Answer

(b) In case of isoelectronic species ionic radii $\propto \frac{1}{\text { atomic number }}$

The ionic radii increases as the positive charge decreases or the negative charge increases.

• Option (a) $\mathrm{F}^{-}<\mathrm{O}^{2-}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}$:

  • This option is incorrect because it suggests that $\mathrm{F}^{-}$ has a smaller radius than $\mathrm{O}^{2-}$, which is not true. $\mathrm{O}^{2-}$ has a larger radius than $\mathrm{F}^{-}$ due to its higher negative charge, which results in greater electron-electron repulsion and thus a larger ionic radius.

• Option (c) $\mathrm{O}^{2-}<\mathrm{F}^{-}<\mathrm{Na}^{+}<\mathrm{Mg}^{2+}$:

  • This option is incorrect because it suggests that $\mathrm{O}^{2-}$ has a smaller radius than $\mathrm{F}^{-}$, which is not true. $\mathrm{O}^{2-}$ has a larger radius than $\mathrm{F}^{-}$ due to its higher negative charge, which results in greater electron-electron repulsion and thus a larger ionic radius.
  • Additionally, it places $\mathrm{Na}^{+}$ before $\mathrm{Mg}^{2+}$, which is incorrect. $\mathrm{Mg}^{2+}$ has a smaller radius than $\mathrm{Na}^{+}$ due to its higher positive charge, which results in a stronger attraction between the nucleus and the electrons, leading to a smaller ionic radius.

• Option (d) $\mathrm{O}^{2-}<\mathrm{F}^{-}<\mathrm{Mg}^{2+}<\mathrm{Na}^{+}$:

  • This option is incorrect because it suggests that $\mathrm{O}^{2-}$ has a smaller radius than $\mathrm{F}^{-}$, which is not true. $\mathrm{O}^{2-}$ has a larger radius than $\mathrm{F}^{-}$ due to its higher negative charge, which results in greater electron-electron repulsion and thus a larger ionic radius.
  • Additionally, it places $\mathrm{Na}^{+}$ after $\mathrm{Mg}^{2+}$, which is incorrect. $\mathrm{Mg}^{2+}$ has a smaller radius than $\mathrm{Na}^{+}$ due to its higher positive charge, which results in a stronger attraction between the nucleus and the electrons, leading to a smaller ionic radius.

Answer

(d) Elements with atomic number, $Z=90$ to 103 are called actinoids. Thus, terbium $(Z=65)$ is not an actinoid. Terbium belong to lanthanoids.

• Curium $(Z=96)$ is an actinoid because it falls within the atomic number range of 90 to 103.
• Californium $(Z=98)$ is an actinoid because it falls within the atomic number range of 90 to 103.
• Uranium $(Z=92)$ is an actinoid because it falls within the atomic number range of 90 to 103.

Thinking Process

To solve this question, keep in mind that shielding effect represent the repulsive force felt by the valence shell from the electrons presents in the inner shells.

Answer

(a) For the same shell screening effect decreases in the order $s>n>d>f$

• Option (b) $f>d>p>s$: This option is incorrect because the screening effect of electrons decreases with increasing azimuthal quantum number (l). The correct order is $s>p>d>f$, not $f>d>p>s$.

• Option (c) $p < d < s > f$: This option is incorrect because it suggests that the screening effect of $s$ orbitals is less than $p$ and $d$ orbitals, which is not true. The correct order is $s>p>d>f$.

• Option (d) $f>p>s>d$: This option is incorrect because it suggests that $f$ orbitals have the highest screening effect, which is not true. The correct order is $s>p>d>f$.

Answer

(a) Follow the following steps to solve out such problems

• Option (b) is incorrect because it suggests that the ionisation enthalpy of sodium (Na) is greater than that of magnesium (Mg), which is not true. Sodium has a lower ionisation enthalpy than magnesium due to its position in the periodic table and its electronic configuration.

• Option (c) is incorrect because it suggests a continuous increase in ionisation enthalpy from sodium (Na) to silicon (Si) without considering the exception. Magnesium (Mg) has a higher ionisation enthalpy than aluminum (Al) due to its completely filled 3s orbital, which makes it more stable and harder to ionize.

• Option (d) is incorrect because it suggests that sodium (Na) has a higher ionisation enthalpy than magnesium (Mg), which is not the case. Sodium has a lower ionisation enthalpy than magnesium due to its electronic configuration and position in the periodic table.

Answer

(c) The electronic configuration of $\mathrm{La}(Z=57)$ is $[\mathrm{Xe}] 5 d^{1} 6 s^{2}$. Therefore, further addition of electrons occurs in the lower energy $4 f$-orbital till it is exactly half-filled at $\mathrm{Eu}(Z=63)$ Thus, the electronic configuration of $\mathrm{Eu}$ is $[\mathrm{Xe}] 4 f^{7} 6 s^{2}$. Thereafter, addition of next electron does not occur in the more stable exactly half-filled $4 f^{7}$ shell but occurs in the little higher energy $5 d$-orbital. Thus, the electronic configuration of $\mathrm{Gd}(Z=64)$ is $[\mathrm{Xe}] 4 f^{7} 5 d^{1} 6 s^{2}$.

• Option (a): The configuration $[\mathrm{Xe}] 4 f^{3} 5 d^{5} 6 s^{2}$ is incorrect because it does not follow the Aufbau principle and Hund’s rule. Gadolinium (Gd) has a more stable configuration with a half-filled $4f$ subshell ($4f^7$) and one electron in the $5d$ subshell.

• Option (b): The configuration $[\mathrm{Xe}] 4 f^{7} 5 d^{2} 6 s^{1}$ is incorrect because it places two electrons in the $5d$ subshell and only one in the $6s$ subshell. This does not align with the observed stability of having two electrons in the $6s$ subshell and only one in the $5d$ subshell for Gadolinium.

• Option (d): The configuration $[\mathrm{Xe}] 4 f^{8} 5 d^{6} 6 s^{2}$ is incorrect because it suggests an overfilled $4f$ subshell and an excessively filled $5d$ subshell, which is not consistent with the actual electron configuration of Gadolinium. The $4f$ subshell should be half-filled ($4f^7$) for stability, and the $5d$ subshell should have only one electron.

Answer

(c) In case of transition elements (or any elements), the order of filling of electrons in various orbital is $3 p<4 s<3 d$. Thus, $3 d$ orbital is filled when $4 s$ orbital gets completely filled.

• (a) The statement is correct because the properties of elements are indeed a periodic function of their atomic numbers, as established by the modern periodic law.

• (b) The statement is correct because non-metallic elements are indeed fewer in number compared to metallic elements in the periodic table.

• (d) The statement is correct because the first ionisation enthalpies of elements generally do increase with an increase in atomic number as we move across a period, due to the increasing nuclear charge which holds the electrons more tightly.

Answer

(c) As we move from $\mathrm{Cl}$ to $\mathrm{I}$ the electron gain enthalpy (i.e., energy released in electron gain) become less and less negative due to a corresponding increase in the atomic size.

However, the electron gain enthalpy of $\mathrm{F}$ is less negative than that of $\mathrm{Cl}$ due to its small size. Thus, the negative electron gain enthalpy follows the order

$$ F < CI < Br < I $$

• Option (a) $\mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$: This option is incorrect because it suggests that fluorine has the most negative electron gain enthalpy, which is not true. Although fluorine is highly electronegative, its small size leads to significant electron-electron repulsion in its compact 2p orbital, making its electron gain enthalpy less negative than that of chlorine.

