The d- and f-Block Elements — JEE Mains Chemistry Formulas and Traps

Electronic Configuration and General Properties

Configurations of d-block atoms and ions

d electrons in a 3d ion

n_d(\mathrm{M^{n+}}) = Z - 18 - n \qquad (n \ge 2)

Ionisation enthalpies across the 3d series

Metal	First IE (kJ/mol)	Second IE (kJ/mol)	Third IE (kJ/mol)	What it shows
Sc	631	1235	2389	Sc³⁺ is d⁰, so +3 is easy and is its only state
Ti	656	1309	2652	A steady rise with the nuclear charge
V	650	1414	2828	A steady rise with the nuclear charge
Cr	653	1592	2987	Low first IE (lone 4s); high second IE (breaks 3d⁵) Highest second IE from Sc to Fe, but its third IE is below Mn's.
Mn	717	1509	3248	High third IE: Mn²⁺ is 3d⁵
Fe	762	1561	2957	Low third IE: Fe²⁺ loses one electron to reach 3d⁵
Co	758	1644	3232	Rises again after the dip at Fe
Ni	736	1752	3393	Rises again after the dip at Fe
Cu	745	1958	3554	Highest second IE of the series: Cu⁺ is 3d¹⁰
Zn	906	1734	3833	Highest first IE: a filled 4s² over a filled 3d¹⁰

Values rounded to the nearest kJ/mol. The kinks, not the exact numbers, decide the questions.

Melting points, atomisation, density, catalysts and interstitial compounds

Metal	Atomisation enthalpy (kJ/mol)	Metallic radius (pm)	Density (g/cm³)	Point tested
Sc	326	164	2.99	Largest atom of the series
Ti	473	147	4.51	Ti⁴⁺ in TiCl₄ is d⁰: the Ziegler–Natta catalyst is diamagnetic
V	515	135	6.11	Highest atomisation enthalpy of the 3d series
Cr	397	129	7.19	Smallest radius among Sc, Ti, V, Cr, Mn and Zn
Mn	281	137	7.21	A dip: 3d⁵ holds its d electrons out of the bonding
Fe	416	126	7.87	Catalyst of the Haber process
Co	425	125	8.90	Dense, high-melting
Ni	430	125	8.91	Catalyst for hydrogenating oils
Cu	339	128	8.96	Densest of the listed 3d metals
Zn	126	137	7.14	Lowest atomisation enthalpy: soft, low-melting Zn, Cd and Hg have filled d subshells; they are the soft end of each series.

The atomisation enthalpy tracks the number of unpaired d electrons that join the metallic bond.

Common traps

Ions lose 4s before 3d

\mathrm{Fe^{2+}}

is

[\mathrm{Ar}]\,3d^{6}

, not

[\mathrm{Ar}]\,3d^{4}4s^{2}

. The 4s electrons fill first but leave first too. Writing the ion as the atom minus 3d electrons gives the wrong count of unpaired electrons and the wrong magnetic moment.

The 4d series has more exceptions than the 3d

In 3d only Cr and Cu take a single s electron. In 4d, Nb, Mo, Ru, Rh and Ag all do, and Pd has none at all. Do not copy the 3d pattern down the group: Nb is

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

although V above it is

3d^{3}4s^{2}

.

Cr beats Mn on the second IE only

Cr's second IE is higher than Mn's, but its third IE is lower. A statement that 'the second and third IEs of Cr are both higher than those of Mn' is false, because

\mathrm{Mn^{2+}}

is the 3d⁵ ion at the third step.

Cr is not the highest second IE of the whole series

Cr has the highest second IE only up to Fe. Copper's second IE (1958) is higher, because it breaks a filled 3d¹⁰. Read which metals the question lists before answering.

A catalyst weakens bonds and uses 4s electrons too

Two false statements recur: that first-row catalysts use only their 3d electrons, and that adsorption strengthens the reactant bonds. The surface bonds through 3d AND 4s electrons, and adsorption weakens the reactant bonds, which is why the activation energy falls.

The group-7/group-8 order flips in the 5d series

Mn melts below Fe and Tc below Ru, but Re melts above Os. Do not extend the 3d and 4d pattern to the 5d pair.

