CBSE Class 11 Chemistry Chapter 3: Classification of Elements and Periodicity in Properties NCERT Solutions

Solution: Isoelectronic species are atoms or ions that contain the same number of electrons. In this case, $Na^+$ (11-1=10 e⁻), $Mg^\{2+\}$ (12-2=10 e⁻), $F^-$ (9+1=10 e⁻), and $O^\{2-\}$ (8+2=10 e⁻) all have 10 electrons.

For isoelectronic species, the ionic radius decreases as the nuclear charge (atomic number) increases. This is because a higher nuclear charge pulls the electrons more strongly, resulting in a smaller radius.

The atomic numbers are:

• $Mg^\{2+\}$: Z = 12
• $Na^\{+\}$: Z = 11
• $F^\{-\}$: Z = 9
• $O^\{2-\}$: Z = 8

Therefore, the order of increasing ionic radii is based on the increasing atomic number:

The correct option is (b).

Solution: The actinoid series includes elements with atomic numbers from 90 (Thorium) to 103 (Lawrencium). These are the elements in which the 5f orbitals are progressively filled.

Let's examine the given options:

• Curium ($Z\; =\; 96$) falls within the range 90-103, so it is an actinoid.
• Californium ($Z\; =\; 98$) falls within the range 90-103, so it is an actinoid.
• Uranium ($Z\; =\; 92$) falls within the range 90-103, so it is an actinoid.
• Terbium ($Z\; =\; 65$) has an atomic number less than 90. Elements with atomic numbers 57-71 are called lanthanoids, and Terbium belongs to this series.

Therefore, Terbium ($Z\; =\; 65$) is not an actinoid.

The correct option is (d).

Solution: The screening effect (or shielding effect) refers to the reduction in the effective nuclear charge experienced by an electron due to the presence of other electrons in the inner shells and the same shell. The extent of screening depends on the shape of the orbital and the distance of the electron from the nucleus.

Electrons in s orbitals are closest to the nucleus and penetrate the nucleus most effectively, providing the strongest screening. Electrons in p orbitals are next, followed by d orbitals, and then f orbitals, which are furthest from the nucleus and penetrate the least.

Therefore, the order of screening effect from strongest to weakest is:

$s\; >\; p\; >\; d\; >\; f$

This order correctly describes how effectively electrons in these orbitals shield the outer shell electrons from the nucleus.

The correct option is (a).

Solution: To determine the order of first ionization enthalpies ($IP\_1$) for Na, Mg, Al, and Si, we need to consider their electronic configurations and positions in the periodic table.

1. Electronic Configurations:
   • Na ($Z=11$): $[Ne]\; 3s^1$
   • Mg ($Z=12$): $[Ne]\; 3s^2$
   • Al ($Z=13$): $[Ne]\; 3s^2\; 3p^1$
   • Si ($Z=14$): $[Ne]\; 3s^2\; 3p^2$
2. General Trend: Ionization enthalpy generally increases across a period from left to right due to an increase in effective nuclear charge. So, we expect .
3. Exceptions: We must consider exceptions. Magnesium ($Mg$) has a completely filled $3s^2$ subshell, which is a stable configuration. Aluminium ($Al$) has a $3p^1$ electron, which is easier to remove than an electron from a filled $3s^2$ subshell.\nTherefore, the first ionization enthalpy of Mg is greater than that of Al ($IP\_1(Mg)\; >\; IP\_1(Al)$).
4. Overall Order:
   • Na has the lowest $IP\_1$ as it has only one valence electron ($3s^1$).
   • Mg has a higher $IP\_1$ than Al due to the stable $3s^2$ configuration.
   • Si has a higher $IP\_1$ than Al because it is further to the right in the period.
   \nCombining these points, the order is and . Thus, the complete order is .

The correct option is (a).

Solution: Gadolinium ($Gd$) has an atomic number $Z\; =\; 64$. Its electronic configuration needs to be determined by considering the filling order of orbitals and the stability of half-filled and fully-filled subshells.

The preceding noble gas is Xenon ($Xe$), with $Z\; =\; 54$. The electronic configuration of $Xe$ is $[Kr]\; 4d^\{10\}\; 5s^2\; 5p^6$. After Xenon, the filling order follows the $\backslash textit\{Aufbau\}$ principle, considering the relative energies of orbitals ($4f,\; 5d,\; 6s$).

