Cl(g) + e^- ⟶Cl^-(g)    AH = 348 kJ mol^-1
For one mole (35.5g) 348 kJ is released.
\(\therefore \text { For } 17.5 \mathrm{~g} \text { chlorine, } \frac{348 \mathrm{~kJ}}{35.5 \mathrm{~\not g}} \times 17.75 \mathrm{~\not g} \text { energy leased. }\)
∴ The amount of energy released =\(\frac { 348 }{ 2 } \) =174 kJ

Atomic number: 120
IUPAC temporary symbol: Unbinilium
IUPAC temporary symbol: Ubn
Possible electronic configuration :[Og] 8s²

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.

Lanthanides: [₅₄Xe] 4f^1-14 5d¹ 6s²
Actinides : [₈₆Rn] 5f^0-146d^0-2 7s²

It is defined as the relative tendency of an element present in a covalently bonded molecule, to attract the shared pair of electrons towards itself.

Mg + I.E₁ ⟶ Mg⁺ +1e^-     .....(i)
Mg⁺ + I.E₂ ⟶ Mg²⁺ + 1e^-  ....(ii)
Mg²⁺ + I.E₃ ⟶ Mg³⁺ + 1e^-  ....(iii)
(i) The step (iii) which involves the formation of Mg³⁺ requires higher ionisation energy.
(ii) Mg²⁺ consist of 10 electrons (2, 8) attaining the stable noble gas configuration of argon (Z = 10).
(iii) Since the valence orbital is completely filled, more energy will be required to remove electrons.

The net nuclear charge experienced by valence electrons in the outermost shell is called the effective nuclear charge.
Z_eff=Z-S
Where Z is the atomic number and 'S' is the screening constant.

Ions of different elements having the same number of electrons are called isoelectronic ions.

The modem periodic law states that, "the physical and chemical properties of the elements are periodic functions of their atomic numbers.

X³⁺,Y²⁺, Z^- are isoelectronic.
∴ Effective nuclear charge is in the order
(Z_eff)_Z-< (Z_eff)_y2+< (Z_eff)_X3+ and hence 
ionic radius should be in the order r_z- > r_y²⁺ >r_X³⁺
∴ The correct values are

(i) ₁₃Al =1s², 2s², 2p⁶, 3s², 3p¹
[Third period and p-block]
(ii) ₂₄Cr = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s¹, 3d⁵
[Fourth period and d-block]
(iii) ₂₉Cu = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s¹, 3d¹⁰
[Fourth period and d-block]
(iv) ₁₁Na = 1s², 2s², 2p⁶, 3s¹
[Third period and s-block]

(i) Lithium and beryllium form more covalent compounds, unlike the alkali and alkali earth metals which predominantly form ionic compounds.
(ii) The elements of the second period have only four orbitals (2s & 2p) in the valence shell and have a maximum co-valence of 4, whereas the other members of the subsequent periods have more orbitals in their valence shell and shows higher valences. For example, boron forms BF₄^- and aluminium forms AIF₆^3-.

In the periodic table vertical columns are called groups.
Elements in the same vertical column possess similar number of electrons in the outer orbitals.
∴ Elements a and c belongs to group 18
Elements b and d belongs to group 13

(i) According to aufbau's principle 5th period has nine orbital (one 5s, five 4d and three 6p) to be filled.
(ii) Nine orbitals can accommodate a maximum of 18 electrons. Hence fifth period of the periodic table should has 18 elements from rubidium (2 = 37) to Xenon (Z = 54).

Z = 118; [₈₆Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶
In the periodic table the element with Z = 118 is located in p-block,
Period no = 7 (as n = 7 for valence shell)
Group no. = 18 (group no = 10+ ns electrons + np electrons) (n - outer most shell)

(i) Electronic configuration is the arrangement of electrons in an atom. The outermost electron shell is often referred to as the "valence shell"determines the chemical properties.
(ii) Ionization energy and electron affinity is the amount of energy released or required in pulling out or adding an electron to a neutral atom. So both depend on electronic configuration of the element.

