Text Book Back Questions and Answers - Chapter 3 Periodic Classification of Elements 11th Chemistry Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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Chapter 3 Periodic Classification of Elements Textual Evaluation Solved

Choose The Correct Answer from The Following

Question 1.
What would be the IUPAC name for an element with atomic number 222?
(a) bibibiium
(b) bididium
(c) didibium
(d) bibibium
Answer:
(d) bibibium

Question 2.
The electronic configuration of the elements A and B are 1s², 2s², 2p⁶, 3s² and 1s², 2s², 2p⁵, respectively. The formula of the ionic compound that can be formed between these elements is ……….
(a) AB
(b) AB₂
(c) A₂B
(d) none of the above.
Answer:
(a) AB₂

Question 3.
The group of elements in which the differentiating electron enters the anti-penultimate shell of atoms are called –
(a) p-block elements
(b) d-block elements
(c) s-block elements
(d) f-block elements
Answer:
(d) f-block elements

Question 4.
In which of the following options the order of arrangement does not agree with the variation of property indicated against it? (NEET 2016 Phase 1)
(a) I < Br < Cl < F (increasing electron gain enthalpy)
(b) Li < Na < K < Rb (increasing metallic radius)
(c) Al³⁺< Mg²⁺< Na⁺ < F^– (increasing ionic size)
(d)  B < C < O < N (increasing first ionization enthalpy)
Answer:
(a) I < Br < Cl < F (increasing electron gain enthalpy)

Question 5.
Which of the following elements will have the highest electro negativity?
(a) Chlorine
(b) Nitrogen
(c) Cesium
(d) Fluorine
Answer:
(d) Fluorine

Question 6.
Various successive ionization enthalpies (in kJ mol^-1) of an element are given below. The element is ………….

(a) phosphorus
(b) sodium
(c) aluminium
(d) silicon table
Answer:
(c) aluminium

Question 7.
In the third period, the first ionization potential is of the order …………..
(a) Na > Al > Mg > Si > P
(b) Na < Al < Mg < Si < P
(c) Mg > Na > Si > P > Al
(d) Na< Al < Mg < Si < P
Answer:
(b) Na < Al < Mg < Si < P

Question 8.
Identify the wrong statement ……………..
(a) Among st the iso electronic species, smaller the positive charge on cation, smaller is the ionic radius
(b) Among-st iso electric species greater the negative charge on the anion, larger is the ionic radius
(c) Atomic radius of the elements increases as one moves down the first group of the periodic table
(d) Atomic radius of the elements decreases as one moves across from left to right in the 2nd period of the periodic table.
Answer:
(a) Among-st the iso electronic species, smaller the positive charge on cation, smaller is the ionic radius

Question 9.
Which one of the following arrangements represent the correct order of least negative to most negative electron gain enthalpy?
(a) Al< O (b) Al < Ca (c) C < F < O < Al < Ca
(d) Ca < Al < C < O < F
Answer:
(d) Ca < Al < C < O < F

Question 10.
The correct order of electron gain enthalpy with negative sign of F, Cl, Br and I having atomic number 9, 17, 35 and 53, respectively is ………..
(a) J > Br > Cl >F
(b) F > Cl > Br >I
(c) Cl > F > Br >I
(d) Br > I > Cl > F
Answer:
(c) Cl > F > Br > I

Question 11.
Which one of the following is the least electro negative element?
(a) Bromine
(b) Chlorine
(c) Iodine
(d) Hydrogen
Answer:
(d) Hydrogen.
Solution:
Hydrogen is the least electro negative element. Since electro negativity increases across the period from left to right. Hydrogen is the first element and it has less electro negativity and down the group electro negativity decreases.

Question 12.
The element with positive electron gain enthalpy is ……….
(a) hydrogen
(b) sodium
(c) argon
(d) fluorine
Answer:
(c) argon
Solution:
Argon has completely filled configuration. So addition of the electron is not possible and has positive electron gain enthalpy.

