Match the solutions given in Column I and the colours given in Column II. Column I (Aqueous solution of salt):

NCERT Exemplar · Q64

Q. Match the solutions given in Column I and the colours given in Column II. Column I (Aqueous solution of salt):

(i)

FeSO_4.7H_2O

(ii)

NiCl_2.4H_2O

(iii)

MnCl_2.4H_2O

(iv)

CoCl_2.6H_2O

(v)

Cu_2Cl_2

Column II (Colour):

(a) Green

(b) Light pink

(d) Pale green

(e) Pink

(f) Colourless

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Concept understanding — Magnetism and Color

Magnetism and Colour: An Intuitive First Look

You've probably noticed that some materials are magnetic (like iron) and others aren't (like wood). And you've seen that objects have different colours — a rose is red, the sky is blue. At first glance, these two properties seem completely unrelated. But at the deepest level, both magnetism and colour come from the same source: how electrons behave inside atoms.

Let's start with a simple picture.

The Intuition: Electrons as Tiny Magnets and Painters

Imagine an electron orbiting the nucleus of an atom. That moving charge is like a tiny loop of electric current — and any loop of current creates a magnetic field. So every electron is a microscopic magnet. In most materials, these tiny magnets point in random directions and cancel out. But in iron, they align, and the material becomes magnetic.

Now, colour. When light hits an atom, electrons can absorb some of its energy and jump to a higher orbit. The colour we see is the light that wasn't absorbed — the leftover wavelengths. Different atoms absorb different colours because their electrons have different "jump sizes" (energy levels).

So both magnetism and colour are about how electrons move and interact with their environment. One is about the direction of electron spin and orbit (magnetism), the other about the energy of electron jumps (colour).

The Precise Statement

Important

Magnetism and colour are both consequences of the electronic structure of atoms, but they arise from different aspects of electron behaviour:

Magnetism originates from the magnetic moments of electrons — their spin and orbital motion. A material is magnetic when these moments align cooperatively.
Colour originates from the absorption of specific wavelengths of light by electrons, which occurs when the photon energy matches the energy difference between two electron states.

How They Connect (and How They Don't)

The two phenomena are linked because they both depend on the arrangement of electrons in orbitals — the so-called electronic configuration. But they are not the same thing, and one does not cause the other.

Here's a table to make the distinction clear:

Property	Origin	What determines it?	Example
Magnetism	Electron spin and orbital motion	Unpaired electrons, crystal structure	Iron is magnetic because it has 4 unpaired electrons per atom
Colour	Electron transitions between energy levels	Energy gap between orbitals	Copper is reddish because its electrons absorb blue-green light

Note

A material can be magnetic and colourless (like pure iron — it's silvery, not colourful). A material can be brilliantly coloured and non-magnetic (like a ruby). The two properties are independent in most everyday cases.

The Deeper Link: Transition Metals

The most interesting connection appears in transition metals (elements like iron, cobalt, nickel, copper). These atoms have partially filled d orbitals. That partial filling does two things:

It leaves unpaired electrons, which can align to produce magnetism. …

Why this formula?

Magnetism and Color: Why the Key Formulas Hold

This is a fascinating intersection of physics and perception. The core idea is that color is not a property of light itself, but of our brain's interpretation of different wavelengths. Magnetism, in turn, can influence how these wavelengths are produced or absorbed.

Let's break down the key formulas and their why.

1. The Fundamental Link: Energy, Frequency, and Color

The most important formula connecting magnetism and color is the Planck-Einstein relation:

E = h \nu

Where:

$E$ = energy of a photon (light particle)
$h$ = Planck's constant ( $6.626 \times 10^{-34} \text{ J·s}$ )
$\nu$ = frequency of the light

Why does this hold?

Quantum nature of light: Light is not a continuous wave, but comes in discrete packets called photons.
Energy quantization: The energy of a photon is directly proportional to its frequency. Higher frequency means higher energy.
Magnetism's role: When an electron in an atom jumps from a higher energy level to a lower one, it emits a photon. The energy difference ( $\Delta E$ ) between these levels determines the photon's frequency:

\Delta E = h \nu

Color perception: Our eyes detect different frequencies as different colors. For example:
- Red light: $\nu \approx 4.3 \times 10^{14} \text{ Hz}$ (lower energy)
- Blue light: $\nu \approx 6.7 \times 10^{14} \text{ Hz}$ (higher energy)

Key insight: The color you see is determined by the energy gap between electron orbits. Magnetism can alter these energy gaps (via the Zeeman effect, see below).

