^-1) and the (∆_egH) electron gain enthalpy (in kJ mol¹) of a few elements are given below :

Which of the above elements is likely to be :
(a) the least reactive element.
(b) the most reactive metal.
(c) the most reactive non-metal.
(d) the least reactive non-metal.
(e) the metal which can form a stable binary halide of the formula MX₂ (X = halogen)
(f) the metal which can form a predominantly stable covalent halide of the formula MX (X = halogen) ?
Answer:
The nature and reactivity of the elements are guided by ionisation enthalpies and electron gain enthalpies. Based upon
the available data :
(a) The least reactive element is ‘V’. It has very high ∆_iH (2372 kJ mol^-1) and ∆_egH is (+ 48 kJ mol^-1) which indicate that it is a noble gas element. The value of ∆_egH suggests it to be helium (He).

(b) The most reactive metal is ‘II’. It has a very low ∆_iH equal to 419 kJ mol^-1 and its negative ∆_eg value is also quite small (- 48 kJ mol^-1). These values correspond to the alkali metal potassium (K).
In general, alkali metals are the most reactive metals in the periodic table.

(c) The most reactive non-metal is ‘III’. Since the element has very high ionization enthalpies (∆_iH₁ as well as ∆_iH₂) and very high negative electron gain enthalpy, it is expected to be a highly reactive non-metal. The values correspond to non-metal fluorine (F).

(d) The least reactive non-metal is ‘IV’. The character is revealed by the moderate values of ∆_iH₁ and ∆_iH₂. The value of electron gain enthalpy indicates that the element is iodine (I).

(e) The metal forming stable binary halide (MX₂) is ‘VI’. The formula MX₂ suggests that the metal belongs to the family of alkaline earth (group 2). The ∆_iH₁ and ∆_iH₂ values suggest it to be Mg (However, ∆_egH does not agree).

(f) The metal which can form a predominantly stable covalent halide of the formula. MX (halogen) is ‘I’. The metal is monovalent in nature and data suggests it to be lithium (Li) because ∆_iH₁ ] value is very small as compared to ∆_iH₂ value. In fact, in alkali metal lithium (Li), the monovalent cation (Li⁺) has the electronic configuration of the noble gas helium (He) The halide is LiX. The covalent character of LiX is due to the reason that the electron cloud of Li⁺ ion attracts the electron cloud of X-ion towards itself. As a result, the halide is of covalent nature.

Question 32.
Predict the formulas of the stable binary compounds that would be formed by the combination of the following
pairs of elements :
(a) Lithium and oxygen
(b) Magnesium and nitrogen
(c) Aluminium and iodine
(d) Silicon and oxy gen
(e) Phosphorus and fluorine
(f) Element 71 and fluorine
Answer:
(a) Li₂O (Lithium oxide)
(b) Mg₃N₂ (Magnesium nitride).
(c) AlI₃ (Aluminium iodide)
(d) SiO₂ (Silicon dioxide)
(e) PF₃ (Phosphorus trifluoride) or PF₅ (Phosphorus pentafluoride)
(f) The element with Z.
= 71 is lutetium (Lu) with electronic configuration [Xe] 4f¹⁴5d¹ 6s². With fluorine, it will form a binary compound of
formula LuF₃.

We hope the NCERT Solutions for Class 11 Chemistry at Work Chapter 3 Classification of Elements and Periodicity in Properties, help you. If you have any query regarding NCERT Solutions for Class 11 Chemistry at Work Chapter 3 Classification of Elements and Periodicity in Properties, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom.

Question 1.
(i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.
Answer:
(i) Mass of an electron = 9.1 × 10^-28 g
9.1 × 10^-28 g is the mass of = 1 electron

(ii) One mole of electrons = 6.022 × 10²³ electrons
Mass of 1 electron = 9.1 × 10^-31 kg
Mass of 6.022 × 10²³ electrons = (9.1 × 10.31kg) × (6.022 × 10²³) = 5.48 × 10^-7 kg
Charge on one electron = 1.602 × 10^-19 coulomb
Charge on one mole electrons = 1.602 × 10^-19 × 6.022 × 10²³ = 9.65 × 10⁴ coulombs