The electronic configuration of Lanthanum ($La$, $Z\; =\; 57$) is $[Xe]\; 5d^1\; 6s^2$. The subsequent elements fill the $4f$ orbitals.

The filling proceeds as follows:

• Europium ($Eu$, $Z\; =\; 63$) has the configuration $[Xe]\; 4f^7\; 6s^2$. This is a stable, half-filled $4f$ subshell.
• Gadolinium ($Gd$, $Z\; =\; 64$) has one more electron than Europium. According to the $\backslash textit\{Aufbau\}$ principle and Hund's rule, the next electron is added to the lowest energy orbital available. While $4f$ is energetically favorable, adding the 64th electron to $4f$ would make it $4f^8$, which is not a particularly stable configuration compared to a half-filled $4f^7$ shell. Instead, the electron occupies the next available higher energy orbital, which is the $5d$ orbital.

Therefore, the electronic configuration of Gadolinium ($Gd$) is:

$[Xe]\; 4f^7\; 5d^1\; 6s^2$

This configuration reflects the stability of the half-filled $4f$ subshell and the filling of the $5d$ orbital.

The correct option is (c).

Solution: Let's analyze each statement regarding the periodic classification of elements:

1. Statement (a): "The properties of elements are periodic function of their atomic numbers." This is the fundamental principle of the modern periodic law, established by Moseley. It is correct.
2. Statement (b): "Non-metallic elements are less in number than metallic elements." In the periodic table, metals occupy the left and central parts, while non-metals are primarily located on the upper right side. The number of metallic elements significantly exceeds the number of non-metallic elements. This statement is correct.
3. Statement (c): "For transition elements, the 3d-orbitals are filled with electrons after 3p-orbitals and before 4s-orbitals." This statement is incorrect. According to the $\backslash textit\{Aufbau\}$ principle, the energy levels of orbitals dictate the order of filling. The $4s$ orbital has a lower energy than the $3d$ orbitals. Therefore, the $4s$ orbital is filled *before* the $3d$ orbitals. For example, in Potassium ($Z=19$) and Calcium ($Z=20$), the $4s$ orbital is filled first ($[Ar]\; 4s^1$ and $[Ar]\; 4s^2$). The filling of $3d$ orbitals begins with Scandium ($Z=21$) with the configuration $[Ar]\; 3d^1\; 4s^2$. So, $3d$ orbitals are filled *after* the $4s$ orbital is filled (or at least partially filled).
4. Statement (d): "The first ionisation enthalpies of elements generally increase with increase in atomic number as we go along a period." This is a general trend observed across a period due to the increasing effective nuclear charge, which holds electrons more tightly. While there are exceptions (like $Mg$ vs $Al$), the general trend is an increase. This statement is correct.

The statement that is not correct is (c).

The correct option is (c).

Solution: Electron gain enthalpy ($\backslash Delta\_\{eg\}H$) is the energy change that occurs when an electron is added to a neutral gaseous atom to form a negative ion.

The general trend for electron gain enthalpy down a group is that it becomes less negative (less energy is released) because the incoming electron is added to shells further from the nucleus, and the atomic size increases, reducing the attraction from the nucleus.

For halogens (Group 17), the electron gain enthalpies are expected to decrease in magnitude (become less negative) from F to I:

• Fluorine ($F$)
• Chlorine ($Cl$)
• Bromine ($Br$)
• Iodine ($I$)

However, there is a significant exception for Fluorine. Although Fluorine is higher up in the group than Chlorine, its electron gain enthalpy is less negative than that of Chlorine. This is because Fluorine is a very small atom, and the addition of an electron to its compact $2p$ subshell results in considerable electron-electron repulsion. Chlorine, being larger, has more diffuse $3p$ orbitals, and the electron-electron repulsion is less significant.

Therefore, the actual order of electron gain enthalpies (energy released, so more negative means more energy released) is:

In terms of the amount of energy released (magnitude), the order is:

$|\; \backslash Delta\_\{eg\}H(Cl)\; |\; >\; |\; \backslash Delta\_\{eg\}H(F)\; |\; >\; |\; \backslash Delta\_\{eg\}H(Br)\; |\; >\; |\; \backslash Delta\_\{eg\}H(I)\; |$

This means Chlorine releases the most energy, followed by Fluorine, then Bromine, and lastly Iodine.

The correct option representing this order is (c) , which implies that the magnitude of energy released follows $CI\; >\; F\; >\; Br\; >\; I$.