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NCERT Exercises

Question 1.
Choose the correct answer. A thermodynamic state function is a quantity
(a) used to determine heat changes
(b) whose value is independent of path
(c) used to determine pressure volume work
(d) whose value depends on temperature only.
Answer.
(b): State function is a property of the system whose value depends only upon the state of the system and is independent of the path or the manner by which the state is reached.

Question 2.
For the process to occur under adiabatic conditions, the correct condition is
(a) ΔT = 0
(b) Δp = 0
(c) g = 0
(d) w=0
Answer..
(c): Adiabatic system does not exchange heat with the surroundings.

Question 3.
The enthalpies of all elements in their standard states are
(a) unity
(b) zero
(c) <0
(d) different for each element.
Answer.
(b) : By convention, the standard enthalpy of formation of every element in its standard state is zero.

Question 4.
ΔU° of combustion of methane is – X kJ mol^-1. The value of ΔH° is
(a) = ΔU°
(b) >ΔU°
(c) <ΔU°
(d) =0
Answer.
(c) : CH_4(g)+2O₂ → CO_{2(g)+2H2Ol
Δn = 1 – 3 = -2. ‘
ΔH° = ΔU° + ΔnRT = -X- 2RT}

Question 5.
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, -890.3 kJ mol^-1,-393.5 kJ mol^-1 and-285.8 kJ mol^-1 respectively. Enthalpy of formation of CH_4(g) will be
(a) -74.8 kJ mol^-1
(b) -52.27 kJ mol^-1
(c) +74.8 kJ mol^-1
(d) +52.26 kJ mol^-1
Solution.

Question 6.
A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be
(a) possible at high temperature
(b) possible only at low temperature
(c) not possible at any temperature
(d) possible at any temperature
Solution.
(d): A + B → C + D + q, ΔS = +ve
Here, ΔH = -ve
ΔG = ΔH – TΔS
For reaction to be spontaneous, ΔG should be -ve. As ΔH = -ve and ΔS is +ve, ΔG will be -ve at any temperature.

Question 7.
In a process, 701 J of heat is absorbed by a system and 394 S of work is done by the system. What is the change in internal energy for the process?
Solution.
Heat absorbed by the system (q) = 701 J
Work done by the system (w) = -394 J
According to first law of thermodynamics,
ΔU = q + w = 701 + (-394) = 701 – 394 = 307 J

Question 8.
The reaction of cyanamide, NH₂CN_(g), with dioxygen was carried out in a bomb calorimeter, and AU was found to be -742.7 kJ mol 1 at 298 K. Calculate enthalpy change for the reaction at 298 K.
NH₂CN_(g) +  → N₂+CO_2(g)+H₂O_l
Solution.

Question 9.
Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol^-1 K^-1.
Solution.
Mass of Al = 60 g
Rise in temperature, ΔT = 55 – 35 = 20°C
Molar heat capacity of Al = 24 J mo^-1 K^-1
Specific heat capacity of Al = 
∴ Energy required = m x c x ΔT
= 
= 1.068kJ or 1.07kJ

Question 10.
Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C. Δ_fusH = 6.03 kJ mol^-1 at 0°C.
C_p[H₂O_(l)] = 75.3 J mol^-1 K^-1,
C_p[H₂O_(s)] = 36.8 J mol^-1 K^-1
Solution.

Question 11.
Enthalpy of combustion of carbon to CO₂ is -393.5 kJ mol^-1. Calculate the heat released upon formation of 35.2 g of CO₂ from carbon and dioxygen gas.
Solution.
C + O₂ → CO₂; ΔT = -393.5 kJ
∵ When 44 g of COz is formed from carbon and dioxygen gas, heat released = 393.5 kj
∴ When 35.2 g of CO₂ is formed from carbon and dioxygen gas, heat released
= 
Thus, ΔH = -314.8 kJ

Question 12.
Enthalpies of formation of CO_(g), CO_2(g), N₂O_(g) and N₂O_4(g) are -110, – 393, 81 and 9.7 kJ mol ^-1 respectively. Find the value of Δ_rf for the reaction
N₂O_4(g) + 3CO_(g) → N₂O_(g) + 3CO_2(g)
Solution.

Question 13.
Given: N_2(g) + 3H_2(g) → 2NH_3(g); Δ_rH° = -92.4 kJ mol^-1 What is the standard enthalpy of formation of NH₃ gas?
Solution.
N_2(g) + 3H_2(g) → 2NH_3(g) ; Δ_rH° = -92.4 kJ mol^-1
∴ Standard enthalpy of formation of NH_3(g)
= 

Question 14.
Calculate the standard enthalpy of formation of CH₃OH_l from the following data :

Solution.

Question 15.
Calculate the enthalpy change for the process
CCl_4(g) → C_(g) + 4Cl_(g)
and calculate bond enthalpy of C – Cl in CCl_4(g).
Δ_vapH°(CCl₄) = 30.5 kJ mol^-1,
Δ_fH°(CCl₄) = -135.5 kJ mol^-1,
Δ_aH°(C) = 715.0 kJ mol^-1, where Δ_aH° is enthalpy of atomisation Δ_aH°(Cl₂) = 242 kJ mol^-1
Solution.

Question 16.
For an isolated system, ΔU = 0, what will be ΔS?
Solution.
When energy factor has no role to play, for the process to be spontaneous ΔS must be +ve i.e., ΔS > 0.

Question 17.
For the reaction at 298 K, 2A + B → C,
ΔH = 400 kJ mol^-1 and ΔS = 0.2 kJ K^-1 mol^-1 At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range.
Solution.
According to Gibbs Helmholtz equation, ΔG = ΔH – TΔS.

Question 18.
For the reaction, 2Cl_(g) → Cl_2(g), what are the signs of ΔH and ΔS?
Solution.
ΔH is negative because bond energy is released and ΔS is negative because there is less randomness among the molecules than among the atoms.