2. The Zeeman Effect: How Magnetic Fields Split Colors

When a magnetic field is applied to an atom, a single spectral line (one color) splits into multiple lines. This is described by:

\Delta E = \mu_B \cdot B \cdot m_l

Where:

$\Delta E$ = energy shift of the spectral line
$\mu_B$ = Bohr magneton ( $9.274 \times 10^{-24} \text{ J/T}$ )
$B$ = magnetic field strength (in Tesla)
$m_l$ = magnetic quantum number (integer: $-l, ..., +l$ )

Why does this hold?

Electron as a tiny magnet: An electron orbiting a nucleus behaves like a tiny current loop, creating a magnetic dipole moment.
Energy in a magnetic field: This dipole moment interacts with an external magnetic field. The interaction energy depends on the orientation of the electron's orbit relative to the field.
Quantized orientations: The magnetic quantum number $m_l$ tells us which orientation is allowed. Each orientation has a slightly different energy.
Result: A single energy level splits into $2l+1$ sub-levels. Transitions between these sub-levels produce photons with slightly different energies — hence different colors appear.

Example: A sodium lamp emits yellow light. In a strong magnetic field, that yellow line splits into three closely spaced lines (normal Zeeman effect).

3. Faraday Rotation: Magnetic Field Twists Light's Color

When polarized light passes through a material in a magnetic field, its plane of polarization rotates. The rotation angle is:

\theta = V \cdot B \cdot d

Where:

$\theta$ = rotation angle (in radians)
$V$ = Verdet constant (material-specific, depends on wavelength)
$B$ = magnetic field strength
$d$ = path length through the material

Why does this hold?

Circular birefringence: In a magnetic field, the material has different refractive indices for left- and right-circularly polarized light.
Phase difference: These two components travel at different speeds, creating a phase difference.
Recombination: When they recombine, the resulting linear polarization is rotated. …

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The colour of a transition-metal salt in solution comes from d-d transitions of electrons in partially filled d-orbitals. Matching the ions to their characteristic colours gives: (i)→(d), (ii)→(a), (iii)→(b), (iv)→(e), (v)→(f).

Why salts have colour — the d-orbital story

Most of us remember that transition metal compounds are colourful, but the reason is beautiful. When a transition metal ion is surrounded by water molecules (or other ligands), its five d-orbitals — which were all equal in energy in the free ion — split into two sets with different energies. Electrons can absorb visible light to jump from the lower set to the higher set. The colour we see is the complement of the light absorbed.

The exact colour depends on:

The metal ion and its oxidation state
The number of d-electrons
The nature of the ligands (here, all are water or chloride)

Let's go salt by salt.

$FeSO_4 \cdot 7H_2O$ — Iron(II) sulphate heptahydrate

Iron in this salt is in the +2 oxidation state. The electron configuration of $Fe^{2+}$ is $[Ar]\,3d^6$ . In an octahedral field of water molecules, the d-orbitals split into $t_{2g}$ (lower) and $e_g$ (higher). With six d-electrons, the lower set is completely filled ( $t_{2g}^6$ ) and the upper set is empty. The d-d transition requires a moderate amount of energy, corresponding to absorption in the red region. The complementary colour is a pale green.

Tip

$Fe^{2+}$ solutions are often described as pale green, but they can appear almost colourless if very dilute. The "pale" is important — it's not the deep green of $Ni^{2+}$ .

So (i) matches (d) Pale green.
$NiCl_2 \cdot 4H_2O$ — Nickel(II) chloride tetrahydrate

Nickel is in the +2 state: $Ni^{2+}$ has $[Ar]\,3d^8$ . In an octahedral environment, the eight d-electrons fill the $t_{2g}$ set completely and put two electrons in the $e_g$ set ( $t_{2g}^6 e_g^2$ ). The d-d transitions absorb light in the blue and red regions, leaving green as the transmitted colour. This is a distinctive, bright green — think of the colour of nickel salts you might have seen in a lab.

So (ii) matches (a) Green.
$MnCl_2 \cdot 4H_2O$ — Manganese(II) chloride tetrahydrate

$Mn^{2+}$ has a $[Ar]\,3d^5$ configuration. This is a special case: with five d-electrons, each orbital gets one electron (Hund's rule), giving a half-filled, highly stable $d^5$ arrangement. The d-d transitions are spin-forbidden — they require an electron to flip its spin, which is much less likely. As a result, the absorption is very weak, and the colour is a very faint light pink (almost colourless in dilute solutions, but distinctly pink when concentrated).

