Classification of elements and periodicity in properties practice · Mixed difficulty

Option D is correct because the ionization energy increases with each successive ionization, meaning ${Z_2}$ (IE2) will always be greater than ${Z_1}$ (IE1).

The correct answer is If both Assertion & Reason are false because Ca2+ is smaller than K+ due to higher effective nuclear charge, and Ca2+ has fewer electrons than K+.

The correct answer is B (Specific heat) because specific heat is an intensive property and does not show a periodic trend like valency, ionization potential, and atomic size.

Option D (If both (A) and (R) are incorrect.) is correct because Cu2+ is paramagnetic due to unpaired electrons, and not all orbitals are doubly occupied.

The correct answer is C (577) because the ionization enthalpy of Al is expected to be between that of Na and Mg, and also less than that of Si due to electronic configuration.

The correct answer is B (Basic oxide) because the electronic configuration indicates an alkali metal, which typically forms basic oxides.

The correct answer is D because after the first ionization, oxygen has a half-filled p orbital, making its second ionization energy higher than fluorine, while nitrogen's stability decreases after the first ionization.

Option B (caesium) is correct because caesium is an alkali metal and has the lowest ionization energy, making it the most electropositive element among the given options.

Option D ($n s^2 n p^{1-6}$) is correct because it accurately represents the general electronic configuration of p-block elements, with 2 electrons in the *ns* orbital and 1-6 electrons in the *np* orbitals.

Option B ($Cl + {e^ - } \to C{l^ - }$) is correct because chlorine has the highest electron affinity among the given options.

Option D (Both (1) and (3)) is correct because transition elements and actinoids both exhibit variable valency due to the involvement of d and f electrons in bonding.

Option A (I and II are true) is incorrect because Statement I is true, but Statement II is false. $\mathrm{Cu}^{+2}$ is paramagnetic, but not all orbitals are doubly occupied.

Option D is correct because the electronic configuration represents a copper(II) ion (Cu2+), which is a metallic cation.

Option C is correct because Manganese (Z = 25) has an electronic configuration of 3d⁵4s², and removing the third electron requires removing an electron from the stable, half-filled 3d orbital, resulting in a significantly higher ionization enthalpy.

Option B (Both Assertion and Reason are true but Reason is not the correct explanation of Assertion) is correct because both statements are true, but the electron configuration is not the direct reason for the less favorable electron affinity.

The correct answer is D (${SO}_3$) because, among the given oxides, sulfur trioxide is the most acidic due to sulfur's high electronegativity and oxidation state.

Option B is correct because Cr has the highest oxidation state (+6) in K2CrO4, leading to the smallest ionic radius due to increased effective nuclear charge.

Option C (CS and Rn) is correct because Cesium (Cs) is the starting element and Radon (Rn) is the last element in the 6th period, which is the largest period in the modern periodic table.

Option B (11) is correct because there are 11 elements that exist as gases under standard conditions.

Option B (acidic, basic) is correct because the acidity or basicity of the solutions depends on the electronegativity difference between M/M' and oxygen.

Option B (Inversely proportional to effective nuclear charge) is correct because as the effective nuclear charge increases, the electrons are pulled closer to the nucleus, resulting in a smaller atomic radius.

The correct answer is 51.40 kJ because the energy required to convert all Al atoms to Al+3 in 270 mg of Al vapor is 51.40 kJ, calculated by summing the ionization energies and multiplying by the number of moles of Al.

Option A (electronegativity) is correct because electronegativity generally increases across a period due to increasing nuclear charge and decreasing atomic size.

The correct answer is D ($\text{Si}<\mathrm{C}< \mathrm{N}< \mathrm{O}< \mathrm{F}$) because electronegativity increases across a period and decreases down a group.

Option B (relative smaller size of their atoms) is correct because lanthanide contraction leads to similar sizes for 4d and 5d series, impacting ionization energy.

Option C (14.5) is correct because nitrogen has a higher ionization energy than its neighbors due to its stable half-filled p-subshell.

Option C is correct because a large jump between the third and fourth ionization energies indicates that the element readily loses three electrons to achieve a stable configuration.

Option D (A $p$-block element) is correct because the last electron enters the p-orbital, making it a p-block element.

Option B (${X_P} < {X_M}$) is correct because electronegativity values on the Mulliken scale are generally larger than those on the Pauling scale.

Option D is correct. The electron affinity of Li is not less than Na. The correct order for E1 is Li > Na > K > Cs.