Question 19.
For the reaction
2A_(g) + B_(g) → 2D_(g), ΔU° = -10.5 kJ and ΔS° = – 44.1 JK^-1.
Calculate ΔG° for the reaction, and predict whether the reaction may occur spontaneously.
Solution.
ΔH° = ΔU° + Δn_(g)RT

Question 20.
The equilibrium constant for a reaction is 10. What will be the value of ΔG° ?
R = 8.314J K^-1 mol^-1, T= 300 K.
Solution.
We know that ΔG° = – 2.303RT logK
= – 2.303 x 8.314 x 300 x log10
= – 5744.14 J/mol

Question 21.
Comment on the thermodynamic stability of NO_(g), given

Solution.
Since Δ_(r)H° is +ve, i.e., enthalpy of formation of NO is positive, therefore NO is unstable. But Δ_(r)H° is negative for the formation of NO₂. So, NO₂ is stable.

Question 22.
Calculate the entropy change in surroundings when 1.00 mol of H₂O_l is formed under standard conditions.
Δ_(r)H° = – 286 kJ mol^-1.
Solution.

NCERT Exercises

Question 1.
What will be the minimum pressure required to compress 500 dm³ of air at 1 bar to 200 dm³ at 30°C?
Solution.
According to Boyle’s law, at constant temperature,

Question 2.
A vessel of 120 mL capacity contains a certain amount of gas of 35°C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35°C.What would be its pressure?
Solution.
Since temperature and amount of gas remain constant, therefore, Boyle’s law is applicable.

Question 3.
Using the equation of state PV = nRT; show that at a given temperature density of a gas is proportional to gas pressure P.
Solution.
According to ideal gas equation for ‘n’ moles of a gas,

Question 4.
At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?
Solution.

Question 5.
Pressure of 1 g of an ideal gas A at 27°C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.
Solution.

Question 6.
The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20°C and one bar will be released when 0.15 g of aluminium reacts?
Solution.
The reaction between aluminium and caustic soda is

Question 7.
What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm³ flask at 27°C?
Solution.

Question 8.
What will be the pressure of the gaseous mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1 L vessel at 27°C?
Solution.
Partial pressure of hydrogen gas

Question 9.
Density of a gas is found to be 5.46 g/dm³ at 27°C and at 2 bar pressure. What will be its density at STP?
Solution.
We know that PV = nRT

Question 10.
34.05 mL of phosphorus vapours weigh 0.0625 g at 546°C and 0.1 bar pressure. What is the molar mass of phosphorus?
Solution.
Molar mass of phosphorus, M

Question 11.
A student forgot to add the reaction mixture to the round bottomed flask at 27°C but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477°C. What fraction of air would have been expelled out?
Solution.
Suppose the number of moles of gas present at 27°C in flask of volume V at pressure P is n₁ then assuming ideal gas behaviour,

Question 12.
Calculate the temperature of 4.0 mol of a gas occupying 5 dm³ at 3.32 bar.
(R – 0.083 bar dm³ K^-1 mol^-1)
Solution.

Question 13.
Calculate the total number of electrons present in 1.4 g of dinitrogen gas.
Solution.
Number of moles of dinitrogen

Question 14.
How much time would it take to distribute one Avogadro number of wheat grains, if 10¹⁰ grains are distributed each second?
Solution.
Time taken to distribute 10¹⁰ grains = 1 sec.
Time taken to distribute 6.023 x 10²³ grains

Question 15.
Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm³ at 27°C.
R = 0.083 bar dm³ K^-1 mol^-1
Solution.
Partial pressure of oxygen gas,

Question 16.
Payload is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the payload when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27°C. (Density of air =1.2 kg m^-3 and R = 0.083 bar dm³ K^-1 mol^-1)
Solution.

Question 17.
Calculate the volume occupied by 8.8 g of CO₂ at 31.1°C and 1 bar pressure.
(R = 0.083 bar L K^-1 mol^-1)
Solution.
According to ideal gas equation, PV= nRT

Question 18.
2.9 g of a gas at 95°C occupied the same volume as 0.184 g of dihydrogen at 17°C, at the same pressure. What is the molar mass of the gas?
Solution.
Let molar mass of gas = M

Question 19.
A mixture of dihydrogen and dioxygen at one bar pressurecontains 20% by weight of dihydgrogen. Calculate the partial pressure of dihydrogen.
Solution.
Let the total mass of the mixture be 100 g

Question 20.
What would be the SI unit for the quantity pV²T²/n?
Solution.

Question 21.
In terms of Charles’ law explain why -273°C is the lowest possible temperature.
Solution.
According to Charles’ law,
Volume of gas at 1°C,
 (∵ V₀ = Volume at 0°C)
Volume of gas at -273°C, 
Thus, -273°C is the lowest possible temperature because below this temperature, the volume will become negative, and that is meaningless. This lowest temperature is called absolute zero temperature.

Question 22.
Critical temperature for carbon dioxide and methane are 31.1°C and -81.9°C respectively. Which of these has stronger intermolecular forces and why?
Solution.
Higher the critical temperature, more easily the gas can be liquefied i.e. greater are the intermolecular forces of attraction, CO₂ has stronger intermolecular forces than methane.

Question 23.
Explain the physical significance of van der Waals parameters.
Solution.
van der Waals constant ‘a’ is a measure of the magnitude of the attractive forces among the molecules of a gas, while constant ‘b’ is a measure of effective size of the gas molecules.

