(a) Answer the following questions: (i) Which element of the first transition series has highest second ionisa

NCERT Exemplar · Q46

Q. (a) Answer the following questions:

(i) Which element of the first transition series has highest second ionisation enthalpy?

(ii) Which element of the first transition series has highest third ionisation enthalpy?

(iii) Which element of the first transition series has lowest enthalpy of atomisation?

(b) Identify the metal and justify your answer.

(i) Carbonyl

M(CO)_5

(ii)

MO_3F

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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 key idea is to use electronic configurations and periodic trends across the first transition series (Sc–Zn). For (a), the highest second ionisation enthalpy belongs to Cu, the highest third to Zn, and the lowest enthalpy of atomisation to Zn. For (b), the metal in $M(CO)_5$ is Fe (as Fe(CO) $_5$ ), and in $MO_3F$ it is Mn (as MnO $_3$ F).

(a) Ionisation enthalpies and enthalpy of atomisation in the first transition series

The first transition series runs from Sc (atomic number 21) to Zn (30). The trends in ionisation enthalpies and enthalpy of atomisation are governed by the stability of half-filled and fully-filled $d$ subshells, and by the strength of metallic bonding.

1. Highest second ionisation enthalpy

Second ionisation enthalpy ( $IE_2$ ) is the energy needed to remove an electron from the $M^+$ ion. For most elements, $IE_2$ is larger than $IE_1$ , but the jump is especially large when the $M^+$ ion has a stable configuration.

Consider Cu ( $3d^{10}4s^1$ ). After losing one electron, Cu $^+$ becomes $3d^{10}$ — a completely filled $d$ subshell, which is very stable. Removing a second electron from this stable $d^{10}$ core requires a lot of energy.
Compare with Zn ( $3d^{10}4s^2$ ): Zn $^+$ is $3d^{10}4s^1$ , not as stable as Cu $^+$ . So Cu has the highest $IE_2$ in the series.

Watch out

A common mistake is to think Zn has the highest $IE_2$ because it has the highest $IE_1$ . But $IE_2$ depends on the stability of the monovalent cation, not the neutral atom. Cu $^+$ ( $d^{10}$ ) is exceptionally stable.

2. Highest third ionisation enthalpy

Third ionisation enthalpy ( $IE_3$ ) is the energy to remove an electron from $M^{2+}$ .

Zn $^{2+}$ has configuration $3d^{10}$ — again a fully filled $d$ subshell. Removing a third electron from this stable $d^{10}$ core is extremely difficult.
No other $M^{2+}$ in the series has a $d^{10}$ configuration (e.g., Cu $^{2+}$ is $d^9$ , Ni $^{2+}$ is $d^8$ ). So Zn has the highest $IE_3$ .

3. Lowest enthalpy of atomisation

Enthalpy of atomisation ( $\Delta H_{\text{atom}}$ ) reflects the strength of metallic bonding. Across the series, metallic bonding depends on the number of unpaired $d$ electrons available for bonding.

Zn has a $3d^{10}4s^2$ configuration — all $d$ electrons are paired. This leads to weak metallic bonding (Zn is relatively volatile, with a low melting point).
In contrast, elements like Cr ( $3d^5 4s^1$ ) or Mn ( $3d^5 4s^2$ ) have many unpaired electrons and stronger bonding.
Hence, Zn has the lowest enthalpy of atomisation.

Tip

A quick way to remember: Zn is the odd one out — it has the highest $IE_3$ (due to $d^{10}$ in $M^{2+}$ ) and the lowest atomisation enthalpy (due to no unpaired $d$ electrons). Cu has the highest $IE_2$ because Cu $^+$ is $d^{10}$ .

(b) Identifying the metal in given compounds

4. Carbonyl $M(CO)_5$ …

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(a) Ionisation Enthalpies & Atomisation Enthalpy

(i) Highest second ionisation enthalpy

Concept:

Second ionisation enthalpy ( $IE_2$ ) is the energy needed to remove an electron from a +1 cation. The highest $IE_2$ occurs when the +1 cation has a stable electronic configuration — removing another electron would break that stability.

