Out of $Fe^{3+}$, $Sc^{3+}$, $Cr^{3+}$ and $Co^{3+}$ ions, the one which is colourless in aqueous solution is : (A) $Sc^{3+}$ (B) $Fe^{3+}$ (C) $Cr^{3+}$ (D) $Co^{3+}$ [Atomic number : Fe = 26, Sc = 21, Cr = 24, Co = 27]

> [!TLDR] > Colour in transition metal ions arises from d–d transitions, which require unpaired electrons in the d‑orbitals. $Sc^{3+}$ has a $3d^0$ configuration (no d‑electrons), so it cannot undergo d–d transitions and is colourless. The correct option is **(A)**. The question asks which of the given trivalent ions is colourless in aqueous solution. Colour in transition metal ions is almost always due to the absorption of visible light by electrons moving between d‑orbitals — the famous d–d transition. For this to happen, the ion must have at least one electron in its d‑orbitals (a partially filled d‑subshell). If the d‑subshell is completely empty ($d^0$) or completely filled ($d^{10}$), no d–d transition is possible, and the ion appears colourless (or white) in solution. Let’s check the electronic configuration of each ion. 1. **$Sc^{3+}$ (Atomic number 21)** Sc ground state: $[Ar]\, 3d^1 4s^2$. Removing three electrons (the two 4s electrons and the one 3d electron) gives $Sc^{3+}$: $[Ar]\, 3d^0$. No d‑electrons at all → no d–d transitions → **colourless**. 2. **$Fe^{3+}$ (Atomic number 26)** Fe ground state: $[Ar]\, 3d^6 4s^2$. Removing three electrons gives $Fe^{3+}$: $[Ar]\, 3d^5$. Five unpaired electrons (half‑filled d‑subshell) → d–d transitions possible → coloured (typically yellow‑brown in aqueous solution). 3. **$Cr^{3+}$ (Atomic number 24)** Cr ground state: $[Ar]\, 3d^5 4s^1$ (exception to the usual filling order). Removing three electrons gives $Cr^{3+}$: $[Ar]\, 3d^3$. Three d‑electrons → d–d transitions possible → coloured (violet or green depending on ligands). 4. **$Co^{3+}$ (Atomic number 27)** Co ground state: $[Ar]\, 3d^7 4s^2$. Removing three electrons gives $Co^{3+}$: $[Ar]\, 3d^6$. Six d‑electrons → d–d transitions possible → coloured (often yellow or brown in solution). > [!WARNING] > A common mistake is to think that all transition metal ions are coloured. That’s false — io

Out of Fe³⁺, Sc³⁺, Cr³⁺ and Co³⁺ ions, the one which is colourless in aqueous solution is : (A) Sc³⁺ (B) Fe³⁺

Question

Fe^{3+}

CBSECBSE Class XII Board 2025MCQ· 1mImportance★★★★★

✓ Free question

Concept understanding — Colour of Transition Metal Ions

Most students first meet colour in transition metals as a striking fact: copper sulphate is blue, potassium dichromate is orange, nickel salts are green. The question is why — after all, most elements form colourless compounds. The answer lives inside the d-orbitals.

The intuition: a window that absorbs some colours

Imagine a white light beam hitting a solution. If the substance absorbs nothing, all colours pass through and you see white (or colourless). If it absorbs only red light, the remaining mixture of colours looks blue-green — that’s the complementary colour. So a coloured compound is simply one that absorbs some part of the visible spectrum and transmits the rest.

For transition metal ions, the absorbing "antenna" is the set of five d-orbitals. In a free ion these orbitals all have the same energy. But when the ion sits inside a crystal or solution, surrounding ligands (water, ammonia, chloride, etc.) push on the d-orbitals unevenly. Some d-orbitals point directly at the ligands and feel strong repulsion; others point between them and feel less. This splits the d-orbital energies into two groups — a lower-energy set and a higher-energy set. The energy gap between them is called $\Delta$ (or $10Dq$ ), and it often falls right in the range of visible light.

The precise mechanism: d–d transition

An electron sitting in a lower d-orbital can absorb a photon whose energy exactly matches $\Delta$ . That photon disappears, and the electron jumps to a higher d-orbital. This is called a d–d transition. The colour you see is white light minus the absorbed wavelength.

