Q.Comprehension given below is followed by some multiple choice questions. Each question has one correct option. Choose the correct option. In the modern periodic table, elements are arranged in order of increasing atomic numbers which is related to the electronic configuration. Depending upon the type of orbitals receiving the last electron, the elements in the periodic table have been divided into four blocks, viz, s, p, d and f. The modern periodic table consists of 7 periods and 18 groups. Each period begins with the filling of a new energy shell. In accordance with the Aufbau principle, the seven periods (1 to 7) have 2, 8, 8, 18, 18, 32 and 32 elements respectively. The seventh period is still incomplete. To avoid the periodic table being too long, the two series of f-block elements, called lanthanoids and actinoids are placed at the bottom of the main body of the periodic table.

Concept

Bohr's quantisation of angular momentum, $mvr_n = n\dfrac{h}{2\pi}$, combined with the de Broglie relation $\lambda = \dfrac{h}{mv}$, gives the standing-wave condition

The orbit radii scale as $r_n = n^2 r_0$, where $r_0$ is the first-orbit radius.

Solution

1. Orbit radius for $n=6$: $r_6 = 6^2 r_0 = 36\,r_0 \text{ Å}.$ …

Given: the middle element $E_2$ has atomic number $50$ (tin, Sn — Group 14, Period 5). The figure fixes $E_1$ and $E_3$ by their positions relative to $E_2$.

Locating $E_1$ (just above $E_2$).

$E_1$ sits directly above $E_2$ in the same group, i.e. one period higher (Period 4, Group 14 = germanium):

Locating $E_3$ (below and to the right of $E_2$). …

The key here is to recall that the group number in the modern (IUPAC) periodic table is determined by the number of valence electrons — but with a twist for transition elements. For main-group elements (s- and p-block), the group number equals the number of valence electrons. For d-block transition elements, the group number is typically the sum of the electrons in the (n−1)d and ns orbitals.

Let’s go element by element.

1. A: Mn, Tc, Re — These are all in Group 7. Manganese (Mn) has the configuration $[\text{Ar}]\,3d^5 4s^2$. The total of d + s electrons is $5 + 2 = 7$, so the group is 7. Thus A matches IV.

2. B: Zn, Cd, Hg — Zinc has $[\text{Ar}]\,3d^{10} 4s^2$. Here the d-subshell is completely filled (d¹⁰), and the s-subshell has 2 electrons. For these elements, the group is determined by the ns² electrons alone — they belong to Group 12. So B matches I. …

The question asks for elements from hydrogen (Z=1) to calcium (Z=20) that have no unpaired electrons in their ground state. That means every electron is paired — the atom is diamagnetic. This happens when all occupied subshells are completely filled.

Let’s walk through the electronic configurations.

1. Hydrogen (Z=1): $1s^1$ — one unpaired electron. Not counted.

2. Helium (Z=2): $1s^2$ — filled shell, no unpaired electrons. Counted.

3. Lithium (Z=3): $1s^2 2s^1$ — one unpaired electron. Not counted.

4. Beryllium (Z=4): $1s^2 2s^2$ — filled $2s$ subshell, no unpaired electrons. Counted.

5. Boron (Z=5): $1s^2 2s^2 2p^1$ — one unpaired electron in $2p$. Not counted.

6. Carbon (Z=6): $1s^2 2s^2 2p^2$ — two unpaired electrons (Hund’s rule). Not counted.

7. Nitrogen (Z=7): $1s^2 2s^2 2p^3$ — three unpaired electrons. Not counted.

8. Oxygen (Z=8): $1s^2 2s^2 2p^4$ — two unpaired electrons. Not counted.

9. Fluorine (Z=9): $1s^2 2s^2 2p^5$ — one unpaired electron. Not counted.

10. Neon (Z=10): $1s^2 2s^2 2p^6$ — all subshells filled, no unpaired electrons. Counted.

11. Sodium (Z=11): $1s^2 2s^2 2p^6 3s^1$ — one unpaired electron. Not counted.

12. Magnesium (Z=12): $1s^2 2s^2 2p^6 3s^2$ — filled $3s$, no unpaired electrons. Counted.

13. Aluminium (Z=13): $1s^2 2s^2 2p^6 3s^2 3p^1$ — one unpaired electron. Not counted.

14. Silicon (Z=14): $1s^2 2s^2 2p^6 3s^2 3p^2$ — two unpaired electrons. Not counted.

15. Phosphorus (Z=15): $1s^2 2s^2 2p^6 3s^2 3p^3$ — three unpaired electrons. Not counted.

16. Sulphur (Z=16): $1s^2 2s^2 2p^6 3s^2 3p^4$ — two unpaired electrons. Not counted.

17. Chlorine (Z=17): $1s^2 2s^2 2p^6 3s^2 3p^5$ — one unpaired electron. Not counted. …

The concept here is significant figures — the digits in a number that carry meaning contributing to its precision. The rules exist to tell us which zeros are placeholders (not significant) and which are measured or known (significant). The trick is that many students misremember or confuse the rules, especially for zeros between digits or trailing zeros.

