Classification of elements and periodicity in properties practice · Easy

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

Option B is correct because elements in the third period have electrons filling the 3s, 3p, and 3d subshells, making them have three complete subshells.

Option C ($Cl$) is correct because Chlorine has the highest electronegativity among the given elements.

Option D (The element forms basic oxide) is correct because the electronic configuration 2,8,1 indicates that the element is a metal, and metals generally form basic oxides.

Option D (All of the above) is correct because the size of the cation is smaller, it's more difficult to remove an electron, and the effective nuclear charge is greater.

Option B ($B{e^ - }$) is correct because Beryllium has a stable electronic configuration, and adding an electron makes it unstable.

Option D (Dependent) is correct because the half-life of a reaction is dependent on the rate constant, which is temperature-dependent according to the Arrhenius equation.

Option A (105) is correct because Unp represents Unnilpentium, where Un = 1, nil = 0, and pent = 5, so the atomic number is 105.

Option B (Li, Na, K) is correct because the atomic weight of Na is approximately the average of the atomic weights of Li and K, satisfying Doberiener's law of triads.

Option B (${MgO}$) is correct because oxides of alkaline earth metals are generally basic, and MgO is more basic than the amphoteric oxides ZnO and Al2O3 and the acidic oxide N2O5.

The correct answer is C (${MX}_3$) because the large difference between the 3rd and 4th ionization enthalpies indicates a stable +3 oxidation state for the metal.

Option C ($K$) is correct because potassium (K) occupies the peak of the Lother Meyer curve due to its large atomic volume.

Option C (Ununquadium) is correct because the IUPAC name for element 114 is Ununquadium.

Option D (Une) is correct because it represents the systematic IUPAC nomenclature for element 109, where Unnilennium is shortened to Une.

Option A (s-orbital) is correct because s-electrons are closest to the nucleus and experience the highest effective nuclear charge, making them the most difficult to remove.

Option A ($s -$subshell) is correct because electrons in s-subshells are closest to the nucleus and experience the highest effective nuclear charge, requiring more energy for removal.

Option D (The cation is smaller than its parent atom) is correct because removing an electron increases the effective nuclear charge, causing the remaining electrons to be pulled closer to the nucleus, thus decreasing the size.

Option C (UuP) is correct because the IUPAC nomenclature rules dictate that the element with atomic number 115 is named Ununpentium, abbreviated as Uup, which is now Moscovium (Mc). However, in the options, UuP is given.

Option A (${O^{2 - }} < {N^{3 - }} < {S^{2 - }} < {P^{3 - }}$) is correct because ionic radii increase down a group and for isoelectronic species, radii decrease with increasing nuclear charge.

Option B is correct because the electron affinity trend is generally increasing across a period and decreasing down a group, but chlorine has a higher electron affinity than fluorine due to its larger size and weaker electron-electron repulsion.

Option C (5) is correct because the naturally occurring diamagnetic p-block elements are Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe), and Radon (Rn).

Option A ($\left( {n - 1} \right){d^{1 - 10}}\,n{s^{0 - 2}}$) correctly represents the outer electronic configuration of transition elements.

Option B (Rb) is correct because ionization energy decreases down the group, and Rb is located furthest down Group 1 compared to the other options.

Option C is correct because the periodic classification is used to examine both the physical and chemical properties of elements.

Option A (BeO) is correct because BeO is amphoteric, but among the given alkaline earth metal oxides, it exhibits the most acidic character due to beryllium's smaller size and higher polarizing power.

Option B (Moseley) is correct because Henry Moseley discovered that the physical and chemical properties of elements are periodic functions of their atomic number.

Option B ($1.12\,\,\,\mathop {\text{A}}\limits^{\text{o}}$) is the most appropriate as the radius of Ne should fall between the radii of $F^-$ and $Na^+$, considering the trend of ionic radii.

Option C (Cation) is correct because when an atom loses an electron, it becomes positively charged, forming a cation.

Option A (${Li}$) is correct because after losing one electron, it attains a stable noble gas configuration, requiring significantly more energy to remove another electron.

Option D (All of the above) is correct because the size of the cation is smaller, it's more difficult to remove an electron, and the effective nuclear charge is greater.