Question 13.
The correct order of decreasing electro negativity values among the elements X, Y, Z and A with atomic numbers 4, 8, 7 and 12 respectively –
(a) Y > Z > X > A
(b) Z > A > Y > X
(c) X > Y > Z > A
(d) X > Y >A >Z
Answer:
(a) Y > Z > X > A

Question 14.
Assertion: Helium has the highest value of ionization energy among all the elements known.
Reason: Helium has the highest value of electron affinity among all the elements known –
(a) Both assertion and reason are true and reason is correct explanation for the assertion
(b) Both assertion and reason are true but the reason is not the correct explanation for the assertion
(c) Assertion is true and the reason is false
(d) Both assertion and the reason are false
Answer:
(c) Assertion is true and the reason is false

Question 15.
The electronic configuration of the atom having maximum difference in first and second ionization energies is ……….
(a) 1s², 2s², 2p⁶, 3s¹
(b) 1s², 2s², 2p⁶, 3s²
(c) 1s², 2s², 2p⁶, 3s², 3s², 3p⁶, 4s¹
(d) 1s², 2s², 2p⁶, 3s², 3p¹
Answer:
(a) 1s², 2s², 2p⁶, 3s¹

Question 16.
Which of the following is second most electro negative element?
(a) Chlorine
(b) Fluorine
(c) Oxygen
(d) Sulphur
Answer:
(a) Chlorine

Question 17.
IE₁  and IE₂ of Mg are 179 and 348 k cal mol^-1 respectively. The energy required for the reaction
Mg → Mg²⁺ + 2e^– is ……..
(a) +169 kcal mol^-1
(b) -169 kcal mol^-1
(c) +527 kcal mol^-1
(d) -527 kcal mol^-1
Answer:
(c) +527 kcal mol^-1

Question 18.
In a given shell the order of screening effect is …………..
(a) s > p > d > f
(b) s > p > f > d
(c) f > d > p > s
(d) f > p > s > d
Answer:
(a) s > p > d > f

Question 19.
Which of the following orders of ionic radii is correct?
(a) H^– > H⁺ > H
(b) Na⁺ > F“ > O^–
(c) F > O^2- > Na⁺
(d) None of these
Answer:
(d) None of these

Question 20.
The first ionization potential of Na, Mg and Si are 496, 737 and 786 kJ mol^-1 respectively. The ionization potential of Al will be closer to
(a) 760 kJ mol^-1
(b) 575 kJ mol^-1
(c) 801 kJ mol^-1
(d) 419 kJ mol^-1
Answer:
(b) 575 kJ mol^-1

Question 21.
Which one of the following is true about metallic character when we move from left to right in a period and top to bottom in a group?
(a) Decreases in a period and increases along the group
(b) Increases in a period and decreases in a group
(c) Increases both in the period and the group
(d) Decreases both in the period and in the group
Answer:
(a) Decreases in a period and increases along the group

Question 22.
How does electron affinity change when we move from left to right in a period in the periodic table?
(a) Generally increases
(b) Generally decreases
(c) Remains unchanged
(d) First increases and then decreases
Answer:
(a) Generally increases.

Question 23.
Which of the following pairs of elements exhibit diagonal relationship?
(a) Be and Mg
(b) Li and Mg
(c) Be and B
(d) Be and Al
Answer:
(d) Be and Al

 II. Write a brief answer to the following questions.

Question 24.
Define modern periodic law.
Answer:
The physical and chemical properties of the elements are periodic functions of their atomic numbers.

Question 25.
What are isoelectronic ions? Give examples.
Answer:
There are some ions of different elements having the same number of electrons are called isoelectronic ions.
Example:
Na⁺, Mg²⁺, Al³⁺, F^– , O^2- and N^3-

Question 26.
What is effective nuclear charge?
Answer:
The net charge experienced by valence electrons in the outermost shell is called the effective nuclear charge. It is approximated by the equation Z_eff = Z – S, where Z is the atomic number and S is the screening constant which can be calculated using Slater’s rules.

Question 27.
Is the definition given below for ionization enthalpy is correct?
“Ionization enthalpy is defined as the energy required to remove the most loosely bound electron from the valence shell of an atom”
Answer:
No. It is not correct. The accurate and absolute definition is as follows:
Ionization energy is defined as the minimum amount of energy required to remove the most loosely bound electron from the valence shell of the isolated neutral gaseous atom in its ground state.

Question 28.
Magnesium loses electrons successively to form Mg⁺, Mg²⁺ and Mg³⁺ ions. Which step will have the highest ionization energy and why?
Answer:

The third step will have the highest ionization energy. I.E₃>I.E₂>I.E₁
Because from a neutral gaseous atom, the electron removal is easy and less amount of energy is required. But from a di positive cation, there will be more number of protons than the electrons and there is more forces of attraction between the nucleus and electron. So the removal of electron in a di positive cation, becomes highly difficult and more energy is required.

Question 29.
Define electro negativity.
Answer:
Electro negativity is the relative tendency of an element present in a covalently bonded molecule, to attract the shared pair of electrons towards itself.

