F^-, O^2- and N^3- are isoelectronic species

The anion with the greater negative charge will have a larger radius because of the lesser attraction of the electrons to the nucleus. Hence the increasing order of ionic radii is, r_N^3- > r_O^2- >r_F^-

Na⁺, Mg²⁺ and Al³⁺ are isoelectronic cations.

The cation with the greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus. Hence the increasing order of ionic radius is, r_Na⁺ > r_Mg²⁺ > r_Al³⁺

(i) It is defined as one half of the distance between the centres of nuclei of the two adjacent atoms in the metallic crystal.
(ii) The metallic radius is always larger than its covalent radius.
(iii) The distance between two adjacent copper atoms in solid copper is 2.56\(\mathring { A } \). Hence, the metallic radius of copper is 1.28 \(\mathring { A } .\)

The distance between the nuclei of two covalently bonded atoms is known as covalent distance or inter-nuclear distance. The one-half of this inter-nuclear distance is called covalent radius.
The covalent distance (Cl-Cl) of Cl₂ molecule is experimentally found as 198 pm (1.98 \(\mathring { A } \)). Its covalent radius is 99 pm (0.99 \(\mathring { A } \)).
Cl-Cl Inter Inter nuclear distance = 1.98 \(\mathring { A } \)
∴ r_Cl =1.98/2 = 0.99 \(\mathring { A } \)

(i) The electronic configuration of the elements changes periodically in a period and group as well.
(ii) We could find a pattern in the physical and chemical properties as we go down in a group or move across a period.
(iii) For example: The chemical reactivity is high at the beginning, lower at the middle and increases to a maximum at group 17 in a period.
(iv) The reactivity increases on moving down the group of alkali metals. But the reactivity decreases on moving down the group of halogens.
(v) Atomic radii increases down the group and decreases across the period.

(i) Non-metals are located at the top right hand side of the periodic table.
(ii) In a period, as we move from left to right the non-metallic character increases while the metallic character increases as we go down a group.
(iii) Non-metals are usually solids or liquids or gases at room temperature with low melting and low boiling points (Exceptions: boron and carbon).
(iv) They are poor conductors of heat and electricity.
(v) Most of the non-metallic solids are brittle and are neither malleable nor ductile.

Electronic configuration of Na and Mg are
Na = Is² 2s²2p⁶ 3s¹
Mg = Is² 2s²2p⁶ 3s²
First electron in both cases has to be removed from 3s-orbital but the nuclear charge of Na (+ 11) is lower than that of Mg (+12) therefore first ionization energy of sodium is lower than that of magnesium.
After the loss of first electron, the electronic configuration of
Na⁺ = Is² 2s² 2p⁶
Mg⁺ = Is² 2s²2p⁶ 3s¹
Here electron is to be removed from inert(neon) gas configuration which is very stable and hence removal of second requires more energy when compared to Mg. Therefore, second ionization enthalpy of sodium is higher of magnesium.

(i) The group 1 elements are extremely reactive because these elements can lose one electron to form cation. Their ionization enthalpy is also least.
(ii) The high reactivity of halogens is due to the ease with which these elements can gain an electron to form anion. Their electron gain enthalpy is most negative.
(iii) The elements at the extreme left (alkalis) exhibit strong reducing behavior, whereas the elements at the extreme right (halogens) exhibit strong oxidizing behaviour.
(iv) The reactivity of elements at the centre of the periodic table becomes low when compared with extreme right and left.

Variation in period: The number of valence electrons increases from 1 to 8 on moving across a period. The valency of the elements with respect to hydrogen and chlorine increases from 1 to 4 and then decreases from 4 to zero.
Variation in group: On moving down a group, the number of valence electrons remains same. All the elements in a group exhibit same valency. For example, all the elements of group 1 have valency equal to 1.
Hence, valency is a periodic property.

Electronegativity \(\alpha\) Non-metallic character
(i) As the electronegativity is directly proportional to the non-metallic character, thus across the period, with an increase in electronegativity, the non-metallic character also increases.
(ii) As we move down the group, decrease in electronegativity is accompanied by a decrease in non-metallic character.

(i) The electron gain enthalpy increases as we move from left to right in a period due to the increase of nuclear charge. However, Be, Mg, N and noble gases have almost zero value of electron gain enthalpy due to extra stability of completely and half filled orbitals.
(ii) When we move in a group of periodic table, the size and nuclear charge increase. But the effect of increase in atomic size is much more pronounced than that of nuclear charge and thus the additional electron feels less attraction by the large atom. Consequently, electron gain enthalpy decreases.