• Option (b) $\mathrm{F}<\mathrm{Cl}<\mathrm{Br}<\mathrm{I}$: This option is incorrect because it implies a continuous increase in electron gain enthalpy from fluorine to iodine. In reality, chlorine has the most negative electron gain enthalpy among the halogens, not fluorine, due to the reasons mentioned above.

• Option (d) $\mathrm{F}<\mathrm{Cl}<\mathrm{Br}<\mathrm{I}$: This option is incorrect for the same reason as option (b). It suggests a continuous increase in electron gain enthalpy from fluorine to iodine, which is not accurate. Chlorine has a more negative electron gain enthalpy than fluorine, and the trend does not follow a simple increasing pattern.

Answer

(c) Since each period starts with the filling of electrons in a new principal quantum number, therefore, the period number in the long form of the periodic table refers to the maximum principal quantum number of any element in the period.

Period number $=$ maximum $n$ of any element (where, $n=$ principal quantum number).

• (a) The magnetic quantum number (m) of an element refers to the orientation of the orbital in space and can have values ranging from -l to +l (where l is the azimuthal quantum number). It does not determine the period number in the periodic table.

• (b) The atomic number of an element is the number of protons in the nucleus of an atom. While elements in a period have consecutive atomic numbers, the period number itself is not equal to the atomic number of any element in that period.

• (d) The azimuthal quantum number (l) defines the shape of the orbital and can have values ranging from 0 to (n-1) for a given principal quantum number (n). The maximum azimuthal quantum number of any element in a period does not determine the period number.

Answer

(c) The elements in which electrons are progressively filled in 4f-orbital are called lanthanoids. Lanthanoids consist of elements from $Z=58$ (cerium) to 71 (lutetium).

• (a) actinoids: Actinoids are elements in which electrons are progressively filled in the 5f-orbital, not the 4f-orbital. They consist of elements from $Z=89$ (actinium) to 103 (lawrencium).

• (b) transition elements: Transition elements are those in which electrons are progressively filled in the d-orbital, specifically the 3d, 4d, 5d, and 6d orbitals. They are not associated with the filling of the 4f-orbital.

• (d) halogens: Halogens are elements in Group 17 of the periodic table and are characterized by having seven electrons in their outermost p-orbital. They do not involve the filling of the 4f-orbital.

Answer

(d) Anion formed after the gain of electron to the neutral atom and cation formed after the lose of electron from outer shell. Hence, cation has smaller size but anion has bigger size than its neutral atom. Thus, $\mathrm{I}^{-}>\mathrm{I}>\mathrm{I}^{+}$.

• Option (a) $\mathrm{I}>\mathrm{I}^{-}>\mathrm{I}^{+}$: This option is incorrect because it suggests that the neutral iodine atom ($\mathrm{I}$) is larger than the iodide anion ($\mathrm{I}^{-}$), which is not true. The iodide anion ($\mathrm{I}^{-}$) is larger than the neutral iodine atom due to the addition of an extra electron, which increases electron-electron repulsion and expands the electron cloud.

• Option (b) $\mathrm{I}^{+}>\mathrm{I}^{-}>\mathrm{I}$: This option is incorrect because it suggests that the iodine cation ($\mathrm{I}^{+}$) is larger than both the iodide anion ($\mathrm{I}^{-}$) and the neutral iodine atom ($\mathrm{I}$). In reality, the iodine cation ($\mathrm{I}^{+}$) is the smallest because it has lost an electron, reducing electron-electron repulsion and resulting in a smaller size.

• Option (c) $\mathrm{I}>\mathrm{I}^{+}>\mathrm{I}^{-}$: This option is incorrect because it suggests that the neutral iodine atom ($\mathrm{I}$) is larger than the iodine cation ($\mathrm{I}^{+}$) but smaller than the iodide anion ($\mathrm{I}^{-}$). However, the correct order is that the iodide anion ($\mathrm{I}^{-}$) is the largest, followed by the neutral iodine atom ($\mathrm{I}$), and the iodine cation ($\mathrm{I}^{+}$) is the smallest.

Answer

(c) Although $\mathrm{O}^{2-}$ has noble gas configuration isoelectronic with neon but its formation is unfavourable due to the strong electronic repulsion between the negatively charged $\mathrm{O}^{-}$ion and the second electron being added.

Hence, the electron repulsion outweighs the stability gained by achieving noble gas configuration.

• (a) Oxygen being more electronegative does not directly explain the unfavourability of forming $\mathrm{O}^{2-}$, as electronegativity refers to the tendency of an atom to attract electrons, not the energy changes involved in adding a second electron to an already negatively charged ion.

• (b) While the addition of an electron does result in a larger size of the ion, this does not directly explain the high endothermic energy change observed. The key issue is the electron-electron repulsion, not just the increase in size.

• (d) The size comparison between $\mathrm{O}^{-}$ and the oxygen atom does not address the energy changes involved in adding a second electron. The main factor is the repulsion between the negatively charged $\mathrm{O}^{-}$ ion and the incoming electron, not the relative sizes of the ions.

Answer

(c) The element with atomic number 57 belongs to $d$-block element as the last electron enters the $5 d$-orbital against the aufbau principle. This anomalous behaviour can be explained on the basis of greater stability of the xenon (inert gas) core.

After barium $(Z=56)$, the addition of the next electron (i.e., 57 th) should occur in $4 f$-orbital in accordance with aufbau principle. This will however, tend to destabilize the xenon core $(Z=54)$, $[\mathrm{Kr}]\left(4 d^{10} 4 f^{0} 5 s^{2} 5 p^{6} 5 d^{0}\right)$ since the $4 f$-orbitals lie inside the core.

Therefore, the 57 th electron prefers to enter $5 d$-orbital which lies outside the xenon core and whose energy is only slightly higher than that of $4 f$-orbital. In doing so, the stability conferred on the atom due to xenon core more than compensates the slight instability caused by the addition of one electron to the higher energy $5 d$ - orbital instead of the lower energy $4 f$ - orbital.

Thus, the outer electronic configuration of $\mathrm{La}(Z=57)$ is $5 d^{1} 6 s^{2}$ rather than the expected $4 f^{1} 6 s^{2}$.

• (a) $s$ - block: The element with atomic number 57 does not belong to the $s$-block because the $s$-block elements have their last electron entering an $s$-orbital. For element 57, the last electron enters a $5d$-orbital, not an $s$-orbital.

• (b) $p$ - block: The element with atomic number 57 does not belong to the $p$-block because the $p$-block elements have their last electron entering a $p$-orbital. For element 57, the last electron enters a $5d$-orbital, not a $p$-orbital.

• (d) $f$ - block: The element with atomic number 57 does not belong to the $f$-block because the $f$-block elements have their last electron entering an $f$-orbital. Although element 57 is closely related to the $f$-block elements (lanthanoids), its last electron enters a $5d$-orbital, not a $4f$-orbital.

Answer

(c) Each period starts with the filling of electrons in a new principal energy shell. Therefore, 6 th period starts with the filling of $6 s$-orbital and ends when $6 p$-orbitals are completely filled.

In between $4 f$ and $5 d$-orbitals are filled in accordance with aufbau principle. Thus, the outmost electronic configuration of the last element of the p-block in the 6th period is $6 s^{2} 4 f^{14} 5 d^{10} 6 p^{6}$ or $4 f^{14} 5 d^{10} 6 s^{2} 6 p^{6}$.

• Option (a): The configuration $7s^2 7p^6$ corresponds to elements in the 7th period, not the 6th period. Therefore, it cannot represent the last element of the p-block in the 6th period.

• Option (b): The configuration $5f^{14} 6d^{10} 7s^2 7p^0$ does not have any electrons in the 7p orbital, indicating it is not a p-block element. Additionally, it corresponds to an element in the 7th period, not the 6th period.

