Easy Practice — Set 1

Quick Recap — Werner's Theory & the Basics

• Coordination sphere is written in square brackets; ions outside are counter ions.
• Primary valence = oxidation state of the metal; secondary valence = coordination number (CN).
• Ligands donate lone pairs: monodentate (Cl$^-$, NH$_3$, H$_2$O, CN$^-$, CO), bidentate (en, oxalate), hexadentate (EDTA).
• Ambidentate ligands can bind through two different atoms (e.g. $NO_2^-$, $SCN^-$).

Beyond-NCERT JEE Formulae

NCERT states most of coordination chemistry qualitatively; JEE Main turns it numerical. This sheet is the calculation toolkit -- each formula carries a when-to-use cue and a [JEE Tip] flagging the trap that usually costs the mark.

1. Effective Atomic Number (EAN)

$$\text{EAN}=Z-(\text{oxidation state})+2\times(\text{coordination number})$$

Here $Z$ is the atomic number of the metal. Use it to test the Sidgwick EAN rule -- whether a complex (most reliably a metal carbonyl) reaches the electron count of the next noble gas.

[JEE Tip] A stable carbonyl lands its EAN on the nearest noble gas: $[Cr(CO)_6]$, $[Fe(CO)_5]$ and $[Ni(CO)_4]$ each give $36$ (Kr). If your EAN misses a noble-gas number, re-check the oxidation state first.

2. Spin-only magnetic moment

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

$n$ is the number of unpaired electrons. Use it to get $\mu$, or to decide paramagnetic vs diamagnetic, once CFT has fixed the spin state. Handy values: $n=1$ gives $1.73$, $n=2$ gives $2.83$, $n=3$ gives $3.87$, $n=4$ gives $4.90$, $n=5$ gives $5.92$ BM.

Read the ligand off the spectrochemical series (weak to strong field):

$$I^-<Br^-<Cl^-<F^-<OH^-<H_2O<NH_3<en<NO_2^-<CN^-<CO$$

[JEE Tip] Fix the oxidation state, then the $d$-count, then the field. Strong-field ligands ($CN^-$, $CO$, $NH_3$) force pairing (low spin, small $\mu$); weak-field ligands ($F^-$, $H_2O$, halides) keep the maximum unpaired electrons (high spin). A quoted $\mu$ near $0$ is a coded message: diamagnetic, hence low spin / strong field.

3. Crystal Field Stabilisation Energy (CFSE)

$$\text{CFSE}_{oct}=(-0.4\,p+0.6\,q)\,\Delta_o\ (+\,mP)$$

$p$ = electrons in $t_{2g}$, $q$ = electrons in $e_g$, $m$ = number of extra electron pairs formed relative to the free ion, and $P$ = pairing energy. For tetrahedral fields,

$$\Delta_t=\dfrac{4}{9}\,\Delta_o$$

Use it to compare stabilities and to justify a spin state.

[JEE Tip] Because $\Delta_t=\dfrac49\Delta_o$ is small (well below $P$), tetrahedral complexes are almost always high-spin -- never write a low-spin tetrahedral in JEE. Octahedral goes low-spin only when $\Delta_o>P$, the strong-field end of the series.

4. Counting isomers

For the standard octahedral and square-planar skeletons:

• Octahedral $MA_4B_2$: 2 geometrical (cis, trans); neither is optically active.
• Octahedral $MA_3B_3$: 2 geometrical (fac, mer); neither is optically active.
• Octahedral $M(AA)_2B_2$ such as $[Co(en)_2Cl_2]^+$: cis and trans, and the cis is chiral ($d$, $l$) while the trans is not, giving 3 stereoisomers.
• Octahedral $M(AA)_3$ such as $[Co(en)_3]^{3+}$: no cis/trans, just a chiral $d$/$l$ pair, giving 2 optical isomers.
• Square planar $MA_2B_2$ such as $[Pt(NH_3)_2Cl_2]$: cis and trans; no optical isomers.
• Tetrahedral $MABCD$: optically active, with no geometrical isomers.

Structural isomerism to keep on the radar: ionisation, hydrate, linkage and coordination isomerism.

[JEE Tip] Square planar can be cis/trans but, with simple monodentate ligands, is never optically active (it keeps a plane of symmetry); tetrahedral turns optically active only when all four groups differ. It is chelation by a symmetric bidentate ($en$) that makes cis-$M(AA)_2B_2$ and $M(AA)_3$ chiral.

5. Werner's theory and conductivity

Primary valence = oxidation state (ionisable, satisfied by counter ions outside the bracket); secondary valence = coordination number (non-ionisable, directional, fixes the geometry). On dissolving, the coordination sphere stays intact, so the number of ions = 1 (complex ion) + number of counter ions.

[JEE Tip] Molar conductivity climbs with the ion count: $[Co(NH_3)_6]Cl_3$ gives 4 ions, $[Co(NH_3)_5Cl]Cl_2$ gives 3, $[Co(NH_3)_4Cl_2]Cl$ gives 2, and $[Co(NH_3)_3Cl_3]$ gives 0 (a non-electrolyte). Only chlorides OUTSIDE the bracket precipitate with $AgNO_3$, so the same series yields 3, 2, 1 and 0 mol of AgCl.

Solved Examples -- Beyond-NCERT Formulae

Example 1 -- EAN of a carbonyl and a cyanido complex. Find the EAN of the metal in (a) $[Ni(CO)_4]$ and (b) $[Fe(CN)_6]^{4-}$, and comment on stability.