Oxidation States and Electrode Potentials

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Why Cu²⁺ is the stable copper ion in water

Iodometry of copper(II)

\mathrm{2Cu^{2+} + 4I^{-} \rightarrow Cu_2I_2 + I_2} \qquad \mathrm{I_2 + 2S_2O_3^{2-} \rightarrow 2I^{-} + S_4O_6^{2-}}

Oxidation states of the 3d metals

Metal	Oxidation states	Most stable in water	Highest fluoride and oxide
Sc	+3	+3	$\mathrm{ScF_3}$ , $\mathrm{Sc_2O_3}$ The only 3d metal with a single oxidation state besides 0.
Ti	+2, +3, +4	+4	$\mathrm{TiF_4}$ , $\mathrm{TiO_2}$
V	+2, +3, +4, +5	+4 (as $\mathrm{VO^{2+}}$ ) and +5	$\mathrm{VF_5}$ , $\mathrm{V_2O_5}$
Cr	+2, +3, +4, +5, +6	+3	$\mathrm{CrF_6}$ , $\mathrm{CrO_3}$
Mn	+2, +3, +4, +5, +6, +7	+2	$\mathrm{MnF_4}$ , $\mathrm{Mn_2O_7}$ Highest oxide (+7) and highest fluoride (+4) differ by 3.
Fe	+2, +3 (+4 and +6 rare)	+3 in air, +2 without it	$\mathrm{FeF_3}$ , $\mathrm{Fe_2O_3}$
Co	+2, +3, +4	+2	$\mathrm{CoF_3}$ , $\mathrm{Co_3O_4}$
Ni	+2, +3, +4	+2	$\mathrm{NiF_2}$ , NiO
Cu	+1, +2	+2	$\mathrm{CuF_2}$ , CuO
Zn	+2	+2	$\mathrm{ZnF_2}$ , ZnO

The number of states peaks at Mn; the ends of the series (Sc, Zn) show one.

E° values: which ions reduce acid and which oxidise

Metal	E° of M²⁺/M (V)	E° of M³⁺/M²⁺ (V)	What it means
Ti	−1.63	−0.37	Ti²⁺ is a reductant and liberates hydrogen
V	−1.18	−0.26	V²⁺ is a reductant and liberates hydrogen
Cr	−0.90	−0.41	Cr²⁺ is a strong reductant: it becomes Cr³⁺, d³
Mn	−1.18	+1.57	Mn³⁺ is a strong oxidant: it becomes Mn²⁺, d⁵
Fe	−0.44	+0.77	Fe³⁺ is a mild oxidant; lower than Mn because Fe³⁺ is d⁵
Co	−0.28	+1.97	Co³⁺ is the strongest oxidant of the series in water
Ni	−0.25	No simple Ni³⁺ in water	Ni²⁺ is the stable ion
Cu	+0.34	No Cu³⁺ in water	The only positive M²⁺/M value: Cu gives no hydrogen with dilute acid Cu has the highest M²⁺/M value of the 3d series.
Zn	−0.76	No Zn³⁺ in water	Zn²⁺ (d¹⁰) is the only ion

Negative M³⁺/M²⁺: the 2+ ion reduces acid. Large positive M³⁺/M²⁺: the 3+ ion is a strong oxidant.

Common traps

The d-block trend runs the other way from the p-block

In group 14 the lower state gets more stable down the group. In a d-block group the HIGHER state does. So 'Cr(VI) is more stable than Mo(VI)' is false, and that is exactly why

\mathrm{CrO_3}

is the stronger oxidant.

Scandium has no +4

Sc loses 3d¹4s² to reach the argon core and stops. A statement giving Sc a +4 state, oxidising or not, is false.

Iron's M³⁺/M²⁺ value is not above manganese's

\mathrm{Fe^{3+}}

is already 3d⁵, so it gains little by taking an electron: +0.77 V.

\mathrm{Mn^{3+}}

reaches 3d⁵ by taking one: +1.57 V. A statement that iron's value is greater is false.

Count the free ion unless a complex is named

When a question pairs an E° clue with a magnetic moment 'in the gaseous state' or with no ligand named, count the free ion.

