The given nom-metals are arranged in the increasing order of non-metallic character as follow

The ionic radii of isoelectronic species decreases with increase in atomic number (as magnitude of the nuclear charge increase with increase in atomic number) 
Therefore, their ionic radii increase in the order.\(\underset { z=13 }{ { Al }^{ 3+ } } <\underset { 12 }{ { Mg }^{ 2+ } } <\underset { 11 }{ { Na }^{ + } } <\underset { 9 }{ { F }^{ - } } <\underset { 8 }{ { O }^{ 2 } } <\underset { 7 }{ { N }^{ 3- } } \)

Since noble gases have a stable electronic configuration (\(ns^{ 2 },np^{ 6 \ }and \ 1s^{ 2 } \ in \ case \ of \ He\)), they have no tendency to attract the bond pair to themselves.
Therefore, their electronegativity is zero. on the other hand, halogens are only one electron short of noble gas configuration. Hence, they have a very high tendency of attracting electron towards itself.
Therefore, their electronegativity values are the highest in the respective periods.

The ionisation enthalpy of the main group elements decreases regularly on moving down the group due to the following two factors.
(i) Atomic size On moving down the group,atomic size increases due to the addition of new higher energy shell.As a result of this, forces of attraction of nucleus for valence electrons decrease and ionisation enthalpy also decreases.
(ii) Screening effect On moving down the group, screening effect or shielding effect increases, so ionisation enthalpy decreases (because forces of attraction between nucleus and electron secreases).

Cations are always smaller in radii than their parent atoms because by the loss of one or two electrons, effective nuclear charge increases. Due to this, forces of attraction of nucleus for electrons increases and hence, ionic radii decreases. On the other hand, anions are always larger in radii than their parent atoms because by the addition of one or two electrons effective nuclear charge decreases.
Due to this, forces of attraction between nucleus and valence shell electrons decreases and hence, ionic radii of anion increases.

Value on Mulliken's scale = 2.8 x 4 = 11.2.

They have fully filled s-orbitals and hence have no tendency to accept an additional electron. Consequently, energy has to be supplied if an extra electron has to be added to the much higher energy p-orbitals of the valence shell. That is why electron gain enthalpies of Be and Mg are positive.

Cations : Na⁺, Mg²⁺ Al³⁺
Anions : N^3- , O^2-,F^-

Se and Te belong to 16 group and fourth and fifth period respectively whereas Br and I belong to 17 group and fourth and fifth period respectively. In a period ionisation energy increase but along a group, it decreases. Thus, the order of ionisation enthalpy is Te < I < Se < Br. i.e. Te has IE of 869 and Se has IE of 1142 kJ mol^-1.

The outer configuration of the element is 3s² 3p² (as it has 4 elements in outermost shell). Thus, the complete configuration is 1s², 2s² 2p ⁶,3s²,3p². So the atomic number is 2 + 8 + 4 = 14. Hence, the element is silicon.

Similarity in outer electronic configuration and gradual addition of an electron into the successive elements,are the most important cause of periodicity.

Mendeleev's periodic law : It states that the properties of the elements are a periodic function of their atomic weights
Modern periodic law : It states that the properties of the elements are a periodic function of their atomic numbers. Thus, change in the base of classification of elements from atomic weight to atomic number is the basic difference between Mendeleev's periodic law and the modern periodic law.

The basic theme of organisation in the periodic table is to simplify and systematise the study of physical and chemical properties of all the elements and their innumerable compounds.

In p -block, when we move from left to right in a period, the acidic character of the oxides increases due to increase in electronegativity. e.g.
( i ) 2nd period
B₂O₃ < CO₂ < N₂O₃ acidic character increases.
( ii ) 3rd period
Al₂O₃ < SiO₂ < P₄O₁₀ < SO₃ < Cl₂O₇ acidic character increases.
on  moving down the group, acidic character decreases and basic character increaseas.e.g.
(a) Nature of oxides of 13 group elements 

