Classification of elements and periodicity in properties practice · Easy

Option B (Uuo) is correct because the IUPAC symbol for element 118 is Ununoctium, with the symbol Uuo.

Option D (decreases from left to right and increases from bottom to top) is correct because the basic character of oxides decreases across a period and increases down a group.

Option A (boron and silicon) is correct because boron and silicon exhibit a diagonal relationship in the periodic table due to their similar electronegativity and charge/size ratio.

Option B ($N{e^ + }$) is correct because removing an electron from a noble gas configuration (Ne+) requires a significantly higher amount of energy due to its stable electronic configuration.

Option C (Dioxygen is paramagnetic) is correct because dioxygen ($\mathrm{O_2}$) has unpaired electrons in its molecular orbital configuration, resulting in its paramagnetic property.

Option D ($2{S^2}2{P^1}$) is correct because adding an electron to the $2p$ orbital will lead to a more stable configuration.

Option B (Both statement-I and II are incorrect) is correct because d-block contains only metals, and non-metals have higher electronegativity and ionization enthalpy than metals.

Option D ($Ti^{4+}$) has the smallest ionic radius because, for isoelectronic species, the ionic radius decreases with increasing nuclear charge (number of protons).

Option C (Vander Waals forces) is correct because in solid argon, the atoms are held together by weak van der Waals forces due to temporary fluctuations in electron distribution.

Option C (Lanthanoids) is correct because lanthanoids are the elements in which the 4f orbitals are progressively filled.

Option D (04:09) is correct because the radius of a Bohr orbit is proportional to n², where n is the principal quantum number.

Option A ($\mathrm{A}_2 \mathrm{O}_3$) is correct because Boron (B) has the smallest atomic radius among the given elements and forms an oxide with the formula $\mathrm{B}_2 \mathrm{O}_3$.

Option D is correct because the ionization energy (IE1) generally increases across a period, but there are exceptions due to electronic configurations; Be has a higher IE1 than B.

Option C (I.E in Al pertains to the removal of p – electron which is relatively easy) is correct because Aluminum's outermost electron is in a p-orbital, which is easier to remove than Magnesium's s-orbital electrons.

Option A (3, 12) is correct because elements with atomic numbers 3 and 12 have their last electron entering the s-orbital, making them s-block elements.

Option C (30) is correct because the number of neutrons is the difference between the mass number and the atomic number (number of protons).

Option C (${\mathrm{N}_2}{\mathrm{O}_3} > \mathrm{C}{\mathrm{O}_2} > {\mathrm{B}_2}{\mathrm{O}_3} > \mathrm{BeO} > \mathrm{L}{\mathrm{i}_2}\mathrm{O}$) is correct because acidic character increases with electronegativity across a period.

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 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 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.

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

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 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 (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.

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