(a) For $Ni$, $Z=28$. CO is neutral, so the oxidation state is $0$ and CN $=4$. $$\text{EAN}=28-0+2\times4=36$$ (b) For $Fe$, $Z=26$. From $x+6(-1)=-4$ the oxidation state is $+2$, and CN $=6$. $$\text{EAN}=26-2+2\times6=36$$ Both reach $36$, the atomic number of krypton, so both obey the EAN (18-electron) rule -- a big part of why they are so stable.

Example 2 -- Magnetic moment and spin state: $[Fe(CN)_6]^{3-}$ vs $[FeF_6]^{3-}$. Both contain $Fe^{3+}$; predict $\mu$ for each.

Oxidation state: $x+6(-1)=-3$ gives $x=+3$, so $Fe^{3+}$ is $3d^5$ in both.

• $[Fe(CN)_6]^{3-}$: $CN^-$ is a strong-field ligand, so $\Delta_o>P$ and the ion is low spin, $t_{2g}^5e_g^0$, with $n=1$. $$\mu=\sqrt{1(1+2)}=\sqrt{3}=1.73\ \text{BM}$$
• $[FeF_6]^{3-}$: $F^-$ is a weak-field ligand, so the ion is high spin, $t_{2g}^3e_g^2$, with $n=5$. $$\mu=\sqrt{5(5+2)}=\sqrt{35}=5.92\ \text{BM}$$ Same metal, same oxidation state -- only the ligand's place in the spectrochemical series flips the answer.

Example 3 -- CFSE of $d^6$: strong field vs weak field. Compare $[Co(NH_3)_6]^{3+}$ and $[CoF_6]^{3-}$, where $Co^{3+}$ is $d^6$.

• Strong field $[Co(NH_3)_6]^{3+}$: low spin $t_{2g}^6e_g^0$. $$\text{CFSE}=(-0.4\times6+0.6\times0)\Delta_o=-2.4\,\Delta_o\ (+\,2P)$$ The three filled $t_{2g}$ pairs are two more than the single pair a free $d^6$ ion already has, hence the $+2P$. With no unpaired electron it is diamagnetic.
• Weak field $[CoF_6]^{3-}$: high spin $t_{2g}^4e_g^2$. $$\text{CFSE}=(-0.4\times4+0.6\times2)\Delta_o=-0.4\,\Delta_o$$ Four unpaired electrons give $\mu=\sqrt{4(4+2)}=\sqrt{24}=4.90\ \text{BM}$. The strong-field ion is far more crystal-field stabilised, which is why $Co^{3+}$ ammines are classic low-spin, inert complexes.

Example 4 -- Same $d$-count, different geometry: $[NiCl_4]^{2-}$ vs $[Ni(CN)_4]^{2-}$. Both are $Ni^{2+}$ ($d^8$); explain the magnetism.

• $[NiCl_4]^{2-}$: $Cl^-$ is weak field. Since $\Delta_t=\dfrac49\Delta_o$ is far too small to pair electrons, the ion is high-spin tetrahedral ($sp^3$) with $n=2$. $$\mu=\sqrt{2(2+2)}=\sqrt{8}=2.83\ \text{BM}$$ so it is paramagnetic.
• $[Ni(CN)_4]^{2-}$: $CN^-$ is strong field, forcing the electrons to pair into a square planar ($dsp^2$) ion with $n=0$, so $\mu=0$ and it is diamagnetic. The pattern JEE tests constantly: weak field with $d^8$ tends to tetrahedral and paramagnetic; strong field with $d^8$ goes square planar and diamagnetic.

Example 5 -- Stereoisomers of $[Co(en)_2Cl_2]^+$. Count all stereoisomers.

This is an octahedral $M(AA)_2B_2$ type. The two chlorides can lie cis or trans, giving 2 geometrical isomers. The trans isomer has a plane of symmetry and is optically inactive. The cis isomer is chiral -- non-superimposable on its mirror image -- so it exists as a $d$/$l$ pair. In total: $$\text{trans}+(\text{cis-}d)+(\text{cis-}l)=3\ \text{stereoisomers}$$

Example 6 -- fac/mer and square-planar isomers. (a) Isomers of octahedral $[Co(NH_3)_3Cl_3]$; (b) isomers of square planar $[Pt(NH_3)_2Cl_2]$.

(a) An $MA_3B_3$ octahedron has 2 geometrical isomers: facial (fac), with the three like ligands on one triangular face, and meridional (mer), with them spread around a meridian. Both keep a plane of symmetry, so neither is optically active. (b) A square planar $MA_2B_2$ has 2 geometrical isomers, cis (this is cisplatin) and trans. Being planar it always retains a mirror plane, so there are no optical isomers.

Example 7 -- Ions from conductivity and precipitation. A cobalt(III) chloride ammine is formulated $[Co(NH_3)_5Cl]Cl_2$. How many ions per formula unit, and how many mol of AgCl does 1 mol give with excess $AgNO_3$?

The coordination sphere stays intact; only the counter ions dissociate: $$[Co(NH_3)_5Cl]Cl_2 \to [Co(NH_3)_5Cl]^{2+}+2Cl^-$$ That is $1+2=3$ ions, so it behaves as a 1:2 electrolyte. Only the two free outer chlorides precipitate, giving 2 mol AgCl; the chloride bonded inside the bracket does not react. For contrast, $[Co(NH_3)_6]Cl_3$ gives 4 ions and 3 mol AgCl, while $[Co(NH_3)_3Cl_3]$ gives 0 ions and 0 mol AgCl.