Which transition metal of 3d series does not show variable oxidation states?

Question

Q. Which transition metal of 3d series does not show variable oxidation states?

CBSECBSE Class XII Board 2020Subjective· 1mImportance★★★★★

✓ Free question

Concept understanding — Transition Element Definition

Transition Elements: From Intuition to Definition

Imagine you're building a house with bricks. Most bricks are identical — you stack them in neat rows. But some bricks are special: they have extra slots on their sides where you can attach hooks, magnets, or other bricks. These special bricks can change the shape of the wall, conduct electricity, or even change colour when you heat them.

In the periodic table, transition elements are those special bricks. They are the metals that sit in the middle block — groups 3 to 12 — and they have a unique ability: they can use their inner electrons (not just the outermost ones) to form bonds, change oxidation states, and create colourful compounds.

The Intuition: Why "Transition"?

The word "transition" comes from the idea that these elements form a bridge between the highly reactive metals on the left (like sodium, magnesium) and the less reactive metals / non-metals on the right (like aluminium, silicon). Their properties are not extreme — they are in-between.

But the real reason they are special lies in their electron configuration.

The Precise Definition (IUPAC)

A transition element is an element whose atom has an incomplete d sub-shell, or which can give rise to cations with an incomplete d sub-shell.

Let's unpack that.

1. The "d" sub-shell

Electrons are arranged in shells (K, L, M, N...) and sub-shells (s, p, d, f). The d sub-shell can hold a maximum of 10 electrons. In transition elements, the d sub-shell is being filled — but not completely.

For example, consider Iron (Fe):

Atomic number 26
Electron configuration: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6$
The 3d sub-shell has 6 electrons — it is incomplete (it can hold 10).

So iron is a transition element.

2. The "or" part — cations matter

Some elements have a complete d sub-shell in their neutral atom, but when they lose electrons to form positive ions (cations), the d sub-shell becomes incomplete.

Example: Zinc (Zn)

Neutral Zn: $[Ar] 3d^{10} 4s^2$ — the 3d sub-shell is full (10 electrons).
But Zn commonly forms $Zn^{2+}$ : $[Ar] 3d^{10}$ — still full.
So zinc is NOT a transition element by the IUPAC definition.

Example: Copper (Cu)

Neutral Cu: $[Ar] 3d^{10} 4s^1$ — 3d is full.
But $Cu^{2+}$ : $[Ar] 3d^9$ — now the 3d sub-shell is incomplete.
So copper IS a transition element.

Watch out

A common mistake: thinking that all elements in the d-block (groups 3–12) are transition elements. They are not. Zinc, cadmium, and mercury are d-block elements but NOT transition elements because their common cations have a full d sub-shell.

The "d-block" vs "Transition Elements"

d-block elements	Transition elements
Groups 3 to 12	Groups 3 to 11 (excluding Zn, Cd, Hg)
All have d electrons	Must have incomplete d sub-shell in atom or common cation
Example: Zn, Cd, Hg are d-block	Example: Sc, Ti, Fe, Cu are transition

Why This Definition Matters

The incomplete d sub-shell gives transition elements their characteristic properties:

Variable oxidation states — iron can be $Fe^{2+}$ or $Fe^{3+}$ , manganese can be $Mn^{2+}$ to $Mn^{7+}$
Coloured compounds — $Cu^{2+}$ is blue, $Cr^{3+}$ is green, $Fe^{3+}$ is yellow-brown
Catalytic activity — iron in Haber process, nickel in hydrogenation
Magnetic properties — unpaired d electrons create paramagnetism

The Final Answer

Important

A transition element is an element whose atom has an incomplete d sub-shell, or which can form a cation with an incomplete d sub-shell. This includes elements from Sc to Cu in the first row (and their heavier counterparts), but excludes Zn, Cd, and Hg because their common cations have a full $d^{10}$ configuration.

The formal definition of a transition element is a must-know fact from the NCERT/CBSE Class 12 Chemistry chapter on d- and f-Block Elements, and ‘why zinc is not a transition element’ is one of the most frequently asked important questions in board exams, JEE Main and NEET. Getting this definition precise is essential for correctly answering competitive-exam MCQs on periodic classification.

Why this formula?

Transition Element Definition: The "Why" Behind the Definition

The Core Definition

A transition element (IUPAC definition) is an element whose atom has an incomplete d-subshell in its ground state or can form stable ions with an incomplete d-subshell.

