Exercises — Coordination number and geometry — 2 (linear), 4 (tetrahedral - square planar), 6 (octahedral)

Step 1 — count donors. Two ${\text{NH}}_3$ molecules, each donating through one nitrogen atom → $2$ donor atoms → ==CN = 2==. Step 2 — name shape. Two electron domains push to opposite poles of the metal → ==linear, $180^{\circ }$==. Answer: CN $=2$, linear.

Step 1. Six ${\text{H}}_2\text{O}$, each donating through one O atom → $6$ donors → CN = 6. Step 2. Six domains at $90^{\circ }$ to their neighbours → octahedral. Answer: CN $=6$, octahedral.

Four ${\text{Cl}}^{-}$, each one hand → $4$ donors → CN = 4. ${\text{Cl}}^{-}$ is bulky and weak-field → the default 4-shape is tetrahedral ($109.5^{\circ }$). Answer: CN $=4$, tetrahedral.

Step 1 — denticity of en. Ethylenediamine grips with 2 nitrogen atoms → bidentate, $d=2$. Step 2 — apply the formula. $n_{\text{en}}=3$, $d_{\text{en}}=2$: $\text{CN}=3\times 2=6.$ Step 3 — geometry. CN $=6$ → octahedral (three en ligands wrap around the six sites). Answer: CN $=6$, octahedral.

Step 1. EDTA donates through 2 amine N + 4 carboxylate O = $6$ donor atoms → $d=6$ (hexadentate). Step 2. $\text{CN}=1\times 6=6$. Step 3. CN $=6$ → octahedral, with a single EDTA molecule wrapping all six positions. Answer: CN $=6$, octahedral.

Mixed denticity — use the sum.

• oxalate: $n=1,d=2\Rightarrow 1\times 2=2$
• water: $n=2,d=1\Rightarrow 2\times 1=2$ $\text{CN}=2+2=4.$ Answer: CN $=4$.

Step 1 — d-count (same metal). $\text{Ni}$ is group 10; ${\text{Ni}}^{2+}$ has lost 2 electrons → $3d^8$ in both complexes. Step 2 — field strength. ${\text{Cl}}^{-}$ = weak field; ${\text{CN}}^{-}$ = strong field. Step 3 — apply the rule.

• $[{\text{NiCl}}_4{]}^{2-}$: $d^8$ + weak field → tetrahedral. Electrons stay spread out with 2 unpaired → paramagnetic.
• $[{\text{Ni(CN)}}_4{]}^{2-}$: $d^8$ + strong field → square planar. All 8 electrons pair into the lower four orbitals → 0 unpaired → diamagnetic. Answer: ${\text{Cl}}^{-}$ complex = tetrahedral, paramagnetic; ${\text{CN}}^{-}$ complex = square planar, diamagnetic.

Step 1 — d-count. ${\text{Zn}}^{2+}$ = $3d^{10}$ (full shell). Step 2 — can it be square planar? Square planar needs an empty high $d_{x^2-y^2}$; but $d^{10}$ has every d-orbital full, so nothing can be left empty — no square-planar stabilisation exists. Step 3 — default. A full, spherical $d^{10}$ shell has no orbital preference → pure VSEPR wins → tetrahedral. Answer: tetrahedral (and diamagnetic — all electrons already paired).

Step 1 — d-count. ${\text{Pt}}^{2+}$ = $5d^8$. Step 2 — why the metal decides here. $\text{Pt}$ is a third-row (5d) transition metal. Its 5d orbitals are large and diffuse, so the field splitting is intrinsically enormous — even a "weak" ligand like ${\text{Cl}}^{-}$ produces a splitting big enough to force square-planar pairing. Step 3 — conclusion. For 4d and 5d $d^8$ ions (Pd(II), Pt(II), Au(III)), square planar is essentially always chosen, regardless of ligand. The "weak → tetrahedral" shortcut is a first-row ($3d$) rule. Answer: square planar; the large 5d splitting overrides ligand-field weakness.

Step 1 — CN. $3\times 2=6$ (en bidentate). Step 2 — d-count. $\text{Co}$ is group 9; ${\text{Co}}^{3+}$ lost 3 electrons → $3d^6$. Step 3 — geometry. CN $=6$ → octahedral. Step 4 — magnetism. en is strong field → low-spin $d^6$: all six electrons pair into the lower three ($t_{2g}$) orbitals → 0 unpaired → diamagnetic. Answer: CN $=6$, octahedral, $d^6$ low-spin, diamagnetic.

Step 1 — CN. en: $2\times 2=4$; ${\text{Cl}}^{-}$: $2\times 1=2$; total $\text{CN}=6$. Step 2 — overall charge. ${\text{Co}}^{3+}$ contributes $+3$; two ${\text{Cl}}^{-}$ contribute $-2$; en is neutral. Net $=+3-2=+1$. Step 3 — geometry. CN $=6$ → octahedral; formula $[{\text{CoCl}}_2(\text{en}{)}_2{]}^{+}$. (It even shows cis–trans isomerism.) Answer: $[{\text{CoCl}}_2(\text{en}{)}_2{]}^{+}$, CN $=6$, octahedral, overall charge $+1$.

Step 1 — oxidation state. CO is neutral and the complex is neutral, so nickel is ${\text{Ni}}^0$ (oxidation state $0$), not ${\text{Ni}}^{2+}$. Step 2 — d-count. ${\text{Ni}}^0$ has configuration $3d^{10}$ (in these complexes all 10 valence electrons act as d-electrons), i.e. a full d-shell — not $d^8$. Step 3 — apply the rule correctly. The square-planar preference needs $d^8$ (one empty high orbital). A $d^{10}$ ion has no empty d-orbital to exploit → falls back to pure VSEPR → tetrahedral. Answer: it is $d^{10}$, not $d^8$, so the square-planar rule never triggers; tetrahedral is correct.

Permanganate $[{\text{MnO}}_4{]}^{-}$. Charge on Mn: $x+4(-2)=-1\Rightarrow x=+7$. ${\text{Mn}}^{7+}$ = $3d^0$ (no d-electrons at all). With $d^0$ there is no orbital to stabilise by square-planar splitting → tetrahedral. $[{\text{Ni(CN)}}_4{]}^{2-}$. ${\text{Ni}}^{2+}$ = $3d^8$ + strong-field ${\text{CN}}^{-}$ → square planar (from L3.1). Answer: $[{\text{MnO}}_4{]}^{-}$ = tetrahedral ($d^0$); $[{\text{Ni(CN)}}_4{]}^{2-}$ = square planar ($d^8$ strong field).

Step 1 — oxidation state & d-count. Charge: $x+2(-1)=-1\Rightarrow x=+1$; ${\text{Cu}}^{+}$ = $3d^{10}$. Step 2 — why only two? A $d^{10}$ shell is completely full and spherical — it gains no crystal-field stabilisation from adding more ligands (there's no empty d-orbital to lower). So the ion just minimises ligand–ligand repulsion by holding the fewest ligands (2), placed as far apart as possible. Step 3 — geometry. Two domains at opposite poles → linear, $180^{\circ }$. Answer: linear; the $d^{10}$ Cu(I) has no field incentive for higher CN, so it settles at CN 2.

• CN from $\sum n_i{d}_i$.
• Oxidation state from charge balance → d-count $=$ group $-$ oxidation state.
• CN 2 → linear ($d^{10}$). CN 6 → octahedral.
• CN 4 → square planar only if $d^8$ AND (strong-field OR 4d/5d metal); otherwise tetrahedral.