Ionisation energy is the amount of energy required to remove the electron from the ground state (EI) to excited state (E^∞)
E₁=-2.18 x 10^-18 J; E_∞=0
ΔE=E_∞-E₁
=0-(-2.18 x 10^-18 J)=2.18 x 10^-18 J
I.E per hydrogen atom = 2.18 x 10^-18 J
I.E per mole of H-atom =2.18 x 10^-18 J x 6.023 x 10²³
=13.13 x 10⁵ J mol^-1

Electronic Configuration of Aluminium
\(\underbrace { { Al }^{ 13 }{ 1s }^{ 2 } }_{ (n-2) } \underbrace { 2s^{ 2 }{ 2p }^{ 2 } }_{ (n-1) } \underbrace { { 3s }^{ 2 }{ 3p }^{ 1 } }_{ n } \)

∴ Effective nuclear charge = Z - S = 13 - 9.5
(Z_eff)_Al =3.5
Electronic Configuration of chlorine
\(\underbrace { { 1s }^{ 2 } }_{ (n-2) } \underbrace { 2s^{ 2 }{ 2p }^{ 2 } }_{ (n-1) } \underbrace { { 3s }^{ 2 }{ 3p }^{ 5 } }_{ n } \)

∴ Effective nuclear charge = Z - S = 17 - 10.9
(Z_eff)_cl = 6.1
(Z_eff)_cl > (Z_eff)_Al and hence r_clAl

(i) ₄Be - 1s² 2s² ; ₅B - 1s² 2s² 2p¹

• The I.E. of Be is more than that of B though the nuclear charge of boron atom (Z = 5) is greater than that of beryllium atom (Z = 4). This can be explained as follows.
• Boron atom (Z = 5; 1s² 2s² 2P_x¹ 2p_y⁰ 2p_z⁰) is having one unpaired electron in the 2p-subshell. Be-atom (Z = 4; 1s² 2s²) is having paired electrons in the 2s-subshell.
• As the fully filled 2s-subshell in Be-atom is more stable than B-atom due to symmetry, more energy would be needed to remove an electron from Be-atom. Hence, Be has high I. E. Hence I.E of Be > B

(ii) ₇N - 1s² ,2s² ,2p_x¹, 2p_y¹, 2p_z¹
₈O -1s² ,2s² ,2p_x², 2p_y¹, 2p_z¹
∴ O has lower I.E. than N.

(i) Variation of Electronegativity in a period: The electronegativity generally increases across a period from left to right. The atomic radius decreases in a period, as the attraction between the valence electron and the nucleus increases. Hence the tendency to attract shared pair of electrons increases. Therefore, electronegativity also increases in a period.
(ii) Variation of Electronegativity in a group: The electronegativity generally 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, the electronegativity decreases.

Screening effect: 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 shielding effect.
Pauling's scale: Pauling, he assigned arbitrary value of electronegativities for hydrogen and fluorine as 2.2 and 4.0 respectively. Based on this the electronegativity values for other elements can be calculated using the following expression.
\(({ X }_{ A }-{ X }_{ B })=0.182\sqrt { E_{ AB } } -({ E }_{ AA }*{ E }_{ BB })^{ 1/2 }\)
Where E_AB' EAA and E_BB are the bond dissociation energies of AB, A₂ and B₂ molecules 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.

(i) Electron configuration of nitrogen
(Z = 7) 1s² 2s² 2p³.
Electron configuration of oxygen
(2= 8) 1s² 2s² 2p⁴.
Nitrogen has a half filled electronic configuration which is much more stable than an incomplete p-orbital of oxygen which would need to give up one of it's electrons to attain the stability of nitrogen. Hence nitrogen would require more ionization energy to remove an electron from it's outer shell than oxygen.
(ii) Electron configuration of carbon
(Z = 6) 1s²2s²2p².
Electron configuration of Boron
(Z = 5) Is²2s²2p¹
The size of a carbon atom is smaller than boron So the valence electron of carbon has greater nuclear charge than that of boron. Hence the first I.E of carbon is greater than that of boron. However, the second ionization enthalpy of boron is higher than that of carbon. This is because after losing electron, Boron has a fully filled orbital (2s²) than carbon (2p¹). Fully filled orbitals have more stability than partially filled orbitals so greater amount of energy will be needed to remove an electron from boron. So in this case, the second I.E of boron is higher than that of carbon.
(iii) The electron affinities of Be, Mg and noble gases are almost zero because both Be (Z = 4; 1s²2s²) and Mg (Z = 12; Is²2s²2p⁶3s²) are having s orbital fully filled in their valence shell. Fully filled orbitals are most stable due to symmetry. Therefore, these elements would be having least tendency to accept electron. Hence, Be and Mg would be having zero electron affinity. Whereas N (Z = 7; 1s²2s²2px¹2py¹2pz¹ and P (Z = 15) Is² 2s² 2p⁶ 3s² 3p³ is having half filled 2p-subshell. Half filled sub shells are most stable due to symmetry (Hund's rule). Thus, nitrogen and phosphorous are having least tendency to accept electron. Hence, have low electron affinity.
(iv) Fluorine is highly electro negative in nature therefore as it gains the electron its octet become stable and releases the energy so exothermic. while in oxygen the addition of first electron is exothermic in nature but addition of second electron experiences high repulsive force. So needs extra external energy to enter outer shell, hence endothermic in nature.