Question 2.
(i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of ¹⁴C. (Assume that mass of a neutron = 1.675 × 10^-27kg).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH₃ at STP.
Will the answer change if the temperature and pressure are changed ?
Answer:
(i) One mole of methane (CH₄) has molecules = 6.022 × 10²³
No. of electrons present in one molecule of CH₄ = 6 + 4 = 10
No. of electrons present in 6.022 × 10²³ molecules of CH₄ = 6.022 × 10²³ × 10
= 6.022 × 10²⁴ electrons

(ii) Step I. Calculation of total number of carbon atoms
Gram atomic mass of carbon (C-14) = 14 g = 14 × 10³ mg
14 × 10³ mg of carbon (C-14) have atoms = 6.022 × 10²³

Step II. Calculation of total number and tatal mass of neutrons
No. of neutrons present in one atom (C-14) of carbon = 14 – 6 = 8
No. of neutrons present in 3-011 × 10²⁰ atoms (C-14) of carbon = 3.011 × 10²⁰ × 8
= 2.408 × 10²¹ neutrons
Mass of one neutron = 1.675 × 10^-27 kg
Mass of 2.408 × 10²¹ neutrons = (1.675X10^-27 kg) × 2.408 × 10²¹
= 4.033 × 10^-6 kg.

(iii) Step I. Calculation of total number ofNH₃ molecules
Gram molecular mass of ammonia (NH₃) = 17 g = 17 × 10³ mg
17 × 10³ mg of NH₃ have molecules = 6.022 × 10²³

Step II. Calculation of total number and mass of protons
No. of protons present in one molecule of NH₃ = 7 + 3 = 10 .
No. of protons present in 12.044 × 10²⁰ molecules of NH₃ = 12.044 × 10²⁰ × 10
= 1.2044 × 10²² protons
Mass of one proton = 1.67 × 10^-27 kg
Mass of 1.2044 × 10²² protons = (1.67 × 10^-27 kg) × 1.2044 × 10²²
= 2.01 × 10^-5 kg.
No, the answer will not change upon changing the temperature and pressure because only the number of protons and mass of protons are involved.

Question 3.
How many protons and neutrons are present in the following nuclei

Answer:

Question 4.
Write the complete symbol for the atom (X) with the given atomic number (Z) and atomic mass (A)
(i) Z = 17,A = 35
(ii) Z = 92, A = 233
(in) Z = 4, A = 9.
Answer:

Question 5.
Yellow light emitted from a sodium lamp has a wavelength (2) of 580 nm. Calculate the frequency (v) and wave number (v) of yellow light.
Answer:

Question 6.
Calculate the energy of each of the photons which
(i) correspond to light of frequency 3 × 1015 Hz
(ii) have wavelength of 0-50 A.
Answer:
(i) Energy of photon (E) = hv
h = 6.626 × 10^-34 J s ; v = 3 × 10¹⁵ Hz = 3 × 10¹⁵s^-1
∴ E = (6.626 × 10^-34 J s) × (3 × 10¹⁵ s^-1) = 1.986 × 10¹⁸ J
Energy of photon (E) = hv = \(\frac { hc }{ \lambda } \)
h = 6.626 × 10 34 J s; c = 3 × 10⁸ m s^-1 ;
λ= 0.50 Å = 0.5 × 10^-10 m.

Question 7.
Calculate the wavelength, frequency and wave number of light wave whose period is 2.0 × 10^-10 s.
Answer:

Question 8.
What is the number of photons of light with wavelength 4000 pm which provide 1 Joule of energy ?
Answer:
Energy of photon (E) = \(\frac { hc }{ \lambda } \)
h = 6.626 × 10^-34 Js, c = 3 × 10⁸ m s^-1, λ = 4000 pm = 4000 × 10^-12 = 4 × 10^-9 m

Question 9.
A photon of wavelength 4 × 10^-7 m strikes on metal surface ; the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon,
(ii) the kinetic energy of the emission
(iii) the velocity of the photoelectron. (Given 1 eV = 1.6020 × 10^-19 J).
Answer:

Question 10.
Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in k-J mol^-1.
Answer:

Question 11.
A 25 watt bulb emits monochromatic yellow light of wavelength 0.57 μm. Calculate the rate of emission of quanta per second.
Answer:

Question 12.
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 A. Calculate threshold frequency (v₀) and work function (W₀) of the metal.
Answer:

Question 13.
What is the wavelength of the light emitted when the electron in a hydrogen atom undergoes transition from the energy level with n = 4 to energy level n = 2 ? What is the colour corresponding to this wavelength ? (Given R_H = 109678 cm^-1)
Answer:

Question 14.
How much energy is required to ionise a hydrogen atom if an electron occupies n = 5 orbit ? Compare your answe r with the ionisation energy of H atom (energy required to remove the electron from n = 1 orbit)
Answer:

Question 15.
What is the maximum number of emission lines when the excited electron of a hydrogen atom in n = 6 drops to the ground state ?
Answer:
The maximum no. of emission lines = \(\frac { n(n – 1) }{ 2 }\) = \(\frac { 6(6 – 1) }{ 2 }\) =3 × 5 = 15

Question 16.
(i) The energy associated with first orbit in hydrogen atom is – 2.17 × 10^-18 J atom^-1. What is the energy associated with the fifth orbit ?
(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.
Answer:

(ii) For hydrogen atom ; r_n = 0.529 x n² Å
r₅ = 0.529 x (5)² = 13.225 Å = 1.3225 nm.

Question 17.
Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.
Answer:

Question 18.
What is the energy in joules required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of light emitted when the electron returns to the ground state ? The ground state electronic energy is – 2.18 × 11^-11 ergs.
Answer:

Question 19.
The electronic energy in hydrogen atom is given by E_n (-2.18 × 10^-18 s) / n²J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition ?
Answer:

Question 20.
Calculate the wavelength of an electron moving with a velocity of 2.05 × 10⁷ m s^-1.
Answer:

Question 21.
The mass of an electron is 9.1 × 10^-31 kg. If its kinetic energy is 3.0 × 10^-25 J, calculate its wavelength.
Answer:

Question 22.
Which of the following are iso-electronic species ?
Na⁺, K⁺, Mg²⁺, Ca²⁺, S^2-, Ar.
Answer:
Na⁺ and Mg²⁺ are iso-electronic species (have 10 electrons) K⁺, Ca²⁺ , S^2- are iso-electronic species (have 18 electrons)

Question 23.
(i) Write the electronic configuration of the following ions : (a) H (b) Na+ (c) 02~ (d) F^–.
(ii) What are the atomic numbers of the elements whose outermost electronic configurations are represented by :
(a) 3s¹ (b) Ip³ and (c) 3d⁶ ?
(iii) Which atoms are indicated by the following configurations ?
(a) [He]2s¹ (b) [Ne] 3s² 3p³ (c) [Ar] 4s² 3d¹.
Answer:
(i) (a) 1s²
(b) 1s² 2s² 2p⁶
(c) 1s²2s²2p⁶
(d) 1s²2s²2p⁶.
(ii) (a) Na (Z = 11) has outermost electronic configuration = 3s¹
(b) N (Z = 7) has outermost electronic configuration = 2p³
(c) Fe (Z = 26) has outermost electronic configuration = 3d⁶
(iii) (a) Li
(b) P
(c) Sc

Question 24.
What is the lowest value of n which allows ‘g’ orbital to exist ?
Answer:
The lowest value of l w’here ‘g’ orbital can be present = 4
The lowest value of n where ‘g’ orbital can be present = 4+1=5.

Question 25.
An electron is in one of the 3d orbitals. Give the possible values of n, l and nil for the electron.
Answer:
For electron in 3d orbital, n = 3, l = 2, mi = -2, -1, 0, +1, +2.

Question 26.
An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.
Answer:
No. of protons in a neutral atom = No. of electrons = 29
Electronic configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹.

Question 27.
Give the number of electrons in the species : H₂⁺, H₂ and 0₂⁺.
Answer:
H₂⁺ = one ; H₂ = two ; 0₂⁺ = 15

Question 28.
(i) An atomic orbital has n = 3. What are the possible values of l and m_l ?
(ii) List the quantum numbers m_l and l of electron in 3rd orbital.
(iii) Which of the following orbitals are possible ?
1p, 2s, 2p and 3f.
Answer:
(i) For n = 3; l = 0, 1 and 2.
For l = 0 ; m_l = 0
For l = 1; m_l = +1, 0, -1
For l = 2 ; m_l = +2, +1,0, +1, + 2
(ii) For an electron in 3rd orbital ; n = 3; l = 2 ; m_l can have any of the values -2, -1, 0,
+ 1, +2.
(iii) 1p and 3f orbitals are not possible.