E_{\text{photon}} = h\nu = \Delta = \frac{hc}{\lambda_{\text{absorbed}}}

The exact colour depends on three things:

The metal ion (more charge → larger $\Delta$ )
The ligand (stronger field → larger $\Delta$ )
The geometry (octahedral, tetrahedral, square planar — each gives a different splitting pattern)

For example, $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ absorbs red-orange light ( $\lambda \approx 600\text{ nm}$ ), so it looks blue. $[\text{Ti}(\text{H}_2\text{O})_6]^{3+}$ absorbs green-yellow and looks violet.

Why not all d-block ions are coloured

A d–d transition is only possible if there is an empty higher d-orbital to jump into. Ions with a full $d^{10}$ configuration (like $\text{Zn}^{2+}$ , $\text{Cu}^+$ , $\text{Ag}^+$ ) have no vacancy — all d-orbitals are filled, so no d–d transition can happen. They are colourless (unless other processes like charge transfer occur, which is a separate topic).

Similarly, $\text{Sc}^{3+}$ has $d^0$ — no d-electrons at all — so there is nothing to excite. Colourless.

Watch out

A common mistake is to think the colour comes from the metal alone. It does not. The same metal ion with different ligands gives different colours. $\text{Ni}^{2+}$ with water is green; with ammonia it turns violet-blue. The ligand changes $\Delta$ , which changes the absorbed colour.

The bigger picture

This explanation — splitting of d-orbitals by ligands, followed by absorption of visible light to promote an electron — is the foundation of crystal field theory. It explains not just colour but also magnetic properties, stability, and even why some complexes are paramagnetic while others are not. For the exam, remember: colour arises from d–d transitions, which require partially filled d-orbitals ( $d^1$ to $d^9$ ) and an appropriate splitting energy in the visible range.

The colour of a transition metal ion is the complementary colour of the light absorbed when an electron jumps between split d-orbitals.

Colour of Transition Metal Ions is a periodic-trend topic that regularly turns up in NCERT/CBSE Class 12 Chemistry as well as in JEE Main and NEET, since questions on colour of Transition Metal Ions periodic trends are a recurring favourite in objective-type tests. Students searching for "Colour of Transition Metal Ions: Definition, Formula & Real-World Examples" or "Colour of Transition Metal Ions class 12 chemistry" will find this explanation aligned with the d- and f-Block Elements unit of the NCERT/CBSE syllabus.

Colour in transition metal ions arises from d–d transitions, which require unpaired electrons in the d‑orbitals. $Sc^{3+}$ has a $3d^0$ configuration (no d‑electrons), so it cannot undergo d–d transitions and is colourless. The correct option is (A).

The question asks which of the given trivalent ions is colourless in aqueous solution. Colour in transition metal ions is almost always due to the absorption of visible light by electrons moving between d‑orbitals — the famous d–d transition. For this to happen, the ion must have at least one electron in its d‑orbitals (a partially filled d‑subshell). If the d‑subshell is completely empty ( $d^0$ ) or completely filled ( $d^{10}$ ), no d–d transition is possible, and the ion appears colourless (or white) in solution.

Let’s check the electronic configuration of each ion.

$Sc^{3+}$ (Atomic number 21)

Sc ground state: $[Ar]\, 3d^1 4s^2$ .

Removing three electrons (the two 4s electrons and the one 3d electron) gives $Sc^{3+}$ : $[Ar]\, 3d^0$ .

No d‑electrons at all → no d–d transitions → colourless.
$Fe^{3+}$ (Atomic number 26)

Fe ground state: $[Ar]\, 3d^6 4s^2$ .

Removing three electrons gives $Fe^{3+}$ : $[Ar]\, 3d^5$ .

Five unpaired electrons (half‑filled d‑subshell) → d–d transitions possible → coloured (typically yellow‑brown in aqueous solution).
$Cr^{3+}$ (Atomic number 24)

Cr ground state: $[Ar]\, 3d^5 4s^1$ (exception to the usual filling order).

Removing three electrons gives $Cr^{3+}$ : $[Ar]\, 3d^3$ .