(i) Size of the atom: The new electron which was added experiences stronger attraction to its nucleus if the atoms are smaller in size.
\(Atomic\ size \ \infty{1\over Electron \ gain \ enthalpy}\)
(ii) Nuclear charge: The new electron which was added experiences stronger attraction to its nucleus if the atom possess greater nuclear charge.
Nuclear charge \(\infty\) Electron gain enthalpy
(iii) Electronic configuration: An atom with stable electronic configuration has no tendency to gain an electron. Such atoms have zero or almost zero electron gain enthalpy.

Mg (Z=12)1s² 2s² 2p⁶ 3s².
Al (Z=13) 1s² 2s² 2p⁶ 3s² 3p¹.
Although the nuclear charge of aluminium is greater than that of magnesium, I.E of Mg is greater than that of AI. It is because Mg atom has more stable configuration than Al atom.
\(\therefore\) I.E₁ of Mg>LE₁ of AI.

(i) The electrons of inner shells form a cloud of negative charge. This cloud shields the outer electron from the nucleus and also acts as a screen between the nucleus and the outer electrons.
(ii) This screen reduces the coulombic attraction between the positive nucleus and the negative outer electrons. Therefore, the outer electrons do not feel the full effect of the positive charge of the nucleus.
(iii) If screening effect increases, ionization energy decreases.

(i) Intermolecular force can affect the value of ionization energy and electron affinity.
(ii) In gaseous state, there is little intermolecular force in a substance and it can be considered negligible in some cases. So the value of I.E and E.A are almost unaffected if they are calculated for gaseous atoms.
(iii) When we are talking about ionization energy and electron affinity, we need to consider atoms and we can find free atoms only when the substance is in gaseous state.

(i) Ionization is always an endothermic process. It absorbs energy.
(ii) Ionization energies always increase in the order, I.E₁ < I.E₂ < I.E_3.
(iii) Ionization energy measurements are carried out with atoms in the gaseous state.

(i) The effective nuclear charge of magnesium is higher than that of sodium. For these reasons, the energy required to remove an electron from magnesium is more than the energy required in sodium. Hence, the first ionization enthalpy of sodium is lower than that of magnesium.
(ii) Noble gases have completely filled electronic configuration and they are more stable. So in Noble gases addition of electron is not possible. Electron gain enthalpy is always the amount of energy released (-ve sign) when an electron is added to an atom. But in noble gases, if an electron is added, they have positive value of electron gain enthalpy.

The 1^st ionization enthalpy of magnesium is higher than that of Na due to higher nuclear charge and slightly smaller atomic radius of Mg than Na. After the loss of first electron, Na⁺ formed has the electronic configuration of neon (2, 8). The higher stability of the completely filled noble gas configuration leads to very high second ionization enthalpy for sodium. On the other hand, Mg⁺ formed after losing first electron still has one more electron in its outermost (3s) orbital. As a result, the second ionization enthalpy of magnesium is much smaller than that of sodium.

(i) F₂ < Cl₂ < O₂ < N₂
(ii) I < Br < F < CI
(iii) N₂ < O₂ < F₂ < Cl₂

The elements of group 1 have only one electron in their respective valence shells and thus have a strong tendency to lose this electron. The tendency to lose electrons in turn, depends upon the ionization enthalpy. Since the ionization enthalpy decreases down the group therefore, the reactivity of group 1 elements increases in the same order Li < Na < K < Rb < Cs. In contrast, the elements of group 17 have seven electrons in their respective valence shells and thus have strong tendency to accept one more electron to make stable configuration. It is linked with electron gain enthalpy and electronegativity. Since both of them decreases down the group, the reactivity also decreases.

For oxygen atom:
\(O_{(g)}+e^-\rightarrow O^-{(g)}\) (e.g.\(\triangle H\) = -141 kJ mol^-1)
\(O^-_{(g)}+e^-\rightarrow O^{2-}_{(g)}\) (e.g.\(\triangle H\)=+780 kJ mol^-1)
The first electron gain enthalpy of oxygen is negative because energy is released when a gaseous atom accepts an electron to form monovalent anion. The second electron gain enthalpy is positive because energy is needed to overcome the force of repulsion between monovalent anion and second incoming electron.

