d-Block Elements

ChemistryThe d- And f-Block ElementsFor JEE aspirants

d-block elements are the elements of groups 3 to 12 in which the last electron enters the penultimate $(n - 1) d$ subshell. Called transition elements because they sit between the s- and p-blocks, they are hard metals with high melting points, variable oxidation states, coloured paramagnetic ions, a strong tendency to form complexes and useful catalytic power. This page explains every trend of the d-block elements with graphs, then covers $K_{2} C r_{2} O_{7}$ , $KMn O_{4}$ and key Cu, Ag and Hg compounds. It is a high-scoring chapter in JEE Main and NEET.

On this page1Position2Configuration3Physical trends4Oxidation states5Colour and magnetism6Complexes, catalysts, alloys7Important compounds8Solved examples

Key Formulas - Quick Reference

★ Must learnConfiguration: $(n - 1) d^{1 - 10} n s^{0 - 2}$ ; exceptions Cr $3 d^{5} 4 s^{1}$ and Cu $3 d^{10} 4 s^{1}$ .
Ions lose $n s$ electrons before $(n - 1) d$ : Fe $[Ar] 3 d^{6} 4 s^{2} \to F e^{2 +}$ $3 d^{6}$ , $F e^{3 +}$ $3 d^{5}$ .
★ Must learnSpin-only magnetic moment: $μ = n (n + 2)$ BM ( $n$ = unpaired electrons); 1.73, 2.83, 3.87, 4.90, 5.92 BM for $n$ = 1 to 5.
★ Must learnHighest oxidation state rises to Mn (+7) then falls; only $C u^{2 +} /Cu$ has positive $E^{\circ}$ (+0.34 V) in the 3d series.
Colour needs $d^{1}$ to $d^{9}$ (a d-d transition); $d^{0}$ and $d^{10}$ ions are colourless.
★ Must learnElectrons gained by $Mn O_{4}^{-}$ : acid 5 ( $\to M n^{2 +}$ ), neutral 3 ( $\to Mn O_{2}$ ), strong alkali 1 ( $\to Mn O_{4}^{2 -}$ ); $C r_{2} O_{7}^{2 -}$ in acid 6.
★ Must learn $C r_{2} O_{7}^{2 -} + H_{2} O ⇌ 2Cr O_{4}^{2 -} + 2 H^{+}$ : orange in acid, yellow in alkali.
Equivalent weight = $M / n$ : $KMn O_{4}$ 31.6 (acid), $K_{2} C r_{2} O_{7}$ 49 (acid).
Lanthanoid contraction makes 4d and 5d radii nearly equal: Zr 160 pm, Hf 159 pm.

1. What Are d-Block Elements?

The d-block elements are formed by filling the 3d, 4d, 5d and 6d subshells. They occupy groups 3 to 12, in the middle of the periodic table. In s- and p-block elements the new electron enters the outermost shell; in the d-block it enters the penultimate shell, $(n - 1) d$ .

They are called transition elements because their position and properties lie between the very reactive, ionic-bond-forming metals of the s-block and the covalent-bond-forming elements of the p-block. There are three complete rows of ten elements (3d: Sc to Zn, 4d: Y to Cd, 5d: La and Hf to Hg) and the 6d row (Ac and Rf to Cn), which is discussed with the f-block.

Figure 1: The d-block fills groups 3-12 between the s- and p-blocks. Its four rows are the 3d, 4d, 5d and 6d series; electrons enter the penultimate

(n - 1) d

shell, which is why these elements are called transition elements.

Definition (IUPAC, NCERT): a transition element is one that has an incompletely filled d subshell either in its ground state atom or in any of its common oxidation states.

By this definition Cu, Ag and Au are transition metals ( $C u^{2 +}$ is $3 d^{9}$ ) even though the atoms have $d^{10}$ . Zn, Cd and Hg are not, because both the atom and the common +2 ion are $d^{10}$ . These three still sit in the d-block and are studied with it.

Transition elements

Partly filled $d$ in the atom or a common ion.
Examples: Sc to Cu, Ag ( $A g^{2 +}$ $4 d^{9}$ ), Au ( $A u^{3 +}$ $5 d^{8}$ ).
Coloured, paramagnetic ions; variable valency.

d-block but not transition

$d^{10}$ in the atom and in the +2 ion.
Examples: Zn, Cd, Hg (the zinc group).
White compounds, fixed +2 state, low melting points.

2. Electronic Configuration

The general configuration is $(n - 1) d^{1 - 10} n s^{0 - 2}$ , where $n$ is the outermost shell. Across a series the outer $n s$ subshell keeps (usually) two electrons, while the penultimate shell grows from 8 to 18 electrons. Palladium is the one element with $n s^{0}$ ( $4 d^{10} 5 s^{0}$ ).

Element	Sc	Ti	V	Cr	Mn	Fe	Co	Ni	Cu	Zn
Z	21	22	23	24	25	26	27	28	29	30
3d	1	2	3	5	5	6	7	8	10	10
4s	2	2	2	1	2	2	2	2	1	2

Figure 2: Box diagrams for Sc to Zn. The 4s pair stays full except in Cr (

3 d^{5} 4 s^{1}

) and Cu (

3 d^{10} 4 s^{1}

), where a half-filled or completely filled

3 d

set gives extra stability.

2.1 Why Cr and Cu Are Anomalous

Chromium is $3 d^{5} 4 s^{1}$ , not $3 d^{4} 4 s^{2}$ , and copper is $3 d^{10} 4 s^{1}$ , not $3 d^{9} 4 s^{2}$ . The $3 d$ and $4 s$ energies are very close, so one electron shifts to give a half-filled ( $d^{5}$ ) or completely filled ( $d^{10}$ ) set. Such sets are extra stable because of their symmetry and their large exchange energy (more pairs of parallel spins).

Exam Trick

Cr and Cu cheat by one. Only these two in the 3d row steal one 4s electron: Cr to reach $d^{5}$ , Cu to reach $d^{10}$ . For ions, 4s fills first and also empties first: $F e^{2 +}$ is $3 d^{6}$ , not $3 d^{4} 4 s^{2}$ .

Key idea

Transition elements differ from main-group elements in one place: the penultimate

(n - 1) d

subshell is being filled.

Quick Recall: tap to check

Configuration of

M n^{2 +}

Mn is

[Ar] 3 d^{5} 4 s^{2}

; remove both 4s electrons:

3 d^{5}

Is Zn a transition element?

No. Zn and

Z n^{2 +}

are both

d^{10}

Which 4d element has an empty 5s subshell?

Palladium,

4 d^{10} 5 s^{0}

3. Physical Properties and Trends

3.1 Metallic Nature, Conductivity and Density

All d-block elements are metals with a typical metallic lustre, high tensile strength, ductility and malleability. Except mercury (a liquid at room temperature) they are solids. They are good conductors of heat and electricity; silver is the best conductor of electricity. Their atoms are small and bonded by strong metallic bonds that use both $n s$ and unpaired $(n - 1) d$ electrons, so their density and hardness are high (osmium and iridium are the densest elements).

3.2 Melting Points and Enthalpy of Atomisation

Melting points and enthalpies of atomisation are high and peak in the middle of each series. The more unpaired d electrons an atom has, the more electrons take part in metallic bonding. Mn and Zn break the pattern: Mn ( $3 d^{5}$ ) holds its half-filled set tightly, and Zn ( $3 d^{10}$ ) has no unpaired d electrons, so it melts at only 693 K. The 4d and 5d metals have higher enthalpies of atomisation than the 3d metals, so heavy transition metals (W, m.p. 3683 K) often form metal-metal bonds.

Figure 3: (A) Melting points peak near V and Cr (about 2180 K) and fall to 693 K for Zn. (B) Ionisation enthalpies rise only slowly, but

IE_{2}

jumps at Cr and Cu and

IE_{3}

at Mn, where a stable

d^{5}

d^{10}

set must be broken.