• Option (d): The configuration $4f^{14} 5d^{10} 6s^2 6p^4$ has only four electrons in the 6p orbital, which means it is not the last element of the p-block in the 6th period. The last element should have a completely filled 6p orbital with six electrons.

Answer

(c) The long form of the periodic table contain element with atomic number 1 to 118.

• (a) 107: This element can be accommodated in the long form of the periodic table as it falls within the range of atomic numbers 1 to 118.

• (b) 118: This element can be accommodated in the long form of the periodic table as it is the highest atomic number currently included in the table.

• (d) 102: This element can be accommodated in the long form of the periodic table as it falls within the range of atomic numbers 1 to 118.

Answer

(a) The fifth period begins with $\mathrm{Rb}(Z=37)$ and ends at $\mathrm{Xe}(Z=54)$. Thus, the element with $Z=43$ lies in the 5 th period. Since, the 4 th period has 18 elements, therefore, the atomic number of the element which lies immediately above the element with atomic number 43 is $43-18=25$.

Now, the electronic configuration of element with $Z=25$ is

$$ 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{5} 4 s^{2} \text { (i.e., } \mathrm{Mn} \text { ). } $$

• Option (b): The configuration $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{5} 4s^{3} 4p^{6}$ is incorrect because the 4s orbital can only hold a maximum of 2 electrons, not 3.

• Option (c): The configuration $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{6} 4s^{2}$ is incorrect because it corresponds to the element with atomic number 26 (Fe), not 25.

• Option (d): The configuration $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{7} 4s^{2}$ is incorrect because it corresponds to the element with atomic number 27 (Co), not 25.

Answer

(b) Each period ends with a noble gas. The atomic number of noble gases (i.e., group 18 elements) are 2, 10, 18, 36, 54 and 86. Therefore, elements with atomic numbers $35(36-1), 53(54-1)$, and $85(86-1)$, lie in a group before noble gases, i.e., halogens (group 17) elements.

Thus, the elements with atomic number 35,53 and 85 are all belongs to halogens.

• (a) noble gases: The elements with atomic numbers 35, 53, and 85 are not noble gases. Noble gases are found in Group 18 of the periodic table, and their atomic numbers are 2, 10, 18, 36, 54, and 86. The given atomic numbers (35, 53, and 85) do not match any of these.

• (c) heavy metals: Heavy metals are typically defined as metals with high atomic weights and densities. The elements with atomic numbers 35 (Bromine), 53 (Iodine), and 85 (Astatine) are non-metals or metalloids, not heavy metals.

• (d) light metals: Light metals are metals with low atomic weights and densities, such as those found in Groups 1 and 2 of the periodic table (e.g., Lithium, Sodium, Magnesium). The elements with atomic numbers 35, 53, and 85 are not metals; they are halogens, which are non-metals.

Answer

(a) Electronic configuration of elements indicate that $A$ is a noble gas (i.e., $\mathrm{Ne}), B$ is oxygen (group 16), $C$ is sodium metal (group 1) and $D$ is fluorine (group 17).

(i) Noble gases have no tendency to gain electrons since all their orbitals are completely filled. Thus, element $A$ has the least electron gain enthalpy.

(ii) Since, element $D$ has one electron less and element $B$ has two electrons less than the corresponding noble gas configuration, hence, element $D$ has the highest electron, gain enthalpy followed by element $B$

(iii) Since, element $C$ has one electron in the $s$-orbital and hence needs one more electron to complete it, therefore, electron gain enthalpy of $C$ is less than that of element $B$. Combining all the facts given above, the electron gain enthalpies of the four elements increase in the order $A<C<B<D$.

• Option (b) $A<B<C<D$: This option is incorrect because it suggests that element $B$ (oxygen) has a lower tendency to gain electrons than element $C$ (sodium). However, oxygen, being a group 16 element, has a higher electron gain enthalpy than sodium, which is a group 1 element. Therefore, $B$ should have a higher tendency to gain electrons than $C$.

• Option (c) $D<B<C<A$: This option is incorrect because it places element $A$ (neon, a noble gas) as having the highest tendency to gain electrons. Noble gases have completely filled orbitals and thus have no tendency to gain electrons. Therefore, $A$ should have the least tendency to gain electrons, not the highest.

• Option (d) $D<A<B<C$: This option is incorrect because it suggests that element $A$ (neon) has a higher tendency to gain electrons than element $B$ (oxygen) and element $C$ (sodium). As a noble gas, $A$ has no tendency to gain electrons, so it should have the least electron gain enthalpy. Therefore, $A$ should not be placed higher than $B$ and $C$ in terms of electron gain tendency.

Answer

$(b, c)$

Elements $\mathrm{Be}$ and $\mathrm{B}$ lie in the 2nd period. They can have a maximum of 8 electrons in the valence shell. In other words, they can have a maximum covalency of $8 / 2=4$.

However, elements $\mathrm{P}$ and $\mathrm{S}$ have vacant $d$ - orbitals in their respective valence shells and hence can accommodate more than 8 electrons in their respective valence shell. In other words, they can show a covalency of more than 4.

• Be: Beryllium (Be) lies in the 2nd period and can have a maximum of 8 electrons in its valence shell. Therefore, it can have a maximum covalency of 4.
• B: Boron (B) also lies in the 2nd period and can have a maximum of 8 electrons in its valence shell. Thus, it can have a maximum covalency of 4.

Answer

$(a, c)$

The elements of group 1 (alkali metals) and group 2 (alkaline earth metals) have 1 and 2 electrons respectively in their valence shells and hence have low ionisation energies. In other words, elements of group 1 and 2 imparts colour to the flame.

• Group 13: The elements in group 13 do not have low ionisation energies compared to groups 1 and 2. They have three electrons in their valence shells, which require more energy to ionize. Therefore, they do not impart colour to the flame.

• Group 17: The elements in group 17 (halogens) have high ionisation energies because they have seven electrons in their valence shells and need only one more electron to achieve a stable octet configuration. As a result, they do not impart colour to the flame.

Answer

$(a, d)$

Elements of $s$ and $p$-block elements are called representative elements. Elements of $d$-block $(Z=21-30 ; 39-48 ; 57$ and $72-80 ; 89$ and 104-112) are called transition elements while those of $f$-block (with $Z=58-71$ and $Z=90-103$ ) are called inner transition elements.

With reference to this division, elements listed under option (a) and option (d) are all representative elements.

• Option (b) is incorrect because the element with atomic number 22 (Titanium) is a transition element, not a representative element.

• Option (c) is incorrect because the elements with atomic numbers 25 (Manganese) and 37 (Rubidium) are not all representative elements. Manganese is a transition element.

Answer

$(a, d)$

Chlorine has the highest tendency to gain one electron because by doing so, it acquires the stable electronic configuration of the nearest noble gas, i.e., argon. Sulphur and oxygen belong to group 16 but the size of oxygen is much smaller than that of sulphur.

As a result, when an electron is added to them, the electron-electron repulsions in the smaller $2 p$ - subshell of oxygen are comparatively stronger than those present in the bigger $3 p$-subshell of sulphur. Therefore, $S$ has a higher tendency to gain an electron than $O$.

$\mathrm{Na}$, on the other hand, has only one electron in the valence shell and hence has a strong tendency to lose rather than gain one electron.

• Sodium (Na): Sodium has only one electron in its valence shell and hence has a strong tendency to lose rather than gain one electron. This makes it less likely to gain an electron compared to other elements in its group.