For first transition series (Sc to Zn), the +1 state is rarely stable.
The element with the highest $IE_2$ is Zinc (Zn).
- Reason: Zn⁺ has configuration $3d^{10}4s^1$ . Removing one more electron gives Zn²⁺ ( $3d^{10}$ ), which is stable. However, the jump from $3d^{10}4s^1$ to $3d^{10}$ is not the highest second ionisation enthalpy in the series.

Correct answer:

Copper (Cu) has the highest $IE_2$ in the first transition series.

Cu⁺ has configuration $3d^{10}$ (completely filled d-subshell — very stable).
Removing an electron from this stable $3d^{10}$ to get Cu²⁺ ( $3d^9$ ) requires a lot of energy.
So $IE_2$ of Cu is the highest.

Common mistake: Students often pick Zn because Zn²⁺ is stable. But Zn⁺ ( $3d^{10}4s^1$ ) is not as stable as Cu⁺ ( $3d^{10}$ ). The stability of the +1 cation determines $IE_2$ .

How to avoid:

Always check the electronic configuration of the +1 ion. The more stable it is, the higher the $IE_2$ .

(ii) Highest third ionisation enthalpy

Concept:

Third ionisation enthalpy ( $IE_3$ ) is the energy to remove an electron from a +2 cation. The highest $IE_3$ occurs when the +2 cation has a stable configuration.

Zinc (Zn) has the highest $IE_3$ $I E_{3}$ .
- Zn²⁺ has configuration $3d^{10}$ (completely filled — very stable).
- Removing one more electron gives Zn³⁺ ( $3d^9$ ), breaking that stable filled subshell.
- This requires maximum energy.

Common mistake: Students sometimes pick Mn (because Mn²⁺ is half-filled stable). But Mn²⁺ ( $3d^5$ ) is stable, so $IE_3$ is not the highest — it’s actually lower than Zn.

How to avoid:

Remember: Filled subshell ( $d^{10}$ ) is more stable than half-filled ( $d^5$ ). So Zn²⁺ > Mn²⁺ in stability → Zn has higher $IE_3$ .

(iii) Lowest enthalpy of atomisation

Concept:

Enthalpy of atomisation ( $\Delta H_{atom}$ ) is the energy required to convert the metal into isolated gaseous atoms. It depends on metallic bond strength.

In the first transition series, Zinc (Zn) has the lowest $\Delta H_{atom}$ $Δ H_{a t o m}$ .
- Zn has no unpaired electrons in its ground state ( $3d^{10}4s^2$ ).
- Metallic bonding is weaker because there are no unpaired d-electrons to contribute to bonding.

Common mistake: Students think Mn (half-filled) has lowest atomisation enthalpy. But Mn has five unpaired electrons → strong metallic bonding → higher $\Delta H_{atom}$ .

How to avoid:

More unpaired d-electrons → stronger metallic bond → higher atomisation enthalpy. Zn has zero unpaired electrons → weakest bonding → lowest $\Delta H_{atom}$ .

(b) Identify the metal and justify

(i) Carbonyl $M(CO)_5$

Concept:

Metal carbonyls follow the 18-electron rule (effective atomic number rule). For a neutral carbonyl $M(CO)_5$ :

Each CO donates 2 electrons.
Total electrons from CO = $5 \times 2 = 10$ .
Metal must contribute 8 electrons to reach 18.
So the metal must have 8 valence electrons in its neutral state.

Which first transition series metal has 8 valence electrons?

Configuration: $3d^6 4s^2$ → Iron (Fe).
Fe(CO) $_5$ is a well-known stable compound.

Common mistake: Students guess Ni or Co. But Ni(CO) $_4$ is tetrahedral (18e⁻), and Co₂(CO)₈ is dimeric. Only Fe gives a stable mononuclear pentacarbonyl.

How to avoid: …

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

COMEDK 2026Set 2026-M1 markMCQ
$\mathrm{Cr}^{3+}$
›Reveal solutionSolution
Colour in aqueous solution arises from d–d transitions in transition-metal ions with partially filled d-orbitals. Among the options, only Cr³⁺ has an incomplete d-subshell (d³), so it is the species that imparts colour.

