Which of the following is not an actinoid? (i) Curium (Z = 96) (ii) Californium (Z = 98) (iii) Uranium (Z = 92

NCERT Exemplar · Q2

Q. Which of the following is not an actinoid?

(i) Curium (Z = 96)

(ii) Californium (Z = 98)

(iii) Uranium (Z = 92)

(iv) Terbium (Z = 65)

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Concept understanding — Periodic Table Blocks

The Intuition: Why "Blocks" at All?

Imagine you're building a house of cards. Each card has a specific shape and a specific place where it fits. The periodic table is like that house — but instead of cards, we have elements, and instead of shapes, we have electron configurations.

The periodic table is arranged in rows (periods) and columns (groups). But if you look closely, you'll notice that the table isn't a perfect rectangle. There's a detached island of elements (the f-block) floating below, and the main body has a strange "staircase" shape. That shape isn't random — it's dictated by which orbital the last electron enters.

That's the core idea: A block is a set of elements whose last electron enters the same type of orbital (s, p, d, or f).

The Precise Statement

Periodic Table Blocks are regions of the periodic table where elements share the same valence subshell — the subshell being filled as you move across that block.

There are four blocks, named after the four types of atomic orbitals:

Block	Orbital being filled	Location in the table	Number of groups
s-block	$ns$	Leftmost 2 columns (Groups 1 & 2)	2
p-block	$np$	Rightmost 6 columns (Groups 13–18)	6
d-block	$(n-1)d$	Middle 10 columns (Groups 3–12)	10
f-block	$(n-2)f$	Two rows below the main table (Lanthanides & Actinides)	14

Note

The "n" in the orbital notation refers to the principal quantum number (the period number). Notice how for d and f blocks, the orbital being filled has a lower n than the period you're in. That's because of the Aufbau principle — orbitals fill in order of increasing energy, and 4s fills before 3d, etc.

How to Read the Blocks

s-block (Groups 1 & 2)

Last electron enters an $s$ orbital.
Examples: Hydrogen ( $1s^1$ ), Lithium ( $2s^1$ ), Beryllium ( $2s^2$ ).
These are highly reactive metals (except H and He). They lose their $s$ electron(s) easily.

p-block (Groups 13–18)

Last electron enters a $p$ orbital.
Examples: Carbon ( $2p^2$ ), Oxygen ( $2p^4$ ), Chlorine ( $3p^5$ ).
This block contains metals, non-metals, and metalloids — the most chemically diverse block.

d-block (Groups 3–12)

Last electron enters a $d$ orbital — specifically, the $(n-1)d$ subshell.
Examples: Iron ( $3d^6$ ), Copper ( $3d^{10}$ ), Zinc ( $3d^{10}$ ).
These are transition metals. They often have variable oxidation states and form coloured compounds.

f-block (Lanthanides & Actinides)

Last electron enters an $f$ orbital — specifically, the $(n-2)f$ subshell.
Examples: Cerium ( $4f^1$ ), Uranium ( $5f^3$ ).
These are inner transition metals. They are placed below to keep the table from being absurdly wide.

Watch out

A common mistake: thinking that the block tells you the group number. It doesn't. The block tells you the orbital type, not the group. For example, both Carbon (Group 14) and Oxygen (Group 16) are in the p-block, but they're in different groups.

Why This Matters

Knowing the block of an element tells you three things instantly:

Which orbital is being filled — the heart of its electron configuration.
General chemical behaviour — s-block elements are electropositive, p-block are diverse, d-block are transition metals, f-block are inner transition metals.
Position in the table — you can locate any element just by knowing its block and period.

Tip

To find an element's block from its electron configuration: look at the last subshell that has electrons. If it ends in $s$ , it's s-block; if $p$ , p-block; if $d$ , d-block; if $f$ , f-block. For example, $[Ar] 4s^2 3d^6$ ends in $3d^6$ → d-block.

The Big Picture

The periodic table is not just a list — it's a map of how electrons fill orbitals. The blocks are the continents on that map. s-block on the left, p-block on the right, d-block in the middle, and f-block as the two islands below.