• Oxygen (O): Oxygen belongs to group 16, but its atomic size is much smaller than that of sulfur. When an electron is added to oxygen, the electron-electron repulsions in the smaller 2p-subshell are comparatively stronger than those present in the bigger 3p-subshell of sulfur. Therefore, oxygen has a lower tendency to gain an electron compared to sulfur.

Answer

$(a, c, d)$

Chlorine has more negative electron gain enthalpy than fluorine. Therefore, all other given statements are correct.

• Chlorine has more negative electron gain enthalpy than fluorine because the added electron in chlorine experiences less electron-electron repulsion compared to fluorine, making it more energetically favorable for chlorine to gain an electron.

Thinking Process

Isoelectronic represents those species which have same number of electrons.

Answer

$(b, c)$

(a) $\mathrm{Zn}^{2+}(30-2=28), \mathrm{Ca}^{2+}(20-2=18), \mathrm{Ga}^{3+}(31-3=28), \mathrm{Al}^{3+}(13-3=10)$.

These species have different number of electrons and hence are not isoelectronic ions.

(b) $\mathrm{K}^{+}(19-1=18), \mathrm{Ca}^{2+}(20-2=18), \mathrm{Sc}^{3+}(21-3=18), \mathrm{Cl}^{-}(17+1=18)$.

These are all isoelectronic ions since each one of them has 18 electrons.

(c) $\mathrm{P}^{3-}(15+3=18), \mathrm{S}^{2-}(16+2=18), \mathrm{Cl}^{-}(17+1=18), \mathrm{K}^{+}(19-1=18)$.

These are all isoelectronic ions since each one of them has 18 electrons.

(d) $\mathrm{Ti}^{4+}(22-4=18), \operatorname{Ar}(18), \mathrm{Cr}^{3+}(24-3=21), \mathrm{V}^{5+}(23-5=18)$.

These have different number of electrons and hence are not isoelectronic ions.

• (a) $\mathrm{Zn}^{2+}(30-2=28), \mathrm{Ca}^{2+}(20-2=18), \mathrm{Ga}^{3+}(31-3=28), \mathrm{Al}^{3+}(13-3=10)$. These species have different number of electrons and hence are not isoelectronic ions.

• (d) $\mathrm{Ti}^{4+}(22-4=18), \operatorname{Ar}(18), \mathrm{Cr}^{3+}(24-3=21), \mathrm{V}^{5+}(23-5=18)$. These have different number of electrons and hence are not isoelectronic ions.

Thinking Process

(i) The ionic size increases as the positive charge on the cation decreases or the negative charge on the anion increases.

(ii) First ionisation enthalpy increases from left to right in the periodic table.

(iii) Electron gain enthalpy increases as the electronegativity of the atom increases.

(iv) The metallic character increases as the size of the metal atom increases.

Answer

$(b, c)$

Due to greater stability of the half filled electronic configuration of nitrogen, its ionisation enthalpy is higher than that of oxygen. Thus, option (b) is incorrect.

Due to stronger electron-electron repulsions in the small size of flourine the negative electron gain enthalpy of flunrinn in Ininan than that af nhlarinn Honnn nntinn ’n ’ is incorrect.

• For option (a): The order of increasing ionic size is incorrect because $\mathrm{Al}^{3+}$ has a smaller ionic radius than $\mathrm{Mg}^{2+}$, $\mathrm{Na}^{+}$, and $\mathrm{F}^{-}$. The correct order should be $\mathrm{Al}^{3+} < \mathrm{Mg}^{2+} < \mathrm{Na}^{+} < \mathrm{F}^{-}$, as given in the question, which is actually correct. Therefore, option (a) is not incorrect.

• For option (d): The order of increasing metallic radius is correct. As we move down the group in the periodic table, the metallic radius increases. Therefore, $\mathrm{Li} < \mathrm{Na} < \mathrm{K} < \mathrm{Rb}$ is the correct order. Hence, option (d) is not incorrect.

Answer

$(a, d)$

Electronegativity and metallic character have no units while electron gain enthalpy and ionisation enthalpy have units of $\mathrm{kJ} \mathrm{mol}^{-1}$.

• Electron gain enthalpy has units because it is the amount of energy released when an electron is added to a neutral atom in the gaseous state, typically measured in $\mathrm{kJ} \mathrm{mol}^{-1}$.
• Ionisation enthalpy has units because it is the amount of energy required to remove an electron from a neutral atom in the gaseous state, also typically measured in $\mathrm{kJ} \mathrm{mol}^{-1}$.

Answer

$(a, c)$

lonic radius decreases as the effective nuclear charge increases.

lonic radius $\propto \frac{1}{\text { effective nuclear charge }}$

Further, ionic radius increases as the screening effect increases.

lonic radius $\propto$ screening effect

• Option (b) is incorrect because ionic radii do not vary in inverse proportion to the square of the effective nuclear charge. The relationship is a direct inverse proportion, not involving the square of the effective nuclear charge.

• Option (d) is incorrect because ionic radii do not vary in direct proportion to the square of the screening effect. The relationship is a direct proportion to the screening effect itself, not its square.

Answer

$(a, c)$

Except boron, all elements of groups 13 are metallic. These exists as solid. Being metallic in nature, aluminium is good conductor of electricity.

• (b) Liquid, metallic: The element in question belongs to the 3rd period and group 13 of the periodic table, which includes elements like aluminum. Aluminum is a solid at room temperature, not a liquid. Therefore, this option is incorrect.

• (d) Solid, non-metallic: While the element is indeed a solid, it is metallic in nature. Group 13 elements, except for boron, are metals. Therefore, this option is incorrect because it describes the element as non-metallic.

Answer

Electron gain enthalply of $\mathrm{F}$ is less negative than that of $\mathrm{Cl}$ because when an electron is added to $F$, the added electron goes to the smaller $n=2$ quantum level and suffers repulsion from other electrons present in this level.

In case of $\mathrm{Cl}$, the added electron goes to the larger $n=3$ quantum level and suffers much less repulsion from other electrons.

Answer

Elements in which the last electron enters in the $d$-orbitals, are called $d$-block elements or transition elements. These elements have the general outer electronic configuration $(n-1) d^{1-10} n s^{0-2} . \mathrm{Zn}, \mathrm{Cd}$ and $\mathrm{Hg}$ having the electronic configuration $(n-1) d^{10} n s^{2}$ do not show most of the properties of transition elements.

The $d$-orbitals in these elements are completely filled in the ground state as well as in their common oxidation states. Therefore, they are not regarded as transition elements.

Thus, on the basis of properties, all transition elements are $d$-block elements but on the basis of electronic configuration, all $d$-block elements are not transition elements.

Answer

The present set up of the Long form of the periodic table can accommodate maximum 118 elements. Thus, in accordance with aufbau principle, the filling of $8 s$-orbital will occur. In other words 119th electron will enter 8s-orbital. As such its outmost electronic configuration will be $8 s^{1}$.

Since, it has only one electron in the valence shell, i.e., $8 s$, therefore, its valency will be 1 and it will lie in the group IA along with alkali metals and the formula of its oxide will be $\mathrm{M}_{2} \mathrm{O}$ where $M$ represents the element.

Answer

To match the correct enthalpy with the elements and to complete the graph the following points are taken into consideration. As we move from left to right across a period, the ionisation enthalpy keeps on increasing due to increased nuclear charge and simultaneous decrease in atomic radius.

