Chapter 3: Classification of Elements and Periodicity in Properties

Ans :

Isoelectronic species refer to atoms and ions that have the same number of electrons.

(i) The ${{\text{F}}^ - }$ ion possesses a total of 10 electrons. As a result, the species that is isoelectronic with it will contain 10 electrons as well. ${\text{N}}{{\text{a}}^ + }$ ion $\left( {11 - 1 = 10} \right)$ electrons, ${\text{Ne}}$ (10 electrons), ${{\text{O}}^{2 - }}$ ion $\left( {8 + 2 = 10} \right)$ electrons, are some of its isoelectronic species.

(ii) ${\text{Ar}}$ possesses a total of 18 electrons. As a result, the species that is isoelectronic with it will contain 18 electrons as well. ${{\text{S}}^{2 - }}$ ion $\left( {16 + 2 = 18} \right)$ electrons, ${\text{C}}{{\text{l}}^ - }$ ion $\left( {17 + 1 = 18} \right)$ electrons, and ${\text{C}}{{\text{a}}^{2 + }}$ ion $\left( {20 - 2 = 18} \right)$electrons are some of its isoelectronic species.

(iii) The electron count of the ${\text{M}}{{\text{g}}^{2 + }}$ ion is 10 electrons. As a result, the species that is isoelectronic with it will contain 10 electrons as well. ${\text{N}}{{\text{a}}^ + }$ ion $\left( {11 - 1 = 10} \right)$ electrons, ${\text{Ne}}$ (10 electrons), ${{\text{O}}^{2 - }}$ ion $\left( {8 + 2 = 10} \right)$ electrons, are some of its isoelectronic species.

(iv) The electron count of the ${\text{R}}{{\text{b}}^ + }$ ion is 36 electrons. As a result, the species that is isoelectronic with it will contain 36 electrons as well. ${\text{B}}{{\text{r}}^ - }$ ion $\left( {35 + 1 = 36} \right)$ electrons, ${\text{Kr}}$ (36 electrons), and ${\text{S}}{{\text{r}}^{2 + }}$ ion $\left( {38 - 2 = 36} \right)$ electrons) are some of its isoelectronic species.

Ans:

The number of electrons in each of the provided species (ions) is the same (10 electrons). As a result, the species in question are isoelectronic.

Ans :

As the magnitudes of nuclear charge fall, the ionic radii of isoelectronic species increases. The following is the order in which the supplied species are arranged in order of increasing nuclear charge.

${{\text{N}}^{3 - }} < {{\text{O}}^{2 - }} < {{\text{F}}^ - } < {\text{N}}{{\text{a}}^ + } < {\text{M}}{{\text{g}}^{2 + }} < {\text{A}}{{\text{l}}^{3 + }}.$

As a result, the following is the order of the supplied species in terms of increasing ionic radii:

\[{\text{A}}{{\text{l}}^{3 + }} < {\text{M}}{{\text{g}}^{2 + }} < {\text{N}}{{\text{a}}^ + } < {{\text{F}}^ - } < {{\text{O}}^{2 - }} < {{\text{N}}^{3 - }}.\]

Ans :

A cation possesses fewer electrons than its parent atom while maintaining the same nuclear charge. As a result, a cation's nucleus attracts electrons more than its parent atom's nucleus. As a result, a cation is smaller than its parent atom in size. An anion, on the other hand, possesses one or more electrons than its parent atom, resulting in higher electron repulsion and a reduction in effective nuclear charge. As a result, anions have a greater distance between their valence electrons and the nucleus than their parent atom. As a result, the radius of an anion is bigger than that of its parent atom.

Ans :

The energy required to remove an electron from an isolated gaseous atom in its ground state is known as ionisation enthalpy. Despite the fact that the atoms are far separated in the gaseous state, there are some attraction forces between them. It is impossible to isolate a single atom to calculate the ionisation enthalpy. However, by lowering the pressure, the power of attraction can be reduced much more. As a result, in the definition of ionisation enthalpy, the term "isolated gaseous atom" is employed.

The most stable state of an atom is called the ground state. If an isolated gaseous atom is in its ground state, removing an electron from it will need less energy. As a result, ionisation enthalpy and electron gain enthalpy for an ‘isolated gaseous atom' and its ‘ground state' must be established for comparison reasons.

Ans :

The energy of an electron in the hydrogen atom's ground state is $ - 2.18 \times {10^{ - 18}}{\text{J}}.$ As a result, the energy required to remove that electron from the hydrogen atom's ground state is $2.18 \times {10^{ - 18}}{\text{J}}.$

Thus, the ionization enthalpy of atomic hydrogen is $2.18 \times {10^{ - 18}}{\text{J}}.$

As a result, the ionization enthalpy of atomic hydrogen in terms of ${\text{J mo}}{{\text{l}}^{ - 1}}$ can be calculated as,

Ionization enthalpy = 2.18 x 10⁻¹⁸ x 6.02 x 10²³J mol⁻¹

= 1.31 x 10⁶ J mol⁻¹

Ans :

(i) The electron to be removed from a beryllium atom during the ionisation process is a $2s$ electron, whereas the electron to be removed from a boron atom is a $2p$-electron. $2s$-electrons are now stronger than $2p$-electrons in their attachment to the nucleus. As a result, removing a $2s$ electron from beryllium requires more energy than removing a $2p$ electron from boron.