\mathrm{Co^{3+}}

is 3d⁶ with 4 unpaired electrons as a free ion, although its complex with water is low spin.

The hydration enthalpy of Cu²⁺ is larger, not smaller

Cu²⁺ is stable in water BECAUSE its hydration enthalpy is much more negative than that of Cu⁺. A reason that says it is 'much less' than that of Cu⁺ is false, even when the assertion beside it is true.

Cu₂I₂ and CuI are one compound

\mathrm{Cu_2I_2}

is only CuI written for two copper atoms. When both appear as options, they name the same white precipitate.

Magnetic Moment and Colour

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The spin-only magnetic moment

Spin-only magnetic moment

\mu = \sqrt{n(n+2)}\ \text{BM}

Colours of the aqueous 3d ions

Ion	d configuration	Unpaired electrons (free ion)	Colour in water
$\mathrm{Sc^{3+}}$	3d⁰	0	Colourless
$\mathrm{Ti^{4+}}$	3d⁰	0	Colourless
$\mathrm{Ti^{3+}}$	3d¹	1	Purple
$\mathrm{V^{4+}}$	3d¹	1	Blue
$\mathrm{V^{3+}}$	3d²	2	Green
$\mathrm{V^{2+}}$	3d³	3	Violet
$\mathrm{Cr^{3+}}$	3d³	3	Violet
$\mathrm{Mn^{3+}}$	3d⁴	4	Violet V²⁺, Cr³⁺ and Mn³⁺ are all violet.
$\mathrm{Cr^{2+}}$	3d⁴	4	Blue
$\mathrm{Mn^{2+}}$	3d⁵	5	Pink
$\mathrm{Fe^{3+}}$	3d⁵	5	Yellow
$\mathrm{Fe^{2+}}$	3d⁶	4	Green
$\mathrm{Co^{3+}}$	3d⁶	4	Blue
$\mathrm{Co^{2+}}$	3d⁷	3	Pink
$\mathrm{Ni^{2+}}$	3d⁸	2	Green
$\mathrm{Cu^{2+}}$	3d⁹	1	Blue
$\mathrm{Zn^{2+}}$	3d¹⁰	0	Colourless

Same colour does not mean same d count: Fe²⁺ (d⁶), Ni²⁺ (d⁸) and V³⁺ (d²) are all green.

Common traps

Past d⁵ the electrons pair up

\mathrm{Cu^{2+}}

is d⁹ with ONE unpaired electron, not nine, and

\mathrm{Ni^{2+}}

is d⁸ with two. Use

10 - x

for d⁶ to d¹⁰.

Take the 4s electrons off first

\mathrm{Mn^{2+}}

is 3d⁵ with five unpaired electrons (5.92 BM). Writing it as 3d³4s² gives three and a wrong answer of 3.87 BM, which is usually one of the options.

Intensely coloured is not paramagnetic

Permanganate is deep purple but Mn(VII) is d⁰, so

\mathrm{MnO_4^{-}}

is diamagnetic. Its colour comes from charge transfer, not from d electrons. The same holds for dichromate and chromate.

Copper is colourless as Cu⁺

\mathrm{Cu^{2+}}

(d⁹) is blue, but

\mathrm{Cu^{+}}

(d¹⁰) is colourless. Always count the d electrons of the ION, not of the element.

Oxides of Transition Metals

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Basic, amphoteric and acidic oxides

Oxide	Metal oxidation state	Character	With acid or alkali
$\mathrm{V_2O_3}$	+3	Basic	Dissolves in acid to give $\mathrm{V^{3+}}$ salts
$\mathrm{V_2O_4}$	+4	Less basic (weakly amphoteric)	Dissolves in acid to give $\mathrm{VO^{2+}}$ salts
$\mathrm{V_2O_5}$	+5	Amphoteric, mainly acidic	$\mathrm{VO_4^{3-}}$ in alkali, $\mathrm{VO_2^{+}}$ in acid The contact-process catalyst, but not a basic oxide.
CrO	+2	Basic	Dissolves in acid to give $\mathrm{Cr^{2+}}$
$\mathrm{Cr_2O_3}$	+3	Amphoteric	Reacts with both acid and alkali
$\mathrm{CrO_3}$	+6	Acidic	With water gives chromic acid, $\mathrm{H_2CrO_4}$
MnO	+2	Basic	Dissolves in acid to give $\mathrm{Mn^{2+}}$
$\mathrm{Mn_2O_7}$	+7	Acidic	With water gives permanganic acid, $\mathrm{HMnO_4}$
ZnO	+2	Amphoteric	Zincate, $\mathrm{[Zn(OH)_4]^{2-}}$ , in excess alkali

Down each metal's column of oxides, the character moves from basic to acidic as the oxidation state rises.

