Classification of elements and periodicity in properties practice · Mixed difficulty

The correct answer is B (All isoelectronic ions belong to same period of the periodic table.) because isoelectronic ions can belong to different periods depending on their charge and electronic configuration.

Option C (p-block) is correct because the element with atomic number 35 (Bromine) has its last electron entering the p-orbital.

Option D is correct because anions are larger than their corresponding neutral atoms, and size decreases across a period, so O2- > F- > O > F.

The correct answer is D ($\mathrm{H}_2 \mathrm{~S}, \mathrm{~K}^{+}, \mathrm{Ar}, \mathrm{Cl}$) because these species do not all have the same number of electrons. $\mathrm{H}_2 \mathrm{~S}, \mathrm{~K}^{+}$, and $\mathrm{Ar}$ have 18 electrons, but $\mathrm{Cl}$ has 17.

Option B (${Na}^{+} < {Mg}^{2+} < {Al}^{3+}$) is correct because for isoelectronic species, the effective nuclear charge increases with increasing nuclear charge, leading to a smaller ionic radius and greater effective nuclear charge.

Option C is correct because representative elements (s and p block elements) have the general electronic configuration of $ns^{1-2}np^{0-5}$.

Option A (Cations derived from metallic elements and anions from non‐metallic elements) is correct because ionic compounds typically consist of cations formed from metals (which lose electrons) and anions formed from non-metals (which gain electrons).

Option A is correct because the electron affinity of Na+ is the energy released when an electron is added, which is numerically equal to the ionization energy of Na.

Option D (4) is correct because an element with atomic number 20 (Calcium) has the electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$, placing it in the 4th period.

Option B (Be) is correct because among the given elements, Be is a metal and has the lowest ionization energy, thus the highest tendency to release electrons.

Option B (Dubnium) is correct because Dubnium is the name of the element with atomic number 105.

Option B ($Be^−$) is the least stable ion because beryllium has a fully filled 2s subshell ($1s^22s^2$), making it less likely to accept an additional electron.

Option B (Increases down the group and decreases across the period) is correct because metallic character increases as you go down a group (due to increased atomic size and easier electron loss) and decreases as you go across a period (due to increased nuclear charge and harder electron loss).

Option C (${\mathrm{K}^ + } < C{\mathrm{a}^{2 + }} < M{\mathrm{g}^{2 + }} < B{\mathrm{e}^{2 + }}$) is correct because the polarizing power increases with increasing charge and decreasing size.

Option A (Decrease in period and increases in a group) is correct because metallic character decreases across a period and increases down a group due to changes in ionization energy and effective nuclear charge.

Option D (Chlorine) is correct because Chlorine has the highest electron affinity among the given options due to its size and electronic configuration.

Option A (Seventh) is correct because the seventh period contains most of the man-made radioactive elements, including transuranic elements.

Option C (Fe) is correct because iron is a transition metal with a partially filled d orbital.

Option C (If Assertion is True but the Reason is False) is correct because ionic bonds are formed between metals and non-metals, but electrons are transferred, not shared.

Option D (Size of atom) is correct because atomic size generally decreases across a period due to increasing nuclear charge.

Option C (Na, Cl) is correct because both sodium (Na) and chlorine (Cl) are in the same period (period 3) of the periodic table.

Option B (O < S < F < Cl) is correct because electron gain enthalpy generally increases across a period and decreases down a group, with chlorine having a higher electron affinity than fluorine due to fluorine's small size.

Option B is correct because removing an electron from P- results in a more stable half-filled p3 configuration, requiring the least energy.

Option B is correct because Fluorine is the most electronegative element, and its outermost electronic configuration is $2s^22p^5$.

Option A is correct because both the assertion and reason are true, and the reason correctly explains why the second ionization potential of oxygen is greater than that of fluorine.

Option A ($Mg, Al^{3+}$) is correct because cations are smaller than their parent atoms, and the size decreases with increasing positive charge. Anions are larger than their parent atoms.

Option D is correct because metals are more abundant than non-metals in the periodic table.

Option A (s – block) is correct because the element with atomic number 55, Cesium (Cs), is an alkali metal and belongs to the s-block of the periodic table.

Option B (p-block) is correct because the last electron enters the p-orbital, indicating that the element belongs to the p-block.

The correct answer is A (${\mathrm{I}^{\mathrm{st}}}\mathrm{\;IE\;of\;C} > {\mathrm{I}^{\mathrm{st}}}\mathrm{\;IE\;of\;B}$) because carbon has a more stable electronic configuration than boron, leading to a higher ionization energy.