Question 29.
Using s, p and d notations, describe the orbitals with follow ing quantum numbers :
(a) n = 1, l = 0
(b) n = 4, l = 3
(c) n = 3, l = 1
(d) n = 4, l = 2
Answer:
(a) 1s orbital
(b) 4f orbital
(c) 3p orbital
(d) 4d orbital

Question 30.
From the following sets of quantum numbers, state which are possible. Explain why the others are not possible.
(i) n = 0, l = 0, m_l = 0, m_s = +1/2
(ii) n = 1, l = 0, m_l = 0, m_s – – 1/2
(iii) n = 1, l = 1, m_l = 0, m_s= +1/2
(iv) n = 1, l = 0, m_l = +1, m_s= +1/2
(v) n = 3, l = 3, m_l = -3, m_s = +1/2
(vi) n = 3, l = 1, m_l = 0, m_s= +1/2
Answer:
(i) The set of quantum numbers is not possible because the minimum value of n can be 1 and not zero.
(ii) The set of quantum numbers is possible.
(iii) The set of quantum numbers is not possible because, for n = 1, l can not be equal to 1. It can have 0 value.
(iv) The set of quantum numbers is not possible because for l = 0. mt cannot be + 1. It must be zero.
(v) The set of quantum numbers is not possible because, for n = 3, l ≠ 3.
(vi) The set of quantum numbers is possible.

Question 31.
How many electrons in an atom may have the following quantum numbers ?
(a) n = 4 ; m_s = -1/2
(b) n = 3, l = 0.
Answer:
(a) For n = 4
Total number of electrons = 2n² = 2 × 16 = 32
Half out of these will have ms = —1/2
∴ Total electrons with ms (-1/2) = 16
(b) For n = 3
l= 0 ; m_l = 0, m_s +1/2, -1/2 (two e^–)

Question 32.
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.
Answer:

Thus, the circumference (2πr) of the Bohr orbit for hydrogen atom is an integral multiple of the de Broglie wavelength.

Question 33.
Calculate the number of atoms present in :
(i) 52 moles of He
(ii) 52 u of He
(iii) 52 g of He.
Answer:

Question 34.
Calculate the energy required for the process :
He⁺fe) → He²⁺(g) + e^–
The ionisation energy’ for the H atom in the ground state is 2.18 × 10^-18  J atom^-1
Answer:

For H atom (Z = 1), E_n =2.18 × 10^-18 × (l)² J atom^-1 (given)
For He⁺ ion (Z = 2), E_n =2.18 × 10^-18 × (2)² = 8.72 × 10^-18 J atom^-1 (one electron species)

Question .35.
If the diameter of carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across a length of a scale of length 20 cm long.
Answer:

Question 36.
2 × 10⁸ atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.
Answer:
The length of the arrangement = 2.4 cm
Total number of carbon atoms present = 2 ×10⁸

Radius of each carbon atom = \(\frac{ 1 }{ 2 }\)(1.2 × 10^-8) = 6.0 × 10^-9cm = 0.06 nm

Question 37.
The diameter of zinc atom is 2.6 Å. Calculate :
(a) the radius of zinc atom in pm
(b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side length wise.
Answer:
(a) Radius of zinc atom =\(\frac { 2.6\AA }{ 2 } \)= 1.3 Å = 1.3 × 10^-10m = 130 × 10^-12m = 130 pm
(b) Length of the scale = 1.6 cm = 1.6 × 10¹⁰ pm
Diameter of zinc atom = 260 pm

Question 38.
A certain particle carries 2.5 x 10^-16 C of static electric charge. Calculate the number of electrons present in it.
Answer:

Question 39.
In Millikan’s experiment, the charge on the oil droplets was found to be – 1.282 x 10^-18C. Calculate the number of electrons present in it.
Answer:

Question 40.
In Rutherford experiment, generally the thin foil of heavy atoms like gold, platinum etc. have been used to be bombarded by the a-particles. If a thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results ?
Answer:
We have studied that in the Rutherford’s experiment by using heavy metals like gold and platinum, a large number of a-particles sufferred deflection while a very few had to retrace their path.

If a thin foil of lighter atoms like aluminium etc. be used in the Rutherford experiment, this means that the obstruction offered to the path of the fast moving a-particles will be comparatively quite less.

As a result, the number of a-particles deflected will be quite less and the particles which are deflected back will be negligible.

