
Aluminium is usually found in +3 oxidation state. In contrast, thallium exists in +1 and +3 oxidation states. This is due to:
A. lanthanide contraction
B. lattice effect
C. diagonal relationship
D. inert pair effect
Answer
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Hint: The term inert pair effect is commonly employed in reference to the increasing stability of oxidation states that are two but the group valency for the heavier elements of groups 13, 14, 15 and 16. The name suggests that the outermost s electrons are more tightly sure to the nucleus in these atoms, and thus harder to ionize or share.
Complete step by step answer:
-Lanthanoid contraction, also called lanthanide contraction, in chemistry, the steady decrease within the size of the atoms and ions of the lanthanide elements with increasing number from lanthanum (atomic number 57) through lutetium (atomic number 71).
-Let's take p-block elements of the 4th, 5th and 6th period come after d-block elements, but the electrons present within the intervening d- (and f-) orbitals don't effectively shield the s-electrons of the valence shell. As a result, the inert pair of ns electrons remains more tightly held by the nucleus and hence participates less in bond formation.
-Diagonal relationships occur thanks to the various ways during which many atomic properties vary down groups and across periods of the table. Additionally to the group and period relationships, the weather of s and p-block elements also shows diagonal relationships.
-The inert pair effect is the tendency of the 2 electrons within the outermost atomic s-orbital to stay unshared in compounds of post-transition metals. As a result, the inert pair of the electrons of ns orbitals remains more tightly held by the nucleus and hence participates less in bond formation.
-Inert pair effect is the prominent character of the p-block element.
In this, the high relative molecular mass element of the group shows a lower oxidation number.
-This is because on taking place the group, the shielding effect increases, but d-subshell and f-subshell show poor shielding effect.
-The high relative molecular mass members of p-block groups contain d-subshell and f-subshell. Because of this, they show a lower number.
Hence, option D is correct.
Note: Aluminum has three oxidation states. The foremost common one is +3. The opposite two are +1 and +2. One +3 number for Aluminum is often found within the compound aluminium oxide, $A{{l}_{2}}{{O}_{3}}$.
Complete step by step answer:
-Lanthanoid contraction, also called lanthanide contraction, in chemistry, the steady decrease within the size of the atoms and ions of the lanthanide elements with increasing number from lanthanum (atomic number 57) through lutetium (atomic number 71).
-Let's take p-block elements of the 4th, 5th and 6th period come after d-block elements, but the electrons present within the intervening d- (and f-) orbitals don't effectively shield the s-electrons of the valence shell. As a result, the inert pair of ns electrons remains more tightly held by the nucleus and hence participates less in bond formation.
-Diagonal relationships occur thanks to the various ways during which many atomic properties vary down groups and across periods of the table. Additionally to the group and period relationships, the weather of s and p-block elements also shows diagonal relationships.
-The inert pair effect is the tendency of the 2 electrons within the outermost atomic s-orbital to stay unshared in compounds of post-transition metals. As a result, the inert pair of the electrons of ns orbitals remains more tightly held by the nucleus and hence participates less in bond formation.
-Inert pair effect is the prominent character of the p-block element.
In this, the high relative molecular mass element of the group shows a lower oxidation number.
-This is because on taking place the group, the shielding effect increases, but d-subshell and f-subshell show poor shielding effect.
-The high relative molecular mass members of p-block groups contain d-subshell and f-subshell. Because of this, they show a lower number.
Hence, option D is correct.
Note: Aluminum has three oxidation states. The foremost common one is +3. The opposite two are +1 and +2. One +3 number for Aluminum is often found within the compound aluminium oxide, $A{{l}_{2}}{{O}_{3}}$.
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