O has smallest radius because it has highest effective nuclear charge because 8 protons attract 8 electrons.
In O^-, 8 protons attract 9 electrons less srongly.
In O^2-, 8 protons attract 10 electrons with least force of atraction among these.

Principal quantum number tells the principal energy level or shell to which the electron belongs. It gives the information about the distance and the energy of the electron.

In a multielectron atom, the electrons present in the inner shells shield the electrons in the valence shell from the attraction of the nucleus or they act as a screen between the nucleus and these electrons. This is known as Sheil ding effect or screening effect. As the screening effect increases, the effective nuclear charge decreases. Consequently, the force of attraction by the nucleus for the valence shell electrons decreases and hence the ionization enthalpy decreases.

Ionization Enthalpy. The minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom so as to convert it into a gaseous cation is called its ionization enthalpy or energy. It is represented by\(\triangle \)_i H. This process may be represented as
M(g) +\(\triangle \)_i H⟶M⁺(g) + e^- (g)
where M (g) is isolated gaseous atom. M⁺ (g) is the resultant cation (a position ion) Variation along a period. Moving from left to right in a period, the ionization enthalpy increases with atomic number.

(i) \(_{ 6 }^{ 13 }{ c }\) ; Atomic no. (Z) = 6 Mass no. (A) = 13
No. of protons (P) = 6 No. of neutrons (n) = 13 - 6 = 7
(ii) ​​​​​​\(_{ 8 }^{ 16 }{ O }\) ; Atomic no. (Z) = 8 Mass no. (A) 16
No. of protons (P) = 8 No. of neutrons (n) = 16 - 8 = 8
(iii) \(_{ 12 }^{ 24 }{ Mg }\) Atomic no. (Z) = 12 Mass no. (A) = 24
No. of protons (P) = 12 No. of neutrons (n) = 24 -12 = 12
(iv) \(_{ 26 }^{ 56 }{ Fe }\) ;Atomic no (Z) = 26 Mass no. (A) = 56
No. of protons (P) = 26 No. of neutrons (n) = 56 - 26 = 30
(v) \(_{ 38 }^{ 88 }{ Sr }\) ;Atomic no (Z) = 38 Mass no. (A) = 88
No. of protons (P) = 38 No. of neutrons (n) = 50.

(i) During the process of ionization, the electron to be removed from beryllium atom is a 2s-electron, whereas the electron to be removed from boron atom is a 2p-electron. Now, 2s-electrons are more strongly attached to the nucleus than 2p-electrons. Therefore, more energy is required to remove a 2s-electron of beryllium than that required to remove a 2p-electron of boron. Hence, beryllium has higher Δ_iH than boron.
(ii) In nitrogen, the three 2p-electrons of nitrogen occupy three different atomic orbitals. However, in oxygen, two of the four 2p-electrons of oxygen occupy the same 2p-orbital. This results in increased electron-electron repulsion in oxygen atom. As a result, the energy required to remove the fourth 2p-electron from oxygen is less as compared to the energy required to remove one of the three 2p-electrons from nitrogen. Hence, oxygen has lower Δ_iH than nitrogen.
Fluorine contains one electron and one proton more than oxygen. As the electron is being added to the same shell, the increase in nuclear attraction (due to the addition of a proton) is more than the increase in electronic repulsion (due to the addition of an electron). Therefore, the valence electrons in fluorine atom experience a more effective nuclear charge than that experienced by the electrons present in oxygen. As a result, more energy is required to remove an electron from fluorine atom than that required to remove an electron from oxygen atom. Hence, oxygen has lower Δ_iH than fluorine.

According to Rutherford's model of atom, the positive charge is concentrated at the centre, that is, at the nucleus. Since, a small fraction of -particles were deflected through small angle and also a few of them bounced back, it proved that the nucleus occupies a very small volume in the atom. The radius of the atom is about 10^-10m while of nucleus is 10^-15 m. The nucleus is composed of protons and neutrons and is surrounded by revolving electrons.