Structure of Mn₂O₇ and the mixed oxides

Oxide	Metal oxidation state	Structure or make-up	Point tested
$\mathrm{Mn_2O_7}$	+7	Two $\mathrm{MnO_4}$ tetrahedra sharing one O	6 terminal Mn=O, 1 bridging O, covalent green oil Mn is tetrahedral, not octahedral, and there is no Mn–Mn bond.
$\mathrm{CrO_3}$	+6	Chains of $\mathrm{CrO_4}$ tetrahedra sharing corners	Acidic, strong oxidant
$\mathrm{Mn_3O_4}$	+2 and +3	MnO· $\mathrm{Mn_2O_3}$	Mixed oxide; paramagnetic
$\mathrm{Fe_3O_4}$	+2 and +3	FeO· $\mathrm{Fe_2O_3}$	Mixed oxide; magnetite, strongly magnetic
$\mathrm{Co_3O_4}$	+2 and +3	CoO· $\mathrm{Co_2O_3}$	Mixed oxide
$\mathrm{Fe_2O_3}$	+3	One oxidation state	Not a mixed oxide

A formula M₃O₄ with an average state of +8/3 hides a +2 and a +3 metal.

Common traps

V₂O₄ with acid gives VO²⁺

Vanadium(IV) in acid is the vanadyl ion,

\mathrm{VO^{2+}}

. The ion

\mathrm{VO_2^{+}}

is vanadium(V), formed from

\mathrm{V_2O_5}

. Match the oxidation state before choosing.

Ionic character falls as the oxidation state rises

A higher oxidation state gives a MORE covalent oxide, which is why

\mathrm{Mn_2O_7}

is a liquid and not an ionic solid. A statement that ionic character increases with oxidation number is false.

Mn₂O₇ is covalent, not ionic

It is a molecular liquid. A statement calling

\mathrm{Mn_2O_7}

an ionic oxide is false; so is one that puts Mn in an octahedron.

An M₂O₃ or M₃O₄ formula alone does not decide 'mixed'

Check for two oxidation states.

\mathrm{Fe_3O_4}

is mixed (+2 and +3);

\mathrm{Fe_2O_3}

is all +3 and

\mathrm{V_2O_4}

all +4.

Potassium Dichromate and Chromium Compounds

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From chromite ore to K₂Cr₂O₇, and chromate against dichromate

The chromate–dichromate equilibrium

\mathrm{2CrO_4^{2-} + 2H^{+} \rightleftharpoons Cr_2O_7^{2-} + H_2O} \qquad \text{Cr stays } +6

Acidified dichromate as an oxidising agent

Electron balance with dichromate

6 \times n\left(\mathrm{Cr_2O_7^{2-}}\right) = (\text{electrons lost per reductant}) \times n(\text{reductant})

The chromyl chloride test and blue CrO₅

Oxidation state of Cr in CrO₅

x + 1(-2) + 4(-1) = 0 \;\Rightarrow\; x = +6

Common traps

Chromate to dichromate is not a redox change

Colour changes from yellow to orange, but Cr is +6 on both sides. A question on 'the change in oxidation state' of Cr between chromate and dichromate has the answer 0.

Only the potassium salt is a primary standard

Sodium dichromate is hygroscopic, so a weighed sample is not pure. Potassium dichromate is less soluble, crystallises pure and is the primary standard.

Electrons per what?

One dichromate takes 6 electrons. But forming one

\mathrm{I_2}

from iodide involves 2 electrons, and forming one S from sulphide also 2. Read whether a question counts per dichromate, per product molecule or per balanced equation before you add.