Question 30.
How would you explain the fact that the second ionization potential is always higher than first ionization potential?
Answer:
The minimum amount of energy required to remove a unipositive cation is called second ionization energy. It is represented by the following equation,
M⁺_(g) + IE₂ – M²⁺_(g) + 1e,
The total number of electrons is less in the cation than the neutral atom while the nuclear charge remains the same. Therefore, the effective nuclear charge of the cation is higher than the corresponding neutral atom. Thus, the successive ionization energies, always increase in the following order I.E₁ < I.E₂. Hence, the second ionization potential is always higher than the first ionization potential.

Question 31.
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 kJ mol^-1.
Answer:
Energy of an electron in the ground state of the hydrogen atom = -2.18 x 10^-18 J
H → H⁺ + e^–
Energy required to ionize 1 mole of hydrogen atoms, we multiply by the Avogadro constant.
E = 2.18 x 10^-18 x 6.023 x 10²³
= 13.123 x 10⁵ J mol^-1
I.E = +1312 K J mol^-1

Question 32.
The electronic configuration of an atom is one of the important factor which affects the value of ionization potential and electron gain enthalpy. Explain.
Answer:
The plot of atomic number vs ionization energy shows that there are two deviations in the trends of ionization energy. It is expected that boron has higher ionization energy than beryllium since it has a higher nuclear charge. However, the actual ionization energies of beryllium and boron are 899 and 800 kJ mol^-1 respectively contrary to the expectation. It is due to the fact that beryllium with completely filled 2s orbital, is more stable than partially filled valence shell electronic configuration of boron. (2s² 2p¹).

Similarly, nitrogen with a 1s² 2s² 2p³ electronic configuration has higher ionization energy (1402 kJ mol^-1) than oxygen (1314 kJ mol^-1). Since the half-filled electronic configuration is more stable, it requires higher energy to remove an electron from 2p orbital of nitrogen. Whereas the removal of one 2p electron from oxygen leads to a stable half-filled configuration. This makes it comparatively easier to remove 2p electron from oxygen.

As we move from alkali metals to halogens in a period, generally electron affinity increases. This is due to an increase in the nuclear charge and a decrease in the size of the atoms. However, in the case of elements such as beryllium (1s² 2s²), nitrogen (1s² 2s² 2p³ ) the addition of extra electrons will disturb their stable electronic configuration and they have almost zero electron affinity.

Noble gases have a stable ns² np⁶ configuration, and the addition of further electrons is unfavourable and requires energy. Halogens having the general electronic configuration of ns² np⁵ readily accept an electron to get the stable noble gas electronic configuration {ns² np⁶) and therefore, in each period the halogen has a high electron affinity.

Question 33.
In what period and group will an element with Z = 118 will be present?
Answer:
The element with atomic number Z = 118 is present in 7^th period and 18^th group.

Question 34.
Justify that the fifth period of the periodic table should have 18 elements on the basis of quantum numbers.
Answer:
In the fifth period, the filling of valence electrons starts with 5s orbital followed by 4d and 6p orbitals. The filling of 4d orbitals starts with Yttribium and ends with cadmium. There are 10 elements present in the second transition series. The period starts with Rubidium (Rb – 5s¹ ) and ends with Xenon (Xe – 5s² 5p⁶).

Question 35.
Elements a, b, c and d have the following electronic configurations:
a : 1s², 2s², 2p⁶
b : 1s², 2s², 2p⁶, 3s², 3p¹
c : 1s², 2s², 2p⁶ 3s²,3p⁶
d : 1s², 2s², 2p¹
Which elements among these will belong to the same group of periodic table?
Answer:

1. In the above elements, Ne and Ar belong to same group (Noble gases – 18th group).
2. Al and B belong to the same group (13^th group).

Question 36.
Give the general electronic configuration of lanthanides and actinides?
Answer:
The general electronic configuration of lanthanides is 4f^1-14 5d^0-1 6s² and for Actinides is 5f^1-14 6d^0-1 7s².

Question 37.
Why halogens act as oxidizing agents?
Answer:
Halogens act as oxidizing agents. Their electronic configuration is ns² np⁵. So all the halogens are ready to gain one electron to attain the nearest inert gas configuration. An oxidizing agent is the one which is ready to gain an electron. So all the halogens act as oxidizing agents. Also halogens are highly electro negative with low dissociation energy and high negative electron gain enthalpies. Therefore, the halogens have a high tendency to gain an electron. Hence they act as oxidizing agents.