3.3 Atomic and Ionic Size

Covalent and metallic radii decrease from left to right across a series. Each new electron enters the inner $(n - 1) d$ subshell, where it shields the outer electrons poorly, so the rising nuclear charge pulls the whole atom in. The decrease slows down in the middle and radii rise slightly at the end (Cu, Zn), because the filled $d$ set adds repulsion.

Down a group, size increases from the 3d to the 4d element (an extra shell: Sc 164 pm, Y 180 pm, La 187 pm). From the 4d to the 5d element, however, there is almost no increase: Zr 160 pm and Hf 159 pm. Before Hf come the 14 lanthanoids, whose $4 f$ electrons shield very poorly. The resulting lanthanoid contraction (studied with the f-block) cancels the effect of the extra shell.

Figure 4: Radii shrink across each series, then level off. The 5d curve sits on top of the 4d curve (Zr 160 pm, Hf 159 pm) because the 14 lanthanoids placed before Hf add poorly shielding

4 f

electrons: lanthanoid contraction.

Exam Trick

Twins of the 4d and 5d rows: Zr-Hf, Nb-Ta and Mo-W have almost the same radius and very similar chemistry, so they occur together in ores and are hard to separate.

3.4 Ionisation Enthalpy

Ionisation enthalpies of transition elements lie between those of the s-block and the p-block. Along a series they rise only gradually: the nuclear charge increases by one each time, but the new $(n - 1) d$ electron adds shielding, so the effective pull grows slowly. The second and third ionisation enthalpies show clear breaks where a stable configuration is disturbed: $I E_{2}$ is high for Cr and Cu (removing an electron from $C r^{+}$ $3 d^{5}$ or $C u^{+}$ $3 d^{10}$ ), and $I E_{3}$ is high for Mn ( $M n^{2 +}$ is $3 d^{5}$ ) but low for Fe ( $F e^{2 +} \to F e^{3 +}$ gives $3 d^{5}$ ).

The 5d elements have higher ionisation enthalpies than the 3d and 4d elements, because after the lanthanoid contraction a much larger nuclear charge acts on electrons at almost the same distance.

Key idea

Small atoms, strong metallic bonds and poor

d

shielding explain nearly every physical trend; lanthanoid contraction explains why 4d and 5d twins look alike.

4. Oxidation States and Electrode Potentials

4.1 Variable Oxidation States

Transition elements show several oxidation states, and the states of one element usually differ by one unit ( $F e^{2 +}$ and $F e^{3 +}$ , $C u^{+}$ and $C u^{2 +}$ ). In p-block elements states usually differ by two (Sn +2/+4, Pb +2/+4). The reason is that the $(n - 1) d$ and $n s$ energies are very close, so a variable number of electrons from both subshells can be used in bonding. Scandium, for example, can in principle be +2 (both 4s electrons used) or +3 (two 4s and one 3d), and +3 is its only common state.

Figure 5: The highest oxidation state climbs with the number of

3 d + 4 s

electrons up to Mn (+7), then falls as the paired

d

electrons hold on. States differ by one unit (

F e^{2 +} /F e^{3 +}

C u^{+} /C u^{2 +}

), unlike p-block elements.

The highest oxidation state equals the total number of $3 d + 4 s$ electrons up to manganese ( $Mn$ +7 in $Mn O_{4}^{-}$ ). After Mn, the d electrons start pairing, are held more firmly and fewer are available, so the maximum falls (Fe +6, Co +4, Ni +4, Cu +2, Zn +2).

4.2 Stability of High and Low Oxidation States

High oxidation states are stabilised by the two most electronegative, small elements, fluorine and oxygen: $V F_{5}$ , $Cr F_{6}$ , $V_{2} O_{5}$ , $Cr O_{3}$ and $M n_{2} O_{7}$ . Oxygen goes further than fluorine because it can form multiple bonds to the metal (Mn is +7 in $M n_{2} O_{7}$ but only +4 in $Mn F_{4}$ ). The ability of $F^{-}$ to stabilise high states comes from its high lattice or bond enthalpy; $I^{-}$ does the opposite and reduces $C u^{2 +}$ , so $Cu I_{2}$ does not exist.

Low oxidation states (0 or +1) appear with ligands that can accept electron density back from the metal, such as CO: $Ni (CO)_{4}$ and $Fe (CO)_{5}$ contain the metal in the zero state.

4.3 Standard Electrode Potentials

The value of $E^{\circ} (M^{2 +} /M)$ reflects three energy steps: atomisation of the metal, ionisation to $M^{2 +}$ and hydration of the ion. Across the 3d series it becomes less negative, because the sum $I E_{1} + I E_{2}$ rises. Only copper has a positive value, so copper does not liberate hydrogen from dilute non-oxidising acids; its high enthalpy of atomisation plus its high $I E_{2}$ is not repaid by its hydration enthalpy.

Figure 6:

E^{\circ} (M^{2 +} /M)

becomes less negative across the series; only Cu is positive (+0.34 V), so Cu cannot liberate

H_{2}

from dilute acids. Mn, Ni and Zn stand out: stable

d^{5}

, very negative hydration enthalpy of

N i^{2 +}

, and stable

d^{10}

Couple	Ti	V	Cr	Mn	Fe	Co	Ni	Cu	Zn
$E^{\circ} (M^{2 +} /M)$ / V	-1.63	-1.18	-0.90	-1.18	-0.44	-0.28	-0.25	+0.34	-0.76
$E^{\circ} (M^{3 +} / M^{2 +})$ / V	-0.37	-0.26	-0.41	+1.57	+0.77	+1.97	-	-	-

The $M^{3 +} / M^{2 +}$ values explain which ions are oxidising or reducing. $E^{\circ} (M n^{3 +} /M n^{2 +}) = + 1.57$ V is high because $M n^{2 +}$ ( $d^{5}$ ) is especially stable, so $M n^{3 +}$ is a strong oxidant. $E^{\circ} (F e^{3 +} /F e^{2 +})$ is only +0.77 V because $F e^{3 +}$ is itself $d^{5}$ . $C r^{2 +}$ is a strong reducing agent, since it changes to $C r^{3 +}$ ( $t_{2 g}^{3}$ , a half-filled lower set). Zinc is stable only as +2 (no second state) because $Z n^{2 +}$ is $d^{10}$ .

JEE Advanced

Why $C u^{+}$ disproportionates in water. Using $E^{\circ} (C u^{+} /Cu) = + 0.52$ V and $E^{\circ} (C u^{2 +} /C u^{+}) = + 0.16$ V, the reaction $2 C u^{+} \to C u^{2 +} + Cu$ has $E_{cell}^{\circ} = 0.52 - 0.16 = + 0.36$ V, so it is spontaneous. The driving force is the much more negative hydration enthalpy of the small, doubly charged $C u^{2 +}$ ion, which outweighs the second ionisation enthalpy of copper. Solid Cu(I) compounds such as CuI and $C u_{2} O$ survive only because they are insoluble.

Exam Trick

$C r^{2 +}$ gives, $M n^{3 +}$ takes. Both are $d^{4}$ . $C r^{2 +}$ gives away an electron to reach $t_{2 g}^{3}$ (reducing), $M n^{3 +}$ takes one to reach $d^{5}$ (oxidising). Every such question is solved by asking: which direction reaches $d^{0}$ , $d^{5}$ , $d^{10}$ or $t_{2 g}^{3}$ ?

Key idea

Oxidation states change in steps of one and peak at Mn; stability follows the

d^{0}

d^{5}

d^{10}

rule, and only Cu has a positive

E^{\circ} (M^{2 +} /M)

5. Colour and Magnetic Properties

5.1 Why Transition Metal Compounds Are Coloured

Many transition metal compounds are coloured, unlike those of the s- and p-blocks. In a free ion the five $d$ orbitals have the same energy (they are degenerate). When ligands such as water surround the ion, the $d$ orbitals split into two groups of different energy ( $t_{2 g}$ and $e_{g}$ in an octahedral complex). An electron can be promoted from the lower to the upper group: this d-d transition needs a small energy that matches visible light. The complex absorbs one colour and we see the complementary colour.