Ans:

The three $2p$-electrons of nitrogen occupy three separate atomic orbitals in nitrogen. In oxygen, however, two of the four $2p$-electrons share the same $2p$-orbital. Increased electron-electron repulsion occurs in the oxygen atom as a result of this. As a result, removing the fourth $2p$-electron from oxygen requires less energy than removing one of the three $2p$-electrons from nitrogen. As a result, oxygen has a lower ionization enthalpy than nitrogen. Fluorine has one extra electron and one more proton than oxygen. The increase in nuclear attraction (due to the addition of a proton) is greater than the rise in electronic repulsion as the electron is added to the same shell (due to the addition of an electron). As a result, the valence electrons in fluorine atoms have a greater effective nuclear charge than the electrons in oxygen atoms. As a result, removing an electron from a fluorine atom requires more energy than removing an electron from an oxygen atom. As a result, oxygen has a lower ionization enthalpy than fluorine.

Ans :

Sodium has a higher initial ionisation enthalpy than magnesium. This is mostly due to the following two factors:

(a) Sodium's atomic size is larger than that of magnesium.

(b) Magnesium has a higher effective nuclear charge than sodium.

As a result, the energy required to take an electron from magnesium is more than that necessary to remove an electron from sodium. As a result, sodium's first ionisation enthalpy is smaller than magnesium. The second ionisation enthalpy of sodium, on the other hand, is larger than that of magnesium. Because sodium loses an electron, it achieves the stable noble gas structure. Magnesium, on the other hand, retains one electron in the 3s orbital after losing an electron. It still has to lose one more electron to achieve the stable noble gas structure. As a result, the energy required to remove the second electron in sodium is significantly more than that required in magnesium. As a result, sodium's second ionisation enthalpy is greater than magnesium's.

Ans :

The following are the causes that cause the ionisation enthalpy of the primary group elements to fall along a group:

(i) Elements' atomic sizes grow larger: The number of shells grows as we proceed down a group. As a result, as you move down a group, the atomic size gradually increases. The valence electrons are not held as tightly when the distance between them and the nucleus grows. As a result, they can be simply removed. As a result, as you move down a group, the ionisation energy diminishes.

(ii) Shielding effect: As you move down a group, the number of electrons in their inner shells grows. As a result, the inner core electrons protect the valence electrons from the nucleus as they go along a group. As a result, the nucleus does not hold the valence electrons particularly tightly. As a result, the energy required to remove a valence electron reduces as the group number increases.

Ans :

The ionization enthalpy reduces with the atomic size and electron shielding lowers as you move down a group. Ionization enthalpy reduces from ${\text{B to Al,}}$ as you move down group 13. ${\text{Ga,}}$ on the other hand, has a larger ionization enthalpy than ${\text{Al}}{\text{.}}$ ${\text{Al}}$ comes after the $s$–block elements, while ${\text{Ga}}$ comes after the $d$–block elements. $d$-electrons do not provide very good shielding. The valence electrons are not well shielded by these electrons. As a result, Ga valence electrons have a higher effective nuclear charge than ${\text{Al}}$ valence electrons. Furthermore, when the atomic size and shielding drop from ${\text{Ga}}$ to ${\text{In,}}$ the ionization enthalpy reduces. However, as the ionization enthalpy increases from ${\text{In to Tl,}}$ the ionization enthalpy decreases. ${\text{Tl}}$ is found after the $4f$ and $5d$ electrons in the periodic table. The electron shielding offered by each of these orbitals is ineffective.

As a result, the nucleus tightly holds the valence electron. As a result, ${\text{Tl's}}$ ionization energy is on the upper side.

Ans :

The elements ${\text{O}}$ and ${\text{F}}$ are found in the same periodic table period. ${\text{F}}$ atom has one more proton and one more electron than an ${\text{O}}$ atom, and because an electron is added to the same shell, ${\text{F}}$ atomic size is lower than ${\text{O}}$'s. Because ${\text{F}}$ has one extra proton than ${\text{O}}$ its nucleus may attract incoming electrons more strongly than the nucleus of the ${\text{O}}$ atom. Furthermore, ${\text{F}}$ just requires one extra electron to achieve the stable noble gas structure. As a result, the electron gain enthalpy of ${\text{F}}$ is lower than that of ${\text{O}}{\text{.}}$

(ii) ${\text{F}}$ and ${\text{Cl}}$ belong to the same periodic table group. Moving down a group, the electron gain enthalpy usually gets less negative. However, the electron gain enthalpy of ${\text{Cl}}$ is less negative than that of ${\text{F}}$ in this scenario. This is due to the fact that ${\text{F}}$ atomic size is smaller than ${\text{Cl}}{\text{.}}$ The electron will be added to quantum level $n = 2$ in ${\text{F}}$ but it will be added to quantum level $n = 3$ in ${\text{Cl}}{\text{.}}$ As a result, there are fewer electron-electron repulsions in ${\text{Cl}}$ and an extra electron can be easily tolerated. As a result, ${\text{Cl}}$ electron gain enthalpy is lower than that of ${\text{F}}{\text{.}}$