Three d‑electrons → d–d transitions possible → coloured (violet or green depending on ligands).
$Co^{3+}$ (Atomic number 27)

Co ground state: $[Ar]\, 3d^7 4s^2$ .

Removing three electrons gives $Co^{3+}$ : $[Ar]\, 3d^6$ .

Six d‑electrons → d–d transitions possible → coloured (often yellow or brown in solution).

Watch out

A common mistake is to think that all transition metal ions are coloured. That’s false — ions with $d^0$ or $d^{10}$ configurations (like $Sc^{3+}$ , $Ti^{4+}$ , $Zn^{2+}$ , $Cu^+$ ) are colourless. Colour requires a partially filled d‑subshell.

Tip

For quick recall: $Sc^{3+}$ is the only trivalent ion among the first‑row transition metals that is colourless. Its $d^0$ configuration is the key.

✓Final answer

Sc^{3+}

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 15 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following ion is colourless in aqueous solution-$
(a)
(i) Fe²⁺
(b)
(ii) Mn²⁺
(c)
(iii) Zn²⁺
(d)
(iv) Cu²⁺
›Reveal solutionSolution
$\text{Zn}^{2+}$ has a fully filled $3d^{10}$ configuration, so it cannot undergo d–d transitions and is colourless. Correct option: (iii).

Concept. A transition-metal ion is coloured when it can absorb visible light by promoting an electron between the split $t_{2g}$ and $e_g$ d-orbitals (a d–d transition). This is only possible if the d-subshell is partially filled ( $d^1$ to $d^9$ ).

Steps (configurations of the ions).

$\text{Fe}^{2+}$ : $3d^{6}$ — partly filled $\Rightarrow$ coloured (pale green).

$\text{Mn}^{2+}$ : $3d^{5}$ — partly filled $\Rightarrow$ coloured (pale pink).

$\text{Zn}^{2+}$ : $3d^{10}$ — completely filled $\Rightarrow$ no d–d transition $\Rightarrow$ colourless.

$\text{Cu}^{2+}$ : $3d^{9}$ — partly filled $\Rightarrow$ coloured (blue).

✓Final answer
$3d^{10}$
CBSE 2025Set 56/4/11 markMCQ
$Fe^{3+}$
›Reveal solutionSolution
Colour in transition metal ions arises from d–d transitions, which require unpaired electrons in the d‑orbitals. $Sc^{3+}$ has a $3d^0$ configuration (no d‑electrons), so it cannot undergo d–d transitions and is colourless. The correct option is (A).

The question asks which of the given trivalent ions is colourless in aqueous solution. Colour in transition metal ions is almost always due to the absorption of visible light by electrons moving between d‑orbitals — the famous d–d transition. For this to happen, the ion must have at least one electron in its d‑orbitals (a partially filled d‑subshell). If the d‑subshell is completely empty ( $d^0$ ) or completely filled ( $d^{10}$ ), no d–d transition is possible, and the ion appears colourless (or white) in solution.

Let’s check the electronic configuration of each ion.

$Sc^{3+}$ (Atomic number 21)

Sc ground state: $[Ar]\, 3d^1 4s^2$ .

Removing three electrons (the two 4s electrons and the one 3d electron) gives $Sc^{3+}$ : $[Ar]\, 3d^0$ .

No d‑electrons at all → no d–d transitions → colourless.

$Fe^{3+}$ (Atomic number 26)

Fe ground state: $[Ar]\, 3d^6 4s^2$ .

Removing three electrons gives $Fe^{3+}$ : $[Ar]\, 3d^5$ .

Five unpaired electrons (half‑filled d‑subshell) → d–d transitions possible → coloured (typically yellow‑brown in aqueous solution).

$Cr^{3+}$ (Atomic number 24)

Cr ground state: $[Ar]\, 3d^5 4s^1$ (exception to the usual filling order).

Removing three electrons gives $Cr^{3+}$ : $[Ar]\, 3d^3$ .

Three d‑electrons → d–d transitions possible → coloured (violet or green depending on ligands).

$Co^{3+}$ (Atomic number 27)

Co ground state: $[Ar]\, 3d^7 4s^2$ .

Removing three electrons gives $Co^{3+}$ : $[Ar]\, 3d^6$ .