Figure 7: Ligands split the

d

set into

t_{2 g}

and

e_{g}

. In

[Ti (H_{2} O)_{6}]^{3 +}

the lone

d

electron absorbs green-yellow light (about 500 nm) to cross

Δ_{o}

, so the solution looks purple, the complementary colour.

Ions with $d^{0}$ ( $S c^{3 +}$ , $T i^{4 +}$ ) or $d^{10}$ ( $Z n^{2 +}$ , $C u^{+}$ ) configurations have no possible d-d transition, so their compounds such as $ZnS O_{4}$ and $Ti O_{2}$ are white. The colour also depends on the ligand: $[Ni (H_{2} O)_{6}]^{2 +}$ is green, $[Ni (N H_{3})_{6}]^{2 +}$ is blue-violet. (Some $d^{0}$ species, such as $Mn O_{4}^{-}$ and $C r_{2} O_{7}^{2 -}$ , are intensely coloured for a different reason: charge transfer from oxygen to the metal.)

5.2 Magnetic Properties

Substances are classified by their behaviour in a magnetic field. Paramagnetic substances are attracted by the field, diamagnetic substances are repelled, and ferromagnetic substances (Fe, Co, Ni) are attracted very strongly and can be permanently magnetised. Paramagnetism is caused by unpaired electrons, each of which behaves like a tiny magnet. So most transition metal ions are paramagnetic, and the more unpaired electrons, the stronger the paramagnetism.

For the 3d ions the magnetic moment comes mostly from the electron spins, and is given by the spin-only formula:

μ = n (n + 2) BM

where $n$ is the number of unpaired electrons and BM is the Bohr magneton. $M n^{2 +}$ ( $d^{5}$ , $n = 5$ ) has the largest value in the series, $35 = 5.92$ BM.

Figure 8: Unpaired electrons rise to 5 at

d^{5}

and fall again, so

μ = n (n + 2)

peaks at 5.92 BM for

M n^{2 +}

. The two ends,

S c^{3 +}

(

d^{0}

) and

Z n^{2 +}

(

d^{10}

), are both colourless and diamagnetic.

Figure 9: One routine answers most colour and magnetism questions.

F e^{3 +}

(

3 d^{5}

n = 5

) gives

μ = 35 = 5.92

BM;

d^{0}

and

d^{10}

ions are colourless and diamagnetic.

Exam Trick

Magic moments: memorise 1.73, 2.83, 3.87, 4.90, 5.92 BM for $n$ = 1 to 5. Reverse questions (' $μ$ = 3.9 BM, how many unpaired electrons?') then take one glance: 3.87, so $n = 3$ .

Key idea

Colour and paramagnetism have the same root: a partly filled

d

set.

d^{0}

and

d^{10}

ions are colourless and diamagnetic.

Quick Recall: tap to check

Why is

[Ti (H_{2} O)_{6}]^{3 +}

purple?

Its single

d

electron absorbs green-yellow light (about 500 nm) in a

t_{2 g} \to e_{g}

jump; purple is the complementary colour.

Spin-only

μ

C r^{3 +}

3 d^{3}

n = 3

μ = 15 = 3.87

BM.

Why is

C u^{+}

colourless?

It is

3 d^{10}

: no d-d transition is possible.

6. Complexes, Catalysis, Interstitial Compounds and Alloys

6.1 Complex Formation

Transition metals have an unmatched tendency to form coordination compounds with Lewis bases called ligands:

C o^{3 +} + 6N H_{3} [Co (N H_{3})_{6}]^{3 +}

F e^{2 +} + 6C N^{-} [Fe (CN)_{6}]^{4 -}

s- and p-block elements form very few complexes. Transition metal ions are good at it because they are small and highly charged (high charge density) and have vacant $d$ orbitals of suitable energy to accept lone pairs from ligands. Other examples are $[Cu (N H_{3})_{4}]^{2 +}$ , $[Ni (CN)_{4}]^{2 -}$ and $[PtC l_{4}]^{2 -}$ .

6.2 Catalytic Properties

Many transition metals and their compounds are catalysts: $V_{2} O_{5}$ (contact process, $S O_{2} \to S O_{3}$ ), finely divided Fe (Haber process), Ni (hydrogenation of oils), Pd (hydrogenation), $FeC l_{3}$ and $TiC l_{4}$ with $Al (C H_{3})_{3}$ (Ziegler-Natta polymerisation). Two features explain this:

Variable oxidation states: the metal forms an unstable intermediate by changing its oxidation state, then returns to its original state. $F e^{3 +}$ catalyses the reaction between iodide and peroxodisulphate this way.
A suitable surface: solid metals adsorb reactant molecules using their free valencies, which weakens bonds and brings molecules together (Ni, Pt, Fe).

2F e^{3 +} + 2 I^{-} 2F e^{2 +} + I_{2}

2F e^{2 +} + S_{2} O_{8}^{2 -} 2F e^{3 +} + 2S O_{4}^{2 -}

Figure 10: (A)

F e^{3 +}

oxidises

I^{-}

, becomes

F e^{2 +}

, and is re-oxidised by

S_{2} O_{8}^{2 -}

, so it is recovered unchanged. (B) Small atoms fit into the holes of the metal lattice, giving hard, high-melting interstitial compounds.

6.3 Interstitial and Non-stoichiometric Compounds

Small atoms such as H, C and N can occupy the holes (interstitial sites) of a transition metal lattice. The products, interstitial compounds such as TiC, $M n_{4} N$ , $F e_{3} H$ , $V H_{0.56}$ and $Ti H_{1.7}$ , keep metallic conductivity, are harder than the pure metal, have very high melting points and are chemically inert. They are usually non-stoichiometric, that is, their composition is not a simple whole-number ratio. Non-stoichiometry also arises from variable valency and lattice defects, as in ferrous oxide, which is really about $F e_{0.94} O$ because some $F e^{2 +}$ is replaced by $F e^{3 +}$ .

6.4 Alloy Formation

An alloy is a homogeneous solid solution of two or more metals made by melting the components together and cooling the melt. Atoms of one metal can take up lattice positions of the other only if their radii differ by not more than about 15%. Transition metals have very similar radii, so they form alloys readily: ferrous alloys with Cr, V, Mo, W and Mn (stainless steel, tool steels), brass (Cu-Zn) and bronze (Cu-Sn).

Applications: iron and steel are the main construction materials; $Ti O_{2}$ is a white pigment; $Mn O_{2}$ is used in dry cells; Zn and Ni/Cd are used in batteries; Cu, Ag and Au are coinage metals.

Key idea

High charge density plus vacant d orbitals gives complexes; variable valency plus a free surface gives catalysis; similar radii give alloys.

7. Important Compounds of Transition Elements

7.1 Copper(II) Sulphate Pentahydrate (Blue Vitriol), $CuS O_{4} \cdot 5 H_{2} O$

Preparation. In the laboratory, cupric oxide, cupric hydroxide or cupric carbonate is dissolved in dilute sulphuric acid:

CuO + H_{2} S O_{4} CuS O_{4} + H_{2} O

Cu (OH)_{2} + H_{2} S O_{4} CuS O_{4} + 2 H_{2} O

CuC O_{3} + H_{2} S O_{4} CuS O_{4} + H_{2} O + C O_{2} ↑

The solution is concentrated and cooled, and blue crystals of $CuS O_{4} \cdot 5 H_{2} O$ separate. Commercially, scrap copper is treated with hot dilute sulphuric acid in the presence of air:

2Cu + 2 H_{2} S O_{4} + O_{2} 2CuS O_{4} + 2 H_{2} O

1. Action of heat. The blue crystals lose water in stages:

CuS O_{4} \cdot 5 H_{2} O 373 K CuS O_{4} \cdot H_{2} O \approx 473 K white CuS O_{4} > 1000 K CuO + S O_{3}

In $CuS O_{4} \cdot 5 H_{2} O$ four water molecules are coordinated to $C u^{2 +}$ , while the fifth is held by hydrogen bonds between a sulphate ion and a coordinated water molecule. This fifth molecule is deep inside the lattice and is the last to be lost.