(a) The given set of quantum numbers is not possible because the value of the principal quantum number (n) cannot be zero.
(b) The given set of quantum numbers is possible.
(c) The given set of quantum numbers is not possible. For a given value of n, ‘l’ can have values from zero to (n – 1). For n = 1, l = 0 and not 1.
(d) The given set of quantum numbers is possible.
(e) The given set of quantum numbers is not possible. For n = 3, l = 0 to (3 – 1) l = 0 to 2 i.e., 0, 1, 2
(f) The given set of quantum numbers is possible.

For n = 3, the possible values of l are 0, 1 and 2. Thus there is one 3s orbital (n = 3, l = 0 and ml = 0); there are three 3p orbitals (n = 3, l = 1 and ml = –1, 0, +1); there are five 3d orbitals (n = 3, l = 2 and ml = –2, –1, 0, +1+, +2).
Therefore, the total number of orbitals is 1+3+5 = 9
The same value can also be obtained by using the relation; number of orbitals = n², i.e. 3² = 9

Isoelectronic species have the same number of electrons but different atomic numbers. Number of positive charge shows the number of electrons lost and number of negative charges shows the number of electrons gained by an atom. Calculation of number of electrons have been shown below.
\(_{ 11 }Na^{ + }=11-1=10{ e }^{ - }, \ _{ 19 }K^{ + }=19-1=18{ e }^{ - },\)
\(_{ 12 }Na^{ 2+ }=12-2=10{ e }^{ - }, \ _{ 20 }Na^{ 2+ }=20-2=18{ e }^{ - },\)
\(_{ 16 }S^{ 2- }=16+2=18{ e }^{ - }, \ _{ 18 }Ar=18{ e }^{ - }\)
Hence, isoelectronic species are
\(Na^{ + }and \ Mg^{ 2+ } \ { k }^{ + },{ Ca }^{ 2+ },{ s }^{ 2- } \ and \ Ar\)

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).

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.

(a) n = 1, l = 0 (Given)
The orbital is 1s.
(b) For n = 3 and l = 1
The orbital is 3p.
(c) For n = 4 and l = 2
The orbital is 4d.
(d) For n = 4 and l = 3
The orbital is 4f.

(a) ₁₅P = 1s², 2s², 2p⁶, 3s², 3p³. 3 unpaired electrons.
(b) ₁₄Si = 1s², 2s², 2p⁶, 3s², 3p². 2 unpaired electrons.
(c) ₁₄Cr = 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁵, 4s¹. 6 unpaired electrons.
(d) ₂₆Fe = 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁶, 4s². 4 unpaired electrons
(e) ₃₆Kr = 1s², 2s², 2p⁶, 3s², 3p⁶, 3d¹⁰, 4s², 4p⁶. No unpaired electrons.

For g-orbitals, l = 4.
As for any value ‘n’ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n – 1).
∴For l = 4, minimum value of n = 5

In this case, \(\overset { 80 }{ 35 } \) Br, Z = 35, A = 80, species is neutral Number of protons = number of electrons = Z = 35 Number of neutrons = 80 – 35 = 45, (mass number (A) = number of protons (Z ) + number of neutrons (n))

According to de Brogile equation \(λ=\frac{h}{mv}= \frac{h}{p}\)
\(λ=\frac{h}{mv}=\frac{6.626×10^{−34}Js}{(0.1 kg)(10 ms^{−1})}\) = 6.626×10⁻³⁴m
[∵ J=Kgm²s⁻²]

\(\mathrm{Cr}(24): 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1} 3 d^{5}\)
Half filled orbitals are more stable because of symmetrical distribution of electrons and exchange energy is maximum.

