Classification of elements and periodicity in properties practice · Mixed difficulty

Option C ($\mathrm{Sn}{\mathrm{O}_2},\mathrm{A}{\mathrm{l}_2}{\mathrm{O}_3},\mathrm{ZnO}$) is correct because all three oxides are known to be amphoteric.

Option C is correct because Cr+ has a stable half-filled d5 configuration, making it difficult to remove another electron.

Option B (Moseley) is correct because Moseley discovered that the properties of elements are periodic functions of their atomic number.

Option A (Alkali metal) is correct because alkali metals have the largest atomic radii in their respective periods, thus occupying the peak positions in Lothar Meyer's curve.

Option B (${F^ - } < {O^{2 - }} < {N^{3 - }}$) is correct because these are isoelectronic species, and size increases with increasing negative charge (decreasing effective nuclear charge).

Option A (Arithmetic mean of the other two elements) is correct because Dobereiner's triads were based on the principle that the atomic weight of the middle element is approximately the arithmetic mean of the other two.

Option D (Siderophiles) is correct because siderophiles are elements that tend to dissolve in iron, which is the primary component of the Earth's core.

Option A (+2, -2) is correct because in OF2, oxygen is bonded to the more electronegative fluorine, giving it a positive oxidation state, while in Na2O, oxygen is bonded to the less electronegative sodium, giving it a negative oxidation state.

Option B (C > Be > B > Li) is correct because it follows the general trend of ionization energy increasing across a period, with the exception of Be being higher than B due to electron configuration.

Option B (18, 7) is correct because IUPAC recognizes 18 groups and 7 periods in the periodic table.

Option A (${N^{ - 3}}$) is correct because it has the lowest nuclear charge among the isoelectronic species, leading to the largest ionic radius.

Option C (6 & 1) is correct because the 6th period is the longest and the 1st period is the shortest in the periodic table.

Option D ($Be,\,Al$) is correct because Beryllium (Be) and Aluminum (Al) exhibit a diagonal relationship due to similarities in charge/size ratio.

Option C (13 and 31) is correct because elements with atomic numbers 13 and 31 belong to the same group (Group 13).

Option A (BeO) is correct because BeO is the most acidic among the given oxides due to Be's high polarizing power.

Option B (Halogens) is correct because after gaining one electron, halogens achieve a stable noble gas configuration, making it energetically unfavorable to add another electron.

Option B is correct because element 104 was named Rutherfordium by the American Society and Kurchatovium by the Soviet Society, leading to a naming dispute.

Option B is correct. Eka-silicon corresponds to Germanium (Ge) and Eka-aluminium corresponds to Gallium (Ga). Eka-Manganese corresponds to Technetium (Tc), not Tellurium (Te).

Option A (${F} > {O} > {Cl} > {Br}$) is correct because it represents the correct order of electronegativity.

The correct answer is D ($6^{\text {th }}$) because the 6th period contains 32 elements, including lanthanides.

Option D (Unniltrium) is correct because it follows the IUPAC nomenclature rules for elements with atomic number 103.

Option D (Number of electrons in the outermost shell of their atoms) is correct because elements in the same group have the same number of valence electrons, which determines their chemical properties.

Option B (B, Al, Ga) is correct because the atomic mass of Al is not the arithmetic mean of B and Ga, thus not forming a triad.

Option D ($\mathrm{Be}{\left( {\mathrm{OH}} \right)_2},\mathrm{Zn}{\left( {\mathrm{OH}} \right)_2}$) is correct because both beryllium hydroxide and zinc hydroxide are amphoteric, reacting with both acids and bases.

The correct answer is (ii) is a d-block element (B) because $[Xe] 4 f^{14} 5 d^1 6 s^2$ is an f-block element (Lanthanide) since the last electron enters the f-orbital after filling the d-orbital.

Option D (118) is correct because there are currently 118 known elements in the periodic table.

The correct answer is A ($6 \mathrm{s}, 4 \mathrm{f}, 5 \mathrm{d}, 6 \mathrm{p}$) because this is the order in which the electronic energy levels are filled according to the Aufbau principle and Hund's rule.

The correct answer is $[Kr]4{d^{10}}5{s^2}5{p^2}$ (D) because the element E is Gallium (Ga) in period 4, group 13. The element diagonally placed to Ga is Germanium (Ge) in period 5, group 14, with the electronic configuration $[Kr]4{d^{10}}5{s^2}5{p^2}$.

Option A is correct because the metallic radius is defined as half the internuclear distance separating the metal cores in the metallic crystal.

Option C (Atomic numbers) is correct because the modern periodic law states that properties vary with atomic numbers.