2. Action of alkalis. NaOH gives a pale blue precipitate of cupric hydroxide; excess aqueous ammonia gives a deep blue solution of tetraamminecopper(II) sulphate:

CuS O_{4} + 2NaOH Cu (OH)_{2} ↓ + N a_{2} S O_{4}

CuS O_{4} + 4N H_{4} OH [Cu (N H_{3})_{4}] S O_{4} + 4 H_{2} O

3. Reaction with KI. Cupric iodide is unstable and breaks down at once into white cuprous iodide and iodine:

CuS O_{4} + 2KI Cu I_{2} + K_{2} S O_{4}

2Cu I_{2} 2CuI ↓ + I_{2}

Iodine is liberated quantitatively, so this reaction (followed by titration of $I_{2}$ with thiosulphate) is used to estimate copper volumetrically.

Uses: electrolyte in electroplating, electrotyping and refining of copper; controlling algae and weeds in reservoirs and swimming pools; fungicide as Bordeaux mixture (copper sulphate with slaked lime, $Ca (OH)_{2}$ ); anhydrous $CuS O_{4}$ (white, turns blue with water) detects moisture in organic liquids such as alcohol and ether.

7.2 Silver Nitrate (Lunar Caustic), $AgN O_{3}$

Preparation. Silver is dissolved in dilute nitric acid and the solution is evaporated to crystallise:

3Ag + 4HN O_{3} 3AgN O_{3} + NO ↑ + 2 H_{2} O

1. Action of heat. It first loses oxygen to give silver nitrite, then silver:

2AgN O_{3} 723 K 2AgN O_{2} + O_{2}

2AgN O_{2} 980 K 2Ag + 2N O_{2}

2. Organic matter such as skin or cloth reduces it to finely divided silver, giving a black stain (hence its use in marking inks).

3. Precipitation reactions. It gives coloured precipitates with many anions, which helps to detect acid radicals:

NaCl + AgN O_{3} AgCl ↓ + NaN O_{3}

N a_{3} P O_{4} + 3AgN O_{3} A g_{3} P O_{4} ↓ + 3NaN O_{3}

K_{2} Cr O_{4} + 2AgN O_{3} A g_{2} Cr O_{4} ↓ + 2KN O_{3}

N a_{2} S + 2AgN O_{3} A g_{2} S ↓ + 2NaN O_{3}

N a_{2} S_{2} O_{3} + 2AgN O_{3} A g_{2} S_{2} O_{3} ↓ + 2NaN O_{3}

N a_{2} C_{2} O_{4} + 2AgN O_{3} A g_{2} C_{2} O_{4} ↓ + 2NaN O_{3}

N a_{3} B O_{3} + 3AgN O_{3} A g_{3} B O_{3} ↓ + 3NaN O_{3}

$AgCl$ is white, $A g_{3} P O_{4}$ yellow, $A g_{2} Cr O_{4}$ brick red, $A g_{2} S$ black, $A g_{2} C_{2} O_{4}$ white. The white $A g_{2} S_{2} O_{3}$ turns yellow, brown and finally black as it hydrolyses to $A g_{2} S$ .

Uses: preparing silver halides for photography; making indelible inks and hair dyes; qualitative and quantitative analysis (Tollens' reagent, Mohr titration of chloride); silvering of glass for mirrors.

7.3 Silver Halides: AgF, AgCl, AgBr and AgI

Preparation. AgCl, AgBr and AgI precipitate when a sodium or potassium halide is added to silver nitrate solution. AgF is soluble, so it is made from silver(I) oxide and HF:

AgN O_{3} + NaX AgX ↓ + NaN O_{3}

A g_{2} O + 2HF 2AgF + H_{2} O

Properties. AgCl is white, AgBr pale yellow and AgI yellow. AgF is soluble in water, the others are insoluble. AgCl dissolves readily in dilute ammonia, AgBr only partly (in concentrated ammonia) and AgI not at all. All silver halides dissolve in potassium cyanide and in sodium thiosulphate (hypo) by forming soluble complexes:

AgCl + 2N H_{4} OH [Ag (N H_{3})_{2}] Cl + 2 H_{2} O

AgCl + 2KCN K [Ag (CN)_{2}] + KCl

AgCl + 2N a_{2} S_{2} O_{3} N a_{3} [Ag (S_{2} O_{3})_{2}] + NaCl

Uses: all silver halides, especially AgBr, are decomposed by light, which is the basis of film photography (section 7.7).

7.4 Mercury Halides

(a) Mercury(I) chloride, mercurous chloride or calomel, $H g_{2} C l_{2}$ . It contains the $H g_{2}^{2 +}$ ion with an Hg-Hg bond. It is prepared by mixing a chloride with a mercury(I) salt, or by heating mercuric chloride with mercury in an iron vessel:

H g_{2} (N O_{3})_{2} + 2NaCl H g_{2} C l_{2} ↓ + 2NaN O_{3}

HgC l_{2} + Hg Δ H g_{2} C l_{2}

It is a white powder, insoluble in water but soluble in chlorine water (which oxidises it); on heating it disproportionates; with ammonia it turns black because of finely divided mercury (a test for $H g_{2}^{2 +}$ ):

H g_{2} C l_{2} + C l_{2} 2HgC l_{2}

H g_{2} C l_{2} Δ HgC l_{2} + Hg

H g_{2} C l_{2} + 2N H_{3} Hg + Hg (N H_{2}) Cl + N H_{4} Cl

Uses: in the standard calomel reference electrode; formerly as a purgative in medicine.

(b) Mercury(II) chloride, mercuric chloride or corrosive sublimate, $HgC l_{2}$ . It is prepared by passing dry chlorine over heated mercury, by dissolving HgO in HCl, or commercially by heating mercuric sulphate with common salt (a little $Mn O_{2}$ oxidises any mercury(I) present):

Hg + C l_{2} Δ HgC l_{2}

HgO + 2HCl HgC l_{2} + H_{2} O

HgS O_{4} + 2NaCl Mn O_{2}, Δ HgC l_{2} + N a_{2} S O_{4}

It is a white crystalline solid, sparingly soluble in cold water but soluble in hot water; adding chloride ions raises its solubility by forming a complex. It dissolves readily in organic solvents, which shows its covalent nature. Stannous chloride reduces it first to white $H g_{2} C l_{2}$ , then to grey mercury (a test for $H g^{2 +}$ ), and copper turnings get coated with a shining grey film of mercury:

HgC l_{2} + 2C l^{-} [HgC l_{4}]^{2 -}

2HgC l_{2} + SnC l_{2} H g_{2} C l_{2} ↓ + SnC l_{4}

H g_{2} C l_{2} + SnC l_{2} 2Hg + SnC l_{4}

HgC l_{2} + Cu Hg + CuC l_{2}

Uses: preserving wood and hides, making fungicides. It is highly poisonous.

(c) Mercury(II) iodide, $Hg I_{2}$ . It is precipitated as a scarlet solid when KI is added to $HgC l_{2}$ . It exists in two forms: red below 400 K and yellow above 400 K. It dissolves in excess KI to form potassium tetraiodomercurate(II):

HgC l_{2} + 2KI Hg I_{2} ↓ + 2KCl

red Hg I_{2} < 400 K ⇌ > 400 K yellow Hg I_{2}

Hg I_{2} + 2KI K_{2} [Hg I_{4}]

An alkaline solution of $K_{2} [Hg I_{4}]$ is Nessler's reagent. With ammonia or ammonium salts it gives a brown precipitate of the iodide of Millon's base, a sensitive test for $N H_{4}^{+}$ :

2 K_{2} [Hg I_{4}] + N H_{3} + 3KOH HgO \cdot Hg (N H_{2}) I ↓ + 7KI + 2 H_{2} O

Uses: preparing Nessler's reagent; ointments for skin infections.