Six d‑electrons → d–d transitions possible → coloured (often yellow or brown in solution).

Watch out

A common mistake is to think that all transition metal ions are coloured. That’s false — ions with $d^0$ or $d^{10}$ configurations (like $Sc^{3+}$ , $Ti^{4+}$ , $Zn^{2+}$ , $Cu^+$ ) are colourless. Colour requires a partially filled d‑subshell.

Tip

For quick recall: $Sc^{3+}$ is the only trivalent ion among the first‑row transition metals that is colourless. Its $d^0$ configuration is the key.

✓Final answer
$Sc^{3+}$
CBSE 2025Set D1 markMCQ
$Q. Which of the following compounds can be coloured?$
(a) Ag2SO4
(b) CuF2
(c) Cu2Cl2
(d) MgF2
›Reveal solutionSolution
A compound is coloured only if the metal ion has a partially filled d subshell; CuF2 (Cu2+ = d9) qualifies.

Colour in transition-metal compounds arises from d-d electronic transitions, which require a partially filled d subshell. Examining each cation:

Ag2SO4: Ag+ is [Kr] 4d10 -> filled d -> colourless.

CuF2: Cu2+ is [Ar] 3d9 -> partially filled d (1 unpaired) -> coloured. (Correct.)

Cu2Cl2: Cu+ is [Ar] 3d10 -> filled d -> colourless.

MgF2: Mg2+ has no d electrons -> colourless.

Hence only CuF2 can be coloured.

✓Final answer
$(B) CuF2.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following transition elements does not give coloured salt?$
(a) Cr
(b) Mn
(c) Cu
(d) Zn
›Reveal solutionSolution
Colour in transition-metal ions arises from d-d electronic transitions, which require partially filled d orbitals; Zn2+ has a full d10 shell with no d-d transition possible, so its salts are colourless.

Transition metal ions are typically coloured because of d-d transitions: an electron in a lower-energy d orbital (in the crystal field of surrounding ligands/anions) absorbs visible light and jumps to a higher-energy d orbital; the colour seen is complementary to the wavelength absorbed. This requires the d subshell to be partially filled (neither completely empty nor completely full), so a d-d transition is possible.

Cr (commonly Cr3+, d3) and Mn (commonly Mn2+, d5) have partially filled d orbitals → coloured salts.

Cu (commonly Cu2+, d9) also partially filled → coloured (blue) salts, e.g., CuSO4·5H2O.

Zn (as Zn2+) has configuration [Ar]3d10 — completely filled d subshell, so no d-d transition is possible, and its salts (e.g., ZnSO4) are colourless/white.

✓Final answer
$(d) Zn does not give a coloured salt.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Colourless metal ion in aqueous solution is -$
(a) Cu2+
(b) Zn2+
(c) Mn2+
(d) V2+
›Reveal solutionSolution
Colour in transition-metal ions arises from d-d transitions of unpaired electrons; Zn2+ (3d10, fully filled) has none, so it is colourless.

Electronic configurations:

Cu2+: [Ar]3d9 - one unpaired electron -> coloured (blue)

Zn2+: [Ar]3d10 - all d orbitals completely filled, no unpaired electron and no vacant d orbital to promote an electron into -> colourless

Mn2+: [Ar]3d5 - unpaired electrons -> coloured (pale pink)

V2+: [Ar]3d3 - unpaired electrons -> coloured (violet)

Since d-d transitions require both a partially filled d subshell (an electron to promote) and a vacancy within it (a place to promote it to), a d0 or d10 ion cannot show d-d transitions, hence is colourless. Zn2+ is d10.