NCERT Exercises

Question 4.1
Explain the formation of a chemical bond.
Answer.
The attractive force which holds various constituents (atoms, ions, etc.) together in different chemical species is called a chemical bond. The formation of chemical compounds takes place as a result of combination of atoms in different ways. The atoms interact with each other on account of following reasons :
(i) Lowering of energy of combining atoms : The process of bonding between the atoms decreases the energy of the combining atoms and gives rise to the formation of a system which has lower energy and hence has greater stability.
(ii) Lewis octet rule : Octet theory of valency or electronic theory of valency states that in the formation of a chemical bond, atoms interact with each other by losing, gaining or sharing of electrons so as to acquire a stable outer shell of eight electrons. The tendency of atoms to achieve eight electrons in their outermost shell is known as Lewis octet rule.

Question 4.2
Write Lewis dot symbols for atoms of the following elements:
Mg, Na, B, O, N, Br.
Answer.
Lewis dot symbols :

Question 4.3
Write Lewis symbols for the following atoms and ions:
S and S^2-; Al and Al³⁺; H and H^–
Answer.
Lewis symbols:

Question 4.4
Draw the Lewis structures for the following molecules and ions:
H²S, SiCl⁴, BeF², , HCOOH.
Answer.
Lewis structures :

Question 4.5
Define octet rule. Write its significance and limitations.
Answer.
Octet rule : The tendency of atoms in molecules to have eight electrons in their valence shells (two for hydrogen atom) is known as the octet rule.
Significance :

1. It explains the formation of most of the compounds.
2. It is quite useful for understanding the structures of most of the organic compounds.

Limitations:

1. In some compounds, the number of electrons surrounding the central atom is less than eight e.g., LiCl, BeH₂ and BCl₃.
2. In molecules with an odd number of electron like nitric oxide (NO) and nitrogen dioxide (NO₂), the octet rule is not satisfied for all the atoms.
   
3. In a number of compounds there are more than eight valence electrons around the central atom. This is termed as the expanded octet. Some of the examples of such compounds are PF₅, SF₆, H₂SO₄ and a number of coordination compounds.

Question 4.6
Write the favourable factors for the formation of ionic bond.
Answer.
The following factors facilitate the formation of an ionic bond between a metal and a non-metal:

1. Ionization energy : Lesser the ionization energy, greater is the ease of formation of a cation.
2. Electron affinity : High electron affinity favours formation of an anion.
3. Lattice Energy : It is defined as the amount of energy released when cations and anions are brought close to each other from infinity to their respective equilibrium sites in the crystal lattice to form one mole of the ionic compound. The higher the magnitude of the lattice energy, the greater is the tendency of the formation of an ionic bond.

Question 4.7
Discuss the shape of the following molecules using the VSEPR model: BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S, PH₃.
Answer.
According to VSEPR theory, the shape of molecule depends upon the number of valence shell electron pairs (bonded or non bonded) around the central atom.
(i) BeCl₂: Presence of two bond pairs and lack of lone pairs gives it linear shape. Cl: Be : Cl
(ii) BCl₃ : Presence of three bond pairs and lack of lone pairs gives it trigonal planar geometry.

(iii) SiCl₄ : Presence of four bond pairs and lack of lone pairs gives it tetrahedral geometry.

(iv) AsF₅ : Presence of five bond pairs and lack of lone pairs gives it trigonal bipyramidal geometry.

(v) H₂S : Presence of two bond pairs and two lone pairs gives it angular or bent shape.

(vi) PH₃ : Presence of three bond pairs and one lone pair gives it trigonal pyramidal shape.

Question 4.8
Although geometries of NH3and H20 molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
Answer.

Bond angle in NH3 is 107.5° and in H₂O, angle is 104.5°. In H₂O, oxygen has two lone pairs and in NH₃, nitrogen has only one lone pair. In H₂O, two lone pairs on oxygen repel each other and the bonded pairs more strongly and cause them to come closer thereby reducing the angle.

Question 4.9
How do you express the bond strength in terms of bond order?
Answer.
Greater the bond order, shorter is the bond length and so, higher is the bond strength.

Question 4.10
Define the bond length.
Answer.
The equilibrium distance between the centres of the nuclei of two atoms bonded together is termed as bond length or bond distance.
Bond length in ionic compounds = 
Bond length in covalent compounds (AB) = r_A+ r_B

Question 4.11
Explain the important aspects of resonance with reference to the  ion.
Answer.
Whenever a single Lewis structure cannot describe a molecule accurately, a number of structures with similar energy, positions of nuclei, bonding and non-bonding pairs of electrons, known as the canonical forms of the hybrid are taken, which describe the molecule accurately. This phenomenon is known as resonance.
Resonance hybrid structures of carbonate ion are shown as follows :

1. The relative position of all the atoms in each of the canonical forms is same.
2. The number of unpaired and paired electrons in each of the canonical forms is same.
3. Negative charge is residing on more electronegative atom.
4. As a result of resonance, bond length and bond order is same.

Question 4.12
H₃PO₃ can be represented by structures (i) and(ii) shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H₃PO₃? If not, give reasons for the same.

Answer.
These two cannot be called as canonical forms as relative position of hydrogen is changed. In resonance, canonical forms should differ only in the position of electrons and not in the position of atoms.

Question 4.13
Write the resonance structures for SO₃, NO₂ and .
Answer.
SO₃

NO₂

Question 4.14
Use Lewis symbols to show electron transfer between the following atoms to form cations and anions :
(a) K and S
(b) Ca and O
(c) Al and N.
Answer.

Question 4.15
Although both CO₂ and H₂O are triatomic molecules, the shape of H20 molecule is bent while that of CO₂ is linear. Explain this on the basis of dipole moment.
Answer.
Even though both CO₂ and H₂O are triatomic, but CO₂, has zero dipole moment. It implies that CO, has a symmetrical structure which is possible only when it is linear. However, H₂0 has a net dipole moment suggesting that the molecule is non-linear.

