Classification of elements and periodicity in properties practice · Easy

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.

Option B ($Pb$) is correct because Lead ($Pb$) is a p-block element and not a transition metal.

Option C (6 & 1) is correct because the 6th period has 32 elements and is the longest, while the 1st period has only 2 elements and is the shortest.

Option D (Bond energies) is correct because Pauling's electronegativity scale is based on bond energies.

Option D (Cs) is correct because ionization potential decreases down the group due to increasing atomic size and weaker nuclear attraction.

The correct answer is S because sulfur has the highest electron affinity among the given elements due to its position in the periodic table.

Option B (S) is correct because electronegativity decreases down the group, and sulfur (S) is below oxygen (O) in the same group.

Option D (Alkali metals) is correct because alkali metals have the lowest ionization potential due to their electronic configuration (ns¹), making it easier to remove an electron.

Option A is correct because elements with atomic numbers 30 (Zn), 48 (Cd), and 80 (Hg) all belong to the same group (group 12) and have 2 electrons in their outermost shell.

Option A (Li) is correct because after losing one electron, Lithium attains a stable noble gas configuration, requiring significantly more energy to remove the second electron.

Option B ((B)) is correct because after losing one electron, element B achieves a stable noble gas configuration, requiring significantly more energy for the second ionization.

Option D (Nitrogen) is correct because nitrogen has a higher ionization energy than boron, carbon, and oxygen due to its stable half-filled p-orbital configuration.

Option A ($X = Na, Y = Mg$) is correct because Na has a significantly higher second ionization energy due to the removal of an electron from a stable noble gas configuration, while Mg's second ionization energy is comparatively lower.

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 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 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 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 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 (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.

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

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 B ($\mathrm{Be}$) is correct because after losing two electrons, Beryllium attains a stable noble gas configuration, making the removal of the third electron require a very high amount of energy.

Option D (Atomic radius) is correct because atomic radius increases down the group due to the addition of new electron shells.