✓Final answer
$(b) Zn2+.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. Direction: In this part of Question No.1, there are two statements labelled as Assertion (A) and Reason (R). From the following options select the correct answer.$
(i) Both A and R are correct and R is the correct explanation of A.
(ii) Both A and R are correct but R is not the correct explanation of A.
(iii) A is correct but R is incorrect.
(iv) Both A and R are incorrect. Assertion (A): The elements of d and f blocks produce coloured ions. Reason (R): Unpaired electrons are present in these elements.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is incorrect.
(d) Both A and R are incorrect.
›Reveal solutionSolution

Unpaired d/f electrons undergo d-d (or f-f) transitions, which is exactly why these elements form coloured ions.

d- and f-block elements have partially filled d or f orbitals in their ions, i.e. unpaired electrons. When such an ion absorbs visible light, an electron is excited from a lower energy d (or f) orbital to a higher energy one within the same subshell (a d-d or f-f transition); the wavelength of visible light absorbed corresponds to the complementary colour observed. Elements/ions with no unpaired d or f electrons (e.g. $Zn^{2+}$ , $Sc^{3+}$ ) are colourless. So both statements A and R are correct, and R is indeed the correct explanation of A.

✓Final answer
$(i) Both A and R are correct and R is the correct explanation of A.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. Which of the following statement about transition element is not correct?$
(a) They show variable oxidation states
(b) They exhibit diamagnetic and paramagnetic properties
(c) All ions are coloured
(d) They exhibit catalytic property
›Reveal solutionSolution

Colour requires a partially filled d-subshell for a d-d transition; ions with an empty (d0) or full (d10) d-subshell are colourless.

Transition elements do show variable oxidation states (a) and both paramagnetic and diamagnetic behaviour depending on the number of unpaired d-electrons (b), and they are well known as catalysts (d) — all true statements. But 'all ions are coloured' is false: colour arises only from a d–d electronic transition, which requires a partially-filled d-subshell. Ions with an empty d-subshell (d⁰, e.g. Sc³⁺, Ti⁴⁺) or a completely filled one (d¹⁰, e.g. Zn²⁺, Cu⁺) have no d–d transition possible and are colourless.

✓Final answer
$(c) All ions are coloured — this statement is NOT correct.$
CBSE 2023Set A1 mark
$Q. Fill in the blank: The colour of Ni^2+ is ______.$
›Reveal solutionSolution

Transition metal ions are typically coloured because of d-d electronic transitions; the hydrated Ni²⁺ ion is green.

Ni²⁺ has the configuration [Ar]3d⁸ — it has partially filled d-orbitals, so it can absorb visible light to promote an electron from a lower-energy d-orbital to a higher-energy one (d-d transition). The colour observed is complementary to the colour of light absorbed.

Aqueous Ni²⁺ salts (e.g. NiSO4, NiCl2 solutions, containing the [Ni(H2O)6]²⁺ ion) characteristically appear green.

✓Final answer
$Green.$
CBSE 2023Set ANNUAL1 markMCQ
$Q. Colours of transition metal ions are due to absorption of light of some wavelength. This results in$
(a) d-d transition
(b) p-p transition
(c) s-s transition
(d) d-f transition
›Reveal solutionSolution

Ligand-field splitting of the d-orbitals creates a small energy gap that visible light can bridge, so absorption/transmission of specific wavelengths gives transition-metal compounds their colour.

In the presence of ligands, the five degenerate d-orbitals of a transition metal ion split into two (or more) sets of different energy (crystal/ligand field splitting, e.g. t2g and eg in an octahedral field). If the ion has partially filled d-orbitals, an electron can absorb a photon of visible light and jump from the lower-energy set to the higher-energy set - a d-d transition. The complementary colour of the absorbed wavelength is what we observe. Ions with completely empty (d0) or completely filled (d10) d-orbitals (e.g. Sc3+, Zn2+) cannot undergo d-d transitions and are typically colourless/white.

✓Final answer
$(a) d-d transition.$
CBSE 2022Set E1 markMCQ
$Q. Which of the following ions is colourless ?$
(a) Cu+
(b) Co2+
(c) Ni2+
(d) Fe3+
›Reveal solutionSolution
Cu+ is 3d10 (no d-d transitions possible) → colourless; Co2+, Ni2+ and Fe3+ have partly filled d subshells → coloured.

Colour in transition-metal ions arises from d-d electronic transitions, which need a partly filled d subshell.

Cu+: Cu is +1 → 3d10 (completely filled) → no d-d transition → colourless.

Co2+: 3d7 → coloured.

Ni2+: 3d8 → coloured.

Fe3+: 3d5 → coloured.

Only Cu+ has a fully filled d subshell, so it is colourless.