However, there are some exceptions given below

(a) In spite of increased nuclear charge, the first ionisation enthalpy of $B$ is lower than that of $\mathrm{Be}$. This is due to the presence of fully filled $2 s$ orbital of $\mathrm{Be}\left[1 s^{2} 2 s^{2}\right]$ which is a stable electronic arrangement. Thus, higher energy is required to knock out the electron from fully filled $2 s$ orbitals. While $B\left[1 s^{2} 2 s^{2} 2 p^{1}\right]$ contains valence electrons in $2 s$ and $2 p$ orbitals. It can easily lose its one $e^{-}$from $2 p$ orbital in order to achieve noble gas configuration. Thus, first ionisation enthalpy of $\mathrm{B}$ is lower than that of $\mathrm{Be}$.

Since, the electrons in $2 s$-orbital are more tightly held by the nucleus than these present in $2 p$-orbital, therefore, ionisation enthalpy of $\mathrm{B}$ is lower than that of $\mathrm{Be}$.

(b) The first ionisation enthalpy of $\mathrm{N}$ is higher than that of $\mathrm{O}$ though the nuclear charge of $\mathrm{O}$ is higher than that of $\mathrm{N}$. This is due to the reason that in case of $\mathrm{N}$, the electron is to be removed from a more stable exactly half-filled electronic configuration $\left(1 s^{2} 2 s^{2} 2 p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1}\right)$ which is not present in $\mathrm{O}\left(1 s^{2} 2 s^{2} 2 p_{x}^{2} 2 p_{y}^{1} 2 p_{z}^{1}\right)$.

Therefore, the first ionisation enthalpy of $\mathrm{N}$ is higher than that of $\mathrm{O}$. The symbols of elements along with their atomic numbers are given in the following graph

Answer

The placing of elements are as

(a) Ionisation enthalpy increases along a period (as we move from left to right in a period) with decrease in atomic size and decreases down the group with increase in atomic size. Hence, carbon has the highest first ionisation enthalpy.

(b) Metallic character decreases across a period but increases on moving down the group. Hence, aluminium has the most metallic character.

Answer

The four important characteristic properties of p-block elements are the following

(a) $p$-Block elements include both metals and non-metals but the number of non-metals is much higher than that of metals. Further, the metallic character increases from top to bottom within a group and non-metallic character increases from left to right along a period in this block.

(b) Their ionisation enthalpies are relatively higher as compared to s-block elements.

(c) They mostly form covalent compounds.

(d) Some of them show more than one (variable) oxidation states in their compounds. Their oxidising character increases from left to right in a period and reducing character increases from top to bottom in a group.

Answer

(a) Atomic radius of $\mathrm{F}$ is expressed in terms of covalent radius while, atomic radius of neon is usually expressed in terms of van der Waals’ radius. van der Waals’ radius of an element is always larger than its covalent radius.

Therefore, atomic radius of $\mathrm{F}$ is smaller than atomic radius of

$\mathrm{Ne}(\mathrm{F}=72 \mathrm{pm}, \mathrm{Ne}=160 \mathrm{pm})$.

Answer

Oxidation state of an element depends upon the electrons present in the outermost shell or eight minus the number of valence shell electrons (outermost shell electrons). e.g.,

Alkali metals (Group 1 elements) General valence shell electronic configuration — $n s^{1}$; Oxidation state $=+1$.

Alkaline earth metals (Group 2 elements) General valence shell electronic configuration $-n s^{2} ;$ Oxidation state $=+2$.

Alkali metals and alkaline earth metals belong to s-block elements and elements of group 13 to group 18 are known as $p$-block elements.

Group 13 elements General valence shell electronic configuration $-n s^{2} n p^{1}$; Oxidation states $=+3$ and +1 .

Group 14 elements General valence shell electronic configuration $-n s^{2} n p^{2}$; Oxidation states $=+4$ and +2 .

Group 15 elements General valence shell electronic configuration $-n s^{2} n p^{3}$; Oxidation states $=-3,+3$ and +5 . Nitrogen shows $+1,+2,+4$ oxidation states also.

Group 16 elements General valence shell electronic configuration $-n s^{2} n p^{4}$; Oxidation states $=-2,+2,+4$ and +6 .

Group 17 elements General valence shell electronic configuration $-n s^{2} n p^{5}$; Oxidation states $=-1 . \mathrm{Cl}, \mathrm{Br}$ and $\mathrm{I}$ also show $+1,+3,+5$ and +7 oxidation states.

Group 18 elements General valence shell configuration $-n s^{2} n p^{6}$. Oxidation state $=$ zero.

Transition elements or d-block elements General electronic configuration $-(n-1) d^{1-10} n s^{1-2}$. These elements show variable oxidation states due to involvement of not only $n s$ electrons but $d$ or $f$-electrons (inner-transition elements) as well. Their most common oxidation states are +2 and +3 .

Answer

Electronic configuration of $ _{7} N = 1 s^2, 2 s^2, 2 p_x^1, 2 p_y^1, 2 p_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 =1 s^2, 2 s^2, 2 p_x^2, 2 p_y^1, 2 p_z^1 $. Oxygen has $2 p^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 $2 p$-orbital, oxygen acquires stable configuration, i.e., $2 p^{3}$. On the other hand, in case of

nitrogen it is not easy to remove one of the three 2 n-electrons due to its stable configuration.

Answer

First member of each group of representative elements (i.e., $s$ - and $p$-block elements) shows anomalous behaviour due to (i) small size (ii) high ionisation enthalpy (iii) high electronegativity and (iv) absence of $d$-orbitals. e.g., in s-block elements, lithium shows anomalous behaviour from rest of the alkali metals.

(a) Compounds of lithium have significant covalent character. While compounds of other alkali metals are predominantly ionic.

(b) Lithium reacts with nitrogen to form lithium nitride while other alkali metals do not form nitrides.

In p-block elements, first member of each group has four orbitals, one $2 s$ - orbital and three $2 p$-orbitals in their valence shell. So, these elements show a maximum covalency of four while other members of the same group or different group show a maximum covalency beyond four due to availability of vacant $d$ - orbitals.

Answer

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 $B_{2} O_{3} <CO_{2} <N_{2} O_{3}$ acidic nature increases.

(ii) 3rd period $ Al_{2} O_{3} <SiO_{2} <P_{4} O_{10} <SO_{3} <Cl_{2} O_{7}$ acidic character increases.

On moving down the group, acidic character decreases and basic character increases. e.g.,

(a) Nature of oxides of 13 group elements

$\underset{\text { Weakly acidic }}{B_2 O_3} \quad \underbrace{Al_2 O_3 \quad Ga_2 O_3}_{\text {Amphoteric }} \quad \underset{\text { Basic}}{In_2 O_3} \quad \underset{\text {Strongly basic}}{TI_2 O} $

(b) Nature of oxides of 15 group elements

$\underset{\text { Strongly acidic }}{N_2 O_5} \quad \underset{\text{Moderately acidic}}{P4 O_{10}} \quad \underset{\text { Amphoteric}}{As_4 O_{10}} \quad \underset{\text {Amphoteric}}{Sb_4 O_{10}} \quad \underset{\text {Basic}}{Bi_2 O_3} $

Among the oxides of same element, higher the oxidation state of the element, stronger is the acid. e.g., $ SO_3 $ is a stronger acid than $ SO_2$.

