Easy Practice — Set 1

Quick Recap — Transition & Inner-transition Elements

• Transition (d-block) elements have partially filled $d$-orbitals; general config $(n-1)d^{1-10}\,ns^{1-2}$; groups 3-12; first series Sc-Zn.
• Characteristics: variable oxidation states, coloured ions (d-d transitions), paramagnetism ($\mu=\sqrt{n(n+2)}$ BM), catalytic activity, alloy and interstitial-compound formation, high melting points.
• Lanthanoids (period 6, 4f) and actinoids (period 7, 5f) are inner-transition; common state $+3$; lanthanoid contraction makes Zr $\approx$ Hf.
• Key oxidisers: $KMnO_4$ (Mn $+7$, purple), $K_2Cr_2O_7$ (Cr $+6$, orange).

Beyond-NCERT JEE Formulae

Use this as a JEE formula sheet for the d- and f-block. The chapter is mostly descriptive, so the ONE genuine calculation is the spin-only magnetic moment; everything else is a systematic trend. Each entry says WHEN to use it, and a [JEE Tip] flags the classic exam trap.

1. Spin-only Magnetic Moment

• Master relation: $\mu=\sqrt{n(n+2)}$ BM, where $n$ is the number of UNPAIRED electrons. Use it whenever a question gives an ion (or its $d^x$ count) and wants the magnetic moment, or gives a measured moment and wants $n$.
• Ready-reckoner: $n=1\to1.73$, $n=2\to2.83$, $n=3\to3.87$, $n=4\to4.90$, $n=5\to5.92$ BM.
• Back-calculation: square the moment, then solve $n(n+2)=\mu^2$. A measured $4.90$ gives $n(n+2)=24$, so $n=4$; a measured $3.87$ gives $15$, so $n=3$.
• [JEE Tip] It is a SQUARE ROOT: for $d^5$, $\mu=\sqrt{35}=5.92$, never $35$. And count $n$ from the ION, not the atom (strip the $4s$ electrons first, see below).

2. Unpaired Electrons from the Ion Configuration

• Golden rule of ionisation: the $ns$ electrons leave BEFORE the $(n-1)d$. Build the ion by writing the neutral atom, removing both $4s$ electrons first, then taking any extra charge from $3d$. Example: Fe is $[Ar]3d^6 4s^2$, so Fe2+ is $3d^6$ and Fe3+ is $3d^5$.
• Unpaired count vs $d^x$ (high-spin): $d^0$ 0, $d^1$ 1, $d^2$ 2, $d^3$ 3, $d^4$ 4, $d^5$ 5, $d^6$ 4, $d^7$ 3, $d^8$ 2, $d^9$ 1, $d^{10}$ 0. The count climbs to a $d^5$ peak, then falls symmetrically.
• Config exceptions (atoms): Cr is $[Ar]3d^5 4s^1$ and Cu is $[Ar]3d^{10} 4s^1$ (extra stability of half- and fully-filled $d$). So a neutral Cr atom has 6 unpaired electrons, and Cu2+ is $3d^9$ (1 unpaired), never $3d^{10}$.
• [JEE Tip] The commonest slip is using the atom's $3d^6 4s^2$ for Fe2+; the ion is $3d^6$ (4 unpaired). A transition-metal ION never keeps its $4s$ electrons.

3. Systematic Trends (with reasoning)

• Variable oxidation states: $(n-1)d$ and $ns$ lie close in energy, so d-electrons count as valence electrons too; successive states then differ by one, e.g. Mn spans $+2$ to $+7$.
• Highest OS peaks at Mn: the maximum oxidation state equals the total $3d+4s$ electrons usable in bonding. It climbs from Sc $(+3)$ up to Mn $(+7)$, then FALLS (Fe tops out at $+6$) because the rising nuclear charge grips the d-electrons. So Mn shows the series' highest OS, $+7$ in KMnO4.
• Lanthanoid contraction: the steady size fall across Ce to Lu, because the diffuse $4f$ electrons shield poorly and the effective nuclear charge rises. Consequences: (i) Zr and Hf have almost equal radii, (ii) the basic strength of Ln(OH)3 falls from La to Lu, (iii) the near-equal sizes make the lanthanoids hard to separate.
• Colour needs a d-d transition: a partly filled $d$-subshell is essential, so $d^0$ (Sc3+, Ti4+) and $d^{10}$ (Zn2+, Cu+) ions are COLOURLESS, while $d^1$ to $d^9$ ions are coloured. An intense colour on a $d^0$ ion such as MnO4- (purple) comes from ligand-to-metal charge transfer, not a d-d jump.
• [JEE Tip] "Colourless" and "diamagnetic" coincide only for $d^0$ and $d^{10}$; a half-filled $d^5$ ion (Mn2+) is faintly coloured and strongly paramagnetic, so do not call it colourless.