Question 4.16
Write the significance/applications of dipole moment.
Answer.
Applications of dipole moment are :

1. To decide polarity of the molecule : Molecules having zero dipole moment are said to be non-polar molecules and those having µ_R ≠ 0 are polar in nature.
2. To determine percentage of ionic character :
   Percentage of ionic character
   
3. To determine geometry of molecules: The values of dipole moments provide valuable information about the structures of molecules.
   • CO₂, CS₂ are linear as values of their dipole moments are zero.
   • H₂O is non-linear as it has a net dipole moment.

Question 4.17
Define electronegativity. How does it differ from electron gain enthalpy?
Answer.
The relative tendency of a bonded atom to attract the shared electron pair towards itself is called electronegativity while electron gain enthalpy is the energy change that occurs for the process of adding an electron to a gaseous isolated atom to convert it into a negative ion i.e., to form, a monovalent anion. Electron gain enthalpy and electronegativity both measure the power of attracting electrons, but electron gain enthalpy is concerned with an isolated gaseous atom while electronegativity is concerned with the atom in combination.

Question 4.18
Explain with the help of suitable example polar covalent bond.
Answer.
In a hetero-atomic molecule the shared pair of electrons between the two atoms gets displaced more towards the electronegative atom. The resultant covalent bond is polar covalent bond e.g., in H – F the electron pair is displaced towards fluorine due to highly electronegative F atom. Hence H develops a partial +ve charge while F develops a partial
-ve charge .

Question 4.19
Arrange the bonds in order of increasing ionic character in the molecules : LiF, K₂O, N₂, SO₂ and CIF3.
Answer.
As per Fajan’s rule, N₂ < SO₂ < CIF₂ < K₂O < LiF

Question 4.20
The skeletal structure of CH₃COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid

Answer.
In the given figure, hydrogen has valency of two, which is not possible and also octet of oxygen is not complete.
So, the correct structure is

Question 4.21.
Apart from tetrahedral geometry, another possible geometry for CH₄ is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH₄ is not square planar?
Answer.
In CH₄ only bond pairs are there and no lone pairs are present so, to attain most stabilized structure it arranges itself in tetrahedral geometry with greater bond angle of 109° rather than in square planar geometry with lesser bond angle of 90°.

Question 4.22
Explain why BeH₂molecule has a zero dipole moment although the Be—H bonds are polar?
Answer.
The dipole moment in case of BeH₂ is zero. This is because the two equal bond dipoles point in opposite directions and cancel the effect of each other.

Question4. 23
Which out of NH₃ and NF₃ has higher dipole moment and why?
Answer.
Both the molecules NH₃ and NF₃ have pyramidal shape with a lone pair of electron on nitrogen atom. Although fluorine is more electronegative than nitrogen, the resultant dipole moment of NH₃ (4.90 x 10^-30 C m) is greater than that of NF₃ (0.8 x 103 the orbital dipole due to lone pair is in the same direction as the resultant dipole moment of the N—H bonds whereas in NF₃ the orbital dipole is in the direction opposite to the resultant dipole moment of the three N—F bonds. The orbital dipole because of lone pair decreases the effect of the resultant N—F bond moment which results in low dipole moment of NF₃ as represented below :

Question 4.24
What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp², sp³ hybrid orbitals.
Answer.
Hybridisation can be defined as the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of new set of orbitals of equivalent energies and shape.
Salient features of hybridisation : The main features of hybridisation are as under :

1. The number of hybrid orbitals is equal to the number of the atomic orbitals that undergo hybridisation.
2. The hybrid orbitals are always equivalent in energy and shape.
3. The hybrid orbitals are more effective in forming stable bonds than the pure atomic orbitals.
4. These hybrid orbitals are directed in space in some preferred directions to have minimum repulsion between electron pairs and thus have a stable arrangement. Therefore, the type of hybridisation indicates the geometry of the molecules.

Important conditions for hybridisation :

1. Only those orbitals that are present in the valence shell of the atom undergo hybridisation.
2. The orbitals undergoing hybridisation should have almost equal energy.
3. Promotion of electron is not essential condition prior to hybridisation.
4. It is not necessary that only half filled orbitals participate in hybridisation. In some cases, even filled orbitals of valence shell take part in hybridisation.

Shapes of hybrid orbitals :

(i) sp-hybridisation : One ‘s’ and one ‘p’ orbital of an atom intermix giving two sp-hybrid orbitals making an angle of 180° with each other. Each hybrid orbital has 50% s and 50% p-orbital character.
The molecules having such type of hybridisation in their central atom have linear geometry with bond angles 180°. Examples : BeCl₂, C₂H₂, CO₂, HgCl₂, CS₂, N₂O, BeH₂, etc.

(ii) sp²-hybridisation : One ‘s’ and two ‘p’ orbitals of an atom intermix giving three sp2-hybrid orbitals, which are directed towards the corners of an equilateral triangle making an angle of 120° with each other. Each hybrid orbital has 33.3% s and 66.7% p-orbital character. The molecules having such type of hybridisation in their central atom are expected to have triangular planar geometry with bond angles 120°.
Examples : BCl₃, BF₃, SO₃, C₂H₄, Graphite, SO₂, etc.

(iii) sp²-hybridisation : One ‘s’ and three ‘p’ orbitals of an atom intermix giving four sp²-hybrid orbitals, which are directed towards the corners of a regular tetrahedron making an angle of 109°28′. Each hybrid orbital has 25% s- and 75% p-orbital character. The molecules having such type of hybridisation in their central atom are expected to have tetrahedral geometry with bond angles 109°28′.
Examples : CH₄, C₂H₆, CCl₄, H₂O, NH₃, diamond, etc.

Question 4.25
Describe the change in hybridisation (if any) of the Al atom in the following reaction.
Alcl₃ + cl^– → 
Answer.
There is change in hybridization from sp² → sp³. In AlCl₃, Al is sp² hybridised. AlCl₃ thus has three sp² hybrid orbitals and a vacant unhybrid p-orbital. When  is formed, Al undergoes sp³ hybridisation and thus Cl^– overlaps with the fourth orbital.

