Classification of elements and periodicity in properties practice · Mixed difficulty

Option A (Atomic weights) is correct because Mendeleev's periodic law stated that the physical and chemical properties of elements are periodic functions of their atomic weights.

The correct answer is B (gadolinium) because the electronic configuration $(n-2) f^7(n-1) d^1 n s^2$ for $n=6$ is $4f^7 5d^1 6s^2$, which corresponds to Gadolinium.

The correct answer is p-block (B) because the electronic configuration of $M^{+3}$ is $[Xe] 4f^{14} 5d^{10} 6s^2$, adding back the 3 electrons gives a configuration ending in $6s^2 6p^1$, indicating a p-block element.

Option A ($\frac{{100}}{{\sqrt 2 }}\,\mathrm{pm}$) is correct because in an fcc structure, $4r = a\sqrt{2}$, where a is the edge length and r is the atomic radius.

Option D (${O^{ - 2}} > {F^ - } > O > F$) is correct because anions are larger than their neutral atoms, and size decreases with increasing nuclear charge for isoelectronic species.

Option A (Unnilunium – Mendelevium) is incorrect because Unnilunium corresponds to atomic number 101, which is indeed Mendelevium.

Option A (${\mathrm{O}^ - }$ ion will tend to resist the addition of another electron) is correct because the negative charge on the O- ion repels the incoming electron, making the process endothermic.

Option D is correct. The general electronic configuration of inner transition elements is ns²(n-1)d⁰⁻¹(n-2)f¹⁻¹⁴.

Option D is correct because $GeO_2$ is an acidic oxide due to Germanium being a non-metal.

Option D (All of the above) is correct because the long form of the periodic table reflects electron filling order, predicts valency, and shows property trends.

Option B (118) is correct because the current long form of the periodic table accommodates 118 elements.

Option D (3) is the correct answer because electronegativity on the Pauling scale is approximately (Ionization Energy + Electron Affinity)/2, divided by a scaling factor.

Option C ($F\mathrm{ } > \mathrm{ }N\mathrm{ } > \mathrm{ }C\mathrm{ } > \mathrm{ }Be\mathrm{ } > \mathrm{ }B$) is correct because it considers the general trend of increasing ionization energy across a period and the exceptions due to stable electronic configurations (half-filled and filled orbitals).

Option A is correct because noble gases have a general outer electronic configuration of ns²np⁶, which represents a completely filled valence shell.

The correct answer is B (Both statement-I and II are incorrect) because metals and non-metals are present in both p and d blocks, but electronegativity and ionization enthalpy of metals are generally lower than non-metals.

Option B (Both (A) & (R) are true but the (R) is not the correct explanation of the (A)) is correct because Mendeleev did base his table on atomic mass, and atomic number equals the number of protons, but the latter doesn't explain the former.

Option A (${M^{2 + }}$) is correct because alkaline earth metals have two valence electrons which they lose to form +2 ions.

Option C (${ZnO}, {Al}_2 {O}_3$) is correct because both Zinc oxide and Aluminum oxide exhibit amphoteric behavior, reacting with both acids and bases.

Option B is correct because elements with atomic numbers 33, 53, and 83 correspond to a metalloid (As), a non-metal (I), and a metal (Bi), respectively.

Option B ($Li > Ne > B > C > Be$) is correct because after the first ionization, Li attains a noble gas configuration, requiring significantly more energy for the second ionization. Ne loses a p electron, B loses an s electron, C loses an s electron, and Be loses an s electron.

Option C (${X_2}{O_3},\; amphoteric) is correct because the element X belongs to group 13 (ns²np¹) and its oxide is amphoteric.$

Option A (${\mathrm{O}^{2 - }} > {\mathrm{F}^ - } > O > F$) is correct because negative ions are larger than their neutral atoms, and among isoelectronic species, the one with fewer protons is larger.

The correct answer is C (Al) because Aluminum (Al) shows a diagonal relationship with Beryllium (Be) in the periodic table.

Option B ($1.12\,\,\,\mathop {\text{A}}\limits^{\text{o}}$) is correct because the ionic radius of Ne should be between the ionic radii of its isoelectronic neighbors, specifically smaller than F- and larger than Na+.

Option D (Chlorine) is correct because chlorine has the highest electron affinity among the halogens due to a balance between its nuclear charge and atomic size.

Option D ($72.23\%$) is correct because it is the ionic character calculated using the Pauling electronegativity difference formula.

Option B (${\mathrm{N}^{3 - }} > {\mathrm{O}^{2 - }} > {\mathrm{F}^ - } > N{\mathrm{a}^ + } > M{\mathrm{g}^{2 + }}$) is correct because for isoelectronic species, ionic radii decrease with increasing nuclear charge.

Option B ($1{s^2}2{s^2}2{p^6}3{s^1}$) is correct because it represents an element with only one valence electron in the 3s orbital, which is easily removed, resulting in low ionization energy.

Option B ($\mathrm{C}{\mathrm{l}^ - }\,\,and\,\,\mathrm{Ar}$) is correct because both the chloride ion and argon have the same number of electrons and thus the same electronic configuration.

Option A (F) is correct because Fluorine is the most electronegative element in the periodic table.