The green paper is not proof of SO₂ alone

Acidified dichromate paper turning green is the standard test for

\mathrm{SO_2}

, but

\mathrm{H_2S}

also reduces dichromate and turns it green. The key follows the named test; the chemistry allows both.

CrO₅ is +6, not +10

Four of its five oxygens are in peroxo groups at −1. Treating all five as −2 gives an impossible +10.

The formula is CrO₂Cl₂

One chromium, two oxygens, two chlorines. Options such as

\mathrm{Cr_2O_2Cl_2}

or a +5 or +3 state are the distractors; there is no redox for chromium in this step.

Potassium Permanganate and Manganese Compounds

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Making KMnO₄: manganate, permanganate and disproportionation

Disproportionation of manganate

\mathrm{3\overset{+6}{Mn}O_4^{2-} + 4H^{+} \rightarrow 2\overset{+7}{Mn}O_4^{-} + \overset{+4}{Mn}O_2 + 2H_2O}

Permanganate as an oxidant: acid against neutral

Two media, two products

\text{acid: } \mathrm{Mn^{+7} \rightarrow Mn^{2+}}\ (5e^{-}) \qquad \text{neutral/faintly alkaline: } \mathrm{Mn^{+7} \rightarrow MnO_2}\ (3e^{-})

Common traps

Manganate +6, permanganate +7

The names are easy to swap. Manganate,

\mathrm{MnO_4^{2-}}

, is green, +6 and paramagnetic. Permanganate,

\mathrm{MnO_4^{-}}

, is purple, +7 and diamagnetic.

Peroxodisulphate goes all the way to permanganate

Oxidising a Mn(II) salt with peroxodisulphate gives

\mathrm{MnO_4^{-}}

, not manganate. A statement that it stops at manganate is false.

Permanganate oxidises; it never reduces

Acidified permanganate OXIDISES oxalate, nitrite and iodide. A statement that it 'reduces oxalate, nitrite and iodide' is false, however familiar the list looks.

Count the water of crystallisation when the question does

If the titration uses ferrous ammonium sulphate HEXAHYDRATE, the 10 formula units bring 60 water molecules of their own. With the 8 formed in the reaction that is 68 per 2

\mathrm{KMnO_4}

.

Lanthanoids and Actinoids

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Lanthanoid configurations and 4f counts

Element (Z)	Atom	M³⁺ ion	Other common ion
La (57)	[Xe]5d¹6s²	4f⁰, colourless	Shows only +3
Ce (58)	[Xe]4f¹5d¹6s²	4f¹	Ce⁴⁺, 4f⁰
Pr (59)	[Xe]4f³6s²	4f²	Pr⁴⁺, 4f¹
Nd (60)	[Xe]4f⁴6s²	4f³	Nd²⁺ 4f⁴; Nd⁴⁺ 4f²
Pm (61)	[Xe]4f⁵6s²	4f⁴	Shows only +3
Sm (62)	[Xe]4f⁶6s²	4f⁵	Sm²⁺, 4f⁶
Eu (63)	[Xe]4f⁷6s²	4f⁶	Eu²⁺, 4f⁷ Eu²⁺ and Gd³⁺ are the two 4f⁷ ions.
Gd (64)	[Xe]4f⁷5d¹6s²	4f⁷	Shows only +3
Tb (65)	[Xe]4f⁹6s²	4f⁸	Tb⁴⁺, 4f⁷
Dy (66)	[Xe]4f¹⁰6s²	4f⁹	Dy⁴⁺, 4f⁸
Ho (67)	[Xe]4f¹¹6s²	4f¹⁰	Shows only +3
Er (68)	[Xe]4f¹²6s²	4f¹¹	Shows only +3
Tm (69)	[Xe]4f¹³6s²	4f¹²	Tm²⁺, 4f¹³
Yb (70)	[Xe]4f¹⁴6s²	4f¹³	Yb²⁺, 4f¹⁴
Lu (71)	[Xe]4f¹⁴5d¹6s²	4f¹⁴, colourless	Shows only +3

The +3 ion always has Z − 57 electrons in 4f; the ions in the last column reach 4f⁰, 4f⁷ or 4f¹⁴, or come close.