Watch out

Don't confuse $Mn^{2+}$ (light pink) with $Mn^{7+}$ in permanganate (deep purple). The oxidation state changes everything — $Mn^{7+}$ has no d-electrons and gets its colour from charge transfer, not d-d transitions.

So (iii) matches (b) Light pink.
$CoCl_2 \cdot 6H_2O$ — Cobalt(II) chloride hexahydrate …

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Method: Colour of Hydrated Transition Metal Ions

This is a memory-based matching problem that relies on knowing the characteristic colours of common hydrated transition metal ions in aqueous solution.

Steps:

Identify the metal ion in each compound (ignore the water of crystallisation and anion for colour purposes — the colour comes from the hydrated cation).
Recall the standard colour of each hydrated transition metal ion in aqueous solution:

Metal Ion	Hydrated Colour
$\text{Fe}^{2+}$	Pale green
$\text{Ni}^{2+}$	Green
$\text{Mn}^{2+}$	Light pink (very pale, almost colourless in dilute solutions)
$\text{Co}^{2+}$	Pink
$\text{Cu}^{+}$	Colourless (unlike $\text{Cu}^{2+}$ which is blue)

Match each compound from Column I to its colour in Column II:

Column I	Metal Ion	Colour	Column II Match
(i) $FeSO_4 \cdot 7H_2O$	$\text{Fe}^{2+}$	Pale green	(d)
(ii) $NiCl_2 \cdot 4H_2O$	$\text{Ni}^{2+}$	Green	(a)

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Here are the most common mistakes students make when matching hydrated salt solutions to their colours, along with how to avoid each.

Mistake 1: Confusing the colours of $Fe^{2+}$ and $Fe^{3+}$ ions

The Error: Students often think all iron compounds are reddish-brown (rust colour). They incorrectly match $FeSO_4 \cdot 7H_2O$ with a brown or dark colour, or confuse it with the yellow/brown of $Fe^{3+}$ .
The Correct Fact: $Fe^{2+}$ (ferrous) in aqueous solution is pale green. $Fe^{3+}$ (ferric) is yellow/brown.
How to Avoid: Memorise the oxidation states. Remember: Ferrous ( $+2$ ) = Pale green; Ferric ( $+3$ ) = Yellow/Brown. For this question, (i) $FeSO_4 \cdot 7H_2O$ matches with (d) Pale green.

Mistake 2: Forgetting that $Cu^+$ is colourless (diamagnetic)

The Error: Students assume all copper compounds are blue. They match $Cu_2Cl_2$ (copper(I) chloride) with blue.
The Correct Fact: $Cu^+$ has a $d^{10}$ configuration (fully filled d-orbital). It has no unpaired electrons and cannot undergo d-d transitions. Therefore, its solutions are colourless (or white if solid). The blue colour comes from $Cu^{2+}$ ( $d^9$ ).
How to Avoid: Remember the diamagnetic rule: Ions with $d^{10}$ or $d^0$ configurations are colourless. $Cu^+$ ( $d^{10}$ ), $Zn^{2+}$ ( $d^{10}$ ), $Sc^{3+}$ ( $d^0$ ) are all colourless. For this question, (v) $Cu_2Cl_2$ matches with (f) Colourless.

Mistake 3: Mixing up the colours of $Ni^{2+}$ and $Co^{2+}$

The Error: Students swap the colours of Nickel and Cobalt salts. They match $NiCl_2$ with pink and $CoCl_2$ with green.
The Correct Fact:
- $Ni^{2+}$ (aqueous) is typically green (or apple green).
- $Co^{2+}$ (aqueous) is typically pink (or rose red).
How to Avoid: Use a simple mnemonic: "Nickel is Nice and Green" ; "Cobalt is Cool and Pink" . For this question:
- (ii) $NiCl_2 \cdot 4H_2O$ → (a) Green
- (iv) $CoCl_2 \cdot 6H_2O$ → (e) Pink

Mistake 4: Ignoring the hydration state or confusing it with the anhydrous form

The Error: Students think $MnCl_2 \cdot 4H_2O$ is colourless (like $KMnO_4$ ) or dark. They also confuse the hydrated $CoCl_2 \cdot 6H_2O$ (pink) with anhydrous $CoCl_2$ (blue).
The Correct Fact: …

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

TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
$Q. The correct order of metallic radius of Al, Ga, In, Tl is (A) Tl > In > Al > Ga (B) Tl > In > Ga > Al (C) In > Ga > Al > Tl (D) Ga > In > Tl > Al$
›Reveal solutionSolution
The metallic radius generally increases down a group, but gallium (Ga) is anomalously small due to its unique crystal structure and the d-block contraction, making the correct order Tl > In > Al > Ga.