Question 41.
Symbols \(_{ 35 }^{ 79 }{ Br }\) and ⁷⁹Br can be written whereas symbols \(_{ 79 }^{ 35 }{ Br }\) and ₃₅Br are not accepted. Answer in brief.
Answer:
In the symbol \(_{ A }^{ B }{ X }\) of an element :
A denotes the atomic number of the element
B denotes the mass number of the element.
The atomic number of the element can be identified from its symbol because no two elements can have the atomic number. However, the mass numbers have to be mentioned in order to identify the elements. Thus,
Symbols \(_{ 35 }^{ 79 }{ Br }\) and ⁷⁹Br are accepted because atomic number of Br will remain 35 even if not mentioned. Symbol \(_{ 79 }^{ 35 }{ Br }\) is not accepted because atomic number of Br cannot be 79 (more than the mass number = 35). Similarly, symbol 35Br cannot be accepted because mass number has to be mentioned. This is needed to differentiate the isotopes of an element.

Question 42.
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the symbol to the element.
Answer:
An element can be identified by its atomic number only. Let us find the atomic number.
Let the number of protons = x
Number of neutrons = x + [\(\frac { x\times 31.7 }{ 100 } \) = (x × 0.317x)
Now, Mass no. of element = no. of protons =no. neutrons
81 = x + x + 0-317 x = 2.317 x or x = \(\frac { 81 }{ 2.317 } \) = 35
∴ No. of protons = 35, No. of neutrons = 81 – 35 =46
Atomic number of element (Z) = No. of protons = 35
The element with atomic number (Z) 35 is bromine \(_{ 35 }^{ 81 }{ Br }\).

Question 43.
An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11 -1% more neutrons than the electrons, find the symbol of the ion.
Answer:
Let the no. of electron in the ion = x
∴ the no. of protons = x – 1 (as the ion has one unit negative charge)
and the no. of neutrons = x + \(\frac { x\times 11.1 }{ 100 } \) = 1.111 x
Mass no. or mass of the ion = No. of protons + No. of neutrons
(x – 1 + 1.111 x)
Given mass of the ion = 37
∴ x- 1 + 1.111 x = 37 or 2.111 x = 37 + 1 = 38
x = \(\frac { 38 }{ 2.111 } \) = 18
No. of electrons = 18 ; No. of protons = 18 – 1 = 17
Atomic no. of the ion = 17 ; Atom corresponding to ion = Cl
Symbol of the ion = \(_{ 17 }^{ 37 }{ Cl }\)^–

Question 44.
An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign symbol to the ion.
Answer:
Let the no. of electrons in the ion = x
∴ the no. of the protons = x + 3 (as the ion has three units positive charge)
and the no. of neutrons = x + \(\frac { x\times 31.7 }{ 100 } \) = xc + 0.304 x
Now, mass no. of ion = No. of protons + No. of neutrons
= (x + 3) + (x + 0.304x)
∴ 56 = (x + 3) + (x + 0.304x) or 2.304x = 56 – 3 = 53
x = \(\frac { 53 }{ 2.304 } \) = 23
Atomic no. of the ion (or element) = 23 + 3 = 26
The element with atomic number 26 is iron (Fe) and the corresponding ion is Fe³⁺.

Question 45.
Arrange the following type of radiations in increasing order of wavelength :
(a) radiation from microwave oven
(b) amber light from traffic signal
(c) radiation from FM radio
(d) cosmic rays from outer space and
(e) X-rays.
Answer:
Cosmic rays < X-rays < amber colour < microwave < FM

Question 46.
Nitrogen laser produces a radiation of wavelength of 337.1 nm. If the number of photons emitted is 5.6 x 10²⁴, calculate the power of this laser.
Answer:

Question 47.
Neon gas is generally used in sign boards. If it emits strongly at 616 nm, calculate :
(a) frequency of emission (b) the distance travelled by this radiation in 30s (c) energy of quantum (d) number of quanta present if it produces 2 J of energy.
Answer:

Question 48.
In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 x 10^-18 J from the radiations of 600 nm, calculate the number of photons received by the detector.
Answer:

Question 49.
Life times of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 x 10¹⁵, calculate the energy of the source.
Answer:
Time duration (t) = 2 ns = 2 x 10^-9 s

Energy of one photon, E = hv = (6.626 x 10^-34 Js) x (10⁹/2 s^-1) = 3.25 x 10^-25J
No. of photons = 2.5 x 10⁵
∴ Energy of source = 3.3125 x 10^-25 J x 2.5 x 10¹⁵ = 8.28 x 10^-10 J

Question 50.
The longest wavelength doublet absorption transition is observed at 589 nm and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.
Answer:

Question 51.
The work function for cesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the cesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron
Answer:

Question 52.
Following results are observed when sodium metal is irradiated with different wavelengths. Calculate threshold wavelength.