✓Final answer
$(a) Cu+.$
CBSE 2019Set HE1 mark
$Q. Answer in one word: The salts of transition metals are generally coloured. Why?$
›Reveal solutionSolution

Coloured salts of transition metals arise because their partially filled d-orbitals allow d–d electronic transitions on absorption of visible light; the observed colour is complementary to the wavelength absorbed.

In a free transition-metal ion, all 5 d-orbitals are degenerate (equal energy). But in a compound/complex, the surrounding ligands (or ions/water molecules) split the d-orbitals into two sets of slightly different energies (crystal field splitting). If the ion has partially filled d-orbitals (i.e. neither completely empty nor completely full), an electron can be promoted from the lower-energy set of d-orbitals to the higher-energy set by absorbing a photon of visible light of just the right energy — this is called a d–d transition.

The wavelengths NOT absorbed (transmitted/reflected) reach our eye and are perceived as the salt's colour — the colour we observe is complementary to the colour actually absorbed.

Ions with completely empty (d⁰, e.g. Sc³⁺) or completely filled (d¹⁰, e.g. Zn²⁺) d-orbitals cannot undergo a d–d transition and their salts are typically colourless — confirming that partially-filled d-orbitals are the reason.

✓Final answer
$Because their partially-filled d-orbitals allow d-d electronic transitions on absorbing visible light (crystal-field splitting of the d-orbitals by ligands makes this transition possible).$
CBSE 2019Set ANNUAL1 mark
$Sc^{3+}$
›Reveal solutionSolution
Colour in transition-metal ions arises from d–d electronic transitions, which require partly-filled d-orbitals; ions with a $d^0$ configuration have no such transition available and are colourless.

Work out each ion's d-electron count:

$Sc^{3+}$ (Z = 21): Sc is $[Ar]3d^1 4s^2$ ; removing 3 electrons ( $4s^2$ , $3d^1$ ) gives $Sc^{3+} = [Ar]3d^0$ — no d electrons.

$Ti^{4+}$ (Z = 22): Ti is $[Ar]3d^2 4s^2$ ; removing all 4 gives $Ti^{4+} = [Ar]3d^0$ — no d electrons.

$Fe^{3+}$ (Z = 26): Fe is $[Ar]3d^6 4s^2$ ; removing 3 electrons ( $4s^2$ , one $3d$ ) gives $Fe^{3+} = [Ar]3d^5$ — 5 unpaired d electrons.

$V^{3+}$ (Z = 23): V is $[Ar]3d^3 4s^2$ ; removing 3 electrons ( $4s^2$ , one $3d$ ) gives $V^{3+} = [Ar]3d^2$ — 2 unpaired d electrons.

Colour in transition-metal ions arises from d–d transitions (an electron absorbing visible light and jumping between split d-orbitals in a ligand field), which is only possible when the d-subshell is partially filled ( $d^1$ to $d^9$ ). Ions with $d^0$ (or $d^{10}$ ) configurations have no such transition available, so they are colourless.

Since $Sc^{3+}$ and $Ti^{4+}$ are $3d^0$ , they are colourless; $Fe^{3+}$ ( $3d^5$ ) and $V^{3+}$ ( $3d^2$ ) have partly-filled d-orbitals and are coloured.

✓Final answer
$Fe^{3+}$

Q.Out of Fe3+Fe^{3+}Fe3+, Sc3+Sc^{3+}Sc3+, Cr3+Cr^{3+}Cr3+ and Co3+Co^{3+}Co3+ ions, the one which is colourless in aqueous solution is : (A) Sc3+Sc^{3+}Sc3+ (B) Fe3+Fe^{3+}Fe3+ (C) Cr3+Cr^{3+}Cr3+ (D) Co3+Co^{3+}Co3+ [Atomic number : Fe = 26, Sc = 21, Cr = 24, Co = 27]

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Q.Out of $Fe^{3+}$ , $Sc^{3+}$ , $Cr^{3+}$ and $Co^{3+}$ ions, the one which is colourless in aqueous solution is : (A) $Sc^{3+}$ (B) $Fe^{3+}$ (C) $Cr^{3+}$ (D) $Co^{3+}$ [Atomic number : Fe = 26, Sc = 21, Cr = 24, Co = 27]

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