r(K⁺)+r(Cl^-) = d(K^+-Cl-) = 3.14 Å.
The effective nuclear charge for K+ and CI- can be calculated as follows.
K+ = (1s²)          (2s²2p⁶)        (3s²3p⁶)
        inner shell    (n-1)th shell   n^th shell
Z*(K-) = Z-S
= 19 - [(0.35 x 7) + (0.85 x 8) + (1 x 2)]
= 19 - 11.25 = 7.75
Z*(Cl-) = 17- [(0.35 x 7) + (0.85 x 8) + (1 x 2)]
= 17-11.25 = 5.75
∴ \(\frac { r({ K }^{ + }) }{ r(Cl^{ - }) } =\frac { Z*(Cl^{ - }) }{ Z*(K^{ + }) } =\frac { 5.75 }{ 7.75 } \)=0.74
∴ r(K⁺) = 0.74 r(Cl^-)
Substitute (2) in (1)
0.74 r(Cl^-) + r(Cl^-) = 3.14 Å.
1.74 r(Cl^-) = 3.14 Å
r(Cl^-) = \(\frac { 3.14\overset { 0 }{ A } }{ 1.74 } \)=1.81.Å.

The electronic configuration of Sodium (Z = 11) Is²2s²2p⁶3s¹.
Magnesium (Z = 12) 1s²2s²2p⁶3s²
Magnesium atom has a smaller radius and higher nuclear charge than a sodium atom, thus more energy will be required to remove the electron from the same orbital (3s), making the first ionisation energy of magnesium higher than that of sodium.
However, the second ionization enthalpy of sodium is higher than that of magnesium. This is because after losing 1 electron, sodium attains the stable noble gas configuration of neon (1s²2s²2p⁶). On the other hand, magnesium, after losing 1 electron still has one electron in the 3s-orbital(1s²2s²2p⁶3s¹). In order to attain the stable noble gas configuration, Thus, the energy required to remove the second electron in case of sodium is much higher than that required in case of magnesium. Hence, the second ionization enthalpy of sodium is higher than that of magnesium.

On moving diagonally across the periodic table, the second and third period elements show certain similarities. It is quite pronounced in the following pair of elements.

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

Variation along a period: Ionisation energy usually increases along a period. This is due to increase of nuclear charge and decrease in size as we move from left to right in a period.
Periodic variation in group: Ionisation 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 ionisation energy also decreases down a group.

(i) Ionic radius of uni-univalent crystal can be calculated using Pauling's method from the inter ionic distance between the nuclei of the cation and anion.
(ii) Pauling assumed that ions present in a crystal lattice are perfect spheres, and they are in contact with each other therefore,
d=rC⁺ + r_A^-   ...(1)
Where d is the distance between the centre of the nucleus of cation C+ and anion A-and rC⁺, rA^- are the radius of the cation and anion respectively.
(iii) Pauling also assumed that the radius of the ion having noble gas electronic configuration is inversely proportional to. the effective nuclear charge.
\({ r }_{ C }^{ + }\alpha \frac { 1 }{ ({ Z }_{ eff }){ C }^{ + } } \) ....(2) and
\({ r }_{ A }^{ - }\alpha \frac { 1 }{ ({ Z }_{ eff }){ A }^{ - } } \)...(3)
Where Z_eff is the effective nuclear charge and Z_eff= Z - S
Dividing the equation 1 by 3
\(\frac { { r }_{ C }^{ + } }{ { r }_{ A }^{ - } } =\frac { ({ Z }_{ eff }){ A }^{ - } }{ ({ Z }_{ eff }){ C }^{ + } } \) ...(4)
On solving equation and (1) and (4) the values of r_C+ and r_A^- can be obtained.