$ B_2 O_3$ is weakly acidic and on dissolution in water, it forms orthoboric acid. Orthoboric acid does not act as a protonic acid (it does not ionise) but acts as a weak Lewis acid.

$$ \begin{aligned} \underset{\text { ron trioxide }}{ B_2 O_3}+3 H_2 O \rightleftharpoons & \underset{\text { Orthoboric acid }}{2 H_3 BO_3} \\ B(OH)_3+ H- OH & \longrightarrow[B OH_4 ]^- + H^+ \end{aligned} $$

$ Al_2 O_3 $ is amphoteric in nature. It is insoluble in water but dissolves in alkalies and reacts with acids.

$$ \underset{\substack{\text { Aluminium } \\ \text { trioxide }}}{ Al_2 O_3}+2 NaHO \xrightarrow[\substack{}]{\Delta} + \underset{\substack{\text { Sodium meta } \\ \text { aluminate }}}{2NaAIO_2}+ H_2 O $$

$$ Al_2 O_3 +6 mHCl \xrightarrow{\Delta} \underset{\begin{array}{c} \text { Aluminium } \\ \text { chloride } \end{array}}{2 AlCl_3}+3 H_2 O $$

$\mathrm{Tl}_{2} \mathrm{O}$ is as basic as $\mathrm{NaOH}$ due to its lower oxidation state (+ 1).

$$ Tl_2 O +2 HCl \longrightarrow 2 TlCl + H_2 O $$

$ P_4 O_{10} $ on reaction with water gives orthophosphoric acid

$\underset{\text{Phosphorus pentoxide}}{P _4O _{10}} + 6H_2O \longrightarrow \underset{\text{Orthophosphoric acid}}{4H_3PO_4} $

$ Cl_2 O_7$ is strongly acidic in nature and on dissolution in water, it gives perchloric acid.

$ \underset{\text{Dichlorine heptoxide }}{Cl_2O_7} + H_2O\longrightarrow \underset{\text{Perchloric acid}}{2HClO_4} $

Thinking Process

The species having exactly half-filled or fully filled orbitals have extra ordinarily high ionisation enthalpies.

Answer

First ionisation enthalpy of sodium $\left(\mathrm{Na}=1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{1}\right)$ is lower than that of magnesium $\left(\mathrm{Mg}=1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2}\right)$ because the electron to be removed in both the cases is from 3s-orbital but the nuclear charge is lower in $\mathrm{Na}$ than that of magnesium.

After the removal of first electron $\mathrm{Na}^{+}$acquires inert gas ( $\mathrm{Ne}$ ) configuration $\left(\mathrm{Na}^{+}=1 s^{2}, 2 s^{2}, 2 p^{6}\right)$ and hence, removal of second electron from sodium is difficult.

While in case of magnesium, after the removal of first electron, the electronic configuration of $\mathrm{Mg}^{+}$is $1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{1}$. In this case $3 s^{1}$ electron is easy to remove in comparison to remove an electron from inert gas configuration. Therefore, $\mathrm{IE}_{2}$ of $\mathrm{Na}$ is higher than that of magnesium.

Answer

Exothermic reactions Reactions which are accompanied by evolution of heat are called exothermic reactions. The quantity of heat produced is shown either along with the products with a ’ + ’ sign or in terms if $\Delta H$ with a ’ - ’ sign. e.g.,

$$ \begin{aligned} C(s)+ O_2(g) & \longrightarrow CO_2(g)+393.5 kJ \\ H_2(g)+\frac{1}{2} O_2(g) & \longrightarrow H_2 O(l) ; \Delta H=-285.8 kJ \quad mol^{-1} \end{aligned} $$

Endothermic reactions Reactions which proceed with absorption of heat are called endothermic reactions. The quantity of heat absorbed is shown either along with the products with a ‘-’ sign or in terms of $\Delta H$ with a ‘+’ sign e.g.,

$$ \begin{aligned} C(s)+ H_2 O(g) & \longrightarrow CO(g)+ H_2(g)-131.4 kJ \\ N_2(g)+3 H_2(g) & \longrightarrow 2 NH_3(g) ; \Delta H=+92.4 kJ mol^{-1} \end{aligned} $$

Answer

The placing of elements are as

(i) Ionisation enthalpy of nitrogen $ ( _{7} N=1 s^2, 2 s^2, 2 p^3 )$ is greater than oxygen ($ _{8} O =1 s^2$, $2 s^2, 2 p^4$ ) due to extra stable exactly half-filled $2 p$-orbitals. Similarly, ionisation enthalpy of phosphorus $ ( _{15} P =1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^3)$ is greater than sulphur $( _16 S=1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^4 )$.

On moving down the group, ionisation enthalpy decreases with increasing atomic size. So, the order is

$$ \mathrm{S}<\mathrm{P}<\mathrm{O}<\mathrm{N} \rightarrow \text { First ionisation enthalpy increases. } $$

(ii) Non-metallic character across a period (left to right) increases but on moving down the group it decreases. So, the order is

$\mathrm{P}<\mathrm{S}<\mathrm{N}<\mathrm{O} \rightarrow$ Non-metallic character increases.

Answer

There is deviation of ionisation enthalpy of some elements from the general trend as shown in figure. The first ionisation enthalpy of $\mathrm{B}$ is lower than that of $\mathrm{Be}$ and in case of nitrogen, the first ionisation enthalpy is higher than that of $O$. (Also, refer to Q. 27)

Answer

(a) Across the period, the nuclear charge increases and the atomic radius decreases. As a result, the tendency of the atom of an element to attract the shared pair of electrons towards itself increases and hence the electronegativity of the element increases. e.g., electronegativity of the elements of the 2nd period increases regularly from left to right as follows $\mathrm{Li}(1.0)$, $\mathrm{Be}(1.5), \mathrm{B}(2.0)$, C (2.5), N (3.0), O (3.5) and F (3.9)

(b) The ionisation enthalpy decreases regularly as we move from top to bottom, as explained below

(i) On moving down a group from top to bottom, the atomic size increases gradually due to the addition of a new principal energy shell at each succeding element. As a result, the distance between the nucleus and the valence shell increases.

In other words, the force of attraction of the nucleus for the valence electrons decreases and hence the ionisation enthalpy should decrease.

(ii) With the addition of new shells, the number of inner shell which shield the valence electrons from the nucleus increases. In other words, the shielding effect or the screening effect increases.

As a result, the force of attraction of the nucleus for the valence electrons further decreases and hence the ionisation enthalpy should decrease.

(iii) Further, in a group from top to bottom nuclear charge increases with increase in atomic number. As a result, the force of attraction of the nucleus for the valence electrons increases and hence the ionisation enthalpy should increase.

The combined effect of the increase in atomic size and screening effect more than compensate the effect of the increased nuclear charge. Consequently, the valence electrons become less and less firmly held by the nucleus and hence the ionisation enthalpy gradually decreases as we move down the group.

Answer

As we move from left to right in a period, the number of valence electrons increases by one at each succeeding element but the number of shells remains same. Due to this effective nuclear charge increases. More is the effective nuclear charge, more is the attraction between nuclei and electron.

Hence, the tendency of the element to lose electrons decreases, this results in decrease in metallic character. Furthermore, the tendency of an element to gain electrons increases with increase in effective nuclear charge, so non-metallic character increases on moving from left to right in a period.

Answer

When an atom loses an electron to form cation, its radius decreases. In a cation, per electron nuclear forces increases due to decrease in number of electrons. As a result of this, effective nuclear charge increases and the radius of cation decreases. e.g., ionic radius of $\mathrm{Na}^{+}$is smaller than the radius of its parent atom $\mathrm{Na}$.

$ \quad \quad \quad \quad \quad \quad \mathrm{Na} \longrightarrow \mathrm{Na}^{+}+1 e^{-}$

Electrons $ \quad \quad \quad 11 \quad 10$

Nuclear charge $\quad 11 \quad 11$

lonic size $\quad \quad \quad 186pm \quad 95pm $

Answer

On moving down the group, electronegativity decreases because atomic size increases. $\mathrm{Fr}$ has the largest size, therefore it is least electronegative.

Thinking Process

(i) All the element given in the question belong to the second period of the periodic table.