Let’s check each option:

1. Option (A): Zeros preceding the first non-zero digit are not significant.

   This is correct. For example, in 0.0025, the zeros before the 2 are just placeholders. They don’t reflect measurement precision. So (A) is a true rule.

2. Option (B): Zeros between two non-zero digits are not significant.

   This is incorrect. In fact, zeros between non-zero digits are always significant. For instance, in 1003, the two zeros are significant because they are “captured” between the 1 and the 3 — they tell us the number is exactly one thousand three, not just roughly a thousand. So (B) is the false statement.

3. Option (C): Zeros at the right end of the number are significant if they are on the right side of the decimal point. …

Concept & Intuition

VSEPR (Valence Shell Electron Pair Repulsion) theory tells us that electron pairs—whether bonding or lone—repel each other and arrange themselves as far apart as possible. The molecular shape is determined by the positions of the atoms only, ignoring lone pairs. So the first step is always: find the number of valence electrons, count the bonds and lone pairs, then predict the geometry.

1. Molecule A: SF₄

   • Sulfur has 6 valence electrons. Each fluorine contributes 1 electron for bonding, and there are 4 fluorines → 4 bonding pairs.
   • Total electrons used in bonds: 4 × 2 = 8. Remaining valence electrons: 6 (from S) + 4 (from F atoms) = 10 total; 10 – 8 = 2 electrons left → 1 lone pair.
   • Electron domains: 5 (4 bonding + 1 lone). The parent geometry is trigonal bipyramidal.
   • Lone pair occupies an equatorial position (less repulsion). The resulting shape from the 4 bonding pairs is see-saw.
   • So A matches III.

2. Molecule B: ClF₃

   • Chlorine has 7 valence electrons. Three fluorines form 3 bonds → 3 bonding pairs.
   • Total valence electrons: 7 (Cl) + 3×1 (F) = 10. Used in bonds: 3×2 = 6. Remaining: 4 → 2 lone pairs.
   • Electron domains: 5 (3 bonding + 2 lone). Again trigonal bipyramidal parent.
   • Two lone pairs occupy two equatorial positions (120° apart). The three bonding pairs then form a T-shaped molecule.
   • So B matches I.

3. Molecule C: BrF₅

   • Bromine has 7 valence electrons. Five fluorines → 5 bonding pairs.
   • Total valence: 7 + 5 = 12. Used: 5×2 = 10. Remaining: 2 → 1 lone pair.
   • Electron domains: 6 (5 bonding + 1 lone). Parent geometry is octahedral.
   • Lone pair can go anywhere (all positions equivalent in octahedron). The 5 bonding pairs give a square pyramidal shape.
   • So C matches IV.

4. Molecule D: XeF₄ …

Concept & Intuition

This is a recall-based question about the physical and thermodynamic properties of elemental substances. Each option describes a property that is either true or false based on well-known chemical facts. The key is to remember the colour, stability, and allotropy of common non-metals. A common pitfall is confusing the thermodynamic stability of ozone with its kinetic instability, or misremembering the colour of chlorine gas.

Step-by-step reasoning

1. Option (A): $N_2$ is a brown coloured gas.

   Nitrogen gas ($N_2$) is colourless, odourless, and makes up most of our atmosphere. Brown fumes are characteristic of nitrogen dioxide ($NO_2$), not dinitrogen. So (A) is false.

2. Option (B): $O_3$ is thermodynamically stable compared to oxygen.

   Ozone ($O_3$) is actually thermodynamically unstable relative to dioxygen ($O_2$). The standard Gibbs free energy of formation of ozone is positive ($\Delta_f G^\circ \approx +163 \, \text{kJ/mol}$), meaning it spontaneously decomposes to $O_2$ under standard conditions. However, it is kinetically stable at room temperature (decomposition is slow). The statement says "thermodynamically stable", which is incorrect. So (B) is false.