Question 4.26.
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 is sp² hybridised and has one vacant p-orbital which gets filled by accepting a lone pair of electron present on the N-atom of NH₃. Nitrogen in this adduct acts as donor atom and BF₃ acts as an acceptor. So, hybridization of B in BF₃ changes from sp² to sp³ whereas there is no change in hybridization of N in NH₃ and in the adduct.

Question 4.27
Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C₂H₄ and C₂H₂ molecules.
Answer.
C₂H₄ molecule

C₂H₂ molecule

Question 4.28.
What is the total number of sigma and pi bonds in the following molecules?
(a) C₂H₂
(b) C₂H₄
Answer.

Question 4.29.
Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why?
(a) 1s and 1s
(b) 1s and 2p_x
(c) 2p_y and 2p_y
(d) 1s and 2s
Answer.
A bond formed between two atoms by the overlap of singly occupied orbitals along their axes (end to end overlap) is called c bond. Hence, 1s and 1s, 1s and 2p_x and 1s and 2s will form σ bonds. π bonds are formed by the sidewise or lateral overlapping of p-orbitals. Hence, p_yand p_y orbitals will form π bond.

Question 4.30.
Which hybrid orbitals are used by carbon atoms in the following molecules?
(a) CH₃CH₃
(b) CH₃CH = CH₂
(c) CH₃CH₂OH
(d) CH₃CHO
(e) CH₃COOH
Answer.
(a) sp³, sp³
(b) sp³, sp², sp²
(c) sp³, sp³
(d) sp³, sp²
(e) sp³, sp²

Question 4.31
What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.
Answer.
Bond pairs are the pairs of electrons taking part in bonding e.g.,

A pair of electrons in a molecule which is not shared by any of the two constituent atoms i.e., does not take part in the direct bonding is called a lone pair e.g.,

Question 4.32.
Distinguish between a sigma and a pi bond.
Answer.
Comparison between σ and π bonds

Question 4.33.
Explain the formation of H2 molecule on the basis of Valence Bond Theory.
Answer.

We can consider two hydrogen atoms A and B approaching each other having nuclei N_A and N_A and electrons present in them are represented by e_A and e_A. When the two atoms are at large distance from each other, there is no interaction between them. As these two atoms begin to approach each other, new attractive and repulsive forces begin to operate.
Attractive forces arise between :

1. nucleus of one atom and its own electron that is N_A – e_A and N_B – e_B.
2. nucleus of one atom and electron of other atom i.e., N_A – e_B, N_B – e_A.

Similarly repulsive forces arise between :

1. electrons of two atoms like e_A – e_B.
2. nuclei of two atoms N_A – N_B.
   Attractive forces tend to bring the two atoms close to each other whereas repulsive forces tend to push them apart.
   
3. Experimentally it has been found that the magnitude of new attractive forces is more than the new repulsive forces. As a result, two atoms approach each other and potential energy decreases. Ultimatelv a stage is reached where the net force of attraction balances the force of repulsion and system acquires minimum energy. At this stage two hydrogen atoms are said to be bonded together to form a stable molecule having the bond length of 74 pm.
   Since the energy gets released when the bond is formed between two hydrogen atoms, the hydrogen molecule is more stable than isolated hydrogen atoms. The energy so released is caljed bond enthalpy, which corresponds to the minimum in the curve depicted in fig. Conversely, 435.8 kj of energy is required to dissociate one mole of H2 molecule.
   H_2(g) + 435.8 kj mol^-1 → H_(g) + H_(g)

Question 4.34
Write the important conditions required for the linear combination of atomic orbitals to
form molecular orbitals.
Answer.
Necessary conditions for linear combination of atomic orbitals :

1. The symmetry and relative energies of combining atomic orbitals _A and _B must be close to each other.
2. The extent of overlap between the charge clouds of _A and _B should be large,
3. Atomic orbitals of the same sign should overlap.

Question 4.35
Use molecular orbital theory to explain why Be₂ molecule does not exist.
Solution.
Beryllium atom has electronic configuration 1s² 2s². So, in Be₂ there are a total of eight electrons which must be filled in four molecular orbitals.
M.O. configuration of Be₂ → σ1s²σ*1s² σ2s² σ*2s²
B.O. = 1/2(4 – 4) = 0. Since the bond order is zero it shows that molecular beryllium will not exist.

Question 4.36
Compare the relative stability of the following species and indicate their magnetic properties; O₂,,  (superoxide),  (peroxide).
Answer.

Question 4.37
Write the significance of a plus and a minus sign shown in representing the orbitals.
Answer.
The plus ancj minus sign indicate the sign of the wave function in the various lobes and not that of the nuclear or electronic charges. The ‘+’ sign (or upward crest) represents one phase and thesign represents second phase of the wave function. Bonding molecular orbital is formed by combination of ‘+’ with ‘+’ and with part of the electron waves whereas antibonding molecular orbital is formed by combination of ‘+’ with part of the electron waves.
As orbitals are represented by wave functions, a plus sign in an orbital represents a +ve wave function and a minus sign represents a -ve wave function.

Question 4.38
Describe the hybridisation in case of PCl₅. Why are the axial bonds longer as compared to equatorial bonds?
Answer.
Formation of PCl₅ (sp³d hybridisation):
The ground state and the excited state outer electronic configurations of phosphorus (Z = 15) are represented here.

Now the five orbitals (i.e., one s, three p and one d orbital) are available for hybridisation to yield a set of five sp³d hybrid orbitals which are directed towards the five corners of a trigonal bipyramid as depicted in the figure.

