Classification of Elements and Periodicity in Properties

In the modern periodic table, the physical and chemical properties are periodic functions of the atomic number of elements. In the modern periodic table, there are 18 groups and 7 periods.

At present, there are 118 elements in the modern periodic table. Out of 118 elements the No. of metals=89, No. of non-metals =22, & metalloids=7.

The properties of elements are function of atomic number ( Z ).
Elements with similar outer electronic configurations are placed in the same group, not the same period.
The number of elements in a period is related to the number of available electrons in subshells, not just orbitals.

Atomic radius is directly related to the size of the atom and the number of shells. it is inversely proportional to the effective nuclear charge. When we move down the group atomic radius increases and across the period atomic radius decreases. Hence, the incorrect order of radii Be >Mg > Al >Si

Atomic radii of Beryllium is smaller than Magnesium.

According to Moseley’s Law, √ν (square root of frequency) vs atomic number (Z) gives a straight line, not frequency (ν) vs Z. Also, √ν vs atomic mass is not linear due to irregular mass variations.

Higher electron gain enthalpy (more negative) means a stronger attraction for electrons, increasing ionic character. Since B (–349 kJ/mol) has the most negative electron gain enthalpy, it forms the most ionic bond with E . The decreasing order of electron gain enthalpy is:
B (–349) > A (–328) > C (–325) > D (–295)
This means EB has the most ionic character, followed by EA, EC, and ED.

The radius of a cation will be smaller than that of the anion as a cation will have a greater positive charge (i.e. number of protons) so it will attract the electrons in the outermost orbital with greater force and hence the smaller size. The more negative charge on the anion more its ionic radius and more charge on the cation , the smaller its ionic radius. Hence, the correct order of ionic radius will be ${\mathrm{O}}^{2-}>{\mathrm{F}}^{-}>{\mathrm{Na}}^{+}>{\mathrm{Mg}}^{2+}$

Down the group, the value of electron gain enthalpy becomes less negative and it should follow the order F>Cl>Br>I
However due to the extremely small size of F, the electron gain enthalpy of F is less negative than Cl. The incoming electron feels repulsion to the electrons of F.
Therefore, the correct order will be F<Cl>Br>I

There is a large jump from 3rd ionization energy to 4th, so, it can be concluded that an element must contain 3 valence electrons, that is it belongs to group 13.

A (Ar) → least negative (or positive)

D (S) → more negative than Ar

B (Br) → more negative than S

C (F) → less negative than Br, due to small size

Ionic radius decreases with increasing nuclear charge for isoelectronic species. Given N³⁻, O²⁻, F⁻, nitrogen has the lowest nuclear charge, so it has the largest radius. Correct order: N³⁻ (1.71 Å) > O²⁻ (1.40 Å) > F⁻ (1.36 Å). Answer: (A).

Boron (B) has the smallest atomic radius among the given elements and forms an oxide of type $B_2{O}_3$, which corresponds to the general formula $A_2{O}_3$.

The size of the anion > the size of the cation

Hence, ${\mathrm{F}}^{-}$ has largest size among all

Electonegativity values acc. to Pauling Scale:

Correct order :

Mg < Al < B < N

Isoelectronic species have the same number of electrons but different nuclear charges. The greater the nuclear charge (Z), the stronger the attraction on electrons, making the radius smaller.
The order of electron gain enthalpy (EGE) for halogens is:

$Cl>F>Br>I$

Palladium -5th period
Iridium, Osmium, Platinum - 6th Period

Be and N having stable electronic configuration. and second EA of neutral element is alwayas negative

The electronegativity values are:

Al ≈ 1.5
Mg ≈ 1.2
B ≈ 2.0
N ≈ 3.0
The order is incorrect because Mg has a lower electronegativity than Al

\(N _2< SO _2< ClF _3< K _2 O < LiF\)

The electronegativity of halogen decreases down the group :

Hence, the correct order is F>Cl>Br>I

The electronegativity of halogen decreases down the group :

Hence, the correct order is F>Cl>Br>I

Elements on extreme left form basic oxides whereas elements on extreme right form acidic oxides.

$$\begin{gathered} {\mathrm{Na}}_2\mathrm{O} + {\mathrm{H}}_2\mathrm{O} \to 2\mathrm{NaOH} \\ {\mathrm{Cl}}_2{\mathrm{O}}_7 + {\mathrm{H}}_2\mathrm{O} \to 2{\mathrm{HClO}}_4 \end{gathered}$$

In the modern periodic table, the physical and chemical properties are periodic functions of the atomic number of elements. In the modern periodic table, there are 18 groups and 7 periods.

At present, there are 118 elements in the modern periodic table. Out of 118 elements the No. of metals=89, No. of non-metals =22, & metalloids=7.

Ionization enthalpy increases across a period (left to right).

Half-filled and full-filled orbitals are more stable → higher ionization energy.

Nitrogen (N) has a half-filled 2p³ configuration → higher IE than Oxygen.

Boron (B) and Be (Beryllium): Be has a filled 2s², so higher IE than B.

Fluorine (F) has the highest IE in this group.

Let’s arrange based on actual ionization energy:

Correct order:
B < Be < O < N < F

Now match this with options:

E (B) < C (Be) < A (O) < B (N) < D (F)

O = 8

F = 9

Na = 11

Mg = 12
Different Z, so this is incorrect

Yes. More electrons than protons = more electron-electron repulsion = larger size → Correct

Yes. Least electrons + highest nuclear charge = strongly pulled inward → Correct

For the isoelectronic species F⁻, Ne, Na⁺, all have 10 electrons.

So, their size cannot depend on:

• (A) Principal quantum number (n) → same n

• (B) Ionic radius → this is what we are trying to compare, not the cause

• (C) Electron–electron interaction → same number of electrons → nearly same e–e repulsion

The only factor that changes is:

More nuclear charge pulls electrons inward → smaller radius.

Order of Z:
F (Z=9) < Ne (Z=10) < Na (Z=11)

Thus radius order:
F⁻ > Ne > Na⁺

Which matches the observed trend.

Ionization Enthalpy depends on:

Atomic size

Nuclear charge

Electronic configuration (stability)

The size of the anion > the size of the cation

Hence, ${\mathrm{F}}^{-}$ has largest size among all