(i) O or F. Both O and F lie in 2^nd period. As we move from O to F the atomic size decreases. Due to smaller size of F nuclear charge increases.
Further, gain of one electron by
\(F\rightarrow F^-\)
ion has inert gas configuration, While tile gain of one electron by
\(O\rightarrow O^-\)
gives O^- ion which does not have stable inert gas configuration. Consequently, the energy released is much higher in going from
\(F\rightarrow F^-\)
than going from\(O\rightarrow O^-\). In other words electron gain enthalpy of F is much more negative than that of oxygen.
(ii) The negative electron gain enthalpy of CI \((e.g.\triangle H=-349\ kJ\ mol^{-1})\) is more than that of \(F(e.g.\triangle H=-328\ kJ\ mol^{-1})\) 
The reason for the deviation is due to the smaller size of F. Due to its small size, the electron repulsions in the relatively compact 2p-subshell are comparatively large and hence the attraction for incoming electron is less as in the case of Cl.

Atomic size: With the increase in atomic size, the number of electron shells increase. Therefore, the force that binds the electrons with the nucleus decreases. The ionization enthalpy thus decreases with the increase in atomic size.
Screening or shielding effect of inner shell electron: With the addition of new shells, the number of inner electron shells which shield the valence electrons increases. As a result, the force of attraction of the nucleus for the valence electrons further decreases and hence the ionization enthalpy decreases.

(i) Metals comprise more than 78% of all known elements.They are present on the left side of the periodic table.
(ii) They are usually solids at room temperature. [Mercury is an exception (Hg-liquid), gallium (303K) and cesium (302K) also have very low melting points].
(iii) Metals usually have high melting and boiling points.
(iv) They are good conductors of heat and electricity.
(v) They are malleable and ductile, and also can be flattened into thin sheets by hammering and drawn into thin wires.

(i) Helium has two electrons. Its electronic configuration is 1s². As per the configuration, it is supposed to be placed in's' block, but actually placed in 18^th group which belongs to 'p' block. Because it has a completely filled valence shell as the other elements present in 18^th group. It also resembles with 18^th group elements in other properties. Hence helium is placed with other noble gases.
(ii) The other exception is hydrogen. It has only one s-electron and hence can be placed in group 1. It can also gain an electron to achieve a noble gas arrangement and hence it can behave as halogens (17^th group elements). Because of these assumptions, position of hydrogen becomes a special case. Finally, it is placed separately at the top of the periodic table.

In sixth period, 32 elements are present. This period starts with the filling of 6^th energy shell, n = 6.There are sixteen orbitals (one 6s, seven 4f, five 4d and three 6p) to be filled.These sixteen orbitals can accommodate 32 (16\(\times\)2 = 32) electrons. Hence, 32 elements are present in sixth period.

In fourth period, 18 elements are present. In this period electrons are entering into fourth energy level, i.e., n = 4. It starts with the filling of 4s-orbitals. However, after the 4s, but before the 4p orbitals, there are five 3d-orbitals also to be filled. Thus, nine orbitals (one 4s, five 3d and three 4p) have to be filled. These nine orbitals can accommodate(9 \(\times\) 2= 18) 18 electrons. Hence, 4^th period contain 18 elements in it.

Electronic configuration of 1^st period:
In 1^st period only two elements are present. This period starts with the filling of electrons in first energy level, n= 1. This level has only one orbital as 1s. Therefore it 'can accommodate two electrons maximum.
Electronic configuration of 2^nd period:
In the 2^nd period 8 elements are present. This period starts with filling of electrons in the second energy level, n = 2. In this level four orbitals (one 2s and three 2p) are present. Hence the second energy level can accommodate 8 electrons. Thus, second period has eight elements.

(i) The distribution of electrons into orbitals, s, p, d and f of an atom is called its electronic configuration. The electronic configuration of an atom is characterized by a set of four quantum numbers, n, 1, m and s. Of these the principal quantum number (n) defines the main energy level known as shells.
(ii) The position of an element in the periodic table is related to the configuration of that element and thus reflects the quantum numbers of the last orbital filled.
(iii) The electronic configuration of elements in the periodic table can be studied along the periods and groups separately for the best classification of elements.
(iv) Elements placed in a horizontal row of a periodic table is called a period. There are seven periods.
(v) A vertical column of the periodic table is called a group. A group consists of a series of elements having similar configuration of the outermost shell. There are 18 groups in periodic table.