Figure 11: Colour clues examiners use: blue vitriol turns white on losing water; silver precipitates are coded by colour (AgCl white, AgBr pale yellow, AgI yellow,

A g_{2} Cr O_{4}

brick red);

SnC l_{2}

reduces

HgC l_{2}

to white

H g_{2} C l_{2}

then grey Hg; Nessler's reagent gives a brown precipitate with ammonia.

7.5 Potassium Dichromate, $K_{2} C r_{2} O_{7}$

Preparation. It is made from chromite ore, $FeC r_{2} O_{4}$ (also written $FeO \cdot C r_{2} O_{3}$ ), in three steps.

Sodium chromate: the powdered ore is fused with sodium carbonate in free access of air.
Sodium dichromate: the yellow chromate solution is filtered and acidified with sulphuric acid.
Potassium dichromate: the sodium salt, which is very soluble, is treated with KCl; less soluble orange $K_{2} C r_{2} O_{7}$ crystallises.

4FeC r_{2} O_{4} + 8N a_{2} C O_{3} + 7 O_{2} 8N a_{2} Cr O_{4} + 2F e_{2} O_{3} + 8C O_{2}

2N a_{2} Cr O_{4} + H_{2} S O_{4} N a_{2} C r_{2} O_{7} + N a_{2} S O_{4} + H_{2} O

N a_{2} C r_{2} O_{7} + 2KCl K_{2} C r_{2} O_{7} + 2NaCl

Properties. It forms orange-red crystals, moderately soluble in cold water and freely soluble in hot water.

1. Action of heat:

4 K_{2} C r_{2} O_{7} Δ 4 K_{2} Cr O_{4} + 2C r_{2} O_{3} + 3 O_{2}

2. Action of alkalis and acids. Alkali converts dichromate to yellow chromate; acid turns it back. Both ions exist in equilibrium in solution, and the pH decides which one dominates:

K_{2} C r_{2} O_{7} + 2KOH 2 K_{2} Cr O_{4} + H_{2} O

2 K_{2} Cr O_{4} + H_{2} S O_{4} K_{2} C r_{2} O_{7} + K_{2} S O_{4} + H_{2} O

C r_{2} O_{7}^{2 -} + H_{2} O ⇌ 2Cr O_{4}^{2 -} + 2 H^{+}

The chromate ion is tetrahedral. The dichromate ion consists of two tetrahedra sharing one corner oxygen, with a Cr-O-Cr bond angle of 126°.

Figure 12:

Cr O_{4}^{2 -}

(yellow) and

C r_{2} O_{7}^{2 -}

(orange) interconvert with pH; the dichromate ion is two tetrahedra sharing one oxygen (Cr-O-Cr 126°). Green

Mn O_{4}^{2 -}

is oxidised to purple

Mn O_{4}^{-}

; all four ions are tetrahedral.

3. Action of concentrated sulphuric acid. In the cold, red crystals of chromic anhydride ( $Cr O_{3}$ ) separate; on heating, oxygen is evolved:

K_{2} C r_{2} O_{7} + 2 H_{2} S O_{4} cold 2Cr O_{3} + 2KHS O_{4} + H_{2} O

2 K_{2} C r_{2} O_{7} + 8 H_{2} S O_{4} hot 2 K_{2} S O_{4} + 2C r_{2} (S O_{4})_{3} + 8 H_{2} O + 3 O_{2}

4. Oxidising properties. In dilute sulphuric acid it is a powerful oxidant and supplies three atoms of available oxygen per formula unit; in ionic terms it gains 6 electrons and turns green ( $C r^{3 +}$ ):

K_{2} C r_{2} O_{7} + 4 H_{2} S O_{4} K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 4 H_{2} O + 3 [O]

C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} 2C r^{3 +} + 7 H_{2} O

Typical oxidations (molecular equations, with the ionic form in the table below):

K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} + 6KI 4 K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 3 I_{2} + 7 H_{2} O

K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} + 6FeS O_{4} K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 3F e_{2} (S O_{4})_{3} + 7 H_{2} O

K_{2} C r_{2} O_{7} + 4 H_{2} S O_{4} + 3 H_{2} S K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 7 H_{2} O + 3S

K_{2} C r_{2} O_{7} + 4 H_{2} S O_{4} + 3NaN O_{2} K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 3NaN O_{3} + 4 H_{2} O

K_{2} C r_{2} O_{7} + H_{2} S O_{4} + 3S O_{2} K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + H_{2} O

Reductant	Ionic half-reaction	Product
iodide	$2 I^{-} \to I_{2} + 2 e^{-}$	iodine (brown)
iron(II)	$F e^{2 +} \to F e^{3 +} + e^{-}$	iron(III)
hydrogen sulphide	$H_{2} S \to S + 2 H^{+} + 2 e^{-}$	sulphur (milky)
nitrite	$N O_{2}^{-} + H_{2} O \to N O_{3}^{-} + 2 H^{+} + 2 e^{-}$	nitrate
sulphur dioxide	$S O_{2} + 2 H_{2} O \to S O_{4}^{2 -} + 4 H^{+} + 2 e^{-}$	sulphate
ethanol	$C H_{3} C H_{2} OH \to C H_{3} CHO \to C H_{3} COOH$	ethanal, then ethanoic acid

5. Chromyl chloride test. When a chloride is heated with solid $K_{2} C r_{2} O_{7}$ and concentrated sulphuric acid, reddish-brown vapours of chromyl chloride are evolved. Passed into NaOH they give yellow sodium chromate, which gives a yellow precipitate of lead chromate with lead acetate. The test detects chloride ions in qualitative analysis (bromides and iodides do not give it):

K_{2} C r_{2} O_{7} + 4KCl + 6 H_{2} S O_{4} 2Cr O_{2} C l_{2} + 6KHS O_{4} + 3 H_{2} O

Uses: volumetric estimation of $F e^{2 +}$ and $I^{-}$ (it is a primary standard); chrome tanning in the leather industry; photography and hardening of gelatin films; the breathalyser test for ethanol.

7.6 Potassium Permanganate, $KMn O_{4}$

Large-scale preparation from the mineral pyrolusite, $Mn O_{2}$ :

$Mn O_{2}$ to potassium manganate: finely powdered $Mn O_{2}$ is fused with KOH in air (or with an oxidant such as $KN O_{3}$ ) to give dark green $K_{2} Mn O_{4}$ .
Manganate to permanganate, chemically: $C O_{2}$ , $C l_{2}$ or ozone is bubbled through the manganate solution; in neutral or acid solution manganate disproportionates.
Manganate to permanganate, electrolytically (the commercial route): the manganate solution is electrolysed between iron electrodes; oxidation at the anode gives permanganate.

2Mn O_{2} + 4KOH + O_{2} 2 K_{2} Mn O_{4} + 2 H_{2} O

3 K_{2} Mn O_{4} + 2C O_{2} 2KMn O_{4} + Mn O_{2} + 2 K_{2} C O_{3}

3Mn O_{4}^{2 -} + 4 H^{+} 2Mn O_{4}^{-} + Mn O_{2} + 2 H_{2} O

2 K_{2} Mn O_{4} + H_{2} O + [O] electrolysis 2KMn O_{4} + 2KOH

Mn O_{4}^{2 -} anode Mn O_{4}^{-} + e^{-}

In the laboratory a manganese(II) salt is oxidised by peroxodisulphate:

2M n^{2 +} + 5 S_{2} O_{8}^{2 -} + 8 H_{2} O 2Mn O_{4}^{-} + 10S O_{4}^{2 -} + 16 H^{+}

Properties. $KMn O_{4}$ forms deep purple (nearly black) prisms, moderately soluble in water at room temperature; solubility increases with temperature. The permanganate ion is tetrahedral and diamagnetic ( $Mn$ +7, $d^{0}$ ); the manganate ion is tetrahedral and paramagnetic ( $d^{1}$ ).