Key exam point: This definition covers both the neutral atom and its common ions.

Why This Definition? The Reasoning

1. The d-orbital filling pattern

In the periodic table, transition elements belong to the d-block (Groups 3–12). As we move across a period, electrons fill the $(n-1)d$ orbitals after the $ns$ orbital.

For example, in Period 4:

Scandium (Sc): $[Ar]\,3d^1\,4s^2$ — has one d-electron → transition element
Zinc (Zn): $[Ar]\,3d^{10}\,4s^2$ — d-subshell is full → not a transition element

2. The "incomplete d-subshell" condition

The definition focuses on incompleteness because:

A full d-subshell ( $d^{10}$ ) is exceptionally stable (like a noble gas configuration for d-orbitals)
Elements with $d^{10}$ configurations do not show the characteristic properties of transition metals (variable oxidation states, coloured compounds, catalytic activity, paramagnetism)

3. Why include ions?

Consider Zinc (Zn):

Ground state: $[Ar]\,3d^{10}\,4s^2$ — d-subshell is full → not a transition element
Common ion: $\text{Zn}^{2+}$ : $[Ar]\,3d^{10}$ — still full → still not a transition element

Now consider Copper (Cu):

Ground state: $[Ar]\,3d^{10}\,4s^1$ — d-subshell is full → by atom definition alone, not a transition element
But $\text{Cu}^{2+}$ : $[Ar]\,3d^9$ — incomplete d-subshell → is a transition element

Therefore: The definition must include ions to correctly classify elements like Cu, which form stable ions with incomplete d-subshells.

The "Formula" — A Decision Tree

The definition can be expressed as a logical condition:

\text{Transition element} \iff \left( \text{Atom has } d^{1-9} \right) \lor \left( \text{Stable ion has } d^{1-9} \right)

Where:

$d^{1-9}$ means incomplete d-subshell (1 to 9 electrons)
$d^{0}$ or $d^{10}$ means complete (empty or full) → not a transition element

Common Exam Exceptions

Element	Ground state	Common ion	Transition element?	Reason
Sc	$3d^1$	$\text{Sc}^{3+}\,(3d^0)$	No	Ion has empty d-subshell
Zn	$3d^{10}$	$\text{Zn}^{2+}\,(3d^{10})$	No	Both have full d-subshell
Cu	$3d^{10}$	$\text{Cu}^{2+}\,(3d^9)$	Yes	Ion has incomplete d-subshell
Cr	$3d^5$	$\text{Cr}^{3+}\,(3d^3)$	Yes	Both have incomplete d-subshell

Why This Matters for Exam Questions

When asked "Is X a transition element?":

First check the ground state electron configuration of the atom
Then check the most stable ion(s)
Apply the OR condition — if either has an incomplete d-subshell, it's a transition element

Remember: The definition exists to capture the chemical behaviour — elements with incomplete d-orbitals show the characteristic properties. Zinc behaves very differently from iron or copper precisely because its d-orbitals are always full.

The key idea is that variable oxidation states in transition metals arise from the availability of unpaired d-electrons for bonding. The 3d series metal that does not show variable oxidation states is Zinc (Zn), because its stable $3d^{10}$ configuration is fully filled, making it chemically inert to further oxidation changes.

Why This Question Matters

Transition metals are famous for their colourful compounds and catalytic properties — both rooted in their ability to exist in multiple oxidation states. But one metal in the 3d series breaks this pattern. Understanding why it breaks the pattern is the real lesson here.

The 3d series runs from Scandium ( $Z=21$ ) to Zinc ( $Z=30$ ). As you move across, electrons fill the 3d orbitals. The ability to show variable oxidation states depends on how easily these d-electrons can be lost or shared.

Step-by-Step Reasoning

1. What causes variable oxidation states?

In transition metals, the $(n-1)d$ and $ns$ orbitals are close in energy. This means electrons from both the 4s and 3d subshells can participate in bonding. The more unpaired d-electrons available, the more oxidation states are possible. For example, Manganese ( $3d^5 4s^2$ ) shows states from +2 to +7 because it can lose up to all seven valence electrons.

2. The critical exception: Zinc

Zinc has the electronic configuration $[Ar] 3d^{10} 4s^2$ . The 3d subshell is completely filled. A fully filled d-subshell is exceptionally stable — it has no unpaired electrons and resists losing or sharing those d-electrons. In chemical reactions, Zinc only loses its two 4s electrons, giving a stable $Zn^{2+}$ ion with a $3d^{10}$ configuration. It never shows +3, +4, or any other oxidation state.