The key concept is crystal field theory and d–d transitions. For a transition-metal ion to appear coloured in solution, it must have at least one unpaired electron in its d-orbitals, allowing it to absorb visible light by promoting an electron from a lower-energy d-orbital to a higher-energy one. Ions with completely filled (d¹⁰) or empty (d⁰) d-subshells cannot undergo such transitions and are typically colourless.

Let’s examine each option:

Cr³⁺ – Chromium in the +3 oxidation state has the electron configuration [Ar] 3d³. The d-subshell is partially filled (three d-electrons). This allows d–d transitions, so Cr³⁺ solutions (e.g., CrCl₃) are typically violet or green. This will impart colour.

Zn²⁺ – Zinc in the +2 state has the configuration [Ar] 3d¹⁰. The d-subshell is completely filled. No d–d transitions are possible, and Zn²⁺ solutions are colourless. No colour.

Ti⁴⁺ – Titanium(IV) has lost all four valence electrons, giving [Ar] (no d-electrons). The d-subshell is empty (d⁰). No d–d transitions possible; Ti⁴⁺ solutions are colourless. No colour. …
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COMEDK 2025Set 2025-M1 markMCQ
$\mathrm{Sc}^{3+}$
›Reveal solutionSolution
The key is to evaluate each statement using principles of transition-metal chemistry: d–d transitions require partially filled d-orbitals, paramagnetism depends on unpaired electrons, ionic character and acidity relate to oxidation state, and Co³⁺ has a 3d⁶ configuration. Only statement (C) is correct.

Concept and Intuition

Transition-metal compounds exhibit colour, paramagnetism, and variable oxidation states due to their partially filled d-orbitals. Each statement here tests a specific concept:

Colour from d–d transitions requires at least one d-electron and an empty d-orbital to allow excitation.

Paramagnetic strength is proportional to the number of unpaired electrons.

Higher oxidation states increase covalent character (Fajan’s rules) and make oxides more acidic.

The electron configuration of an ion is found by removing electrons from the neutral atom’s configuration, starting with the 4s orbital.

Step-by-step analysis

Statement (A):

Sc³⁺ has the configuration [Ar] (no d-electrons). Without any d-electrons, d–d transitions are impossible.

Cu⁺ has configuration [Ar] 3d¹⁰ (full d-subshell). A full d-subshell means no empty d-orbital to accept an excited electron, so d–d transitions cannot occur.

Therefore, neither ion can be coloured via d–d transitions. Statement (A) is false.

Statement (B):

Determine the number of unpaired electrons for each ion (all are 2+ ions of first-row transition metals):

V²⁺: [Ar] 3d³ → 3 unpaired electrons.

Cr²⁺: [Ar] 3d⁴ → 4 unpaired electrons (high-spin, as in aqueous complexes).

Mn²⁺: [Ar] 3d⁵ → 5 unpaired electrons.

Fe²⁺: [Ar] 3d⁶ → 4 unpaired electrons (high-spin).

Paramagnetic nature increases with number of unpaired electrons. So the order should be: V²⁺ (3) < Cr²⁺ (4) = Fe²⁺ (4) < Mn²⁺ (5).

The given order is V²⁺ < Cr²⁺ = Mn²⁺ < Fe²⁺, which incorrectly places Mn²⁺ equal to Cr²⁺ and Fe²⁺ as the most paramagnetic.

Statement (B) is false.

Statement (C): …
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COMEDK 2024Set 2024-E1 markMCQ
$\mathrm{Yb}=70, \quad \mathrm{Lu}=71, \mathrm{Pr}=59, \quad \mathrm{La}=57$
›Reveal solutionSolution
Colour in lanthanoid ions arises from f–f transitions, which require at least one unpaired electron in the 4f subshell. Among the given ions, only Pr³⁺ (4f²) has unpaired electrons, so it forms coloured compounds.

Concept & Intuition

Lanthanoid ions often display beautiful colours in solution or in solids. The colour comes from electronic transitions within the partially filled 4f orbitals. These f–f transitions are Laporte-forbidden but become weakly allowed due to vibronic coupling, giving pale but distinct colours. The key requirement is that the ion must have at least one unpaired electron in its 4f subshell. If the 4f subshell is empty (4f⁰) or completely full (4f¹⁴), no such transitions are possible, and the ion is colourless.