(ii) Atomic radius is the distance from the centre of the nucleus to the point upto which the density of the electron cloud (i.e., probability of finding the electron) is maximum.

Answer

All the given elements are of same period and along a period, atomic radii decreases because effective nuclear charge increases. Thus, the order of atomic radii is $\mathrm{O}<\mathrm{N}<\mathrm{C}<\mathrm{B}<\mathrm{Be}$ or, $\mathrm{Be}=111 \mathrm{pm}, \mathrm{O}=66 \mathrm{pm}, \mathrm{C}=77 \mathrm{pm}, \mathrm{B}=88 \mathrm{pm}, \mathrm{N}=74 \mathrm{pm}$.

Answer

(i) Most reactive non-metal has high $\Delta_{i} H_{1}$ and $\Delta_{i} H_{2}$ and most negative $\Delta_{\mathrm{eg}} H$. Therefore, the element is $B$.

(ii) Most reactive metal has low $\Delta_{i} H_{1}$ and high $\Delta_{i} H_{2}$ (because the second electron has to be lost from noble gas configuration) and has small negative $\Delta_{\mathrm{eg}} H$. Therefore, the element is $A$.

(iii) Noble gases are the least reactive elements. They have very high $\Delta_{i} H_{1}$ and $\Delta_{i} H_{2}$ and have positive $\Delta_{\mathrm{eg}} H$ values. Thus, the element is $D$.

(iv) Metal forming binary halides are alkaline earth metals. They have $\Delta_{i} H_{1}$ and $\Delta_{i} H_{2}$ values little higher than those of most reactive metals (such as $A$ ) and have comparatively slightly less negative $\Delta_{\mathrm{eg}} H$ values. Thus, the element is $C$.

Answer

A. $\rightarrow(4)$

B. $\rightarrow(1)$

C. $\rightarrow(2)$

D. $\rightarrow$ (3)

A. This electronic configuration corresponds to the noble gas i.e., neon. Since, noble gases have $+\Delta_{\mathrm{eg}} H$ values, therefore, electronic configuration (A) corresponds to the $\Delta_{\text {eg }} H=+48 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

B. This electronic configuration corresponds to the alkali metal i.e., potassium. Alkali metals have small negative $\Delta_{\mathrm{eq}} H$ values, hence, electronic configuration (B) corresponds to $\Delta_{\mathrm{eg}} H=-53 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

C. This electronic configuration corresponds to the halogen i.e., fluorine. Since, halogens have high negative $\Delta_{\mathrm{eg}} H$ values, therefore, electronic configuration (C) corresponds to $\Delta_{\text {eg }} H=328 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

D. This electronic configuration corresponds to the chalcogen i.e., oxygen. Since, chalcogens have $\Delta_{\mathrm{eg}} H$ values less negative than those of halogens, therefore, electronic configuration (D) corresponds to $\Delta_{\mathrm{eg}} H=-141 \mathrm{kJmol}^{-1}$.

Answer

(b) Assertion and reason both are correct statements and reason is correct explanation of assertion. Ionisation enthalpy increases along a period because effective nuclear charge increases and atomic size decreases.

Answer

(c) Assertion and reason both are correct statements and reason is correct explanation for assertion.

Boron has a smaller first ionisation enthalpy than beryllium because the penetration of a $2 s$ electron to the nucleus is more than the $2 p$ electron. Hence, $2 p$ electron is more shielded by the inner core of electron than the $2 s$ electron.

Answer

(b) Assertion and reason both are correct statements and reason is correct explanation for assertion.

Electron gain enthalpy becomes less negative as the size of an atom increases down the group. This is because within a group screening effect increases on going downward and the added electron would be farther away from the nucleus.

Answer

Electron gain enthalpy of an element is equal to the energy released when an electron is added to valence shell of an isolated gaseous atom.

$A(g)+e^{-} \longrightarrow A^{-}(g) ; \Delta_{e g} H=$ negative

Factors affecting electron gain enthalpy

(i) Effective nuclear charge Electron gain enthalpy increases with increase in effective nuclear charge because attraction of nucleus towards incoming electron increases. (ii) Size of an atom Electron gain enthalpy decreases with increase in the size of valence shell.

(iii) Type of subshell More closer is the subshell to the nucleus, easier is the addition of electron in that subshell.

Electron gain enthalpy (in decreasing order) for addition of electron in different subshell ( $n$-same) is $s>p>d>f$

(iv) Nature of configuration Half-filled and completely-filled subshell have stable configuration, so addition of electron in them is not energetically favourable.

Variation in the periodic table As a general rule, electron gain enthalpy becomes more and more negative with increase in the atomic number across a period. The effective nuclear charge increases from left to right across a period and consequently it will be easier to add an electron to a smaller atom.

Electron gain enthalpy becomes less negative as we go down a group because the size of the atom increases and the added electron would be farther from the nucleus.

Electron gain enthalpy of $\mathrm{O}$ or $\mathrm{F}$ is less than that of the succeeding element $(\mathrm{S}$ or $\mathrm{Cl})$ because the added electron goes to the smaller $n=2$ level and suffers repulsion from other electrons present in this level. For the $n=3$ level $(\mathrm{S}$ or $\mathrm{Cl}$ ), the added electron occupies a larger region of space and suffers much less repulsion from electrons present in this level.

Answer

Ionisation 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 ionisation enthalpy. It is represented by $\Delta_{i} H$.

Factors affecting ionisation enthalpy of the elements

lonisation enthalpy depends upon the following factors

(i) Nuclear charge The ionisation enthalpy increases with increase in nuclear charge. This is due to the fact that with increase in nuclear charge, the electrons of the outer shell are more firmly held by the nucleus and thus greater energy is required to pull out an electron from the atom.

e.g., the ionisation enthalpy increases as we move along a period from left to right due to increased nuclear charge.

(ii) Atomic size or radius lonisation enthalpy decreases as the atomic size increases. As the distance of the outer electrons from the nucleus increases with increase in atomic radius, the attractive force on the outer electron decreases.

As a result, outer electrons are held less firmly and hence lesser amount of energy is required to knock them out. Thus, ionisation enthalpy decreases with increase in atomic size. Ionisation enthalpy is found to decrease on moving down a group

(iii) Penetration effect of the electrons lonisation enthalpy increases as the penetration effect of the electrons increases. It is well known fact that in case of multielectron atoms, the electrons of the s-orbital has the maximum probability of being found near the nucleus and this probability goes on decreasing in case of $p, d$ and $f$-orbitals of the same shell.

In other words, s-electrons of any shell are more penetrating towards the nucleus than $p$-electrons the same shell. Thus, within the same shell, the penetration effect decreases in the order $s>p>d>f$

e.g., First ionisation enthalpy of aluminium is lower than that of magnesium. This is due to the fact that in case of aluminium $\left(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p_{x}^{1}\right)$, we have to pull out a $p$-electron to form $\mathrm{Al}^{+}$ion whereas in case of magnesium $\left(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}\right)$ we have to remove an $s$-electron of the same energy shell to produce $\mathrm{Mg}^{+}$ion.

(iv) Shielding or screening effect of inner shell electrons As the shielding or the screening effect of the inner electrons increases, the ionisation enthalpy decreases. Consequently, the force of attraction by the nucleus for the valence shell electrons decreases and hence the ionisation enthalpy decreases.