Watch out

A common mistake is to think Scandium ( $3d^1 4s^2$ ) also lacks variable states. While Sc mostly shows +3, it can show +2 in some compounds (e.g., ScCl₂). So it does have variable oxidation states, just very limited ones. Zinc is the only one that shows only one oxidation state (+2) under normal conditions.

3. Check the other candidates

Copper ( $3d^{10} 4s^1$ ): Shows +1 and +2 (and rarely +3).
Nickel ( $3d^8 4s^2$ ): Shows +2, +3, and +4.
Iron ( $3d^6 4s^2$ ): Shows +2 and +3 (and +6 in ferrates).

All have partially filled d-orbitals, allowing multiple oxidation states.

4. The deeper reason

The stability of a half-filled ( $d^5$ ) or fully-filled ( $d^{10}$ ) subshell is a key concept in transition metal chemistry. For Zinc, the $3d^{10}$ configuration is so stable that removing even one d-electron requires too much energy — it's not observed in normal chemical reactions. This is why Zinc is placed at the end of the 3d series and behaves more like a main-group metal in its oxidation behaviour.

Tip

A quick memory aid: The only 3d metal with a completely filled d-subshell in its ground state is Zinc ( $3d^{10}$ ). All others have partially filled d-orbitals. If the d-subshell is full, variable oxidation states are impossible.

✓Final answer

The transition metal of the 3d series that does not show variable oxidation states is Zinc (Zn) .

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

Showing the 12 most recent of 37 on this concept.

CBSE 2026Set 56/1/11 markMCQ
$Q. Which of the following is not a transition metal ? (A) Sc (B) Ag (C) Hg (D) Cu$
›Reveal solutionSolution
Transition metals are defined by having partially filled d-orbitals in their common oxidation states. Mercury (Hg) has a full d¹⁰ configuration in both its elemental and common +2 state, so it is not a transition metal. The correct answer is (C).

Why This Definition Matters

The classification of transition metals isn't about being a metal or having d-electrons — it's about partially filled d-orbitals in the atom or in any common ion. This is the IUPAC definition. If an element's d-subshell is completely full (d¹⁰) in all its common forms, it doesn't qualify, even if it sits in the d-block of the periodic table.

Let's check each option against this rule.

Scandium (Sc, Z=21)

Electronic configuration: $[\text{Ar}]\,3d^1 4s^2$ .

Its common ion is Sc³⁺: $[\text{Ar}]$ — the 3d orbital is empty.

But the atom has a partially filled d-orbital (3d¹). The definition says "in the atom or in any common oxidation state." Since the atom itself has an incomplete d-subshell, Sc is a transition metal.

Silver (Ag, Z=47)

Configuration: $[\text{Kr}]\,4d^{10} 5s^1$ .

Common ion: Ag⁺ → $[\text{Kr}]\,4d^{10}$ — full d-subshell.

However, silver also forms Ag²⁺ (e.g., in AgF₂), which has $4d^9$ — a partially filled d-orbital. Because at least one common oxidation state (Ag²⁺) has an incomplete d-subshell, Ag qualifies as a transition metal.

Mercury (Hg, Z=80)

Configuration: $[\text{Xe}]\,4f^{14} 5d^{10} 6s^2$ .

Common ions: Hg²⁺ → $[\text{Xe}]\,4f^{14} 5d^{10}$ (full d¹⁰); Hg₂²⁺ (dimeric) also has each Hg with a full d¹⁰ core.

Mercury does not form any stable ion with a partially filled d-orbital. Its d-subshell is always full. Therefore, Hg is not a transition metal.

Copper (Cu, Z=29)

Configuration: $[\text{Ar}]\,3d^{10} 4s^1$ .

Common ions: Cu⁺ → $3d^{10}$ (full); Cu²⁺ → $3d^9$ (partially filled).

Since Cu²⁺ has an incomplete d-subshell, copper is a transition metal.

Watch out

A common mistake is to think that because Hg is a d-block element and a metal, it must be a transition metal. But the d-block includes two "exceptions": zinc, cadmium, and mercury — their d-orbitals are always full, so they are not transition metals. They are often called "post-transition metals" or simply d-block metals.

