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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.

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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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