(v) Effect of arrangement of electrons If an atom contains exactly half filled or completely filled orbitals then such an arrangement has extra stability. Therefore, the removal of an electron from such an atom requires more energy than expected.

e.g., $\mathrm{Be}\left(1 s^{2} 2 s^{2}\right)$ has higher ionisation enthalpy than $\mathrm{B}\left(1 s^{2} 2 s^{2} 2 p^{1}\right)$ and $\mathrm{N}\left(1 s^{2} 2 s^{2} 2 p_{x}^{6} 2 p_{y}^{1} 2 p_{z}^{1}\right)$ has higher ionisation enthalpy than $\mathrm{O}\left(1 s^{2} 2 s^{2} 2 p_{x}^{2} 2 p_{y}^{1} 2 p_{z}^{1}\right)$. In general, as we move from left to right in a period, the ionisation enthalpy increases with increasing atomic numbers.

The ionisation enthalpies keep on decreasing regularly as we move down a group from one element to the other.

$\left(\Delta_{i} H\right)$

Answer

There are numerous physical properties of elements such as melting points, boiling points, heats of fusion and vaporisation, energy of atomisation, etc., which show periodic variations.

The cause of periodicity in properties is the repetition of similar outer electronic configurations after certain regular intervals. e.g., all the elements of 1st group (alkali metals) have similar outer electronic configuration, i.e., $n s^{1}$.

$$ \begin{aligned} & _{3} Li =1 s^2, 2 s^1 \\ & _{11} Na =1 s^2, 2 s^2, 2 p^6, 3 s^1 \\ & _{19} K =1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 4 s^1 \\ & _{37} Rb =1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^{10}, 4 s^2, 4 p^6, 5 s^1 \\ & _{55} Cs =1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^{10}, 4 s^2, 4 p^6, 4 d^{10}, 5 s^2, 5 p^6, 6 s^1 \\ & _{87} Fr =1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^{10}, 4 s^2, 4 p^6, 4 d^{10}, 4 f^14 \\ & 5 s^2, 5 p^6, 5 d^{10}, 6 s^2, 6 p^6, 7 s^1 \end{aligned} $$

Therefore, due to similar outermost shell electronic configuration all alkali metals have similar properties. e.g., sodium and potassium both are soft and reactive metals. They all form basic oxides and their basic character increases down the group. They all form unipositive ion by the lose of one electron.

Similarly, all the elements of 17th group (halogens) have similar outermost shell electronic configuration, i.e., $n s^{2} n p^{5}$ and thus possess similar properties.

$$ \begin{aligned} & _{9} F =1 s^2, 2 s^2, 2 p^5 \\ & _{17} Cl =1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^5 \\ & _{35} Br =1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^{10}, 4 s^2, 4 p^5 \\ & _{53} I=1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{6}, 3 d^{10}, 4 s^{2}, 4 p^{6}, 4 d^{10}, 5 s^{2}, 5 p^{5} \\ & _{85} At =1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{6}, 3 d^{10}, 4 s^{2}, 4 p^{6}, 4 d^{10}, \\ & 4 f^{14}, 5 s^{2}, 5 p^{6}, 5 d^{10}, 6 s^{2}, 6 p^{5} \end{aligned} $$

Answer

All the elements of group IA (or I), i.e., alkali metals have the similar outer electronic configuration, i.e., $n s^{1}$ where $n$ refers to the number of principal shell. These electronic configurations are given below

Hence, placement of all these elements in group 1 of the periodic table because of similarity in electronic configuration and all the elements have similar properties.

Answer

The main drawbacks of Mendeleev’s periodic table are

(i) Some elements having similar properties were placed in different groups whereas some elements having dissimilar properties were placed in the same group.

e.g., alkali metals such as $\mathrm{Li}, \mathrm{Na}, \mathrm{K}$, etc., (IA group) are grouped together with coinage metals such as $\mathrm{Cu}, \mathrm{Ag}, \mathrm{Au}$ (IB group) though their properties are quite different. Chemically similar elements such as $\mathrm{Cu}(\mathrm{IB}$ group) and $\mathrm{Hg}$ (IIB group) have been placed in different groups.

(ii) Some elements with higher atomic weights are placed before the elements with lower atomic weights in order to maintain the similar chemical nature of elements.

i.e., $ _{18}^{39.9} Ar \quad \text{and} \quad$ $ _{19}^{39.1} K ; _{27}^{58.9} Co \quad and \quad$ $ _{28}^{58.7} Ni $, etc.

(iii) Isotopes did not find any place in the periodic table. However, according to Mendeleev’s classification, these should be placed at different places in the periodic table.

(All the above three defects were however removed when modern periodic law based on atomic number was given).

(iv) Position of hydrogen in the periodic table is not fixed but is controversial.

(v) Position of elements of group VIII could not be made clear which have been arranged in three triads without any justification.

(vi) It could not explain the even and odd series in IV, V and VI long periods.

(vii) Lanthanides and actinides which were discovered later on have not been given proper positions in the main frame of periodic table.

Answer

The long form of the periodic table is better than Mendeleev’s periodic table because it classifies the elements on the basis of electronic configurations of their atoms.

The characteristics of this table are

(i) The table consists of 9 vertical columns, called the groups and 7 horizontal rows, called the periods.

(ii) The groups are marked 0 to VIII out of which group I to VII are subdivided into subgroups A and B.

(iii) The group IA elements ( $\mathrm{Li}, \mathrm{Na}, \mathrm{K}, \mathrm{Rb}, \mathrm{Cs}$ and $\mathrm{Fr}$ ) are known as alkali metals and the group IIA elements ( $\mathrm{Be}, \mathrm{Mg}, \mathrm{Ca}, \mathrm{Sr}, \mathrm{Ba}$ and $\mathrm{Ra}$ ) are known as alkaline earth metals. Elements in group VIII A ( $\mathrm{F}, \mathrm{Cl}, \mathrm{Br}, \mathrm{I}$ and $\mathrm{At}$ ) are called halogens and elements in group VIII $(\mathrm{He}, \mathrm{Ne}, \mathrm{Ar}, \mathrm{Kr}, \mathrm{Xe}$ and $\mathrm{Rn}$ ) are called noble gases or rare gases.

(iv) The group VIII has there similar elements placed together in one place. These are called transition triads, e.g., $\mathrm{Fe}, \mathrm{Co}$ and $\mathrm{Ni}$, $\mathrm{Ru}, \mathrm{Rh}$ and $\mathrm{Pd}$; $\mathrm{Os}$, Ir and Pt etc.

(v) In the 6th and 7th period, 14 elements present called as lanthanides and actinides respectively.

(vi) Based on their electronic configuration, elements have been grouped into $s-, p-, d-$ and $f$-blocks. This has helped us to understand their properties more easily.

(vii) There is gradual change in properties seen from one end to the other.

Answer

The ionisation enthalpies decreases regularly as we move down a group from one element to the other. This is evident from the values of the first ionisation enthalpies of the elements of group 1 (alkali metals) and group 17 elements as given in table and figure.

Given trend can be easily explained on the basis of increasing atomic size and screening effect as follows

(i) On moving down the group, the atomic size increases gradually due to the addition of one new principal energy shell at each succeeding element. Hence, the distance of the valence electrons from the nucleus increases.

Consequently, the force of attraction by the nucleus for the valence electrons decreases and hence the ionisation enthalpy should decrease.

(ii) With the addition of new shells, the shielding or the screening effect increases. As a result, the force of attraction of the nucleus for the valence electrons further decreases and hence the ionisation enthalpy should decrease.

(iii) Nuclear charge increases with increase in atomic number. As a result, the force of attraction by the nucleus for the valence electrons should increase and accordingly the ionisation enthalpy should increase.

The combined effect of the increase in the atomic size and the screening effect more than compensates the effect of the increased nuclear charges. Consequently, the valence electrons become less and less firmly held by the nucleus and hence the ionisation enthalpies gradually decrease as move down the group.