Tip

To quickly spot non-transition metals in the d-block, look for elements whose common ions have a d¹⁰ configuration. Zn²⁺ (3d¹⁰), Cd²⁺ (4d¹⁰), and Hg²⁺ (5d¹⁰) are the classic trio. If an element only forms d¹⁰ ions, it's out.

✓Final answer
$The element that is not a transition metal is (C) Hg .$
CBSE 2026Set 56/2/11 markMCQ
$Q. Assertion (A) : Zinc, cadmium and mercury are not considered as transition elements. Reason (R) : These elements have completely filled orbitals in their ground state as well as in their common oxidation states.$
›Reveal solutionSolution
The key idea is that transition elements are defined by having partially filled d-orbitals in their ground state or common oxidation states. Since Zn, Cd, and Hg have completely filled d¹⁰ configurations in both, they fail this criterion. Thus, both Assertion (A) and Reason (R) are true, and (R) correctly explains (A).

Why this approach works

The definition of a transition element is not arbitrary — it stems from the unique chemistry of d-block elements. The IUPAC defines a transition element as an element whose atom has a partially filled d-subshell, or which can give rise to cations with an incomplete d-subshell. This means we must check two things: the ground state electron configuration of the neutral atom, and the configurations of its common oxidation states. If either has a partially filled d-orbital, the element qualifies. If both are completely filled, it does not.

Zinc, cadmium, and mercury sit at the end of their respective d-block series (Group 12). Their neutral atoms have the configuration $(n-1)d^{10} ns^2$ — the d-subshell is full. Their common oxidation state is +2, formed by losing the two s-electrons, leaving $(n-1)d^{10}$ — still full. No partially filled d-orbital appears anywhere. Hence, they are not transition elements.

Let’s walk through the reasoning step by step.

Recall the defining criterion for transition elements.

A transition element must have an atom or a common ion with an incomplete d-subshell. This is the official IUPAC definition. Elements that have completely filled d-orbitals in both the ground state and all common oxidation states are excluded.

Examine the ground state configurations of Zn, Cd, and Hg.

Zinc ( $Z = 30$ ): $[Ar] 3d^{10} 4s^2$

Cadmium ( $Z = 48$ ): $[Kr] 4d^{10} 5s^2$

Mercury ( $Z = 80$ ): $[Xe] 4f^{14} 5d^{10} 6s^2$ In each case, the d-subshell is completely filled ( $d^{10}$ ). No partially filled d-orbital exists in the neutral atom.

Check their common oxidation states.

The most stable and common oxidation state for all three is +2. For example:

Zn loses its two 4s electrons to form $Zn^{2+}$ : $[Ar] 3d^{10}$

Cd loses its two 5s electrons to form $Cd^{2+}$ : $[Kr] 4d^{10}$

Hg loses its two 6s electrons to form $Hg^{2+}$ : $[Xe] 4f^{14} 5d^{10}$ In every case, the d-subshell remains completely filled. No partially filled d-orbital appears in any common oxidation state.

Watch out

A common mistake is to think that because these elements are in the d-block, they must be transition elements. But the d-block is a broader category based on the last electron entering a d-orbital. Transition elements are a subset of the d-block, defined by the incompleteness of the d-subshell. Group 12 elements are d-block but not transition elements.

Consider any other oxidation states.

Zinc and cadmium rarely show oxidation states other than +2. Mercury can show +1 (as in $Hg_2^{2+}$ ), but even there, the configuration is $5d^{10} 6s^1$ for each Hg atom — the d-subshell is still full. No common oxidation state of any of these three elements has a partially filled d-orbital.

Connect Assertion (A) and Reason (R).

Assertion (A) states that Zn, Cd, and Hg are not considered transition elements. This is true.

Reason (R) states that these elements have completely filled d-orbitals in their ground state as well as in their common oxidation states. This is also true.

Moreover, (R) is the correct explanation for (A): because the d-orbitals are always full, the defining condition for a transition element is never met.

Tip

A quick way to remember: For Group 12 elements, the d-subshell is always $d^{10}$ — think of them as "full-house" elements. Transition elements require a "vacancy" (at least one empty d-orbital) in the atom or a common ion. No vacancy here, so no transition element status.

✓Final answer
$Both Assertion (A) and Reason (R) are true, and (R) is the correct explanation of (A).$
CBSE 2026Set DZ1 markMCQ
$Q. Which of the following is not a transition metal?$
(a) Cu
(b) Cr
(c) Fe
(d) Na
›Reveal solutionSolution
A transition metal has a partially filled d-subshell (in the atom or a common ion). Cu, Cr and Fe qualify; Na does not — so option (d).