Once you see the blocks, the periodic table stops being a random grid and becomes a logical, predictable structure. Every element's position tells you its electron configuration, and every configuration tells you its block.

The s, p, d, and f block classification of the periodic table is a foundational topic in the NCERT Class 11 Chemistry chapter on Classification of Elements and Periodicity, and "periodic table blocks explained with examples" is a commonly searched revision topic for CBSE boards and JEE Main/NEET. Quickly identifying an element's block from its electron configuration is also a frequently tested skill in "periodic table important questions" for competitive chemistry.

Actinoids are the fourteen elements from thorium ( $Z = 90$ ) to lawrencium ( $Z = 103$ ) in which the $5f$ subshell is progressively filled. Terbium ( $Z = 65$ ) lies in the lanthanoid series, not the actinoid series. The answer is (D).

The actinoid series is defined by the progressive filling of the $5f$ orbitals, just as the lanthanoid series is characterized by the filling of $4f$ orbitals. Understanding where these series begin and end in the periodic table immediately tells us which elements belong to each family.

The actinoids span from thorium ( $Z = 90$ ) to lawrencium ( $Z = 103$ ), occupying the bottom row of the f-block. These elements follow actinium ( $Z = 89$ ) and are characterized by electrons entering the $5f$ subshell (though there are some irregularities in electron configurations due to the close energy levels of $5f$ , $6d$ , and $7s$ orbitals).

Let me examine each option:

Curium ( $Z = 96$ ): This element falls squarely in the middle of the actinoid series. With 96 protons, it lies between americium ( $Z = 95$ ) and berkelium ( $Z = 97$ ), well within the $Z = 90$ to $103$ range. Curium is definitely an actinoid.
Californium ( $Z = 98$ ): Similarly, californium sits comfortably within the actinoid series. Named after California (where it was first synthesized), it continues the $5f$ filling pattern. This is an actinoid.
Uranium ( $Z = 92$ ): The most famous actinoid, uranium is the heaviest naturally occurring element in significant quantities. At $Z = 92$ , it's the third member of the actinoid series (after thorium and protactinium). Uranium is certainly an actinoid.
Terbium ( $Z = 65$ ): Here's the outlier. With only 65 protons, terbium falls far short of the actinoid range. In fact, $Z = 65$ places it in the lanthanoid series (also called the rare earth elements), which runs from cerium ( $Z = 58$ ) to lutetium ( $Z = 71$ ). Terbium belongs to the $4f$ block, not the $5f$ block.

Note

The lanthanoids occupy the first row of the f-block (elements $58–71$ ), while the actinoids occupy the second row (elements $90–103$ ). The two series are often placed below the main periodic table to keep the table compact.

✓Final answer

The correct option is (D) — Terbium is a lanthanoid, not an actinoid.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
$r_0$
›Reveal solutionSolution
By de Broglie, the circumference of the $n$ th Bohr orbit holds a whole number of electron wavelengths: $2\pi r_n = n\lambda$ . With $r_n = n^2 r_0$ , the wavelength for $n=6$ is $\lambda = 2\pi r_6 / 6 = 12\pi r_0$ Å — option (D).

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
$2\pi r_n = n\lambda \quad\Longrightarrow\quad \lambda = \frac{2\pi r_n}{n}.$
The orbit radii scale as $r_n = n^2 r_0$ , where $r_0$ is the first-orbit radius.

Solution

Orbit radius for $n=6$ :

$r_6 = 6^2 r_0 = 36\,r_0 \text{ Å}.$

Electron wavelength in that orbit:

$\lambda = \frac{2\pi r_6}{6} = \frac{2\pi(36\,r_0)}{6} = 12\pi r_0 \text{ Å}.$
Equivalently, since $\lambda_n = \dfrac{2\pi r_n}{n} = 2\pi n\,r_0$ , we get $\lambda_6 = 12\pi r_0$ Å directly.