(i) The number of electrons increases by the same number as the increase in the atomic number.
(ii) As the number of electrons increases, the electronic structure of the atom changes.
(iii) Electrons in the outermost shell of an atom (valence shell electrons) determine the chemical properties of the elements.

(i) Henry Moseley studied the X-ray spectra of several elements and determined their atomic numbers (Z).
(ii) He discovered a correlation between atomic number and the frequency of X-rays generated by bombarding an element with high energy of electrons.
(iii) Moseley correlated the frequency of the X-ray emitted by an equation as, \(\sqrt { \upsilon } \) = a(Z-b)
Where ሀ = Frequency of the X-rays emitted by the elements.a and b = Constants.
(iv) From the square root of the measured frequency of the X-rays emitted, he determined the atomic number of the element.

(i) This law is quite well for lighter elements but not supported to heavier elements.
(ii) Elements were arranged in increasing atomic masses without taking an account on the properties of elements.
(iii) This law was seemed to be applicable only for elements upto calcium.

When the halogens, chlorine, bromine and iodine are placed on below the others, they had similar properties, The atomic weight of bromine was close to the average of the atomic weights of chlorine and iodine.

(i) Classification is a fundamental and essential process in our day-to-day life for the effective utilization of resources, daily events and materials
(ii) In such a way, for the effective utilization of discovered elements becomes fundamentally essential process.
(iii) The periodic classification of the elements is one of the outstanding contributions to the progress of chemistry.

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

Li (2, 1) = 2 x 0.85 = 1.7
Be (2, 2) = 2 x 0.35 + 2 x 0.85 = 2.05
B (2, 3) = 2 x 0.35 + 2 x 0.85 = 2.40
C (2, 4) = 3 x 0.35 + 2 x 0.85 = 2.75
N (2, 5) = 4 x 0.35 + 2 x 0.85 = 3.10
0(2,6) = 5 x 0.35 + 2 x 0.85 = 3.45
F (2, 7) = 6 x 0.35 + 2 x 0.85 = 3.80
Ne (2, 8) = 7 x 0.35 + 2 x 0.85 = 4.15

Li (2, 1) = 2 x 0.85 = 1.7
Na (2,8, 1) = 8 x 0.85+ 2 x 1 = 8.8
K (2,8,8, 1) =8 x 0.85 +- 10 x 1 = 16.8
Rb (2, 8, 18, 8, 1) = 8 x 0.85 + 28 x 1 = 34.8
CS (2, 8, 18, 18,8, 1) = 8 x 0.85 + 46 x 1 = 52.8

IE = E_{\(\infty \)} -E_n
= 0-(-2.18 x 10^-18)
= 2.18 x 10^-18J
Ionisation energy per mole = 2.18 x 10^-18 x 6.023 x 10²³
= 1.313 x 10⁶J
(2.18 x 10^-18J is IE for one electron. I.E permole of electron, it has to be multiplied by Avagadro Number).

The second electron gain enthalpy of oxygen is positive. When an electron is added to neutral oxygen atom, a monovalent anion (O^-) is formed and energy is released. i.e. first electron gain enthalpy is negative. When the electron is added to form 0^-2 anion, there is a lot of electrostatic repulsion as both 0^- anion and the electron. To overcome their repulsion energy is absorbed. Thus, the second IE of oxygen is positive \(\triangle\)H
O(g) + e \(\rightarrow\) o^- (g) - 141.0 kj mol^-1
O^-(g) + e \(\rightarrow\) 0^-2 (g) + 780 kj mol^-1

(i) Atomic size: Atomic size increases due to addition of a new energy level. This decreases ,the force of attraction and consequently ionisation enthalpy decreases.
(ii) Screening effect: Due to increase of inner electrons, the shielding effect increases which reduces force of attraction towards nucleus and hence IE decreases.

(i) The electronic configuration of Be and B are Be = Is² 2s² ; B = 1s² 2s² 2pl
In Be, the electron is removed from completes filled orbital (2s) while in boron, from 2p sub shell. 2s electrons are more strongly bond to the nucleus than 2p electron and therefore more energy is required to remove a 2s electron than 2p electron. Consequently, Be has higher first IE than boron.
(ii) In nitrogen, all the 'p' orbitals are singly occupied. Thus it is a stable arrangement compared to oxygen. As a result, the removal of an electron from nitrogen requires more energy than the removal of an electron from oxygen atom. Then, nitrogen has higher IE then oxygen

(i) Isolated atom means a single gaseous atom free from other atoms. No energy should be utilised to separate it from other atoms. It is an ideal situation which is difficult to achieve. However, the inter atomic distances are reduced to minimum by carrying out the measurement of IE at a low pressure.
(ii) 'Ground state' means that the atom must be present in the most stable state. i.e., in the lowest energy state possible for the atom.

