Classification of elements and periodicity in properties practice · Mixed difficulty

Option B (Lanthanoid and actinoid series) is correct because the 4f inner-transition series is called the Lanthanoid series, and the 5f inner-transition series is called the Actinoid series.

Option C is correct because statement I is false, as reactivity doesn't regularly increase from group 1 to 18, and statement II is true, as group 1 oxides are basic and group 17 oxides are acidic.

Option B ($B{r_2}$) is correct because bromine is the only non-metal that exists as a liquid at room temperature.

Option A (Smaller size of 2p – subshell) is correct because the smaller size of the 2p subshell in fluorine leads to greater interelectronic repulsion, reducing the electron gain enthalpy.

The correct answer is B ($1{s^2}2{s^2}2{p^6}3{s^1}$) because removing the second electron from this configuration requires breaking a stable noble gas configuration, leading to a very high second ionization energy.

Option C (20) is correct because elements from atomic number 91 to 110 have electrons in the 5f subshell, totaling 20 elements.

Option C ($\mathrm{B} < C < N < O$ increasing first ionisation enthalphy) is correct because the ionization enthalpy of Nitrogen is higher than Oxygen due to the stable half-filled electronic configuration of Nitrogen.

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

Option D is correct because Beryllium has a higher ionization potential than Boron due to its stable, fully-filled 2s² electron configuration.

The correct answer is $Na > Mg > Li$ because atomic size decreases across a period (Na > Mg) and increases down a group (Na > Li).

Option B ($Li < Be < B < C -$First IP) is wrong because the first ionization potential generally increases across a period, but there are exceptions due to electronic configurations.

Option B (Electronic configuration) is correct because electronic configuration describes the arrangement of electrons in the different orbitals of an atom.

Option D is correct. Metals lose electrons to form positive ions, resulting in a smaller ionic radius due to increased effective nuclear charge.

Option D (90 to 103) is correct because the actinide series consists of the elements with atomic numbers from 90 (Thorium) to 103 (Lawrencium).

Option C (Na) is correct because among the given species, Na is the only cation and neutral atom, while others are isoelectronic anions. Cations are smaller than their corresponding neutral atoms.

Option D ($M_{\left( g \right)}^{2 + } \to M_{\left( g \right)}^{3 + }\,$) is correct because ionization enthalpy increases with each successive ionization.

Option D (Cobalt – Nickel) is correct because Cobalt and Nickel are adjacent in the periodic table and have very similar properties, so they are not considered an anomalous pair.

Option B (Be & He) is correct because beryllium (Be) and helium (He) have stable electronic configurations, resulting in very low or zero electron affinity.

Option C (18) is correct because the 5th period of the periodic table contains 18 elements.

Option C (3) is correct because there is a large jump in ionization enthalpy between the 3rd and 4th ionization, indicating 3 valence electrons.

Option C (Cr) is correct because Chromium achieves a stable electronic configuration ($4s^1 3d^5$) after losing one electron, requiring more energy for the second ionization.

Option A is correct because the first ionization potential generally increases across a period, with exceptions due to electron configuration stability.

Option A (A > B > C) is correct because ionization potential decreases down the group as the outermost electron is further from the nucleus.

Option B ($\mathrm{Na}^{-} > \mathrm{Na} > \mathrm{Na}^+$) is correct because anions are larger than their neutral atoms, and cations are smaller.

The correct answer is C ([Ne] 3s² 3p³) because half-filled p orbitals provide extra stability, leading to a higher ionization energy.

Option A (Electron gain enthalpy) is correct because the given reaction represents the addition of an electron to a gaseous atom, which is the definition of electron gain enthalpy.

The correct answer is B (Rb) because ionization energy decreases down a group due to increasing atomic size and shielding effect.

Option B ($\mathrm{CO}$) is correct because carbon monoxide is a neutral oxide, meaning it does not react with acids or bases to form salts.

Option D (Both (1) & (3)) is correct because the electronic configurations of Ac and Uup place them in the d-block and p-block, respectively.

Option B (Size decreases and increases respectively) is correct because removing an electron increases the effective nuclear charge, decreasing size, while adding an electron increases electron-electron repulsion, increasing size.