Lanthanoid ions outside the +3 state

Ion	4f configuration	Why it exists	Behaviour
Ce⁴⁺	4f⁰	Noble-gas (Xe) core	Strong oxidant; E° = +1.74 V back to Ce³⁺ The noble-gas core favours forming Ce⁴⁺, but Ce³⁺ is still the more stable state in water.
Tb⁴⁺	4f⁷	Half-filled 4f	Stronger oxidant than Ce⁴⁺; found in $\mathrm{TbO_2}$
Pr⁴⁺, Nd⁴⁺, Dy⁴⁺	4f¹, 4f², 4f⁸	Stabilised only in the solid oxide	Found only as $\mathrm{MO_2}$ ; oxidants
Eu²⁺	4f⁷	Half-filled 4f after losing 6s²	Strong reductant; turns into Eu³⁺
Yb²⁺	4f¹⁴	Full 4f after losing 6s²	Reductant; diamagnetic
Sm²⁺	4f⁶	Close to 4f⁷	Reductant
Ln³⁺ (all)	4f¹ to 4f¹⁴	Loss of 6s² and one more electron	The stable state of every lanthanoid

+4 ions are oxidants and +2 ions are reductants, because each tends to return to +3.

Actinoids compared with lanthanoids

Property	Lanthanoids	Actinoids
Subshell being filled	4f, deeply buried	5f, less buried, reaches further out
f electrons in bonding	Very little	To a far greater extent
Oxidation states	Mostly +3; a few +2 and +4	+3 common; up to +7 (Np) in the first half
Contraction along the series	Lanthanoid contraction	Actinoid contraction: larger from element to element
Radioactivity	Only Pm	All of them
Example configuration	Gd [Xe]4f⁷5d¹6s²	Cm [Rn]5f⁷6d¹7s²

Almost every actinoid difference traces back to 5f orbitals being less buried than 4f.

Common traps

The 5d electron in Gd and Lu does not change the ion

Gd is 4f⁷5d¹6s², but Gd³⁺ is still 4f⁷: the three electrons lost are 6s², 5d¹. Use Z − 57 for every +3 ion and the atom's quirks drop out.

Isoelectronic means the same total, Z minus charge

Compare Z − charge, not the charge.

\mathrm{Tb^{2+}}

has 63 electrons and

\mathrm{Tm^{4+}}

has 65, so they are not isoelectronic even though both look like 'lanthanoid ions near 4f⁷'.

A noble-gas core does not make Ce⁴⁺ the stable state

The Xe core favours FORMING Ce⁴⁺, but in water Ce⁴⁺ returns to Ce³⁺, which is why it is an oxidant. 'Ce is more stable as Ce⁴⁺ than as Ce³⁺' is false.

4f⁷ does not stop Eu²⁺ reducing

Eu²⁺ has a half-filled 4f⁷, yet it is a strong reductant. The +3 state is still preferred, so the special configuration does not protect it.

Name the right contraction

The shrinking of

\mathrm{M^{3+}}

across Th to Lr is the ACTINOID contraction. A reason that calls it the lanthanoid contraction is wrong for actinoid ions such as

\mathrm{Bk^{3+}}

and

\mathrm{Np^{3+}}

.

Cm has eight unpaired electrons

Am and Cm both have 5f⁷, but curium adds a 6d electron. 'Cm and Am have seven unpaired electrons' is false.

Qualitative Analysis of Ions

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Cation groups and their group reagents