Nature of oxides of 15 group elements
N₂O₅   P₄O₁₀    As₄O₁₀    Sb₄O₁₀   Bi₂O3
Strongly acidic   Moderately acidic  Amphoteric  Amphoteric Basic
Among the oxides of same element, higher the oxidation state of the element, stronger is the acid. e.g. SO₃ is a stronger is the acid than SO₂.
B₂O₃ is weakly acidic and on dissolution in water, ti forms orthoboric acid. Orthoboric acid does not act as a protonic acid ( it does not ionise ) but acts as a weak Lewis acid.
B₂O₃  + 3H₂O  \(\rightleftharpoons\)  2H₃BO₃
Boron trioxide                       Orthoboric acid
B ( OH )₃  +  H-----OH \(\longrightarrow\) [ B ( OH )₄ ]^- + H⁺
Al₂O₃ is amphoteric in nature. It is insoluble in water bur dissolves in alkalies and react with acids.
Al₂O₃  + 2NaOH \(\overset { \triangle }{ \longrightarrow } \) 2NaAlO₂       +              H₂O
Aluminiun trioxide                     Sodium meta ailuminate
Al₂O₃ + 6HCl \(\overset { \triangle }{ \longrightarrow }\) 2AlCl₃  + 3H₂O
Aluminium chloride
Tl₂O is as basic as NaOH due to its lower oxidation state ( +1 )
Tl₂O + 2HCl \(\longrightarrow \) 2TlCl + H₂O
P₄O₁₀ on reaction with water gives orthophosphoric acid.
P₄O10  +  6H₂O  \(\longrightarrow\) 4H₃PO₄
Phosphorus pentaxide                   Orthophosphoric acid
Cl₂O₇ is strongly acidic in nature and on dissolution in water, it gives perchloric acid.
Cl₂O₇  +  H₂O \(\longrightarrow\) 2HClO₄
Dichlorine heptoxide       Perchloric acid

(a) Element V is likely to be the least reactive element. This is because it has the highest first ionization enthalpy (Δ_iH₁) and a positive electron gain enthalpy (Δ_egH).
(b) Element II is likely to be the most reactive metal as it has the lowest first ionization enthalpy (Δ_iH₁) and a low negative electron gain enthalpy (Δ_egH).
(c) Element III is likely to be the most reactive non–metal as it has a high first ionization enthalpy (Δ_iH₁) and the highest negative electron gain enthalpy (Δ_egH).
(d) Element V is likely to be the least reactive non–metal since it has a very high first ionization enthalpy (Δ_iH₂) and a positive electron gain enthalpy (Δ_egH).
(e) Element VI has a low negative electron gain enthalpy (Δ_egH). Thus, it is a metal. Further, it has the lowest second ionization enthalpy (Δ_iH₂). Hence, it can form a stable binary halide of the formula MX₂ (X=halogen).
(f) Element V has the highest first ionization energy and high second ionization energy. Therefore, it can form a predominantly stable covalent halide of the formula MX (X=halogen).

(i) Potassium, K (Z = 19)
(ii) Strontium, Sr (Z = 38)
(iii) Ruthenium, Ru (Z = 44)
(iv) Praseodymium, Pr (Z = 59 ) 
(v) Xenon, Xe(Z = 54 )

Non-metallic character across a period (left to right) increase but on moving down the group it decreases. So, the increasing order of non0metalic character is P.

Electronic on configuration of \(_{ 7 }N\) = \({ 1s }^{ 2 },{ 2s }^{ 2 },{ 2p }_{ x }^{ 1 },{ 2p }_{ y }^{ 1 },{ 2p }_{ z }^{ 1 }\)
Nitrogen has stable configuration because p-orbital is half-filled. Therefore, addition of extra electron to any of the p-orbital; requires energy.
Electronic configuration of \(_{ 8 }O\) = \({ 1s }^{ 2 },{ 2s }^{ 2 },{ 2p }_{ x }^{ 2 },{ 2p }_{ y }^{ 1 },{ 2p }_{ z }^{ 1 }\)
Oxygen has  \(2p^{ 4 }\)electrons, so process of adding an electron to the p-orbital is exothermic.
Oxygen has lower ionisation enthalpy than nitrogen because by removing one electron from 2p-orbital, oxygen acquires stable configuration, i.e., \(2p^{ 3 }\). On the other hand, in case of nitrogen it is not easy to remove one of the three 2p-electrons due to its stable configuration.

(i) ns² np⁴ for n = 3
n = 3 means element belongs to third period. Since, last electron enters in the p-orbital, it belongs to p-block. For p-block elements, the group number = 10+valence shell electrons = 10 + (2 + 4) = 16.
(ii) (n - 1) d² ns² for n = 4
n = 4 means the element belongs to fourth period. Since, last electorn enters in d-orbital, the given element belongs to d-block. For d-block elements, group number = number of d-electrons + number of ns electrons = 2 + 2 = 4
(iii) (n - 2) f⁷ (n-1) d¹ ns² for n = 6
n = 6 means the element belongs to sixth period. Since, last electron enters in f-orbital, the given elements are the members of third group. Hence, the element belongs to third group.