Let’s check each option.

Yb²⁺ (Ytterbium, atomic number 70)

Neutral Yb: [Xe] 4f¹⁴ 6s²

Yb²⁺ loses the two 6s electrons: [Xe] 4f¹⁴

4f subshell is completely full → no unpaired electrons → colourless.

La³⁺ (Lanthanum, atomic number 57)

Neutral La: [Xe] 5d¹ 6s²

La³⁺ loses the 5d and both 6s electrons: [Xe] 4f⁰

4f subshell is empty → no unpaired electrons → colourless.

Lu³⁺ (Lutetium, atomic number 71)

Neutral Lu: [Xe] 4f¹⁴ 5d¹ 6s²

Lu³⁺ loses the 5d and both 6s electrons: [Xe] 4f¹⁴

Again, 4f subshell is full → no unpaired electrons → colourless.

Pr³⁺ (Praseodymium, atomic number 59)

Neutral Pr: [Xe] 4f³ 6s²

Pr³⁺ loses the two 6s electrons and one 4f electron: [Xe] 4f² …
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KCET 2023Set D-21 markMCQ
$Q. Which of the following is CORRECT with respect to melting point of a transition element? (A) V > Cr (B) Cr > Mn (C) Mn > Fe (D) Ti > V$
›Reveal solutionSolution
Recall the NCERT melting-point trend across the 3d series and test each inequality; Mn's anomalously low melting point makes Cr > Mn the robust correct statement.

Step 1 — The concept behind the trend

In a transition metal, the strength of metallic bonding depends on how many unpaired d-electrons are available to take part in interatomic (covalent-like) bonding on top of the metallic sea. Bonding strength — and hence melting point — therefore rises to a maximum near the middle of the series and falls off towards the end as d-electrons begin to pair up.

Step 2 — The data (NCERT, 3d series, $^\circ$ C)
$\mathrm{Ti}\ 1677 \quad \mathrm{V}\ 1917 \quad \mathrm{Cr}\ 1903 \quad \mathrm{Mn}\ 1244 \quad \mathrm{Fe}\ 1535$
Step 3 — The Mn anomaly (the point of the question)

Manganese has the configuration $3d^5 4s^2$ . Its half-filled d-subshell is exceptionally stable, so those five electrons are reluctant to delocalise into the metallic bond. With fewer electrons effectively contributing to bonding, Mn's metallic bonding is weak and its melting point ( $1244^\circ$ C) is far below both its neighbours Cr ( $1903^\circ$ C) and Fe ( $1535^\circ$ C). This dip is the famous irregularity in the melting-point curve.

Step 4 — Test each option

(B) Cr > Mn: $1903 > 1244$ — TRUE, and by a very large margin (this is the Mn anomaly). …
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COMEDK 2023Set 2023-M1 markMCQ
$\mathrm{Ti}^{2+}$
›Reveal solutionSolution
Colour in transition-metal ions comes from $d$ – $d$ electronic transitions, which require partially filled $d$ orbitals. $\mathrm{Ti^{2+}}$ ( $3d^2$ ) has $d$ electrons and is coloured (purple); $\mathrm{Ti^{4+}}$ ( $3d^0$ ) has none and is colourless.

Titanium: $\mathrm{Ti}$ ( $Z=22$ ) $= [\mathrm{Ar}]3d^2 4s^2$ .

$\mathrm{Ti^{2+}}$ : remove the two $4s$ electrons $\Rightarrow [\mathrm{Ar}]3d^2$ . With two $d$ electrons, $d$ – $d$ transitions are possible, absorbing visible light and imparting the purple colour. …
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KCET 2021Set B-21 markMCQ
$+2$
›Reveal solutionSolution

Every $+2$ ion from Sc $^{2+}$ ( $3d^1$ ) to Cu $^{2+}$ ( $3d^9$ ) carries unpaired $3d$ electrons, so all are paramagnetic — option (D).