Group	Cations	Group reagent	Precipitated as
Zero	$\mathrm{NH_4^{+}}$	No group reagent; heat with NaOH	Ammonia gas, confirmed with Nessler's reagent
I	$\mathrm{Pb^{2+}}$	Dilute HCl	White $\mathrm{PbCl_2}$
II	$\mathrm{Pb^{2+}}$ , $\mathrm{Cu^{2+}}$ , $\mathrm{Cd^{2+}}$ , $\mathrm{As^{3+}}$	$\mathrm{H_2S}$ in dilute HCl	Sulphides: PbS and CuS black, CdS and $\mathrm{As_2S_3}$ yellow Pb²⁺ shows up in group I and again in group II, because PbCl₂ is partly soluble.
III	$\mathrm{Fe^{3+}}$ , $\mathrm{Al^{3+}}$ , $\mathrm{Cr^{3+}}$	$\mathrm{NH_4OH}$ with $\mathrm{NH_4Cl}$	Hydroxides: $\mathrm{Fe(OH)_3}$ reddish-brown, $\mathrm{Al(OH)_3}$ white, $\mathrm{Cr(OH)_3}$ green
IV	$\mathrm{Zn^{2+}}$ , $\mathrm{Mn^{2+}}$ , $\mathrm{Co^{2+}}$ , $\mathrm{Ni^{2+}}$	$\mathrm{H_2S}$ in $\mathrm{NH_4OH}$	Sulphides: ZnS white, MnS buff, CoS and NiS black
V	$\mathrm{Ba^{2+}}$ , $\mathrm{Sr^{2+}}$ , $\mathrm{Ca^{2+}}$	$\mathrm{(NH_4)_2CO_3}$ in $\mathrm{NH_4OH}$	White carbonates
VI	$\mathrm{Mg^{2+}}$	No group reagent; ammonium phosphate	White $\mathrm{MgNH_4PO_4}$

Acidic H₂S catches only group II; alkaline H₂S catches group IV as well, which is why group II must be removed first.

Confirmatory tests and the colours they give

Ion	Reagent	Observation	Product
$\mathrm{Cu^{2+}}$	$\mathrm{K_4[Fe(CN)_6]}$ in acetic acid	Chocolate-brown precipitate	$\mathrm{Cu_2[Fe(CN)_6]}$
$\mathrm{Fe^{3+}}$	$\mathrm{K_4[Fe(CN)_6]}$	Prussian blue precipitate	$\mathrm{Fe_4[Fe(CN)_6]_3}$
$\mathrm{Fe^{3+}}$	KSCN	Blood-red colour	$\mathrm{[Fe(SCN)]^{2+}}$
$\mathrm{Zn^{2+}}$	$\mathrm{K_4[Fe(CN)_6]}$ , after neutralising	White or bluish-white precipitate	$\mathrm{K_2Zn_3[Fe(CN)_6]_2}$
$\mathrm{Ni^{2+}}$	Dimethylglyoxime in $\mathrm{NH_4OH}$	Brilliant red precipitate	$\mathrm{[Ni(dmg)_2]}$ , five-membered chelate rings
$\mathrm{Co^{2+}}$	$\mathrm{KNO_2}$ in acetic acid	Yellow precipitate	$\mathrm{K_3[Co(NO_2)_6]}$
$\mathrm{Mn^{2+}}$	NaOH, then left in air	White precipitate turning brown	$\mathrm{MnO(OH)_2}$
$\mathrm{Mg^{2+}}$	Ammonium phosphate in $\mathrm{NH_4OH}$	White crystalline precipitate	$\mathrm{MgNH_4PO_4}$
$\mathrm{NH_4^{+}}$	Nessler's reagent, $\mathrm{K_2[HgI_4]}$ in KOH	Brown precipitate	Iodide of Millon's base
$\mathrm{PO_4^{3-}}$	Ammonium molybdate in $\mathrm{HNO_3}$	Canary-yellow precipitate	$\mathrm{(NH_4)_3PO_4 \cdot 12MoO_3}$
$\mathrm{S^{2-}}$	Sodium nitroprusside	Violet colour	$\mathrm{Na_4[Fe(CN)_5NOS]}$

Ferrocyanide alone confirms three cations: brown for copper, blue for iron(III), white for zinc.

