Classification of Elements and Periodicity in Properties — Important Questions

The very small size of fluorine (2p subshell) causes high electron-electron repulsion in the compact 2p orbital when an electron is added, making the process less exothermic than for chlorine (3p subshell).

Step 1: The general trend in a group is that electron gain enthalpy becomes less negative going down. So F should have more negative EGE than Cl — but it doesn't. Step 2: Fluorine is extremely small (atomic radius = 0.72 Å). When an incoming electron enters the 2p subshell of fluorine, it encounters intense inter-electronic repulsion from the existing 2p electrons in this very compact subshell. Step 3: In chlorine, the added electron goes into the larger, more diffuse 3p subshell, where electron-electron repulsion is much smaller. Step 4: This reduced repulsion in Cl makes the electron additio

Ununpentium

Step 1: The IUPAC systematic name is built from numerical roots for each digit of the atomic number. Step 2: Atomic number 115 has digits 1, 1, and 5. Step 3: Numerical roots: 1 = 'un', 1 = 'un', 5 = 'pent'. Step 4: Combining: un + un + pent + ium = Ununpentium. The symbol would be Uup. Step 5: This element has now been officially named Moscovium (Mc) by IUPAC. Option B 'Unquadpentium' incorrectly uses 'quad' for digit 1. Option C 'Unnilpentium' would be for Z=105 (digits 1,0,5). Option D 'Unununpentium' would suggest Z=111+5 = wrong interpretation. The IUPAC official name is Moscovium (Mc).

Mg²⁺ has a higher nuclear charge (+12) compared to Na⁺ (+11), so the same 10 electrons experience a stronger nuclear pull, reducing the ionic radius.

Step 1: Both Na⁺ and Mg²⁺ have 10 electrons (isoelectronic — same as Ne). Step 2: Na has Z=11, so Na⁺ has nuclear charge +11 pulling 10 electrons. Z/e ratio = 11/10 = 1.1. Step 3: Mg has Z=12, so Mg²⁺ has nuclear charge +12 pulling 10 electrons. Z/e ratio = 12/10 = 1.2. Step 4: Higher Z/e for Mg²⁺ means each electron is pulled more strongly toward the nucleus → smaller size. Step 5: Option A is wrong — electron-electron repulsion plays a minor role; the key factor is nuclear charge. Option C confuses the number of electrons lost with radius determination. Option D is completely wrong — nuclear

Group 13

Step 1: The key to identifying the group is to find the large jump in successive ionisation energies — this jump occurs when we go from removing valence electrons to removing core (inner shell) electrons. Step 2: Looking at the data: IE₁=577, IE₂=1817, IE₃=2745 — these increase gradually (removing valence electrons). Step 3: From IE₃ to IE₄ there is a HUGE jump: 2745 → 11,578 kJ/mol. This means the 4th electron is extremely difficult to remove. Step 4: This indicates the element has 3 valence electrons (IE₁, IE₂, IE₃ involve valence electrons; IE₄ involves a core electron). Step 5: An element