(i) Action of heat:

2KMn O_{4} 513 K K_{2} Mn O_{4} + Mn O_{2} + O_{2}

(ii) Action of concentrated sulphuric acid. Cold acid gives the explosive oily oxide $M n_{2} O_{7}$ , which decomposes on warming; hot acid liberates oxygen:

2KMn O_{4} + 2 H_{2} S O_{4} cold M n_{2} O_{7} + 2KHS O_{4} + H_{2} O

2M n_{2} O_{7} Δ 4Mn O_{2} + 3 O_{2}

4KMn O_{4} + 6 H_{2} S O_{4} hot 2 K_{2} S O_{4} + 4MnS O_{4} + 6 H_{2} O + 5 O_{2}

(iii) Oxidising properties. $KMn O_{4}$ is a powerful oxidant; what it becomes depends on the medium.

(a) Neutral or weakly alkaline medium (moderate oxidant, 3 electrons, brown $Mn O_{2}$ ):

2KMn O_{4} + H_{2} O 2KOH + 2Mn O_{2} + 3 [O]

Mn O_{4}^{-} + 2 H_{2} O + 3 e^{-} Mn O_{2} + 4O H^{-}

8KMn O_{4} + 3N a_{2} S_{2} O_{3} + H_{2} O 3 K_{2} S O_{4} + 8Mn O_{2} + 3N a_{2} S O_{4} + 2KOH

2KMn O_{4} + 4 H_{2} S 2MnS + S + K_{2} S O_{4} + 4 H_{2} O

2KMn O_{4} + H_{2} O + KI 2Mn O_{2} + 2KOH + KI O_{3}

(b) Strongly alkaline medium (1 electron, green manganate):

2KMn O_{4} + 2KOH 2 K_{2} Mn O_{4} + H_{2} O + [O]

Mn O_{4}^{-} + e^{-} Mn O_{4}^{2 -}

(c) Acidic medium (strongest, 5 electrons, $M n^{+ 7} \to M n^{2 +}$ , the purple colour disappears):

2KMn O_{4} + 3 H_{2} S O_{4} K_{2} S O_{4} + 2MnS O_{4} + 3 H_{2} O + 5 [O]

Mn O_{4}^{-} + 8 H^{+} + 5 e^{-} M n^{2 +} + 4 H_{2} O

Important oxidations in acid:

2KMn O_{4} + 3 H_{2} S O_{4} + 5 H_{2} S K_{2} S O_{4} + 2MnS O_{4} + 8 H_{2} O + 5S

2KMn O_{4} + 5S O_{2} + 2 H_{2} O K_{2} S O_{4} + 2MnS O_{4} + 2 H_{2} S O_{4}

2KMn O_{4} + 3 H_{2} S O_{4} + 5KN O_{2} K_{2} S O_{4} + 2MnS O_{4} + 3 H_{2} O + 5KN O_{3}

2KMn O_{4} + 3 H_{2} S O_{4} + 5 C_{2} H_{2} O_{4} K_{2} S O_{4} + 2MnS O_{4} + 8 H_{2} O + 10C O_{2}

2KMn O_{4} + 8 H_{2} S O_{4} + 10FeS O_{4} K_{2} S O_{4} + 2MnS O_{4} + 5F e_{2} (S O_{4})_{3} + 8 H_{2} O

2KMn O_{4} + 8 H_{2} S O_{4} + 10KI 6 K_{2} S O_{4} + 2MnS O_{4} + 8 H_{2} O + 5 I_{2}

Figure 13: The medium fixes the product and the electron count: acid 5, neutral 3 and strong alkali 1 for

Mn O_{4}^{-}

; 6 for

C r_{2} O_{7}^{2 -}

in acid. Equivalent weight

= M / n

, so

KMn O_{4}

is 31.6 in acid and

K_{2} C r_{2} O_{7}

is 49.

Sulphuric acid, not hydrochloric acid, is used to acidify permanganate in titrations, because $KMn O_{4}$ would oxidise HCl to chlorine. No indicator is needed: the first excess drop colours the solution pink (self-indicator).

Uses: volumetric estimation of ferrous salts, oxalates, iodides and $H_{2} O_{2}$ ; oxidising agent in the laboratory and industry (Baeyer's reagent is cold, dilute alkaline $KMn O_{4}$ ); disinfectant and germicide.

K_{2} C r_{2} O_{7}

Orange; ore: chromite.
Acid: $C r_{2} O_{7}^{2 -} \to 2C r^{3 +}$ (green), 6 $e^{-}$ .
Primary standard; does not oxidise $C l^{-}$ in dilute acid; needs an indicator (diphenylamine).

KMn O_{4}

Purple; ore: pyrolusite.
Acid: $Mn O_{4}^{-} \to M n^{2 +}$ (colourless), 5 $e^{-}$ .
Not a primary standard; oxidises HCl; acts as its own indicator.

7.7 Silver Halides in Film Photography

Black-and-white film photography, now largely replaced by digital sensors, is a classic application of silver bromide, which decomposes and blackens in light:

2AgBr light 2Ag + B r_{2}

Preparing the film. Ammoniacal $AgN O_{3}$ is added to $N H_{4} Br$ solution containing gelatin, giving an emulsion of AgBr in gelatin. It is allowed to stand so that the AgBr grains grow (ripening), solidified, washed free of $N H_{4} N O_{3}$ , melted and spread on a glass plate or celluloid film: $N H_{4} Br + AgN O_{3} \to AgBr + N H_{4} N O_{3}$ .
Exposure. Light from the object reaches the film for a fraction of a second. AgBr grains that receive light are partly reduced to silver specks, forming an invisible latent image.
Developing. In a dark room, a reducing developer (quinol or pyrogallol) reduces the exposed grains completely to black silver. Bright parts of the object become dark: this is the negative.
Fixing. The negative is dipped in hypo, which dissolves the unexposed AgBr, making the image permanent; the film can now be taken into light.
Printing. Less sensitive printing paper (printing-out paper coated with AgCl and $AgN O_{3}$ , or bromide paper) is exposed through the negative. The negative of the negative is a positive print.
Toning. For a golden tint the print is dipped in dilute gold chloride; for a grey tint, in potassium chloroplatinite, $K_{2} PtC l_{4}$ . Silver on the print is replaced by gold or platinum.

Reactions in developing, fixing and toning:

C_{6} H_{4} (OH)_{2} + 2AgBr 2Ag + C_{6} H_{4} O_{2} + 2HBr

AgBr + 2N a_{2} S_{2} O_{3} N a_{3} [Ag (S_{2} O_{3})_{2}] + NaBr

AuC l_{3} + 3Ag 3AgCl + Au

K_{2} PtC l_{4} + 2Ag 2AgCl + Pt + 2KCl

Key idea

Remember each compound by its colour clue and its electron count: blue vitriol goes white, silver precipitates are colour-coded, dichromate takes

6 e^{-}

, permanganate takes 5, 3 or

1 e^{-}

Quick Recall: tap to check

What colour change marks the end point of a

KMn O_{4}

titration?

The first extra drop leaves a permanent pale pink colour; permanganate is its own indicator.

Which red-brown vapours prove chloride in the chromyl chloride test?

Reddish-brown

Cr O_{2} C l_{2}

; in NaOH it gives yellow chromate.

Why does

CuS O_{4}

with KI give a white precipitate?

Cu I_{2}

is unstable; it gives white CuI and brown

I_{2}

8. Revision Map

Use this map to revise the whole chapter in two minutes before a test.

Figure 14: The whole page on one map: configuration and trends explain the properties, and the properties explain the chemistry of

K_{2} C r_{2} O_{7}

KMn O_{4}

and the Cu, Ag and Hg salts.

9. Solved Examples

Solved Example 1

Although copper, silver and gold have completely filled d orbitals in their ground state, they are considered transition metals. Why?

Solution:

A transition metal needs a partly filled d subshell in the atom or in a common oxidation state. $C u^{2 +}$ is $3 d^{9}$ , silver in $Ag F_{2}$ is $A g^{2 +}$ , $4 d^{9}$ , and $A u^{3 +}$ is $5 d^{8}$ . So all three qualify.

Solved Example 2

Zinc, cadmium and mercury are generally not regarded as transition metals. Give reasons.