Borax beads, anion tests and Mohr's salt

Test	Conditions	Observation	Reason
Borax bead: Cu	Oxidising flame	Green when hot, blue when cold	Copper metaborate; red and opaque in the reducing flame
Borax bead: Fe	Oxidising and reducing flame	Yellowish-brown hot, yellow cold (oxidising); green (reducing)	Iron(III) metaborate; iron(II) in the reducing flame
Borax bead: Ni	Oxidising flame	Violet when hot, reddish-brown when cold	Nickel metaborate
Borax bead: Mn	Oxidising flame	Violet (amethyst), hot and cold	Manganese metaborate; colourless in the reducing flame
Borax bead: Co	Either flame	Blue, hot and cold	Cobalt metaborate
Borax bead: Cr	Either flame	Green, hot and cold	Chromium metaborate
Brown ring ( $\mathrm{NO_3^{-}}$ )	Fresh $\mathrm{FeSO_4}$ , then conc. $\mathrm{H_2SO_4}$ down the side	Brown ring where the layers meet	$\mathrm{[Fe(H_2O)_5(NO)]^{2+}}$ , Fe +1 The complex is nitrosoferrous sulphate.
Acetate ( $\mathrm{CH_3COO^{-}}$ )	Neutral $\mathrm{FeCl_3}$ , then boil	Deep red colour, then a brown-red precipitate	Basic ferric acetate, Fe +3
Chloride ( $\mathrm{Cl^{-}}$ )	$\mathrm{AgNO_3}$ in dilute $\mathrm{HNO_3}$ , then $\mathrm{NH_4OH}$	Curdy white precipitate that dissolves	AgCl, then $\mathrm{[Ag(NH_3)_2]Cl}$
Mohr's salt preparation	Dilute $\mathrm{H_2SO_4}$ added; no prolonged heating	Pale green crystals	Acid stops hydrolysis; heating would oxidise $\mathrm{Fe^{2+}}$

The bead colour depends on the metal AND on the part of the flame used.

Common traps

Mn²⁺ is group IV, Fe³⁺ is group III

Both are d⁵ ions, but

\mathrm{Fe(OH)_3}

precipitates in group III while

\mathrm{Mn^{2+}}

waits for alkaline

\mathrm{H_2S}

as MnS. Do not group transition metal ions by their configuration.

Acid decides which sulphides come down

In dilute HCl the sulphide ion concentration is tiny, so only group II sulphides precipitate. Adding a group IV cation to that step catches nothing; it needs the ammonia step.

The dimethylglyoxime rings are five-membered

Each glyoxime binds nickel through two nitrogens, closing a five-membered Ni–N–C–C–N ring. A statement calling it a six-membered chelate is false.

Nessler's reagent has no nitrogen

The reagent is

\mathrm{K_2[HgI_4]}

in KOH: potassium, mercury, iodine, and oxygen and hydrogen from the alkali. It detects nitrogen as

\mathrm{NH_4^{+}}

; it does not contain any.

Iron in the brown ring is +1

The ring is

\mathrm{[Fe(H_2O)_5(NO)]^{2+}}

. NO bonds as

\mathrm{NO^{+}}

, so

x + 1 = +2

and

x = +1

. Treating NO as neutral gives +2, which is the common wrong option.

The acetate test needs NEUTRAL ferric chloride

Acid destroys the red iron acetate complex. The red colour, and the brown-red precipitate on boiling, appear only with neutral

\mathrm{FeCl_3}

.

The d- and f-Block Elements formulas

Electronic Configuration and General Properties

Configurations of d-block atoms and ions

Ionisation enthalpies across the 3d series

Melting points, atomisation, density, catalysts and interstitial compounds

Common traps

Oxidation States and Electrode Potentials

Why Cu²⁺ is the stable copper ion in water

Oxidation states of the 3d metals

E° values: which ions reduce acid and which oxidise

Common traps

Magnetic Moment and Colour

The spin-only magnetic moment

Colours of the aqueous 3d ions

Common traps

Oxides of Transition Metals

Basic, amphoteric and acidic oxides

Structure of Mn₂O₇ and the mixed oxides

Common traps

Potassium Dichromate and Chromium Compounds

From chromite ore to K₂Cr₂O₇, and chromate against dichromate

Acidified dichromate as an oxidising agent

The chromyl chloride test and blue CrO₅

Common traps

Potassium Permanganate and Manganese Compounds

Making KMnO₄: manganate, permanganate and disproportionation

Permanganate as an oxidant: acid against neutral

Common traps

Lanthanoids and Actinoids

Lanthanoid configurations and 4f counts

Lanthanoid ions outside the +3 state

Actinoids compared with lanthanoids

Common traps

Qualitative Analysis of Ions

Cation groups and their group reagents

Confirmatory tests and the colours they give

Borax beads, anion tests and Mohr's salt

Common traps

More JEE Mains Chemistry formula sheets