Chapter 3: Classification of Elements and Periodicity in Properties

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Energy is released when an electron is added to the ${\text{O}}$ atom to form the ${{\text{O}}^ - }$ ion. As a result, ${\text{O}}$ first electron gain enthalpy is negative.

${\text{O}} + {{\text{e}}^ - } \to {{\text{O}}^ - }$

When an electron is added to an ${{\text{O}}^ - }$ ion to generate an ${{\text{O}}^{2 - }}$ ion, however, energy must be given out to overcome the strong electronic repulsions. As a result, ${\text{O's}}$ second electron gain enthalpy is positive.

${{\text{O}}^ - } + {{\text{e}}^ - } \to {{\text{O}}^{2 - }}$

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The tendency of an isolated gaseous atom to accept an electron is measured by electron gain enthalpy, whereas the tendency of an atom in a chemical compound to attract a shared pair of electrons is measured by electronegativity.

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An element's electronegativity is a changeable property. It varies depending on the chemical. As a result, the statement that all nitrogen compounds have an electronegativity of 3.0 on the Pauling scale is erroneous. The electronegativity of ${\text{N}}$ in ${\text{N}}{{\text{H}}_{\text{3}}}$ and ${\text{N}}{{\text{O}}_{\text{2}}}$ is different.

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(a) An atom's size increases when it gains an electron. The number of electrons increases by one when one is added. The electrons' repulsion increases as a result of this. The number of protons, however, remains constant. As a result, the atom's effective nuclear charge drops while its radius increases.

(b) When an atom loses an electron, the number of electrons in the atom reduces by one, but the nuclear charge does not change. As a result, the atom's interelectronic repulsions decrease. The effective nuclear charge rises as a result. As a result, the atom's radius shrinks.

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The number of electrons and protons (nuclear charge) in an atom determines its ionisation enthalpy. The protons and electrons in an element's isotopes are now the same. As a result, for two isotopes of the same element, the first ionisation enthalpy should be the same.

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An element with 5 electrons in its outermost subshell should have the electrical configuration $n{s^2}n{p^5}.$ The electronic configuration of the halogen group is as follows. The element can therefore be ${\text{F, Cl, Br or I}}{\text{.}}$

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To achieve the stable noble gas configuration, an element with two valence electrons can simply lose two electrons. The element's overall electrical configuration will be $n{s^2}.$ The electrical configuration of group 2 components is shown here. ${\text{Be, Mg, Ca, Sr and Ba}}$ are the elements found in group 2.

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If an element only requires two electrons to achieve the stable noble gas state, it is likely to gain two electrons. As a result, such an element's general electrical configuration should be $n{s^2}n{p^4}.$ The electrical configuration of the oxygen family is as follows.

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At room temperature, Group 17 contains metals, nonmetals, liquids, and gases.

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Group 1 elements have only one valence electron, which they frequently lose. The noble gas structure, on the other hand, requires only one electron for Group 17 elements. The ionization enthalpies drop as you move along group 1. This reduces the energy required to remove the valence electron. As a result, on moving down a group, the reactivity increases.

${\text{Li}} < {\text{Na}} < {\text{K}} < {\text{Rb}} < {\text{Cs}}$

As a result, the following is the increasing order of reactivity among group 1 elements:

The electron gain enthalpy decreases as we proceed down the group from ${\text{Cl}}$ to ${\text{I}}$ in group 17. Its tendency to gain electrons reduces as it moves along group 17. As a result, responsiveness reduces the size of a group. ${\text{F}}$ has a lower negative electron gain enthalpy than ${\text{Cl}}{\text{.}}$ Even yet, it is the most reactive of the halogens. This is due to the fact that it has a low bond dissociation energy. As a result, the following is the decreasing order of reactivity among group 17 elements: ${\text{F}} > {\text{Cl}} > {\text{Br}} > {\text{I}}{\text{.}}$

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$s$ block element: $n{s^{1 - 2}},$ where, $n = 2 - 7$

$p$ block element: $n{s^2}n{p^{1 - 6}},$ where, $n = 2 - 6$

$d$ block element: $\left( {n - 1} \right){d^{1 - 10}}n{s^{0 - 2}},$ where, $n = 4 - 7$

$f$ block element: \[\left( {n - 2} \right){f^{1 - 14}}\left( {n - 1} \right){d^{1 - 10}}n{s^2},\] where, $n = 6 - 7$

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The element belongs to the 3rd period since $n = 3.$ Because the last electron occupies the $p$ orbital, it is a $p$–block element. In the $p$–orbital, there are four electrons. As a result, the element's matching group.

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The element belongs to the 4th period since $n = 4.$ Because the $d$– orbitals are not completely filled, it is a $d$–block element. In the $d$–orbital, there are two electrons. As a result, the element's matching group.

corresponding group of element = number of elements in (d - block + s - block)  

= 2 + 2  

= 4

As a result, it belongs to the fourth period and fourth group. As a result, the element's name is Titanium.

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The element is present in the 6th period because $n = 6.$ The last electron fills the $f$–orbital, making it a $f$–block element. It belongs to group 3 of the periodic table, which includes all $f$-block elements. The electronic configuration is as follows: $\left[ {{\text{Xe}}} \right]4{f^7}5{d^1}6{s^2}.$ As a result,

atomic number = 54 + 7 + 2 + 1

= 64

As a result, the element's name is Gadolinium.