A transition element is defined as one whose atom or one of its stable ions has an incompletely filled d-orbital.

Cu ( $[Ar]3d^{10}4s^1$ ): forms $Cu^{2+}$ ( $3d^9$ ) — partially filled d → transition metal.

Cr ( $[Ar]3d^{5}4s^1$ ) — partially filled d → transition metal.

Fe ( $[Ar]3d^{6}4s^2$ ) — partially filled d → transition metal.

Na ( $[Ne]3s^1$ ) — an s-block element with no d-electrons → not a transition metal.

✓Final answer
$(d) Na .$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Transition element among the following is$
(a) Zinc
(b) Cadmium
(c) Cerium
(d) Rutherfordium
›Reveal solutionSolution
A transition element is defined as one having a partially (incompletely) filled d-subshell in the ground state or in any of its common oxidation states.

Zinc (3d10 4s2) and Cadmium (4d10 5s2) have a completely filled d-subshell in the atom AND in their only common +2 ion, so they are NOT classified as transition elements (they are 'post-transition'/group 12 metals).

Cerium is an f-block (lanthanide) element, not d-block.

Rutherfordium (Z = 104, [Rn]5f14 6d2 7s2) is a d-block transactinide with an incompletely filled d-subshell - it is classified as a genuine (4th-series) transition element.

✓Final answer
$(d) Rutherfordium.$
CBSE 2026Set ANNUAL1 mark
$Q. Why is Zn not considered as transition element? (Z = 30)$
›Reveal solutionSolution

Transition elements must have partly filled d-orbitals in the elemental or common ionic state; zinc's d-subshell is always completely full, so it does not qualify.

Zinc (Z = 30) has the electronic configuration [Ar] 3d10 4s2. When it forms its common oxidation state, Zn2+, it loses the two 4s electrons, giving [Ar] 3d10 - the d-subshell remains completely filled (all 10 electrons present) in both the metal and its ion.

By the standard definition, a transition element must have an incompletely (partially) filled d-subshell in at least one of its common oxidation states. Since Zn never shows a partially filled d-subshell (it is always d10 or, when all electrons are removed, d0), it does not meet this criterion and is therefore excluded from the transition elements, despite being placed in the d-block of the periodic table.

✓Final answer
$Zn (and Zn2+) has a completely filled 3d10 configuration, never a partially filled d-subshell, so it is not classified as a transition element.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Chromium belongs to which block?$
(a) s-block
(b) p-block
(c) d-block
(d) f-block
›Reveal solutionSolution

Chromium (Z=24) has the configuration [Ar] 3d⁵ 4s¹, placing it in the d-block.

Chromium has atomic number 24 and electronic configuration [Ar] 3d⁵ 4s¹ (an exception to the expected 3d⁴4s², arising from the extra stability of a half-filled d-subshell). Since its last-entering electron occupies a (n−1)d orbital, it belongs to the d-block (transition elements), specifically the 3d transition series.

✓Final answer
$(c) d-block.$
CBSE 2026Set ANNUAL1 mark
$Q.______ block elements are called transition elements.$
›Reveal solutionSolution

d-block elements are called transition elements.

The periodic table is divided into four blocks (s, p, d, f) based on which subshell receives the last (differentiating) electron. Elements in which the last electron enters a d-orbital form the d-block, spanning groups 3 to 12. These are called transition elements because most of them have partially filled d-orbitals in their elemental or common ionic states, giving them properties 'transitional' between the highly reactive s-block metals and the covalent p-block elements — e.g. variable oxidation states, coloured ions, catalytic activity, and complex-forming ability.

✓Final answer
$The d-block elements are called transition elements.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Transition metals have incomplete$
(a) s-orbital
(b) p-orbital
(c) d-orbital
(d) f-orbital
›Reveal solutionSolution

Transition metals have incomplete d-orbitals.

Transition elements are defined as those whose atoms or stable ions have partially filled (incomplete) d-orbitals. This partly filled d-subshell is responsible for their characteristic properties: variable oxidation states, coloured ions, catalytic behaviour and complex formation.

✓Final answer
$(C) d-orbital.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Total number of elements are in d-block is ______$
(a) 39
(b) 40
(c) 38
(d) 36
›Reveal solutionSolution
Four transition series of 10 elements each -> 40 d-block elements.