The key concept here is periodic trends in atomic (metallic) radius down Group 13. Normally, as you go down a group, each element adds a new electron shell, so the atomic radius increases. However, Group 13 has a famous exception: gallium is actually smaller than aluminum. This happens because gallium has a filled d-subshell (3d¹⁰) that doesn’t shield the nuclear charge well, pulling the outer electrons in tighter. Additionally, gallium’s crystal structure is less metallic and more molecular, leading to a smaller metallic radius than expected.

Let’s work through the order step by step.

General trend down Group 13 From aluminum (Al) to indium (In) to thallium (Tl), each element is in a higher period, so the atomic radius should increase:

$\text{Al} < \text{In} < \text{Tl}$
This is the baseline expectation.

The gallium anomaly Gallium (Ga) sits directly below Al in the periodic table. However, its metallic radius is actually smaller than Al’s. Why?

Ga has the electron configuration [Ar] 3d¹⁰ 4s² 4p¹. The 3d electrons are poor at shielding the 4p electron from the nuclear charge, so the effective nuclear charge felt by the outer electron is higher.

Also, gallium’s crystal structure is orthorhombic (not close-packed like Al), leading to shorter interatomic distances. The result:

$r_{\text{Ga}} < r_{\text{Al}}$

Comparing Ga with In and Tl …
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TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
$\ce{Cl^-}$
›Reveal solutionSolution
The key idea is that beryllium and aluminium show diagonal similarity in the periodic table, which leads to shared properties like forming bridged chlorides and a strong tendency to form complexes, but their maximum covalency differs (4 for Be, 6 for Al). Thus only statements I and II are correct.

Concept and Intuition

Beryllium (Be, group 2) and aluminium (Al, group 13) are diagonal neighbours in the periodic table. This diagonal relationship arises because the increase in nuclear charge down a group is offset by the increase in atomic size across a period, leading to similar charge-to-size ratios and thus similar chemical behaviour. However, covalency limits depend on the number of available empty orbitals: Be has only 2s and 2p orbitals (max 4), while Al has 3s, 3p, and 3d orbitals (max 6). This distinction is crucial.

Step-by-step reasoning

Statement I: Chlorides with $\ce{Cl^-}$ bridged structure in vapour phase

Beryllium chloride ( $\ce{BeCl2}$ ) in the vapour phase exists as a dimer $\ce{Be2Cl4}$ , where each Be is surrounded by four Cl atoms, with two Cl atoms bridging the two Be atoms.

Aluminium chloride ( $\ce{AlCl3}$ ) also dimerises in the vapour phase to $\ce{Al2Cl6}$ , with chlorine bridges.

Both exhibit electron-deficient character and form bridged structures to complete their octet.

Conclusion: Statement I is correct.

Statement II: Strong tendency to form complexes

Be²⁺ has a high charge density (small size, +2 charge), making it a strong Lewis acid that readily accepts electron pairs from ligands (e.g., $\ce{[Be(H2O)4]^{2+}}$ , $\ce{[BeF4]^{2-}}$ ).

Al³⁺ has an even higher charge density and forms many complexes (e.g., $\ce{[AlF6]^{3-}}$ , $\ce{[Al(H2O)6]^{3+}}$ ).

Both show a strong tendency to form coordination compounds.

Conclusion: Statement II is correct. …
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TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
$>$
›Reveal solutionSolution
Atomic radii in group 13 do not increase smoothly down the group due to d- and f-block contractions; the correct order is Tl > In > Al > Ga, so option (D) is correct.

The key concept here is periodic trends in atomic radii, but with a twist. Normally, as you go down a group, atomic radius increases because each new element has an additional electron shell. However, group 13 (B, Al, Ga, In, Tl) shows an anomaly: gallium (Ga) is actually smaller than aluminium (Al), and thallium (Tl) is only slightly larger than indium (In). This happens because of the d-block contraction (for Ga and In) and the f-block contraction (for Tl). These contractions occur when electrons fill inner d or f orbitals, which are poor at shielding the nuclear charge, so the outer electrons are pulled in more tightly.

Let’s work through the reasoning step by step.

Start with the general trend.

Down a group, atomic radius should increase: B < Al < Ga < In < Tl. But experimental data show this is not true. The first break occurs at Ga.