(i) The electronic configuration of the element is 3s² 3p⁴. Thus the element belongs to 3rd period and group 16.
(ii) The electronic configuration of the element is 3d² 4s². Thus the element belongs to 4th period and group 4.
(iii) The electronic configuration of the element' is 4f⁷5d¹ 6s² i.e it belongs to 6th period and group 3 (lanthanides).

(i) III is a non-metal because 1st and II_nd ionization enthalpies are high.
(ii) I is an alkali metal because the second ionization enthalpy is high. In alkali metals, the s.econd electron is removed from complete octet.
(iii) II is an alkaline earth metal because these elements have two electrons in the outermost shell. First and second IE's are in almost 1: 2 ratio.

(i) The size of fluorine atom is small compared to chlorine and the electron density around the nucleus is high. This resists the addition of electron, hence the electron affinity of fluorine' is less than that of chlorine atom
(ii) The electronic configuration of nitrogen and oxygen are as follows:
N = \({ 1s }^{ 2 }{ 2s }^{ 2 }{ 29 }_{ x }^{ 1 }2p_{ y }^{ 1 }{ 2p }_{ z }^{ 1 };0=1s^{ 2 }{ 2s }^{ 2 }{ 2p }_{ x }^{ 2 }{ 2p }_{ y }^{ 1 }{ 2p }_{ z }^{ 1 }\)
In, nitrogen, 'p' orbitals are half filled and hence possess a stable electronic configuration. It requires more energy to remove an electron. Hence IE of nitrogen is higher than that of oxygen which has a less stable electronic configuration.
(iii) The electronic configuration ofMg and Al are as follows:
1s² 2s² 2p⁶ 3s² ; 1s²,2s² 2p⁶,3s² 3p¹
It is more difficult to remove an electron from 3s orbital in comparison to 3p orbital because 3s electrons are closer to the nucleus. Further 3s electrons are paired and 3p is simply occupied. Hence IE of Mg is higher than that of AI.

(i) Fluorine is expected to have higher first IE than chlorine because, it is smaller in size than chlorine and IE decreases down the group.
(ii) Chlorine is expected to have higher IE than sulphur, because it has a smaller size than sulphur. IE decreases along the period.
(iii) Neon is expected to have higher IE because it has completely filled orbitals from which removal of electron is difficult.
(iv) Argan is expected to have higher IE, because it has completely filled orbitals.

Iso electronic species are those which have the same number of electrons.
(i) Ar has 18 electrons. Therefore, the species p^-3, S^-2, Cl^-, K+, Ca⁺² are isoelectronic with Ar.
(ii) F^- has 10 electrons. Therefore the species, N^-3, 0^-2, Ne, Na⁺, Mg⁺² are isoelectronic with F^-
(iii) K⁺ has 18 electrons. Therefore, the species p^-3, S^-2, CI^-, Ar, Ca⁺²are isoelectronic with K⁺
(iv) Sr⁺²has 36 electrons. Therefore Br^-, Kr, Rb⁺ etc are isoelectronic with Sr⁺²

Mg, Mg⁺², AI, Al⁺³
Mg and Al belong to the same period

Atomic size decreases from left to right across a period. Thus Mg is a larger atom than AI. Cation
is smaller than the neutral atom. Mg⁺² is smaller than Mg and Al⁺³ is smaller than AI. Thus, Al⁺³ ion the smallest and Mg is the largest in size among the given species.

(i) If the difference, X_A- X_B= 0, i.e., X_A= X_B' the bond is purely covalent and non polar.
eg: H₂, Cl₂,0₂ and N₂ molecules.
(ii) The difference X_A- X_Bis small i.e., XA > XB' the bond is polar covalent.
(iii) The difference X_A - X_B = 1.9, the bond is 50% covalent and 50% ionic.
(iv) The difference X_A  - X_B.is very high the bond is more ionic and less covalent.
The percentage ionic character may be calculated by using the formula. % ionic character = 16 (X_A- X_B)+ 3.5 (X_A- X_B)²