Test each statement across Sc–Cu (the $3d$ series).

(A) False — Sc's lowest common state is $+3$ and Cu shows $+1$ , so $+2$ is not the lowest for every element.

(B) False — in Cr $[\text{Ar}]3d^5 4s^1$ and Cu $[\text{Ar}]3d^{10}4s^1$ the $4s$ orbital holds only one electron.

(C) False — Cu is $[\text{Ar}]3d^{10}4s^1$ , so its $3d$ subshell is completely filled.

(D) True — the two $4s$ electrons are lost first, so each $+2$ ion is $3d^n$ : …
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KCET 2021Set B-21 markMCQ
$Q. The property of the alkaline earth metals that increases with their atomic number is (A) Ionisation enthalpy (B) Electronegativity (C) Solubility of their hydroxide in water (D) Solubility of their sulphate in water$
›Reveal solutionSolution
Only hydroxide solubility rises down group 2; ionisation enthalpy, electronegativity and sulphate solubility all fall.

Step 1 — The governing principle for solubility trends

Whether an ionic solid dissolves depends on the competition between two quantities:
$\Delta H_{\text{soln}} = \Delta H_{\text{hydration}} - \Delta H_{\text{lattice}}$
Both lattice and hydration enthalpies decrease as the cation gets bigger down the group. Which one falls faster decides the trend — and that depends on the size of the anion:

With a small anion ( $OH^-$ , $F^-$ ), the lattice enthalpy is very sensitive to the cation size, so it falls faster than the hydration enthalpy $\Rightarrow$ solubility increases down the group.

With a large anion ( $SO_4^{2-}$ , $CO_3^{2-}$ ), the lattice enthalpy is dominated by the big anion and barely changes; the hydration enthalpy of the cation falls faster $\Rightarrow$ solubility decreases down the group.

Step 2 — Apply it to each option

(C) Hydroxides. $OH^-$ is a small anion, so the first case applies:
$Be(OH)_2 < Mg(OH)_2 < Ca(OH)_2 < Sr(OH)_2 < Ba(OH)_2$
Solubility (and hence basic strength) increases with atomic number. $Mg(OH)_2$ is only sparingly soluble (milk of magnesia), while $Ba(OH)_2$ is appreciably soluble and a strong base. ✓ This is the property that increases. …
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←Previous questionQ45 — On the basis of Lanthanoid contraction, explain the following: (i) Nat…→Next questionQ47 — Mention the type of compounds formed when small atoms like H, C and N …

Q.(a) Answer the following questions:

Concept understanding — Magnetism and Color

Magnetism and Colour: An Intuitive First Look

The Intuition: Electrons as Tiny Magnets and Painters

The Precise Statement

How They Connect (and How They Don't)

The Deeper Link: Transition Metals

Why this formula?

Magnetism and Color: Why the Key Formulas Hold

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

Why does this hold?

2. The Zeeman Effect: How Magnetic Fields Split Colors

Why does this hold?

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

Why does this hold?

(a) Ionisation enthalpies and enthalpy of atomisation in the first transition series

(b) Identifying the metal in given compounds

Method: Electronic Configuration & Periodic Trends Analysis

(a) Step-by-step reasoning

(i) Highest second ionisation enthalpy

(ii) Highest third ionisation enthalpy

(iii) Lowest enthalpy of atomisation

(b) Identify the metal and justify

(i) Carbonyl $M(CO)_5$

(a) Ionisation Enthalpies & Atomisation Enthalpy

(i) Highest second ionisation enthalpy

(ii) Highest third ionisation enthalpy

(iii) Lowest enthalpy of atomisation

(b) Identify the metal and justify

(i) Carbonyl $M(CO)_5$

Step 1 — The concept behind the trend

Step 2 — The data (NCERT, 3d series, $^\circ$ C)

Step 3 — The Mn anomaly (the point of the question)

Step 4 — Test each option

Step 1 — The governing principle for solubility trends

Step 2 — Apply it to each option

Q.(a) Answer the following questions:

(a) Ionisation enthalpies and enthalpy of atomisation in the first transition series

(b) Identifying the metal in given compounds

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