✓Final answer
$\lambda = 12\pi r_0$
TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
$E_1$
›Reveal solutionSolution

$(Z_2 - Z_1) = 52$ — option (A).

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):
$Z_1 = 32.$
Locating $E_3$ (below and to the right of $E_2$ ).

Moving from $E_2$ to $E_3$ 's position in the figure lands on the element with
$Z_2 = 84 \quad(\text{Group 16, Period 6}).$
Difference:
$Z_2 - Z_1 = 84 - 32 = 52.$
✓Final answer
$52$
TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
$Q. Match the following List - 1 (Elements) List - 2 (Group) A Mn, Tc, Re I 12 B Zn, Cd, Hg II 4 C Ti, Zr, Hf III 17 D Ga, In, Tl IV 7 V 13 The correct answer is (A) A - III, B - V, C - I, D - IV (B) A - IV, B - I, C - II, D - V (C) A - IV, B - II, C - I, D - V (D) A - III, B - I, C - II, D - V$
›Reveal solutionSolution
This question tests your knowledge of group numbers in the modern periodic table based on the elements’ outer electronic configurations. The correct matching is A–IV, B–I, C–II, D–V.

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.

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.

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.

Watch out

A common mistake is to count the d¹⁰ electrons and think the group is 12 anyway — which is correct, but the reasoning matters. For Zn, Cd, Hg, the d-subshell is full and does not contribute to variable oxidation states; they are placed in Group 12 precisely because of the ns² configuration.

C: Ti, Zr, Hf — Titanium has $[\text{Ar}]\,3d^2 4s^2$ . The sum of d + s electrons is $2 + 2 = 4$ , placing it in Group 4. So C matches II.

D: Ga, In, Tl — Gallium is in Group 13 (the boron family). Its outer configuration is $4s^2 4p^1$ , giving 3 valence electrons. So D matches V.

Now check the options. The matching we have is: A–IV, B–I, C–II, D–V. That corresponds exactly to option (B).

✓Final answer
$The correct option is (B) .$
TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
$Q. Observe the elements from H(Z=1) to Ca(Z=20). How many elements have no unpaired electrons in their ground state? (A) 6 (B) 7 (C) 8 (D) 9$
›Reveal solutionSolution
The key is to identify elements with completely filled subshells (no unpaired electrons) from H to Ca. Counting these gives 6 elements: He, Be, Ne, Mg, Ar, and Ca.

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Argon (Z=18): $1s^2 2s^2 2p^6 3s^2 3p^6$ — all filled, no unpaired electrons. Counted.

Potassium (Z=19): $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$ — one unpaired electron. Not counted.

Calcium (Z=20): $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$ — filled $4s$ , no unpaired electrons. Counted.

So the counted elements are: He, Be, Ne, Mg, Ar, Ca — that’s 6 elements.

Watch out

A common mistake is to include zinc (Zn, Z=30) because it has a filled $3d^{10}$ and $4s^2$ , but the question explicitly limits to Z=20 (calcium). Also, some students think that elements like carbon or oxygen have no unpaired electrons — but they do, due to Hund’s rule.

Tip

For quick recall: noble gases (He, Ne, Ar) and alkaline earth metals (Be, Mg, Ca) are the diamagnetic ones in this range. That’s 3 + 3 = 6.

✓Final answer
$The number of elements from H to Ca with no unpaired electrons in their ground state is 6, which corresponds to option (A) .$
TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
$Q. The incorrect rule regarding the determination of significant figures is (A) Zeros preceeding to first non zero digit are not significant (B) Zeros between two non-zero digits are not significant (C) Zeros at the right end of the number are significant if they are on the right side of decimal point. (D) All non-zero digits are significant$
›Reveal solutionSolution
The question asks which rule about significant figures is incorrect. The wrong rule is (B), because zeros between two non-zero digits are always significant.

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:

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.

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.

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

This is correct. For example, in 2.50, the zero is significant — it tells us the measurement was precise to the hundredths place. So (C) is true.

Option (D): All non-zero digits are significant.