All the bond angles in trigonal bipyramidal geometry are not equivalent. In PCl₅ the five sp³d orbitals of phosphorus overlap with the singly occupied p orbitals of chlorine atoms to form five P-Cl sigma bonds. Three P-Cl bonds lie in one plane and make angle of 120° with each other; these bonds are termed as equatorial bonds. The remaining two P-Cl bonds, one lying above and the other lying below the equatorial plane, make an angle of 90° with the plane. These bonds are called axial bonds. As the axial bond pairs suffer more repulsive interaction from the equatorial bond pairs, therefore axial bonds have been found to be slightly longer and hence slightly weaker than the equatorial bonds.

Question 4.39
Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?
Answer.
Hydrogen bond can be defined as an attractive force which binds hydrogen atom of one molecule with the electronegative atom (F, O or N) of another molecule. Following conditions should be fulfilled by a molecule for the formation of hydrogen bond.
(a) Hydrogen atom should be linked to a highly electronegative atom, such as F, O or N through a covalent bond.
(b) The size of the electronegative atom bonded covalently to the hydrogen atom should be small.
Types of H-bonding:
Intermolecular hydrogen bond : When hydrogen bond is formed between the hydrogen atom of one molecule and the electronegative atom of another molecule of the same compound (such as HF) or different compounds (such as water and alcohol), it is known as intermolecular hydrogen bond.
>
Intermolecular Hydrogen bond : When hydrogen bond is formed between the hydrogen atom and an electronegative atom, present in the same molecule, it is known as intramolecular hydrogen
bond.

H-bonding is stronger than van der Waals forces.

Question 4.40
What is meant by the term bond order?
Calculate the bond order of: N₂,O₂, and 
Answer.
Bond order of a molecule or an ion is a measure of the strength or stability of the bond. Numerically, it is half of the difference between number of electrons in bonding (N_b) and in anti-bonding (N_a) molecular orbitals.
Bond order = 
If B.O. is zero or negative, the bond is highly unstable, i.e. the molecule does not exist. Higher bond order means high bond dissociation energy and greater stability. Greater the bond order, lesser is the bond length.
M.O. configuration of N₂ :

NCERT Exercises

Question 3.1
What is the basic theme of organisation in the periodic table?

Answer.
By 1865, number of identified elements was 63 and with such a large number of elements it was very difficult to study individually the chemistry of all these elements and their innumerable compounds individually. To ease out this problem, scientists searched for a systematic way to organize their knowledge by classifying the elements. Here comes the basis of periodic table. Various elements have now been divided into different groups on the basis of similarities in chemical properties. This has made the study simpler as now the properties of elements are studied in groups rather than individually.

Question 3.2
Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to it?

Answer.
Mendeleev arranged elements in horizontal rows and vertical columns of a table in order of their increasing atomic weights in such a way that the elements with similar properties occupied the same vertical column or group. He realized that some of the elements did not fit in with his scheme of classification if the order of atomic weight was strictly followed. So, he did not stick to his criteria, he ignored the order of atomic weights, thinking that the atomic measurements might be incorrect, and placed the elements with similar properties together.

Question 3.3
What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law?
Answer.
According to Mendeleev’s periodic law, the physical and chemical properties of the elements are periodic functions of their atomic masses, but according to modern periodic law, the physical and chemical properties of the elements are periodic functions of their atomic numbers.

Question 3.4
On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
Answer.
In the sixth period, the orbitals to be filled are 6s, 4f, 5d and 6p. The complete filling of these orbitals require 2 + 14 + 10 + 6 = 32 electrons, hence, the sixth period of the periodic table should have 32 elements.

Question 3.5
In terms of period and group where would you locate the element with Z= 114?
Answer.
The outermost electronic configuration of element (₁₁₄Z) is
[Rn] 5f¹⁴ 6d¹⁰7s²7p². It has n = 7, so period → 7 It belongs to p-block so,
group number = 10 + 4 = 14.

Question 3.6
Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
Answer.
₁₇Cl → It belongs to the third period. So, outermost shell is n = 3. Its configuration is [Ne] 3s²3p⁵.
Therefore, its atomic number = 17.

Question 3.7
Which element do you think would have been named by
(i) Lawrence Berkeley Laboratory
(ii) Seaborg’s group?
Answer.
(i) Lawrencium  →  ₁₀₃Lr
(ii) Seaborgium  →  ₁₀₆Sg

Question 3.8
Why do elements in the same group have similar physical and chemical properties?
Answer.
In a group, the chemical properties of the elements remain nearly the same due to same valence shell configuration.

Question 3.9
What does atomic radius and ionic radius really mean to you?
Answer.
Atomic radius: It is the distance between the centre of the nucleus and outermost shell where electrons are present.
Ionic radius : It is the distance between the nucleus and outermost shell of an ion.

Question 3.10
How do atomic radius vary in a period and in a group? How do you explain the variation?
Answer.
Variation of atomic radii in a period :
As we move from left to right across a period, there is regular decrease in atomic radii of representative elements. This can be explained on the basis of effective nuclear charge which increases gradually in a period, i.c, electron cloud is attracted more strongly towards nucleus as the effective nuclear charge becomes more and more along the period. The increased force of attraction brings contraction in size.

Variation of atomic radii in a group : Atomic radii in a group increase as the atomic number increases. The increase in size is due to extra energy shell which outweighs the effect of increased nuclear charge.