Solution:

Their atoms are $(n - 1) d^{10} n s^{2}$ and their only common ions, $M^{2 +}$ (and $H g_{2}^{2 +}$ ), are also $d^{10}$ . The d subshell is never partly filled, so they do not meet the definition; this is why their compounds are white and diamagnetic.

Solved Example 3

In what way is the electronic configuration of transition elements different from that of non-transition elements?

Solution:

Transition elements have an incompletely filled penultimate d subshell, $(n - 1) d^{1 - 10} n s^{0 - 2}$ . Non-transition (main-group) elements have either no d subshell or a completely filled one, and their outer shell is $n s^{1 - 2}$ or $n s^{2} n p^{1 - 6}$ .

Solved Example 4

K_{2} PtC l_{6}

is a well-known compound, whereas the corresponding nickel compound is not known. State a reason.

Solution:

Both would need the metal in the +4 state. The sum of the first four ionisation enthalpies is about $11.29 \times 1 0^{3}$ kJ $mol^{- 1}$ for Ni but only about $9.30 \times 1 0^{3}$ kJ $mol^{- 1}$ for Pt (NCERT data). It is far easier to form $P t^{4 +}$ than $N i^{4 +}$ , so Pt(IV) compounds are stable and Ni(IV) compounds are not.

Solved Example 5

Why do transition elements show variable oxidation states?

Solution:

The energies of the $(n - 1) d$ and $n s$ orbitals are very close, so electrons from both can take part in bonding. Different numbers of electrons can be lost or shared, giving several oxidation states that usually differ by one.

Solved Example 6

Explain briefly how the +2 state becomes more and more stable in the first half of the first-row transition elements with increasing atomic number.

Solution:

Going from Sc to Mn, the third ionisation enthalpy rises steeply as the $3 d$ set fills towards $d^{5}$ , so it becomes harder and harder to oxidise $M^{2 +}$ to $M^{3 +}$ . This shows in $E^{\circ} (M^{3 +} / M^{2 +})$ : about $- 0.4$ to $- 0.3$ V for Ti, V and Cr, but $+ 1.57$ V for Mn, where $M n^{2 +}$ is a stable half-filled $d^{5}$ ion. Beyond Mn, $Z n^{2 +}$ is stabilised by its full $d^{10}$ set and $N i^{2 +}$ by its highly negative enthalpy of hydration.

Solved Example 7

Why are

Z n^{2 +}

salts white while

N i^{2 +}

salts are green?

Solution:

$Z n^{2 +}$ is $3 d^{10}$ : no d-d transition is possible, so no visible light is absorbed. $N i^{2 +}$ is $3 d^{8}$ with two unpaired electrons; d-d transitions absorb red light, so hydrated nickel salts look green.

Solved Example 8

Why are

Z n^{2 +}

salts white while

C u^{2 +}

salts are blue?

Solution:

Same reasoning: $Z n^{2 +}$ ( $3 d^{10}$ ) cannot undergo a d-d transition. $C u^{2 +}$ ( $3 d^{9}$ ) has one vacancy in the upper $e_{g}$ set; the hydrated ion absorbs orange-red light (around 600-800 nm) and appears blue.

Solved Example 9

Giving reasons, indicate which of the following would be coloured:

C u^{+}

V O^{2 +}

S c^{3 +}

N i^{2 +}

(At. Nos: Cu = 29, V = 23, Sc = 21, Ni = 28).

Solution:

$C u^{+}$ : $3 d^{10}$ , colourless. $S c^{3 +}$ : $3 d^{0}$ , colourless. $V O^{2 +}$ : V is +4, $3 d^{1}$ , coloured (blue). $N i^{2 +}$ : $3 d^{8}$ , coloured (green). So $V O^{2 +}$ and $N i^{2 +}$ are coloured, because both have partly filled d orbitals.

Solved Example 10

Why does Mn(II) show the maximum paramagnetic character among the bivalent ions of the first transition series?

Solution:

$M n^{2 +}$ is $3 d^{5}$ with five unpaired electrons, the largest possible number for a d subshell. So $μ = 5 \times 7 = 5.92$ BM, the highest among the $M^{2 +}$ ions.

Solved Example 11

A substance is found to have a magnetic moment of 3.9 BM. How many unpaired electrons does it contain?

Solution:

$n (n + 2) = 3.9 \Rightarrow n^{2} + 2 n = 15.21 \Rightarrow n = - 1 + 16.21 = 3.03$ . So $n = 3$ unpaired electrons (spin-only value 3.87 BM).

Solved Example 12

Which of the following ions has the highest spin-only magnetic moment?
(A)

C r^{3 +}

(B)

F e^{2 +}

(C)

F e^{3 +}

(D)

C o^{2 +}

Solution:

Answer: (C). Unpaired electrons: $C r^{3 +}$ $3 d^{3}$ (3), $F e^{2 +}$ $3 d^{6}$ (4), $F e^{3 +}$ $3 d^{5}$ (5), $C o^{2 +}$ $3 d^{7}$ (3). $F e^{3 +}$ : $μ = 35 = 5.92$ BM.

Solved Example 13

The equivalent weight of

KMn O_{4}

(molar mass 158 g

mol^{- 1}

) when it oxidises in acidic medium is
(A) 158
(B) 52.7
(C) 31.6
(D) 79

Solution:

Answer: (C). In acid, $Mn$ goes from +7 to +2, a gain of 5 electrons. Equivalent weight = 158/5 = 31.6. (B) is the value in neutral medium (158/3).

Solved Example 14

What volume of 0.02 M

KMn O_{4}

is needed to oxidise 20 mL of 0.1 M

FeS O_{4}

in acidic medium?

Solution:

Mn O_{4}^{-} + 5F e^{2 +} + 8 H^{+} M n^{2 +} + 5F e^{3 +} + 4 H_{2} O

Moles of $F e^{2 +}$ = $0.1 \times 0.020 = 2.0 \times 1 0^{- 3}$ . Moles of $Mn O_{4}^{-}$ = $2.0 \times 1 0^{- 3} /5 = 4.0 \times 1 0^{- 4}$ . Volume = $4.0 \times 1 0^{- 4} /0.02 = 0.020$ L = 20 mL.

Solved Example 15

C r^{2 +}

is strongly reducing whereas

M n^{3 +}

is strongly oxidising, although both are

d^{4}

. Explain.

Solution:

$C r^{2 +}$ is oxidised to $C r^{3 +}$ ( $d^{3}$ , half-filled $t_{2 g}$ set), which is very stable; $E^{\circ} (C r^{3 +} /C r^{2 +}) = - 0.41$ V. $M n^{3 +}$ is reduced to $M n^{2 +}$ ( $d^{5}$ , half-filled), also very stable; $E^{\circ} (M n^{3 +} /M n^{2 +}) = + 1.57$ V. Each ion moves towards the more stable configuration.

Solved Example 16

An orange crystalline compound A, heated with NaCl and concentrated

H_{2} S O_{4}

, gives red-brown vapours B. B dissolves in NaOH to give a yellow solution C, which gives a yellow precipitate D with lead acetate and acetic acid. Identify A to D.

Solution:

A = $K_{2} C r_{2} O_{7}$ , B = $Cr O_{2} C l_{2}$ (chromyl chloride), C = $N a_{2} Cr O_{4}$ , D = $PbCr O_{4}$ .