Why is Ga smaller than Al?

Gallium comes after the first row of transition metals (Sc to Zn). In those elements, the 3d orbitals are filled. The 3d electrons are poor at shielding the 4s and 4p electrons from the nuclear charge. This d-block contraction (also called scandide contraction) makes the effective nuclear charge on the outer electrons higher, shrinking the atom. So Ga (atomic radius ~135 pm) is actually smaller than Al (~143 pm).

What about indium?

Indium is below Ga, so it has an extra shell (n=5). Despite the d-block contraction from the 4d electrons, the addition of a shell wins out, so In (~167 pm) is larger than both Al and Ga.

Now thallium — the second twist.

Thallium comes after the lanthanides (Ce to Lu). The 4f orbitals are filled in the lanthanide series, and these f-electrons are even worse at shielding than d-electrons. This f-block contraction (lanthanide contraction) is so strong that Tl (~170 pm) is only slightly larger than In (~167 pm), despite having an extra shell. In fact, Tl is smaller than you’d expect from a simple periodic trend. …
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TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
$Q. Match the following A Mn, Tc, Re B Zn, Cd, Hg C Ti, Zr, Hf D Ga, In, Tl I 12 II 4 III 17 IV 7 V 13 The correct answer is (A) A - IV, B - I, C - II, D - V (B) A - IV, B - II, C - I, D - V (C) A - III, B - I, C - II, D - V (D) A - III, B - V, C - I, D - IV$
›Reveal solutionSolution
The problem matches groups of elements (A–D) to their periodic‑table group numbers (I–V).

Mn, Tc, Re are in group 7 (IV); Zn, Cd, Hg in group 12 (I); Ti, Zr, Hf in group 4 (II); Ga, In, Tl in group 13 (V).

The correct match is A–IV, B–I, C–II, D–V, which corresponds to option (A).

The key idea is that each set of elements belongs to the same vertical column (group) in the modern periodic table. The group number tells you the number of valence electrons (for main‑group elements) or the typical oxidation states (for transition metals). Instead of memorizing, you can reason from the element’s position: for example, Mn is directly above Tc and Re, so they share group 7. Similarly, Zn, Cd, Hg are the “volatile” group 12 metals; Ti, Zr, Hf are group 4; Ga, In, Tl are group 13 (the boron family). The Roman numerals I–V are just the group numbers 12, 4, 17, 7, 13 respectively.

Identify the group of Mn, Tc, Re (Set A).

Manganese (Mn) is in the first transition series, atomic number 25. Its electron configuration ends in $3d^5 4s^2$ , placing it in group 7. Technetium (Tc) and rhenium (Re) are directly below Mn in the same column. So Set A belongs to group 7, which is listed as IV (since IV = 7).

Thus A → IV.

Identify the group of Zn, Cd, Hg (Set B).

Zinc (Zn) ends in $3d^{10}4s^2$ , a full d‑subshell, placing it in group 12. Cadmium (Cd) and mercury (Hg) are below Zn. Group 12 is listed as I (since I = 12).

Thus B → I.

Identify the group of Ti, Zr, Hf (Set C).

Titanium (Ti) has configuration $3d^2 4s^2$ , making it group 4. Zirconium (Zr) and hafnium (Hf) are directly below. Group 4 is listed as II (since II = 4).

Thus C → II.

Identify the group of Ga, In, Tl (Set D). …
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TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
$Q. The correct order of atomic radii of Al, Ga, In, Tl is (A) Al < Ga < In < Tl (B) Ga < Al < In < Tl (C) Ga < Al < Tl < In (D) Al < In < Ga < Tl$
›Reveal solutionSolution
Atomic radii generally increase down a group, but gallium (Ga) is an exception due to the d-block contraction, making it smaller than aluminium (Al). The correct order is Ga < Al < In < Tl.

The question asks for the correct order of atomic radii among the group 13 elements: Al, Ga, In, and Tl. This is a classic case where the simple periodic trend — radius increases down a group — hits a snag. The reason lies in the electronic configurations and the effect of inner electron shells.

Let’s recall the trend: as you go down a group, new electron shells are added, so atomic radius should increase. For group 13, that would predict Al < Ga < In < Tl. But experiment shows otherwise. Why?

The key is the d-block contraction. Gallium sits right after the first transition series (elements Sc to Zn). The 3d electrons in Ga are poor at shielding the nuclear charge, so the effective nuclear charge felt by the outermost electrons is higher than expected. This pulls the electron cloud inward, making Ga’s atomic radius smaller than Al’s, despite Ga being one period below.