This is correct by definition. Every digit 1–9 always counts as significant. So (D) is true.

Watch out

A common mistake is to think zeros between digits are just placeholders — they are not. They are as significant as any non-zero digit because they are part of the measured value.

Tip

A quick memory aid: “Sandwich zeros count” — any zero trapped between non-zero digits is significant.

✓Final answer
$The correct option is (B). ANSWER: B$
TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
$_4$
›Reveal solutionSolution
The key is to apply VSEPR theory: count the total electron domains (bonding + lone pairs) around the central atom, then deduce the molecular geometry from the arrangement of bonding pairs only. The correct matches are A–III, B–I, C–IV, D–II, so option (D) is correct.

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.

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.

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.

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.

Molecule D: XeF₄

Xenon has 8 valence electrons. Four fluorines → 4 bonding pairs.

Total valence: 8 + 4 = 12. Used: 4×2 = 8. Remaining: 4 → 2 lone pairs.

Electron domains: 6 (4 bonding + 2 lone). Octahedral parent.

Two lone pairs occupy opposite positions (trans) to minimize repulsion. The 4 bonding pairs lie in a plane → square planar.

So D matches II.

Watch out

A common mistake is to forget that lone pairs occupy more space than bonding pairs. For example, in ClF₃, if you place lone pairs axially, you get a different (wrong) shape. Always put lone pairs in equatorial positions first in trigonal bipyramidal cases.

Tip

For molecules with 5 or 6 electron domains, remember the "lone pair placement" rule: in a trigonal bipyramid, lone pairs go equatorial; in an octahedron, they go opposite each other (trans) if there are two.

Final matching:

A → III, B → I, C → IV, D → II. This corresponds to option (D).

✓Final answer
$The correct option is (D). ANSWER: D$
TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
$N_2$
›Reveal solutionSolution
The question tests basic chemical properties of common elements. The only correct statement is that rhombic sulphur is the stable allotrope at room temperature; the others are false. The correct option is (C).

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

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.

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.

Watch out

Do not confuse thermodynamic stability (spontaneity) with kinetic stability (rate). Ozone is thermodynamically unstable but kinetically stable at room temperature.

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

Sulphur has several allotropes. The most stable form below 95.5°C is rhombic sulphur ( $\alpha$ -sulphur). Above that temperature, it converts to monoclinic sulphur ( $\beta$ -sulphur). At room temperature (≈25°C), rhombic sulphur is indeed the stable allotrope. So (C) is true.

Option (D): $Cl_2$ is a colourless gas.

Chlorine gas ( $Cl_2$ ) is actually greenish-yellow in colour. It is famously a coloured gas, unlike fluorine (pale yellow) or the colourless noble gases. So (D) is false.

Tip

A handy mnemonic: "Chlorine is greenish-yellow, bromine is red-brown, iodine is violet." Colourless gases include $H_2$ , $O_2$ , $N_2$ , and noble gases.

Conclusion: Only statement (C) is correct.

✓Final answer
$The correct option is (C) . ANSWER: C$
TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
$Q. Assertion (A): First ionisation enthalpy of oxygen is less than that of nitrogen Reason (R): Atoms with half-filled or completely filled orbitals are less stable The correct option among the following is (A) (A) and (R) are true. (R) is the correct explanation of (A) (B) (A) and (R) are true, but (R) is not correct explanation of (A) (C) (A) is true but (R) is false (D) (A) is false but (R) is true$
›Reveal solutionSolution
The assertion is true (oxygen’s first ionisation enthalpy is lower than nitrogen’s), but the reason given is false (half-filled orbitals are more stable, not less). So the correct choice is (C).

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:

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.

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.

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.

Check the Assertion (A)

“First ionisation enthalpy of oxygen is less than that of nitrogen” — this is a well‑known experimental fact (≈1314 kJ/mol for N vs. ≈1314? Actually N: 1402 kJ/mol, O: 1314 kJ/mol). So (A) is true.

Check the Reason (R)

“Atoms with half‑filled or completely filled orbitals are less stable” — this is the opposite of the truth. Such atoms are more stable. So (R) is false.