Cr O_{2} C l_{2} + 4NaOH N a_{2} Cr O_{4} + 2NaCl + 2 H_{2} O

N a_{2} Cr O_{4} + (C H_{3} COO)_{2} Pb PbCr O_{4} ↓ + 2C H_{3} COONa

Practice Questions

Most of the transition metals do not displace hydrogen from dilute acids. Why?Answer: Only metals with positive $E^{\circ} (M^{2 +} /M)$ , such as Cu (+0.34 V), Ag, Au and Pt, cannot. Many 3d metals (Mn, Fe, Zn) do give $H_{2}$ , though Cr, Ti and Ni react slowly because a protective oxide film forms.
Why are the ionisation energies of 5d elements greater than those of 3d elements?Answer: After the 14 lanthanoids the nuclear charge rises by 32 while the 4f electrons shield poorly (lanthanoid contraction), so 5d electrons feel a much larger effective nuclear charge.
Why are $M n^{2 +}$ compounds more stable than $F e^{2 +}$ towards oxidation to the +3 state?Answer: $M n^{2 +}$ is $3 d^{5}$ (half-filled, stable), so losing an electron is hard ( $I E_{3}$ of Mn is high). $F e^{2 +}$ ( $3 d^{6}$ ) readily loses one electron to become $3 d^{5}$ .
Why do the transition elements have high enthalpies of hydration?Answer: Their ions are small and highly charged, so their charge density is high and they attract water dipoles strongly.
$[Ti (H_{2} O)_{6}]^{3 +}$ is coloured while $[Sc (H_{2} O)_{6}]^{3 +}$ is colourless. Explain.Answer: $T i^{3 +}$ is $3 d^{1}$ : a d-d transition absorbs visible light (purple). $S c^{3 +}$ is $3 d^{0}$ : no d-d transition.
Why is copper sulphate pentahydrate coloured?Answer: $C u^{2 +}$ ( $3 d^{9}$ ) surrounded by water ligands has split d levels; a d-d transition absorbs orange-red light, so it looks blue. Anhydrous $CuS O_{4}$ is white.
Why is $M n_{2} O_{7}$ acidic while MnO is basic?Answer: In +7 the metal is small and highly charged, so the oxide is covalent and acidic (gives permanganic acid); in +2 it is ionic and basic.

Common Mistakes to Avoid

Watch out

Removing 3d electrons before 4s: $F e^{2 +}$ is $[Ar] 3 d^{6}$ , not $3 d^{4} 4 s^{2}$ .
Writing Cr as $3 d^{4} 4 s^{2}$ or Cu as $3 d^{9} 4 s^{2}$ ; both have a single 4s electron.
Calling Zn, Cd and Hg transition elements: they are d-block elements but always $d^{10}$ .
Putting the total number of d electrons into $μ = n (n + 2)$ ; $n$ is the number of unpaired electrons.
Assuming $KMn O_{4}$ always takes 5 electrons: it takes 3 in neutral and 1 in strongly alkaline medium.
Acidifying permanganate with HCl in titrations: HCl is oxidised to chlorine and the result is wrong; use dilute sulphuric acid.
Saying acidified dichromate becomes colourless: it turns green ( $C r^{3 +}$ ). Permanganate becomes nearly colourless ( $M n^{2 +}$ ).
Expecting 5d atoms to be larger than 4d atoms: lanthanoid contraction makes them almost equal (Zr 160 pm, Hf 159 pm).

Frequently Asked Questions

What are d-block elements and why are they called transition elements?

d-Block elements are the elements of groups 3 to 12 in which the last electron enters the penultimate (n-1)d subshell. They are called transition elements because they lie between the s-block and p-block and their properties are transitional between the reactive ionic s-block metals and the covalent p-block elements.

Why are zinc, cadmium and mercury not considered transition elements?

A transition element must have a partly filled d subshell in the atom or a common ion. Zn, Cd and Hg have a full $d^{10}$ set both as atoms and as their common +2 ions, so their compounds are white, diamagnetic and show only one main oxidation state. They are still placed in the d-block.

Why do chromium and copper have anomalous electronic configurations?

In Cr and Cu one 4s electron moves into the 3d subshell, giving $3 d^{5} 4 s^{1}$ and $3 d^{10} 4 s^{1}$ . The 3d and 4s energies are very close, and half-filled or completely filled d sets are extra stable because of their symmetry and large exchange energy.

How do you calculate the spin-only magnetic moment of a transition metal ion?

Write the ion's configuration (remove 4s electrons first), count the unpaired d electrons n by Hund's rule and use $μ = n (n + 2)$ BM. For example, $F e^{3 +}$ is $3 d^{5}$ , $n = 5$ , so $μ = 5.92$ BM; $C u^{2 +}$ is $3 d^{9}$ , $n = 1$ , so $μ = 1.73$ BM.

Why are most transition metal compounds coloured?

Ligands split the d orbitals into two groups of slightly different energy. An electron can jump from the lower to the upper group by absorbing visible light (a d-d transition), and we see the complementary colour. Ions with $d^{0}$ or $d^{10}$ configurations cannot do this, so they are colourless.

What is the difference between KMnO4 in acidic, neutral and alkaline media?

In acid, permanganate gains 5 electrons and becomes colourless $M n^{2 +}$ ; in neutral or weakly alkaline solution it gains 3 electrons and gives a brown precipitate of $Mn O_{2}$ ; in strongly alkaline solution it gains 1 electron and becomes green manganate. Its equivalent weight is 31.6, 52.7 and 158 respectively.

Which d-block topics are most important for NEET?

For NEET, focus on NCERT statements: configurations with the Cr and Cu exceptions, magnetic moment calculations, colour of ions, trends in oxidation states and E° values, lanthanoid contraction, and the preparation, structures and reactions of $K_{2} C r_{2} O_{7}$ and $KMn O_{4}$ , especially their ionic equations.

How is the d-block chapter tested in JEE Main and Advanced?

JEE Main asks direct questions on magnetic moments, oxidation states, electrode potential anomalies and $KMn O_{4}$ or $K_{2} C r_{2} O_{7}$ reactions. JEE Advanced adds reasoning with E° values (disproportionation of $C u^{+}$ , oxidising $M n^{3 +}$ versus reducing $C r^{2 +}$ ), identification puzzles such as the chromyl chloride test, and redox titration calculations.

Previous year questions on d-Block Elements

26 questions from past papers, each with a step-by-step solution.

Show all 26 questions

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1. What Are d-Block Elements?

2. Electronic Configuration

2.1 Why Cr and Cu Are Anomalous

3. Physical Properties and Trends

3.1 Metallic Nature, Conductivity and Density

3.2 Melting Points and Enthalpy of Atomisation

3.3 Atomic and Ionic Size

3.4 Ionisation Enthalpy

4. Oxidation States and Electrode Potentials

4.1 Variable Oxidation States

4.2 Stability of High and Low Oxidation States

4.3 Standard Electrode Potentials

5. Colour and Magnetic Properties

5.1 Why Transition Metal Compounds Are Coloured

5.2 Magnetic Properties

6. Complexes, Catalysis, Interstitial Compounds and Alloys

6.1 Complex Formation

6.2 Catalytic Properties

6.3 Interstitial and Non-stoichiometric Compounds

6.4 Alloy Formation

7. Important Compounds of Transition Elements

7.1 Copper(II) Sulphate Pentahydrate (Blue Vitriol), CuSO4​⋅5H2​O

7.2 Silver Nitrate (Lunar Caustic), AgNO3​

7.3 Silver Halides: AgF, AgCl, AgBr and AgI

7.4 Mercury Halides

7.5 Potassium Dichromate, K2​Cr2​O7​

7.6 Potassium Permanganate, KMnO4​

7.7 Silver Halides in Film Photography

8. Revision Map

9. Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

What are d-block elements and why are they called transition elements?

Why are zinc, cadmium and mercury not considered transition elements?

Why do chromium and copper have anomalous electronic configurations?

How do you calculate the spin-only magnetic moment of a transition metal ion?

Why are most transition metal compounds coloured?

What is the difference between KMnO4 in acidic, neutral and alkaline media?

Which d-block topics are most important for NEET?

How is the d-block chapter tested in JEE Main and Advanced?

Previous year questions on d-Block Elements

Ready to master The d- And f-Block Elements?

7.1 Copper(II) Sulphate Pentahydrate (Blue Vitriol), $CuS O_{4} \cdot 5 H_{2} O$

7.2 Silver Nitrate (Lunar Caustic), $AgN O_{3}$

7.5 Potassium Dichromate, $K_{2} C r_{2} O_{7}$

7.6 Potassium Permanganate, $KMn O_{4}$