As we move down from top to bottom in a group, electronegativity decreases due to increased atomic radius. Fluorine has the highest value of electronegativity among all the elements.

Noble gases: Ioniation 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.

It is defined as the amount of energy released (required in the case noble gases) when an electron is added to the valence shell of an isolated neutral gaseous atom in its ground state to form its anion. It is expressed in kJ mol^-1.
A  + e^- \(\rightarrow\)A^- + E_A

Galium and Germanium

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.

Transition metals form a bridge between the chemically active metals of s-block and the less active elements of groups of 13th and 14 and thus are called as Transition elements.

In periodic table the elements placed beyond uranium (Z = 92) are all synthetic elements and are called as transuranium elements.

(i) Sodium (Z = 11)
(ii) krypton (Z = 36)
(iii) Bromine (Z = 35)
(iv) Scandium (Z = 21)

(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-.

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) 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.

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.