Indium and thallium follow after the 4d and 5d series, but the effect is less dramatic for In (4d) and Tl (5d) because the 4f and 5f contractions also play a role, but the overall trend of increasing size resumes after Ga.

Let’s break it down step by step.

Identify the group and period positions.

Al is in period 3, Ga in period 4, In in period 5, Tl in period 6. Normally, period 4 elements are larger than period 3 elements. But Ga is an exception.

Understand the d-block contraction.

Ga has the configuration $[\text{Ar}] 3d^{10} 4s^2 4p^1$ . The 3d electrons are in a full d-subshell, which is diffuse and does not shield the 4p electron effectively from the nuclear charge. This increases the effective nuclear charge ( $Z_{\text{eff}}$ ) on the outermost electron, shrinking the atomic radius. Al has $[\text{Ne}] 3s^2 3p^1$ — no d-electrons, so its $Z_{\text{eff}}$ is lower, and its radius is larger than Ga’s.

Compare Al and Ga.

Experimental atomic radii (in picometers, typical values):

Al: 143 pm

Ga: 135 pm So Ga < Al.

Compare In and Tl. …
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TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
$Q. The atomic radius of gallium is less than that of aluminium. This is due to (A) Greater shielding power of s-electrons of gallium atom (B) poor shielding power of s-electrons of gallium (C) Poor shielding power of d-electrons of gallium (D) Greater shielding power of d-electrons of gallium$
›Reveal solutionSolution
The atomic radius of gallium is smaller than that of aluminium because the d-electrons in gallium provide poor shielding, allowing the nucleus to pull the outer electrons inward more strongly. The correct option is (C).

The key concept here is shielding (or screening) effect and how different types of orbitals (s, p, d) contribute to it. In the periodic table, atomic radius generally increases down a group because new electron shells are added. However, gallium (Ga) is an exception: it is smaller than aluminium (Al), even though it lies directly below Al in Group 13. This anomaly is due to the presence of a filled set of 3d orbitals in gallium, which are absent in aluminium.

Let’s break down why this happens.

Recall the trend and the anomaly

Aluminium (atomic number 13) has the electron configuration $[Ne] 3s^2 3p^1$ . Gallium (atomic number 31) has $[Ar] 3d^{10} 4s^2 4p^1$ . Normally, adding a shell (from n=3 to n=4) should increase the radius. But gallium’s radius is actually about 135 pm vs. aluminium’s 143 pm — a clear contraction.

Understand shielding and effective nuclear charge

Electrons in inner shells “shield” outer electrons from the full positive charge of the nucleus. The effective nuclear charge ( $Z_{\text{eff}}$ ) felt by an outer electron is the actual nuclear charge minus the shielding constant. If shielding is poor, $Z_{\text{eff}}$ is higher, pulling electrons closer and shrinking the atom.

Identify the role of d-electrons

In gallium, the 3d subshell is completely filled. d-orbitals are diffuse and penetrate poorly toward the nucleus compared to s- or p-orbitals. This means d-electrons are not very effective at shielding the outer 4s and 4p electrons from the nuclear charge. As a result, the 4p electron in gallium experiences a higher $Z_{\text{eff}}$ than expected, contracting the atomic radius.

Compare with aluminium

Aluminium has no d-electrons; its inner electrons are 1s, 2s, and 2p — all of which are better shielders. So aluminium’s outer electron feels a lower $Z_{\text{eff}}$ relative to its nuclear charge, giving a larger radius.

Evaluate the options …
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TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
$\mathrm{Mn^{3+}}$
›Reveal solutionSolution
The question asks for the incorrect statement about transition metals. The key is to recall that Mn²⁺ is more stable than Mn³⁺ due to the half-filled d⁵ configuration, making option (C) false. The correct answer is (C).

The concept here is stability of oxidation states in transition metals, governed by electronic configuration and the extra stability of half-filled and fully-filled d-subshells. While melting points generally increase with more unpaired electrons (due to stronger metallic bonding), and Cr, Mo, W indeed have very high melting points, the stability of Mn²⁺ over Mn³⁺ is a classic exception. Let’s check each statement.

Statement (A): Cr, Mo and W have high melting points

These elements belong to Group 6. They have a high number of unpaired d-electrons (Cr: d⁵, Mo: d⁵, W: d⁴ in their metallic states), leading to strong metallic bonding. This results in exceptionally high melting points (e.g., W melts at ~3422°C). This statement is true.