3. Option (C): Rhombic sulphur is stable at room temperature. …

Why this works — the core concept:

Ionisation enthalpy measures the energy needed to remove the most loosely bound electron. The stability of an atom’s electron configuration directly affects this energy: a more stable configuration means a higher ionisation enthalpy. Nitrogen has a half-filled 2p subshell (three electrons, each in a separate orbital), which is exceptionally stable. Oxygen has one extra electron that must pair up in an already-occupied orbital, causing electron‑electron repulsion and making that electron easier to remove. The “Reason” statement gets the stability direction exactly backwards.

Step‑by‑step reasoning:

1. Recall the electron configurations

   • Nitrogen (Z = 7): $1s^2\,2s^2\,2p_x^1\,2p_y^1\,2p_z^1$ — the 2p subshell is half‑filled (three orbitals, one electron each).
   • Oxygen (Z = 8): $1s^2\,2s^2\,2p_x^2\,2p_y^1\,2p_z^1$ — the 2p subshell has four electrons; one orbital now contains a pair.

2. Understand the stability of half‑filled subshells

   A half‑filled subshell (like nitrogen’s 2p³) has all electrons unpaired and parallel spins. This minimises electron‑electron repulsion and gives extra exchange energy, making the atom more stable. A completely filled subshell (like noble gas configurations) is even more stable. The “Reason” claims the opposite — that such atoms are less stable — which is false.

3. Connect stability to ionisation enthalpy

   Because nitrogen’s 2p subshell is half‑filled and extra stable, removing one electron disrupts that stability, requiring more energy. Oxygen’s 2p subshell has a paired electron; the repulsion between the two electrons in the same orbital makes one of them easier to remove. Hence oxygen’s first ionisation enthalpy is lower than nitrogen’s.

4. Check the Assertion (A) …

The concept here is electronic configuration anomalies in transition metals. Normally, you’d expect the 5s orbital to fill before the 4d (since 5s is lower in energy for neutral atoms). But for elements in the 4d series, especially around Nb and Mo, the energy difference between 4d and 5s is very small. A half-filled d-subshell (five electrons) or a completely filled d-subshell (ten electrons) has extra stability due to exchange energy and symmetry. So the atom “borrows” an electron from the 5s orbital to achieve that stable configuration — even if it means leaving the 5s orbital only half-filled or empty.

Let’s work through it step by step.

1. Identify the atomic numbers.

   Niobium (Nb) is element 41. Molybdenum (Mo) is element 42. Their ground-state configurations follow the pattern of the 4d transition series.

2. Recall the expected order for the 4d series.

   For elements before Nb, the 5s fills first: e.g., Y (39) is $4d^1 5s^2$, Zr (40) is $4d^2 5s^2$. So you’d naively predict Nb (41) as $4d^3 5s^2$ and Mo (42) as $4d^4 5s^2$.

3. Apply the half-filled stability rule.

   A half-filled d-subshell ($d^5$) is especially stable. For Nb, moving one electron from 5s to 4d gives $4d^4 5s^1$ — not yet half-filled. But for Mo, moving one electron gives $4d^5 5s^1$, which is exactly half-filled in the d-subshell. That extra stability makes this the actual ground state.

4. Check Nb more carefully.

   Why doesn’t Nb go all the way to $4d^5 5s^0$? That would also give a half-filled d-subshell, but it costs more energy to remove both 5s electrons than to keep one. The compromise is $4d^4 5s^1$ — the d-subshell is one electron short of half-filled, but the s-orbital retains one electron. This is the experimentally observed configuration for Nb.

5. Confirm Mo. …

Concept and Intuition

VSEPR (Valence Shell Electron Pair Repulsion) theory arranges the electron domains (bonds and lone pairs) around a central atom so that repulsion is minimised, then reads off the molecular shape from the positions of the atoms only. For $\mathrm{BrF_3}$, bromine (Group 17, 7 valence electrons) forms three Br-F bonds and keeps two lone pairs, giving five electron domains - a trigonal-bipyramidal electron geometry. Because lone pairs repel more strongly than bonding pairs, both lone pairs take equatorial positions, which forces the three fluorine atoms into a T-shape.

Step-by-Step Reasoning

1. Valence electrons and bonding.

   Bromine has 7 valence electrons; three are used in three Br-F single bonds, leaving $7 - 3 = 4$ electrons as two lone pairs.

2. Count electron domains.

   3 bonding pairs + 2 lone pairs = 5 electron domains.

3. Electron-group geometry.

   Five domains arrange as a trigonal bipyramid - two axial positions ($180^\circ$ apart) and three equatorial positions ($120^\circ$ apart).

4. Place the lone pairs.

   Equatorial sites have only two close ($90^\circ$) neighbours versus three for axial sites, so both lone pairs occupy equatorial positions to minimise repulsion.

5. Read off the molecular shape. …