Chapter 3: Classification of Elements and Periodicity in Properties

Ans:

The least reactive element is likely to be element V. This is due to the fact that it has the highest initial ionization enthalpy $\left( {{\Delta _i}{H_1}} \right)$ and the highest positive electron gain enthalpy $\left( {{\Delta _{eg}}H} \right).$

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Because it has the lowest first ionization enthalpy $\left( {{\Delta _i}{H_1}} \right)$ and a low negative electron gain enthalpy $\left( {{\Delta _{eg}}H} \right).$ Element II is predicted to be the most reactive metal.

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Element III, with a high first ionization enthalpy $\left( {{\Delta _i}{H_1}} \right)$ and the largest negative electron gain enthalpy $\left( {{\Delta _{eg}}H} \right).$ is anticipated to be the most reactive nonmetal.

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Because it has a very high first ionization enthalpy $\left( {{\Delta _i}{H_2}} \right)$ and a positive electron gain enthalpy $\left( {{\Delta _{eg}}H} \right).$ element V is predicted to be the least reactive nonmetal.

(e) The metal which can form a stable binary halide of the formula ${\text{M}}{{\text{X}}_{\text{2}}}$ ($X$=halogen).

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The negative electron gain enthalpy of element VI is low$\left( {{\Delta _{eg}}H} \right).$ As a result, it is a metal. It also has the smallest second ionization enthalpy, $\left( {{\Delta _i}{H_2}} \right).$ As a result, a stable binary halide with the formula ${\text{M}}{{\text{X}}_{\text{2}}}$ ($X$=halogen) can be formed.

(f) The metal which can form a predominantly stable covalent halide of the formula ${\text{MX}}$ ($X$=halogen)?

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The first ionization energy of element I is low, whereas the second ionization energy is large. As a result, a primarily stable covalent halide with the formula ${\text{MX}}$ ($X$=halogen) can be formed.

Ans: ${\text{L}}{{\text{i}}_{\text{2}}}{\text{O}}$

Ans : ${\text{M}}{{\text{g}}_3}{{\text{N}}_2}$

Ans: ${\text{Al}}{{\text{I}}_3}$

Ans: ${\text{Si}}{{\text{O}}_2}$

(e) Phosphorus and fluorine

Ans: ${\text{P}}{{\text{F}}_{\text{3}}}{\text{ or P}}{{\text{F}}_{\text{5}}}$

(f) Element 71 and fluorine

Ans: Lutetium is an element with the atomic number 71. $\left( {{\text{Lu}}} \right){\text{.}}$ It has a valency of three. As a result, the compound's formula is ${\text{Lu}}{{\text{F}}_{\text{3}}}{\text{.}}$

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In the Modern periodic table, the value of the primary quantum number $\left( n \right)$ for the outermost shell or the valence shell denotes a period.

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Because a maximum of 10 electrons can fill all of the orbitals in a $d$ subshell, the $d$-block contains ten columns.

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The valence electrons are unaffected by nuclear mass.

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The valence electrons are unaffected by nuclear mass.

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The nucleus is more attracted to electrons in orbitals with a lower n value than electrons in orbitals with a higher $n$ value. As a result, removing electrons from orbitals with a larger $n$ value is easier than removing electrons from orbitals with a smaller $n$ value.

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Over time, the metallic character of components reduces from left to right. As a result, ${\text{Mg}}$ has a more metallic quality than ${\text{Al}}{\text{.}}$ Elements get more metallic as they progress through the group. As a result, ${\text{Al}}$ has a more metallic character than ${\text{B}}{\text{.}}$

${\text{K}} > {\text{Mg}}$ is the result of the foregoing statements.

As a result, the right metallic character order is ${\text{K}} > {\text{Mg}} > {\text{Al}} > {\text{B}}$

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Over time, the non-metallic nature of elements grows from left to right. ${\text{F}} > {\text{N}} > {\text{C}} > {\text{B}}$ is the decreasing order of non-metallic nature.

The non-metallic nature of elements reduces with each successive group. As a result, the non-metallic characteristics of ${\text{C}}$ and ${\text{Si}}$ are in decreasing order: ${\text{C}} > {\text{Si}}$. ${\text{Si,}}$ on the other hand, is less non-metallic than ${\text{B,}}$ hence ${\text{B}} > {\text{Si}}{\text{.}}$

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Over time, the oxidizing property of elements rises from left to right. As a result, we have ${\text{F}} > {\text{O}} > {\text{N}}$ as the decreasing order of oxidizing properties.

The ability of elements to oxidize reduces as they progress through a group. As a result,${\text{F}} > {\text{Cl}}{\text{.}}$ However, ${\text{O}}$ has a higher oxidizing character than ${\text{Cl,}}$ i.e., ${\text{O}} > {\text{Cl}}{\text{.}}$ As a result, in terms of their oxidizing property, the right sequence of chemical reactivity of \[{\text{F, Cl, O and N,}}\] is ${\text{F}} > {\text{O}} > {\text{Cl}} > {\text{N}}{\text{.}}$