Statement (B): With increase in number of unpaired electrons melting point increases

In transition metals, metallic bonding strength depends on the number of delocalized electrons contributed by d-orbitals. More unpaired electrons mean more electrons available for bonding, raising the melting point. This trend generally holds across a period (e.g., from Sc to Cr). This statement is true.

Statement (C): Mn³⁺ is more stable than Mn²⁺ …
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TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
$Q. Assertion (A) : In general, transition metals have high melting points. Reason (R) : More number of electrons from '(n-1)d' and 'ns' are involved in interatomic metallic bonding. The correct option among the following is (A) (A) is true, (R) is true and (R) is the correct explanation for (A) (B) (A) is true, (R) is true but (R) is not the correct explanation for (A) (C) (A) is true but (R) is false (D) (A) is false but (R) is true$
›Reveal solutionSolution
Transition metals do have high melting points, and the reason is exactly the one stated — participation of both $(n-1)d$ and $ns$ electrons in metallic bonding. (R) explains (A) — option (A).

The concept first

In a metallic lattice, positive kernels sit in a "sea" of delocalised electrons. The strength of the metallic bond — and hence the melting point, boiling point and enthalpy of atomisation — grows with the number of electrons each atom contributes to that sea, particularly the number of unpaired electrons available for bonding.

For $s$ -block metals only the $ns$ electrons (one or two) are available. For a transition metal, the $(n-1)d$ and $ns$ orbitals lie very close in energy, so the $d$ electrons can also take part. That gives a much larger pool of bonding electrons.

Step-by-step

Is the Assertion true? Yes. Compare (approximate melting points):

$\mathrm{Na}\ 371\ \mathrm{K}, \quad \mathrm{Ca}\ 1112\ \mathrm{K} \qquad\text{vs}\qquad \mathrm{Fe}\ 1808\ \mathrm{K}, \quad \mathrm{Cr}\ 2130\ \mathrm{K}, \quad \mathrm{W}\ 3683\ \mathrm{K}$
Transition metals are, in general, hard, dense and high-melting. (A) is true.

2. Is the Reason true? Yes — the near-degeneracy of $(n-1)d$ and $ns$ means both sets of electrons enter the delocalised bonding, so more electrons per atom participate in interatomic metallic bonding. (R) is true.

3. Does (R) explain (A)? Yes, and there are two beautiful pieces of supporting evidence within the block itself: …
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TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ

4d^5\ 5s^1

›Reveal solution

Molybdenum adopts the $4d^5\ 5s^1$ configuration (rather than $4d^4\ 5s^2$ ) to maximize exchange energy from having six unpaired electrons in half-filled plus one orbitals, and this configuration indeed provides the highest exchange energy in the second transition series. Both statements are true and causally connected.

Understanding Electronic Configuration Anomalies

Molybdenum (atomic number 42) presents a classic case of an anomalous electron configuration. We'd naively expect it to follow the Aufbau principle strictly and have $4d^4\ 5s^2$ , but nature has other plans.

The key concept here is exchange energy — the quantum mechanical stabilization that occurs when electrons with parallel spins can exchange positions. The more unpaired electrons with the same spin, the greater the exchange energy stabilization. This is a manifestation of the Pauli exclusion principle and electron correlation effects.

Evaluating Assertion (A)

Let's verify Mo's ground state configuration:

Expected configuration: Following the simple filling order, we'd predict $[\text{Kr}]\ 4d^4\ 5s^2$
Actual configuration: Molybdenum actually has $[\text{Kr}]\ 4d^5\ 5s^1$
Why the promotion? Moving one electron from the $5s$ to the $4d$ orbital creates:
- Five unpaired electrons in the $4d$ subshell (half-filled)
- One unpaired electron in the $5s$ orbital
- Total: six unpaired electrons
Energy balance: The exchange energy gained from having six unpaired electrons (instead of four) more than compensates for the small energy cost of promoting an electron from $5s$ to $4d$

Assertion (A) is TRUE.