Question 3.11
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
(iii) Mg²⁺
(iv) Rb⁺
Answer.
Isoelectronic species are those which have same number of electrons.
(i) F ^– has 10 electrons. Therefore, the species N³⁺, O^2-, Ne, Na⁺, Mg²⁺, etc., are isoelectronic with F ^–.
(ii) Ar has 18 electrons. Therefore, the species P^3-, S^2-, Cl^–, K⁺, Ca²⁺, etc., are isoelectronic with Ar
(iii) Mg²⁺ has 10 electrons. Therefore, the species N^3-, O^2-, Ne, Na⁺, etc., are isoelectronic with Mg²⁺.
(iv) Rb⁺ has 36 electrons. Therefore, the species Br^–, Kr, Sr²+, etc., are isoelectronic with Rb⁺

Question 3.12
Consider the following species :
N^3-, O^2-, F^–, Na⁺, Mg²⁺ and Al³⁺
(a) What is common in them?
(b) Arrange them in the order of increasing ionic radii.
Answer.
(a) All these are isoelectronic species as they are having same number of electrons i.c., 10.
(b) As Z/e decreases, size increases so, order should be,
N^3- > O^2- > F ^– > Na⁺ > Mg²⁺ > Al³⁺.

Question 3.13
Explain why cations are smaller and anions larger in radii than their parent atoms?
Answer.
A cation is smaller than its parent atom because it has fewer electrons while its nuclear charge remains the same. The size of an anion will be larger than that of the parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge.

Question 3.14
What is the significance of the terms – ‘isolated gaseous atom’and’ground state’while defining the ionization enthalpy and electron gain enthalpy? [Hint: Requirements for comparison purposes.]
Answer.
In the definition of ionization enthalpy and electron gain enthalpy, isolated gaseous atom is required for comparison purposes. Ionization energy is the minimum amount of energy required to remove most loosely bound electron from an isolated atom in the gaseous state of an element so as to convert it into gaseous monovalent positive ion. Electron gain enthalpy is the energy change accompanying the process of adding an electron to a gaseous isolated atom to convert it into a negative ion, i.e., a monovalent anion.

Both the above mentioned processes are carried out on an isolated gaseous atom, which in turn is obtained from either the excitation of a ground state atom (in case the element is monoatomic) or atomisation of polyatomic elements.

The force with which an electron is attracted by the nucleus is appreciably affected by presence of other atoms in the neighbourhood. Since in the gaseous state the atoms are widely separated, therefore these interatomic forces are minimum.
The term ground state means that the atom must be present in the most stable state.

Question 3.15
Energy of an electron in the ground state of the hydrogen atom is -2.18 x 10^-18 J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol^-1. [Hint: Apply the idea of mole concept to derive the answer]
Answer.
Energy of the electron in the ground state of H-atom, E₁ = -2.18 x 10^-18 J
Ionisation energy = E_∞ – E_n
Ionisation enthalpy per mole of atomic hydrogen = (E_∞ – E₁)NA
= [0 – (- 2.18 x 10^-18)] x 6.023 x 10²³
= 2.18 x 6.023 x 10⁵ J/mol = 13.13 x 10⁵ j/mol
= 1.313 x 10⁶ J/mol

Question 3.16
Among the second period elements the actual ionization enthalpies are in the order Li <B <Be< C< 0 < N< F < Ne. Explain why?
(i) Be has higher ∆_iH than B
(ii) O has lower ∆_iH than N and F?
Answer.
(i) An s-electron is attracted to the nucleus more than a p-electron. In beryllium, the electron removed during the ionization is an s-electron whereas the electron removed during ionization of boron is a p-electron. The penetration of a 2s-electron to the nucleus is more than that of a 2p-electron; hence the 2 p electron of boron is more shielded from the nucleus by the inner core of electrons than the 2s electrons of beryllium. Therefore, it is easier to remove the 2p-electron from boron as compared to the removal of a 2s-electron from beryllium. Thus, boron has a smaller first ionization enthalpy than beryllium.
(ii) O has lower ionisation energy than N because N (1s² 2s² ) has extra stable electronic configuration whereas O (1s² 2s² ) does not. O has lower ionisation energy than F because O has larger size than F.

Question 3.17
How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
Answer.
The electronic configurations of Na and Mg are :
Na : 1s² 2s² 2p⁶ 3s¹
Na⁺ : 1s² 2s² 2p⁶
Mg : 1s² 2s² 2p⁶ 3s²
Mg⁺ : 1s² 2s² 2p⁶ 3s¹
Mg²⁺ : 1s² 2s² 2p⁶

The 1st ionization enthalpy of Na is lesser than that of Mg because Mg has an extra stable configuration and smaller size, so, a larger amount of energy would be required to remove an electron from the 3s orbital, which has a pair of electrons.

The 2^nd ionization enthalpy of Na is more than that of Mg because Na⁺ has an extra stable configuration (complete octet), whereas Mg⁺ does not have an extra stable configuration.

Question 3.18
What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
Answer.

We have to consider two factors :
(i) the attraction of electrons towards the nucleus, and (ii) the repulsion of electrons from each other. The effective nuclear charge experienced by a valence electron in an atom will be less than the actual charge on the nucleus because of shielding or screening of the valence electron from the nucleus by the intervening core electrons. As we descend the group, the outermost electron being increasingly farther from the nucleus, there is an increased shielding of the nuclear charge by the electrons in the inner levels. The increase in shielding outweighs the increasing nuclear charge and the removal of the outermost electron requires less energy down a group.

Question 3.19
The first ionization enthalpy values (in kJ mol^-1) of group 13 elements are :
B         Al      Ga    In       Tl
801    577    579   558   589
How would you explain this deviation from the general trend?
Answer.
(i) Al has lower ionization enthalpy than B due to larger size.
(ii) Ga has slightly higher ionization enthalpy than Al due to ineffective shielding by 3d electrons.
(iii) In has lower ionization enthalpy than Ga due to larger size.
(iv) Tl has higher ionization enthalpy than In due to ineffective shielding by 4f electrons.

Question 3.20
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 : F has more negative electron gain enthalpy than O due to smaller size, higher nuclear charge and greater possibility of attaining the nearest stable noble gas configuration by gaining one electron.
(ii) F or Cl : Cl has more negative electron gain enthalpy because in F the incoming electron is added to the smaller n = 2 quantum level and suffers significant repulsion from the other electrons present in this level. In Cl, the added electron goes to n = 3 quantum level and occupies a larger region of space and electron-electron repulsion experienced is far less.