4. Electrode Potentials, Disproportionation and Observed Moments

• $E^\circ(M^{2+}/M)$ trend: generally NEGATIVE across the 3d series (the metals are reducing); copper alone is positive at $+0.34$ V, because its large $\Delta_{sub}H+\Delta_iH$ is not repaid by the hydration enthalpy.
• Irregularities: the extra-stable half-filled $d^5$ (Mn2+) and filled $d^{10}$ (Zn2+) give unusually negative $E^\circ(M^{2+}/M)$; for the $M^{3+}/M^{2+}$ couple, Cr2+ is a strong reductant (it forms the stable $d^3$ ion Cr3+) whereas Co3+ is a strong oxidant.
• Disproportionation: $3MnO_4^{2-}+4H^+\to 2MnO_4^-+MnO_2+2H_2O$ (green manganate to purple permanganate plus brown MnO2); and $2Cu^+\to Cu^{2+}+Cu$ in aqueous solution.
• Spin-only vs observed: the formula ignores orbital motion, so ions with an unquenched orbital contribution (Co2+, Ni2+) show observed moments a little ABOVE the spin-only value; JEE writes "spin-only" when it wants the formula answer.
• [JEE Tip] Only copper has a positive $E^\circ(M^{2+}/M)$ in the 3d series; do not generalise it, since the rest are negative.

Solved Examples — Beyond-NCERT Formulae

Example 1 - Spin-only moment of Fe2+. Find the spin-only magnetic moment of the iron(II) ion.

• Fe is $[Ar]3d^6 4s^2$; strip the $4s^2$ pair first, so Fe2+ is $3d^6$.
• High-spin $d^6$ is $t_{2g}^4 e_g^2$: one orbital is doubly filled and four are singly filled, so $n=4$.
• $\mu=\sqrt{4(4+2)}=\sqrt{24}=4.90$ BM.

Example 2 - Spin-only moment of Mn2+. Find the spin-only moment of the manganese(II) ion.

• Mn is $[Ar]3d^5 4s^2$; remove the $4s^2$ pair, so Mn2+ is $3d^5$.
• Half-filled $d^5$ has every d-orbital singly occupied, so $n=5$, the maximum for the 3d series.
• $\mu=\sqrt{5(5+2)}=\sqrt{35}=5.92$ BM. Fe3+ (also $3d^5$) gives the same 5.92 BM.

Example 3 - Unpaired electrons from a measured moment. A first-series ion has spin-only $\mu=2.83$ BM. Find the number of unpaired electrons and name a possible ion.

• Square the moment: $n(n+2)=\mu^2=(2.83)^2\approx 8$.
• Solve $n^2+2n-8=0$, i.e. $(n+4)(n-2)=0$, giving $n=2$.
• A $d^8$ ion fits, e.g. Ni2+ ($3d^8$). Trap: reading 2.83 as 3 unpaired; $n=3$ would give $3.87$, not 2.83.

Example 4 - Order ions by magnetic moment. Arrange Ti3+, Ni2+, Cr3+ and Mn2+ in increasing spin-only moment.

• Ion configs: Ti3+ is $3d^1$, Cr3+ is $3d^3$, Mn2+ is $3d^5$, Ni2+ is $3d^8$.
• Unpaired electrons: Ti3+ 1, Ni2+ 2, Cr3+ 3, Mn2+ 5.
• More unpaired electrons means a larger $\mu$, so the order is Ti3+ (1.73) < Ni2+ (2.83) < Cr3+ (3.87) < Mn2+ (5.92) BM.

Example 5 - Coloured vs colourless ions. Which of Sc3+, Ti3+, Cu2+, Zn2+ and Cu+ are colourless in dilute aqueous solution?

• Colour needs a partly filled d-subshell for a d-d transition.
• Sc3+ is $3d^0$, Zn2+ is $3d^{10}$ and Cu+ is $3d^{10}$: no d-d transition, so all three are COLOURLESS.
• Ti3+ ($3d^1$, violet) and Cu2+ ($3d^9$, blue) have partly filled d-subshells, so they are COLOURED.
• Colourless count $=3$.

Example 6 - Highest number of unpaired electrons. Among Fe2+, Fe3+, Co2+ and Ni2+, which ion has the most unpaired electrons?

• Configs: Fe2+ is $3d^6$ (4), Fe3+ is $3d^5$ (5), Co2+ is $3d^7$ (3), Ni2+ is $3d^8$ (2).
• The half-filled $d^5$ of Fe3+ wins with $n=5$, so Fe3+ has the most ($\mu=\sqrt{35}=5.92$ BM).
• Note Fe3+ beats Fe2+ despite having FEWER d-electrons, because $d^5$ maximises the unpaired spins.

Example 7 - The copper anomaly. Find the spin-only moment of Cu2+.

• Cu is anomalous: $[Ar]3d^{10} 4s^1$, not $3d^9 4s^2$.
• Remove the $4s^1$ electron (Cu+ is $3d^{10}$), then one $3d$ electron, so Cu2+ is $3d^9$.
• Nine electrons in five d-orbitals leave one unpaired, $n=1$, so $\mu=\sqrt{1(1+2)}=\sqrt{3}=1.73$ BM.

Example 8 - The chromium anomaly (neutral atom). How many unpaired electrons, and what spin-only moment, does a neutral Cr atom have?

• Cr adopts the extra-stable half-filled ground state $[Ar]3d^5 4s^1$, not $3d^4 4s^2$.
• All five $3d$ orbitals are singly filled (5) and the lone $4s$ electron adds one more, so $n=6$.
• $\mu=\sqrt{6(6+2)}=\sqrt{48}=6.93$ BM, higher than any first-series ION because both subshells contribute unpaired electrons.