Evaluating Reason (R)

Now let's examine the exchange energy claim across the second transition series (Y through Cd):

Element	Config	Unpaired e⁻	Exchange pairs
Y	$4d^1\ 5s^2$	1	0
Zr	$4d^2\ 5s^2$	2	1
Nb	$4d^4\ 5s^1$	5	10
Mo	$4d^5\ 5s^1$	6	15
Tc	$4d^5\ 5s^2$	5	10
Ru	$4d^7\ 5s^1$	4	6

n

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TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
$Q. Assertion (A): The ionic radii of the alkaline earth metals are smaller than those of alkali metals in the same period. Reason (R): Alkali metals have higher nuclear charge than that of the alkaline earth metals. The correct option among the following is (A) (A) is true, (R) is true and (R) is the correct explanation for (A) (B) (A) is true, (R) is true but (R) is not the correct explanation for (A) (C) (A) is true but (R) is false (D) (A) is false but (R) is true$
›Reveal solutionSolution
The assertion is true — alkaline earth metals have smaller ionic radii than alkali metals in the same period — but the reason given is false, because alkali metals actually have lower nuclear charge than alkaline earth metals in the same period.

The concept: periodic trends in ionic radii and nuclear charge.

Ionic radius depends on two competing factors: the number of protons in the nucleus (nuclear charge) and the number of electrons. In the same period, as you move from left to right, the nuclear charge increases while electrons are added to the same principal shell. A higher nuclear charge pulls the electron cloud inward more strongly, shrinking the ionic radius. So the key idea is: greater nuclear charge → smaller ionic radius, for ions with the same number of electron shells.

Now let’s apply this to the given statements.

Check the Assertion (A): “The ionic radii of the alkaline earth metals are smaller than those of alkali metals in the same period.”

In the same period (e.g., Period 3), the alkali metal is sodium (Na) and the alkaline earth metal is magnesium (Mg). The common ions are Na⁺ and Mg²⁺. Both have the same electron configuration as neon (2,8), but Mg²⁺ has 12 protons pulling on 10 electrons, while Na⁺ has only 11 protons. The stronger pull in Mg²⁺ makes its ionic radius smaller. This trend holds across all periods: Li⁺ > Be²⁺, Na⁺ > Mg²⁺, K⁺ > Ca²⁺, etc. So Assertion (A) is true.

Check the Reason (R): “Alkali metals have higher nuclear charge than that of the alkaline earth metals.” …
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TG EAPCET 2021Set ap-2021-08-09-FN1 markMCQ
$Q. Which of the following statements are correct? A) The separation of actinoid elements from each other is difficult B) The covalency of the compounds of actinoid metals decrease from left to right along actinoid series C) The compounds of Lawrencium are most covalent D) U and Th occur naturally in substantial quantities (A) A and B only (B) B and C only (C) A, C and D only (D) A, B, C and D$
›Reveal solutionSolution
The key idea is that actinoid chemistry is dominated by the 5f orbitals, which are less shielded and more involved in bonding than 4f orbitals, leading to complex separation chemistry and a decrease in covalency across the series. The correct statements are A, C, and D, so the answer is option (C).

The question tests your understanding of the actinoid series (elements 89–103). Unlike lanthanoids, where 4f electrons are deeply buried, actinoid 5f orbitals are more extended and can participate in covalent bonding, especially early in the series. This leads to several distinctive trends.

Statement A: "The separation of actinoid elements from each other is difficult"

This is correct. Actinoids have very similar chemical properties (e.g., all form +3 ions, and early ones also show +4, +5, +6 states). Their ionic radii decrease slowly across the series (actinoid contraction), making separation by ion exchange or solvent extraction challenging — much like lanthanoids, but with added complexity from multiple oxidation states.

Statement B: "The covalency of the compounds of actinoid metals decrease from left to right along actinoid series"

This is incorrect. Covalency increases from left to right. Early actinoids (Th, Pa, U) have more diffuse 5f orbitals that overlap poorly with ligands, so bonding is more ionic. As you move right, the 5f orbitals contract and become more like 4f orbitals, but they also become more available for covalent interaction due to better energy matching with ligand orbitals. Actually, the trend is: covalency is highest in the middle of the series (around Pu, Am) and then decreases again? Let’s be precise: For actinoids, the 5f orbitals are more extended than 4f, so early members show some covalency, but the maximum covalency occurs around the middle (e.g., in Np, Pu, Am) because the 5f orbitals are still relatively diffuse and the energy gap to ligand orbitals is smallest. However, the statement says "decrease from left to right" — that is false; it generally increases then decreases, but certainly not a monotonic decrease. The correct trend is that covalency increases initially, then decreases. So B is wrong.

Statement C: "The compounds of Lawrencium are most covalent" …
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Q.Match the solutions given in Column I and the colours given in Column II. Column I (Aqueous solution of salt):

Why salts have colour — the d-orbital story

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Q.Match the solutions given in Column I and the colours given in Column II.
Column I (Aqueous solution of salt):

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