Question 38.
Mention any two anomalous properties of second period elements.
Answer:
The anomalous properties of second-period elements are as follows. The ionization energy of Boron is greater than that of Beryllium due to the fact that Be has completely filled 2s orbital which is more stable than the partially filled valence shell electronic configuration of Boron. Similarly, nitrogen with a half-filled electronic configuration has higher ionization energy than oxygen because the half-filled electronic configuration is more stable.

Question 39.
Explain the Pauling’s method for the determination of ionic radius.
Answer:
1. Ionic radius is defined as the distance from the center of the nucleus of the ion up-to which it exerts its influence on the electron cloud of the ion.
2. Ionic radius of uni-univalent crystal can be calculated from the inter-ionic distance between the nuclei of the cation and anion.
3. Pauling assumed that ions present in a crystal lattice are perfect spheres and they are in contact with each other, therefore
d = r_C⁺ + r_A^– ………(1)
Where, d = distance between the center of the nucleus of cation C+ and the anion A-
r_C⁺ = radius of cation
r_A^– = radius of anion.
4. Pauling assumed that the radius of the ion having noble gas configuration (Na⁺ and F^– having 1s², 25², 2p⁶ configuration) is inversely proportional to the effective nuclear charge felt at the periphery of the ion.

Where Z_eff is the effective nuclear charge
Z_eff = Z – S
5. Dividing the equation (2) by (3)

On solving equation (1) and (4), the values of r_C⁺ and r_A^– can be obtained.

Question 40.
Explain the periodic trend of ionization potential.
Answer:
Variation along the period:
The ionization energy usually increases along a period with few exceptions. When we move from left to right along a period, the valence electrons are added to the same shell, at the same time protons arc added to the nucleus. This successive increase of nuclear charge increases the electrostatic attractive force on the valence electron and more energy is required to remove the valence electron resulting in high ionization energy.

Consider the variation in ionization energy of second-period elements. The plot of atomic number vs ionization energy is given below. The plot of atomic number vs ionization energy shows that there are two deviations in the trends of ionization energy. It is expected that boron has higher ionization energy than beryllium since it has a higher nuclear charge.

However, the actual ionization energies of beryllium and boron are 899 and 800 kJ mol^-1 respectively contrary to the expectation. It is due to the fact that beryllium with completely filled 2s orbital, is more stable than partially filled valence shell electronic configuration of boron. (2s², 2p¹).

Similarly, nitrogen with 1s², 2s², 2p³ electronic configuration has higher ionization energy (1402 kJ mol^-1) than oxygen (1314 kJ mol^-1). Since the half-filled electronic configuration is more stable, it requires higher energy to remove an electron from the 2p orbital of nitrogen. Whereas the removal of one 2p electron from oxygen leads to a stable half-filled configuration. This makes it comparatively easier to remove 2p electron from oxygen.

Variation along with the group:
The ionization energy decreases down a group. As we move down a group, the valence electron occupies new shells, the distance between the nucleus and the valence electron increases. So, the nuclear forces of attraction on valence electron decreases, and hence, ionization energy also decreases down a group.

As we move down a group, the number of inner-shell electrons increases which in turn increases the repulsive force exerted by them on the valence electrons, i.e., the increased shielding effect caused by the inner electrons decreases the attractive force acting on the valence electron by the nucleus. Therefore, the ionization energy decreases.

Question 41.
Explain the diagonal relationship.
Answer:

• On moving diagonally across the periodic table, the second and the third period elements show certain similarities.
• Even though the similarity is not same as we see in a group, it is quite pronounced in the following pair of elements.

• The similarity in properties existing between the diagonally placed elements is called “diagonal relationship”.

Question 42.
Why the first ionization enthalpy of sodium is lower than that of magnesium while its second ionization enthalpy is higher than that of magnesium?
Answer:
The first ionization enthalpy of sodium is lower than that of magnesium.
Na(1s², 2s², 2p⁶, 3s¹)+ IE₁ → Na⁺ (1s², 2s², 2p⁶) + e

Mg(1s², 2s², 2p⁶, 3s²) + IE₁ → Mg⁺(1s², 2s², 2p⁶, 3s¹) + e

Magnesium has completely filled 3s orbital (1s², 2s², 2p⁶, 3s²) , is more stable than partially filled valence shell electronic configuration of sodium (1s², 2s², 2p⁶, 3s¹).