Question 2.21
Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.
Answer.
The second electron gain enthalpy of O is +ve. This is because energy has to be supplied to convert O^–_(g) to O^2-_(g) in order to overcome the repulsive forces.

Question 3.22
What is the basic difference between the terms electron gain enthalpy and electronegativity?
Answer.

Question 3.23
How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?
Answer.
The statement is not correct because electronegativity of an element varies with the state of hybridisation and oxidation state of the element.

Question 3.24
Describe the theory associated with the radius of an atom as it
(a) gains an electron
(b) loses an electron.
Answer.
The distance between the nucleus and the outermost shell of an ion is known as ionic radius.
(a) The gain of an electron leads to the formation of an anion. The radius of the anion is larger than the atomic radius of its parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge which acts on more electrons so, each electron is held less tightly and thereby the electron cloud expands.

(b) The removal of an electron from an atom results in the formation of a cation. A cation is smaller than its parent atom because it has fewer electrons while its nuclear charge remains the same and since it is now acting on lesser number of electrons and pulls them closer, the ion is smaller.

Question 3.25
Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.
Answer.
Two isotopes of the same element have the same first ionization enthalpies because of same effective nuclear charges.
e.g.,  and  have same ionization enthalpies.

Question 3.26
What are the major differences between metals and non-metals?
Answer.

Question 3.27

Use the periodic table to answer the following questions.
(a) Identify an element with five electrons in the outer subshell.
(b) Identify an element that would tend to lose two electrons.
(c) Identify an element that would tend to gain two electrons.
(d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature.
Answer.
(a) Fluorine
(b) Magnesium
(c) Oxygen
(d) Group 17 (Halogens) :

• F, Cl – Non metals and gases
• Br – Non metal and liquid
• I – Shows metallic lustre.

Question 3.28
The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas that among group 17 elements is F > Cl > Br > I. Explain.
Answer.
The trend Li < Na < K < Rb < Cs is observed for chemical reactivity because upon descending the group the ionization energy of alkali metals decreases i.e., it is easy for them to lose an electron from their valence shell and attain the nearest stable noble gas configuration. The trend F > Cl > Br > I is observed for chemical reactivity in halogens because the standard reduction potential decreases as we descend the group. F being the most electronegative readily accepts an e~ to complete its octet and hence the trend.

Question 3.29
Write the general outer electronic configuration of s-, p-, d- and f- block elements.
Answer.
General outer electronic configuration :

Question 3.30
Assign the position of the element having outer electronic configuration
(i) ns² np⁴ for n = 3
(ii) (n – 1)d²ns² for n = 4, and
(iii) (n-2)f ⁷(n-1)d¹ns² for n = 6, in the periodic table.
Answer.

Question 3.31
The first (∆_iH₁) and the second (∆_iH₁) ionization enthalpies (in kJ mol^-1) and the (∆H₁) electron gain enthalpy (in kJ mol-1) 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.
(a) V : The element V has highest first ionization enthalpy and positive electron gain enthalpy hence, it is least reactive.
(b) II : The element II has the least first ionization enthalpy hence, it is most reactive metal.
(c) III; The element III has very high negative electron gain enthalpy hence, it is most reactive non-metal.
(d) IV : Element IV has high negative electron gain enthalpy but ionization energy is not that high hence, it is least reactive non-metal.
(e) VI : The first and second ionization energies of element VI indicate that it can form a stable binary halide.
(f) I : The element 1 has very low value of first ionization energy but very high second ionization energy. Hence, it will form a stable covalent halide of the formula MX.

Question 3.32
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 oxygen
(e) Phosphorus and fluorine
(f) Element 71 and fluorine
Answer.
(a) Li₂O
(b) Mg₃N₂
(c) AlI₃
(d) SiO₂
(e) PF₅
(f) LuF₃

Question 3.33
In the modern periodic table, the period indicates the value of
(a) atomic number
(b) atomic mass
(c) principal quantum number
(d) azimuthal quantum number.
Answer.
(c) Principal quantum number

Question 3.34
Which of the following statements related to the modern periodic table is incorrect?
(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell.
(b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.
(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
(d) The block indicates value of azimuthal quantum number (/) for the last subshell that received electrons in building up the electronic configuration.
Answer.
(b): The d-block has 10 columns, because a maximum of 10 electrons can occupy all the orbitals in a d-subshell.

Question 3.35
Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?
(a) Valence principal quantum number (n)
(b) Nuclear charge (Z)
(c) Nuclear mass
(d) Number of core electrons.
Answer.
(c): Nuclear mass

Question 3.36
The size of isoelectronic species – F^–, Ne and Na⁺ is affected by
(a) nuclear charge (Z)
(b) valence principal quantum number (n)
(c) electron-electron interaction in the outer orbitals
(d) none of the factors because their size is the same.
Answer.
(a): Nuclear charge (Z).

Question 3.37
Which one of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.
Answer.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.

Question 3.38
Considering the elements B, Al, Mg, and K, the correct order of their metallic character is
(a) B > Al > Mg > K
(b) Al > Mg > B > K
(c) Mg > Al > K > B
(d) K > Mg > Al > B
Answer.
(d) K > Mg > Al > B

Question 3.39
Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is
(a) B>C>Si>N>F
(b) Si>C>B>N>F
(c) F>N>C>B>Si
(d) F > N > C > Si > B
Answer.
(c) F>N>C>B>Si

Question 3.40
Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is
(a) F > Cl > O > N
(b) F > O > Cl > N
(c) CI>F>0>N
(d) O > F > N > Cl
Answer.
(b) F > O > Cl > N