Answer:

Question 53.
The ejection of the photoelectrons from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.
Answer:

Question 54.
If the photon of the wavelength 150 pm strikes an atom, one of its inner bound electrons is ejected out with a velocity of 1.5 x 10⁷ m s^-1. Calculate the energy with which it is bound to the nucleus.
Answer:

Question 55.
Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 x 10¹⁵ (Hz) [1/3² – 1 /n²]
Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.
Answer:

Question 56.
Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
Answer:

Question 57.
Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 x 10⁶ m s^-1, calculate de Brogile wavelength associated with this electron.
Answer:

Question 58.
Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.
Answer:

Question 59.
If the velocity of the electron in Bohr’s first orbit is 2.19 x 10⁶ m s^-1, calculate the de Brogile wavelength associated with it.
Answer:

Question 60.
The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 x 10⁵ m s^-1. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.
Answer:

Question 61.
If the position of the electron is measured within an accuracy of ± 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is \(\frac { h }{ 4\pi } \) x 0.05 nm. Is there any problem in defining this value ?
Answer:

Since actual momentum is smaller than the uncertainty in measuring momentum, therefore, the momentum of electron can not be defined.

Question 62.
The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. List if any of these combination(s) has/have the same energy
(i) n = 4, l = 2, m_l = -2, m_s = -1/2
(ii) n = 3, l = 2, m_l = 1, m_s = +1/2
(iii) n = 4, l = 1, m_l = 0, m_s = +1/2
(iv) n = 3, l = 2, m_l = -2, m_s = -111
(v) n = 3, l = l, m_l = -1, m_s = +1/2
(vi) n = 4, l = 1, m_l = 0, m_s = +1/2
Answer:
The electrons may be assigned to the following orbitals :
(i) 4d
(ii) 3d
(iii) 4p
(iv) 3d
(v) 3p
(vi) 4p.
The increasing order of energy is :
(v) < (ii) = (iv) < (vi), = (iii) < (i)

Question 63.
The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electrons in 4p orbital. Which of these electrons experiences lowest effective nuclear charge ?
Answer:
4p electron experiences lowest effective nuclear charge because of the maximum magnitude of screening or shielding effect. It is farthest from the nucleus.

Question 64.
Among the following pairs of orbitals, which orbital will experience more effective nuclear charge (i) 2s and 3s (ii) 4d and 4f (iii) 3d and 3p ?
Answer:
Please note that greater the penetration of the electron present in a particular orbital towards the nucleus, more will be the magnitude of the effective nuclear charge. Based upon this,
(i) 2s electron will experience more effective nuclear charge.
(ii) 4d electron will experience more effective nuclear charge.
(iii) 3p electron will experience more effective nuclear charge.

Question 65.
The unpaired electrons in A1 and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus ?
Answer:
Configuration of the two elements are :
A1 (Z = 13) : [Ne]¹⁰3s²3p¹ ; Si (Z = 14) : [Ne] 103s23p2
The unpaired electrons in silicon (Si) will experience more effective nuclear charge because the atomic number of the element Si is more than that of A1.

Question 66.
Indicate the number of unpaired electrons in :
(a) P (b) Si (c) Cr (d) Fe and (e) Kr.
Answer:
(a) P (z=15) : [Ne]¹⁰3s²3p³ ; No. of unpaired electrons = 3
(b) Si (z=14) : [Ne]¹⁰3s²3p² ; No. of unpaired electrons = 2
(c) Cr (z=24): [Ar]¹⁸4s¹3d⁵ ; No. of unpaired electrons = 6
(d) Fe (z=26): [Ar]¹⁸4s²3d⁶ ; No. of unpaired electrons = 4
(e) Kr (z=36) : [Ar]¹⁸4s²3d¹⁰4p⁶ ; No. of unpaired electrons = Nil.

Question 67.
(a) How many sub-shells are associated with n = 4 ?
(b) How many electrons will be present in the sub-shells having ms value of -1/2 for n = 4 ?
Answer:
(a) For n = 4 ; No. of sub-shells = (l = 0, l = 1, l = 2, l = 3) = 4.
(b) Total number of orbitals which can be present = n² = 4² = 16.
Each orbital can have an electron with m_s = – 1/2 -‘. Total no. of electrons with m, = – 1/2 is 16.