Na⁺ (1s², 2s², 2p⁶) + IE₂ → Na²⁺ (1s², 2s², 2p⁵) + e
Mg⁺ (1s², 2s², 2p⁶, 3s¹) + IE₂ → Mg²⁺ (1s², 2s², 2p⁶, 3s²) + e

Na⁺ has completely filled 2p orbital (1s², 2s², 2p⁶), is more stable than partially filled valence shell electronic configuration of Mg⁺ (1s², 2s², 2p⁶). Hence, the second ionization energy of sodium is higher than that of magnesium.

Question 43.
By using Pauling’s method calculate the ionic radii of K⁺ and Cl^– ions in the potassium chloride crystal. Given that dK+−cl− = 3.14 Å
Answer:
Given

We know that,

(Z_eff)_Cl^– = Z – S
= 17 – [(0.35 x 7) + (0.85 x 8) + (1 x 2)]
= 17 – 11.25 = 5.75
(Z_eff)_K⁺ = Z – S
= 19 – [(0.35 x 7) + (0.85 x 8) + (1 x 2)]
= 19 – 11.25 = 7.75

r_(K⁺) = 0.74 r_(Cl^–)
Substitute the value of r_(K⁺) in equation (1)
0.74 r_(Cl^–) + r_(Cl^–) = 3.14 Å
1.74 r_(Cl^–) = 3.14Å1.74 = 1.81 Å.

Question 44.
Explain the following, give appropriate reasons.

1. Ionization potential of N is greater than that of O.
2. First ionization potential of C-atom is greater than that of B-atom, where as the reverse is true for second ionization potential.
3. The electron affinity values of Be, Mg and noble gases are zero and those of N (0.02 eV) and P (0.80 eV) are very low
4. The formation of F^– (g) from F(g) is exothermic while that of O^2-(g) from O (g) is endothermic.

Answer:
1. N (Z = 7) 1s² 2s² 2p_x¹ 12p_y¹ 2p_z¹. It has exactly half-filled electronic configuration and it is more stable. Due to stability, ionization energy of nitrogen is high.
O (Z = 8) 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹. It has incomplete electronic configuration and it requires less ionization energy.
I.E₁ N > I.E₁O

2. C (Z = 6) 1s² 2s² 2p_x¹ 2p_y¹. The electron removal from p orbital is very difficult. So carbon has highest first ionization potential. B (Z = 5) 1s² 2s² 2p¹. In boron nuclear charge is less than that of carbon, so boron has lowest first ionization potential.
I.E₁ C > I.E₁ B
But it is reverse in the case of second ionization energy. Because in case of B+ the electronic configuration is 1s² 2s², which is completely filled and it has high ionization energy. But in C+ the electronic configuration is 1s² 2s² 2p¹, one electron removal is easy so it has low ionization energy.
I.E₂ B > I.E₂ C

3. Be (Z = 4) 1s² 2s²
Mg (Z = 12) 1s² 2s² 2p⁶ 3s²
Noble gases has the electronic configuration of ns² np⁶. All these are completely filled and are more stable. For all these elements Be, Mg and noble gases, addition of electron is unfavorable and so they have zero electron affinity.

Nitrogen (Z = 7) 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹. It has half filled electronic configuration. So addition of electron is unfavorable and it has very low electron affinity value of 0.02 eV. Phosphorus (Z = 15) 1s² 2s² 2p⁶ 3s² 3p_x¹ 3p_y¹ 3p_z¹. It also has half filled electronic configuration. Due to the symmetry and more stability, it has very low electron affinity value of 0.80 eV.

4. F_(g) + e^– → F_(g)^– exothermic
F (Z = 9) 1s² 2s² 2p⁵. It is ready to gain one electron to attain the nearest inert gas configuration. By gaining one electron, energy is released, so it is an exothermic reaction.
O_(g) + 2e^– → O^2-_(g) endothermic
O (Z = 8) 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹. It is the small atom with high electron density. The first electron affinity is negative because energy is released in the process of adding one electron to the neutral oxygen atom. Second electron affinity is always endothermic (positive) because the electron is added to an ion which is already negative, therefore it must overcome the repulsion.

Question 45.
What is screening effect? Briefly give the basis for Pauling’s scale of electro negativity. Screening effect:
Answer:
In addition to the electrostatic forces of attraction between the nucleus and the electrons, there exists repulsive forces among the electrons. The repulsive force between the inner shell electrons and the valence electrons leads to a decrease in the electrostatic attractive forces acting on the valence electrons by the nucleus. Thus, the inner shell electrons act as a shield between the nucleus and the valence electrons. This effect is called the shielding effect.