Determine the correct option

(A) true, (R) false → matches option (C).

Watch out

A common mistake is to think that pairing electrons always lowers energy. In fact, pairing increases repulsion; the extra stability of half‑filled shells comes from exchange energy and reduced repulsion. Always check the direction of stability.

Tip

A quick mnemonic: “Half‑filled = happy, full‑filled = fuller happiness.” Nitrogen is “happy” (harder to ionise), oxygen is “less happy” (easier to ionise).

✓Final answer
$The correct option is (C) . ANSWER: C$
TG EAPCET 2022Set ap-2022-07-31-FN1 markMCQ
$4d^5 5s^0$
›Reveal solutionSolution
The key idea is that Nb and Mo are exceptions to the usual Aufbau order because half-filled and completely filled d-subshells are extra stable. Nb ends up as $4d^4 5s^1$ and Mo as $4d^5 5s^1$ , so the correct option is (D).

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.

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.

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 $4d^4 5s^2$ .

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.

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.

Confirm Mo.

Mo’s actual configuration is $4d^5 5s^1$ , not $4d^4 5s^2$ or $4d^5 5s^0$ . The half-filled d-subshell provides enough stabilization to justify promoting one 5s electron, but not both (again, removing both 5s electrons costs too much).

Match to the options.

(A) says Nb $4d^5 5s^0$ and Mo $4d^4 5s^2$ — both wrong.

(B) says Nb $4d^4 5s^1$ and Mo $4d^4 5s^2$ — Nb is right, Mo is wrong.

(C) says Nb $4d^5 5s^0$ and Mo $4d^5 5s^1$ — Nb wrong, Mo right.

(D) says Nb $4d^4 5s^1$ and Mo $4d^5 5s^1$ — both correct.

Watch out

A common mistake is to think that every transition metal near the middle of a series will have a half-filled d-subshell. For example, Tc (43) is $4d^5 5s^2$ , not $4d^6 5s^1$ , because the energy cost of promoting an electron outweighs the stability gain once you pass the half-filled point.

Tip

A quick memory aid: For the 4d series, the two big exceptions are Nb ( $4d^4 5s^1$ ) and Mo ( $4d^5 5s^1$ ). For the 5d series, the analogous exceptions are W ( $5d^4 6s^2$ is actually normal, but Pt and Au have anomalies — different story). So just remember “Nb and Mo are the odd ones in row 5.”

✓Final answer
$The correct option is (D) . ANSWER: D$
TG EAPCET 2021Set ap-2021-08-10-FN1 markMCQ
$\mathrm{BrF_3}$
›Reveal solutionSolution
VSEPR predicts the shape of $\mathrm{BrF_3}$ from the electron domains around bromine: 3 bonding pairs + 2 lone pairs = 5 domains (trigonal-bipyramidal arrangement) with both lone pairs equatorial, giving a T-shape. The correct option is (C).

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

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.

Count electron domains.

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

Electron-group geometry.

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

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.

Read off the molecular shape.

The three fluorines then occupy the two axial positions and the one remaining equatorial position. With the two equatorial lone pairs squeezing the axial bonds slightly inward, the atoms trace a (bent) T-shape.

Confirm with a known example.

$\mathrm{BrF_3}$ (like $\mathrm{ClF_3}$ ) is the textbook example of a T-shaped $\mathrm{AB_3E_2}$ molecule.

Watch out

Five electron domains give a trigonal-bipyramidal electron geometry, but the molecular shape is T-shaped only because two of the domains are lone pairs. An all-bonding $\mathrm{AB_5}$ (e.g. $\mathrm{PCl_5}$ ) would be trigonal bipyramidal.

Tip

For $\mathrm{AB_nE_m}$ with $n+m=5$ : no lone pairs $\to$ trigonal bipyramid, one lone pair $\to$ see-saw, two lone pairs $\to$ T-shape, three lone pairs $\to$ linear.

✓Final answer
$The correct option is (C) . ANSWER: C$

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