The d-block (transition elements) consists of elements in which the last electron enters a d-subshell. There are four such series:

3d series (Sc to Zn),

4d series (Y to Cd),

5d series (La/Hf to Hg),

6d series (Ac/Rf onward).

Each series contains 10 elements (d-subshell holds up to 10 electrons), so the total number of d-block elements = 4 x 10 = 40.

✓Final answer
$(b) 40.$
CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: Zn, Cd and Hg generally do not considered as a ______ elements.$
›Reveal solutionSolution

Zn, Cd and Hg are not true transition elements because their d-orbitals are completely filled.

A transition element is defined as one that has a partially filled d-subshell in its atomic state or in one of its common oxidation (ionic) states. Zinc, cadmium and mercury have the configuration (n-1)d10 ns2, and even in their common +2 ions they are (n-1)d10 - the d-subshell is completely filled. Since they never show a partly filled d-subshell, they are generally not considered transition elements (they are placed in the d-block only by position).

✓Final answer
$Transition elements.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Choose the correct order of atomic and ionic radii of chlorine atom and chloride (Cl-) ion (in pm) is$
(a) a) 136 and 90
(b) b) 167 and 99
(c) c) 99 and 181
(d) d) 186 and 90
›Reveal solutionSolution
[!TLDR]

c) 99 and 181

Why

The covalent/atomic radius of Cl is about 99 pm, while the Cl- ion is larger (extra electron, same nuclear charge) at about 181 pm.

[!ANSWER]

c) 99 and 181
CBSE 2025Set 56/6/11 markMCQ
$Q. Which of the following is the softest metal ? (A) Zn (B) Sc (C) Cu (D) Fe$
›Reveal solutionSolution
Softness in metals is determined by how easily their atoms slide past one another; among the transition elements listed, scandium (Sc) has the weakest metallic bonding due to having only one $d$ -electron available for bonding, making it the softest. The answer is (B).

Understanding Metallic Softness

Hardness in metals arises from the strength of metallic bonding—the electrostatic attraction between the "sea" of delocalized electrons and the positive metal ions. The more electrons available for delocalization (especially from $d$ -orbitals in transition metals), and the smaller the atomic radius, the stronger the bonding and the harder the metal.

Conversely, a soft metal has weaker metallic bonds. Its atoms can slide past one another more easily under stress, making it malleable and easy to deform.

Comparing the Four Metals

Let's examine each candidate by looking at their electronic configurations and the number of electrons contributing to metallic bonding:

Zinc (Zn): Electronic configuration $[\text{Ar}] \, 3d^{10} \, 4s^2$

All ten $d$ -electrons are paired in filled orbitals. While the two $4s$ electrons participate in bonding, the filled $d^{10}$ subshell contributes relatively little to directional bonding. Zinc is moderately soft but not the softest here.

Scandium (Sc): Electronic configuration $[\text{Ar}] \, 3d^1 \, 4s^2$

Only one $d$ -electron is available. This is the first transition metal in the series, with minimal $d$ -orbital participation in bonding. The metallic bond is weak because there are fewer electrons to delocalize and the $d$ -orbitals are only just beginning to fill. Scandium is notably soft and can be cut with a knife.

Copper (Cu): Electronic configuration $[\text{Ar}] \, 3d^{10} \, 4s^1$

Although it has a filled $d^{10}$ subshell, copper exhibits strong metallic bonding due to effective overlap of orbitals and high electron density. Copper is relatively hard and tough.

Iron (Fe): Electronic configuration $[\text{Ar}] \, 3d^6 \, 4s^2$

Six $d$ -electrons contribute to bonding, and iron forms a body-centered cubic structure at room temperature with strong metallic bonds. Iron is hard and strong—far from soft.

Tip

Across the first transition series (Sc → Zn), hardness generally increases from Sc to the middle elements (like Cr, Fe) as more $d$ -electrons participate in bonding, then decreases slightly toward Zn as the $d$ -subshell fills completely.

The Verdict

Scandium, with only one $d$ -electron and the weakest metallic bonding among the four, is the softest metal in this list. It is light, silvery, and can be easily deformed—characteristic of metals with minimal $d$ -electron participation.

✓Final answer
$The correct option is (B) Sc .$

Q.Which transition metal of 3d series does not show variable oxidation states?

Why This Question Matters

Step-by-Step Reasoning

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