Electronegativity is defined as the relative tendency of an element present in a covalently bonded molecule, to attract the shared pair of electrons towards itself. Electronegativity is not a measurable quantity. In the Pauling scale, an arbitrary values of electronegativities for hydrogen and fluorine are assigned as 2.2 and 4.0 respectively. Based on this, the electronegativity values for other elements can be calculated using the following expression,
(χ_A – χ_B) = 0.182 EAB−−−−√ – (E_AA × E_BB)^1/2
where E_AB, E_AA, and E_BB are the bond dissociation energies of AB, A_2, and B₂molecules respectively.

The electronegativity of any given element is not a constant and its value depends on the element to which it is covalently bound. The electronegativity values play an important role in predicting the nature of the bond.

Question 46.
State the trends in the variation of electro negativity in period and group.
Answer:
Variation of electron negativity in a period:
The electro negativity increases across a period from left to right. Since the atomic radius decreases in a period, the attraction between the valence electron and the nucleus increases. Hence the tendency to attract shared pair of electrons increases. Therefore, electro negativity increases in a period.

Variation of electro negativity in a group:
The electro negativity decreases down a group. As we move down a group, the atomic radius increases and the nuclear attractive force on the valence electron decreases. Hence electro negativity decreases in a group.

Text Question – Evaluate your self

Question 1.
What is the basic difference in approach between Mendeleev’s periodic table and modern periodic table?
Answer:
The main basic difference between Mendeleev’s periodic table and modem periodic table is that first one is constructed on the basis of atomic weight and the later is constructed on the basis of atomic number.

Question 2.
The element with atomic number 120 has not been discovered so far. What would be the IUPAC name and the symbol for this element? Predict the possible electronic configuration of this element.
Answer:
Atomic number : 120
IUPAC temporary symbol : Unbinilium
IUPAC temporary symbol : Ubn
Possible electronic configuration : [Og] 8s²

Question 3.
Predict the position of the element in periodic table satisfying the electronic configuration (n – 1 )d² ns² where n = 5?
Answer:
Electronic Configuration : (n – 1 )d² ns²
for n = 5, the electronic configuration is,
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d² 5s²
Atomic number : 40
4th group 5th period (d block element) = Zirconium

Question 4.
Using Slater’s rule calculate the effective nuclear charge on a 3p electron in aluminium and chlorine. Explain how these results relate to the atomic radii of the two atoms.
Answer:
Electronic Configuration of Aluminium

Effective nuclear charge = Z – S = 13 – 9.5
(Z_eff)_Al = 3.5
Electronic Configuration of chlorine

Effective nuclear charge = Z- S = 17 – 10.9
(Z_eff)_Cl = 6.1
(Z_eff)_Cl > (Z_eff)_Cl and hence r_Cl< r_Al

Question 5.
A student reported the ionic radii of iso electronic species X³⁺ , Y²⁺ and Z^– as 136 pm, 64 pm and 49 pm respectively. Is that order correct? Comment.
Answer:
X³⁺, Y²⁺, Z^– are iso electronic.
∴ Effective nuclear charge is in the order
(Z_eff)_Cl^– < (Z_eff)Y_Y²⁺ < (Z_eff)_X³⁺ and hcnce, ionic radii should be in the order r_Z^– > r_Y²⁺ > r_X³⁺
∴ The correct values are:

Question 6.
The first ionisation energy (IE₁) and second ionisation energy (IE₂) of elements X, Y and Z are given below.

Which one of the above elements is the most reactive metal, the least reactive metal and a noble gas?
Answer:
Noble gases:
Ionisation energy ranging from 2372 KJmol^-1 to 1037 kJ mol^-1. For element X, the IE₁ value is in the range of noble gas, moreover for this element both IE₁ and IE₂ are higher and hence X is the noble gas. For Y, the first ionisation energy is low and second ionisation energy is very high and hence Y is most reactive metal.
For Z, both IE₁ and IE₂ are higher and hence it is least reactive.

Question 7.
The electron gain enthalpy of chlorine is 348 kJ mol^-1. How much energy in kJ is released when 17.5 g of chlorine is completely converted into Cl^– ions in the gaseous state?
Cl_(g)+ e^– → Cl^–_(g)
∆H = 348 kJ mol^-1
For one mole (35.5g) 348 kJ is released.
∴ For 17.5g chlorine,  energy leased.
∴ The amount of energy released = 3482 = 174 kJ.