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The Address NCERT Solutions for Class 11 English Snapshots Chapter 2

The Address NCERT Text Book Questions and Answers

Question 1.
‘Have you come back?’ said the woman. ‘I thought that no one had come back.’ Does this statement give some clue about the story? If yes, what is it?
Answer:
The narrator went back after the war to Marconi Street—Number 46, in search of her mother’s belongings. The belongings were with a non-Jewish lady, Mrs Dorling, who had stayed close by. This woman had often come to their place and regularly taken heavy boxes of silverware, cutlery, crockery etc. to keep with her till after the war. Not only did she volunteer to keep them, but had also insisted that they leave the things with her for safety.

However, when the narrator returned, Mrs Dorling at first refused to recognise her and then she expressed her surprise and said that she had not expected anyone to return. This is the first vital clue that the people who had once left the country were unwelcome. The people, who had volunteered to keep their belongings safely, were using their things as their own and had no intention of returning them. The political animosity had seeped into personal lives as well.

Question 2.
The story is divided into pre-War and post-War times. What hardships do you think the girl underwent during these times?
Answer:
The story is divided into pre-War and post-War times. The situations lend a direct contrast to each other. During the pre¬War times, the narrator and her mother lived a comfortable life, where there was bonding between the people in their neighbourhood. When during the first half of the War, the narrator visited her home she noticed that various things were missing. Her mother told her about Mrs Dorling, an old acquaintance, who had suddenly turned up and renewed their contact.

Since then, she had come regularly. She had insisted on taking their things to “save” all the “nice things”. The narrator’s mother also censured her daughter for not trusting the lady.As the narrator feared, when she went back after the war, Mrs Dorling stood at the door and “wanted to prevent it opening any further. Her face gave absolutely no sign of recognition.” She was wearing her mother’s green knitted cardigan but refused to talk to the narrator. She said that it was “not convenient” for her to talk. It was a betrayal of trust and sentiment.

The girl, like anyone who goes through war, must have undergone a traumatic experience. They were uprooted and insecure. They had to leave the country that they thought to be their own, leave their house and belongings that were not merely things but held memories and had sentiments attached to them. The basic necessities of life were not available. Moreover, they left behind the people who they thought were their friends. Above all, they lived a threatened life. This is evident through the observation of the narrator—“But gradually everything became more normal again. Bread was getting to be a lighter colour, there was a bed one could sleep in unthreatened, a room with a view one was more used to glancing at each day.”

Question 3.
Why did the narrator of the story want to forget the address?
Answer:
The narrator’s visit to Mrs Dorling’s house horrified her. She had come out of curiosity to see her possessions to see them, touch them, and relive the memories attached with them. But she felt oppressed in the strange atmosphere. Her eyes fell on the woollen tablecloth. The memories came flooding back to her. She followed the lines of the pattern and knew that somewhere on the edge there should be a bum mark that had never been repaired.

The cups on the tea table, the white pot, the spoons, all were so familiar and yet so strange. She recalled how as a child she had always fancied the apple on the pewter plate. She said one gets so used to touching all the lovely things in the house that one ceases to notice them unless something is missing. She recalled the time her mother had asked her to polish the silver. It was then that she had realised the spoons, forks and knives, they ate off every day were silver.

The objects were linked in her memory with the familiar life of earlier times but now they had lost their value because with the passage of time she felt cut off from them as they were now in unfamiliar surroundings. Moreover, she now lived in a small rented room where no more than a handful of cutlery could fit in the narrow table drawer. Hence, she thought of an easier way out to forget the address.

Question 4.
‘The Address’ is a story of human predicament that follows war. Comment.
Answer:
Although, this is apparently a very sad story about loss and regret emanating from the persecution of the Dutch Jews during the Second World War, it also speaks, more intimately, of the personal challenges we all must face as individuals in resolving crisis in our own lives. The story relays events before and after the war as the female narrator attempts to confront her past as she visits “the address” where her family’s past belongings were ‘stored,’ at a non- Jewish neighbour’s house.

She felt the urge to see them, touch and recall memories. On a deeper level, the story is a commentary on memories and remembering on what is worth remembering and what is worth forgetting: things “lose their value when you see them again, tom out of context…”