Number of radial nodes = (n - l - 1)
For 3s orbital, n = 3, l = 0
When 'n' is principal quantum
number, 'l' is azimuthal quantum number
l = 0 for s-orbital
l = 1 for p-orbltal
Hence number of radial nodes = (3 - 0 - 1) = 2
For 2p orbital, n = 2, l = 1
∴ Number of radial nodes = (2 - 1- 1) = 0

(i) Aufbau Principle: In the ground state of the atoms, the orbitals are filled in the order of their increasing energies. In other words, electrons first occupy the lowest-energy orbital available to them and enter into higher energy orbitals only after the lower energy orbitals are filled.
The order in which the energies of the orbitals increase and hence the order in which the orbitals are filled is as follows:
15, 25, 2p, 3s, 3p, 4s, 3d, 4p, 55, 4d, 5p, 6s, 4f, 5d, 6p, 75, Sf, 6d, 7p .......
(ii) Pauli Exclusion Principle: An orbital can have maximum of two electrons and
these must have opposite signs.
For example: Two electrons in an orbital can be represented by

The two electrons have opposite spin, if one is revolving clockwise, the other is revolving anticlockwise or vice versa.
(iii) Hund's Rule of Maximum Multiplicity: Electron pairing in p, d and f orbitals cannot occur until each orbital of a given subshell contains one electron each or is single occupied.
For example: For the element nitrogen which contains 7 electrons, the following configuration can be written.

Total spin of unpaired electrons \(=\frac { 1 }{ 2 } +\frac { 1 }{ 2 } +\frac { 1 }{ 2 } =1\frac { 1 }{ 2 } \)

Electron gain enthalpy is the energy released when an isolated gaseous atom is converted into a negative ion by adding an extra electron. Electron gain enthalpy is denoted by the sign \(\triangle\)_eg H.
The process may be represented by
M(g) + e^-  ⟶ M^- (g)
neutral gaseous anion atom
\(\triangle\) H = \(\triangle\)_eg H
electron gain enthalpy is negative or positive it depends upon the nature of the element. For example. For halogens it is highly negative, because they can acquire the noble gas configuration by accepting an extra electron. In contrast. For noble gases have positive electron gain enthalpy because energy has to be supplied to the element.
Factors on which electron gain enthalpy depends:
(i) Atomic size. As the size of an atom increases, the distance between its nucleus and the incoming electron also increases. Therefore, the force of attraction between the nucleus and the incoming electron decreases and hence the electron gain enthalpy becomes less negative.
(ii) Nuclear charge. As the nuclear charge increases force of attraction for the incoming electron increases and thus electron gain enthalpy becomes more negative.
(iii) Symmetry of electronic configuration. Elements having symmetrical configuration (Either half filled or fully filled orbitals in the same subshell) having no attraction for electron because by accepting electron their configuration becomes less stable. In that case energy has to be supplied to accept electron. Thus electron gain enthalpy will be positive.

Electronegativity:
1. The tendency of an atom in a molecule to attract the shared pair of electrons towards itself is known as electronegativity.
2. There is no specific unit for electronegativity. In the modern periodic table:
3. In the modern periodic table: as we move left to right, across a period the nuclear charge increases, atomic size decreases, and electronegativity value increases.
4. As we move down the group there is an increase in the atomic number also nuclear charge and a decrease in the electronegativity value.
5. In general, metal shows a lower electronegativity value compared to on-metals.

We know that, mass of electron = 9.11 x 10^-31 kg
h = 6.626 x 10^-34 Js
Wavelength, 
\(\lambda =\frac { h }{ mv } =\frac { 6.626\times{ 10 }^{ -34 }kg \ { m }^{ 2 }{ s }^{ -1 } }{ 9.11\times{ 10 }^{ -31 }kg\times2.19\times{ 10 }^{ 6 }\times m{ s }^{ -1 } }\)
\( \lambda =3.32\times{ 10 }^{ -10 }m=332pm\)

In the modern periodic table, each period begins with the filling of a new shell. Therefore, the period indicates the value of principal quantum number. Thus, option (c) is correct.

(c) Assertion is true but Reason is false.

(a) Number of orbitals = n ² = 3 ² = 9 orbitals, 3s, 3 p_x, 3 p_y , 3 p_z, 3 d _{x² - y²,} 3 dz², 3 dxy, 3 dyz and 3 dzx.
(b) (i) 2p (ii) 4s
(c) Orbital can have maximum two electron which must be of opposite spin.
(d) 's' orbitals are spherical and 'p' orbitals have dumb-bell shaped.
(e) d _{x² - y²} , dz².