Q.Draw the geometrical isomers of $[Co(NO_2)_3(NH_3)_3]$. (½+½=1)

[Co(NH3)4Cl2]+ is an octahedral complex of the type [MA4B2], which shows cis-trans geometrical isomerism depending on the relative positions of the two identical Cl ligands.

The complex [Co(NH3)4Cl2]+ has an octahedral geometry with 4 NH3 and 2 Cl- ligands around the central Co(III) ion. For an [MA4B2] type octahedral complex, two arrangements of the two B (Cl) ligands are possible:

• cis-isomer: the two Cl- ligands occupy adjacent positions on the octahedron, with a Cl-Co-Cl bond angle of 90 degrees. (Structure: picture an octahedron with NH3 on four positions and the two Cl ligands on two adjacent corners.) …

Geometrical (cis/trans, fac/mer) isomerism requires ligand positions that are not all equivalent/adjacent; a tetrahedral geometry has no such distinction, so it alone among these never shows geometrical isomerism.

• (a) Square planar complexes (e.g. $[Pt(NH_3)_2Cl_2]$, type $MA_2B_2$) do show cis–trans geometrical isomerism, since two positions can be adjacent (cis, $90^\circ$) or opposite (trans, $180^\circ$).
• (c) Octahedral complexes (types $MA_4B_2$, $MA_3B_3$, etc.) do show both cis–trans and facial–meridional (fac/mer) geometrical isomerism, since some positions are adjacent and some are directly opposite. …

This octahedral $MA_2B_4$ complex can arrange its two identical $NH_3$ ligands either adjacent to each other (cis) or directly opposite each other (trans), giving two geometrical isomers.

Identifying the isomerism

$[Fe(NH_3)_2(CN)_4]^-$ is an octahedral complex of the general type $[MA_2B_4]$, where $A = NH_3$ (2 ligands) and $B = CN^-$ (4 ligands). This type of complex shows cis–trans (geometrical) isomerism depending on the relative positions of the two $A$ ($NH_3$) ligands.

cis-isomer: Picture an octahedron with the six positions labelled $+x,-x,+y,-y,+z,-z$. In the cis isomer, the two $NH_3$ ligands occupy adjacent positions, i.e. at $90^\circ$ to each other (e.g. one $NH_3$ at $+z$ and the other at $+x$), with the four $CN^-$ ligands occupying the remaining four positions ($-z, -x, +y, -y$).

trans-isomer: In the trans isomer, the two $NH_3$ ligands occupy diametrically opposite positions, i.e. at $180^\circ$ to each other (e.g. one at $+z$ and the other at $-z$), with all four $CN^-$ ligands occupying the equatorial plane ($+x, -x, +y, -y$).

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Co(NH3)4Br2Cl is written as [Co(NH3)4Br2]Cl; being an octahedral MA4B2-type complex it shows cis-trans (geometrical) isomerism, and because Cl and Br can swap places inside/outside the coordination sphere it also shows ionization isomerism.

Formula analysis: cobalt is in the +3 oxidation state; 4 NH3 (neutral) and 2 Br- occupy the coordination sphere (charge = 3 - 2 = +1), balanced by one Cl- as the counter ion outside the sphere: [Co(NH3)4Br2]+ Cl-.

Geometrical isomerism: this is an octahedral complex of type MA4B2 (4 identical NH3 and 2 identical Br in the sphere). The two Br ligands can be mutually cis (adjacent, 90 degrees apart) or trans (opposite, 180 degrees apart), giving cis- and trans-tetraamminedibromidocobalt(III) chloride. (Note: MA4B2 does not show optical isomerism, because both cis and trans forms possess a plane of symmetry.)

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$MX_6$ and $MX_5L$ complexes genuinely show no geometrical isomerism, but that is NOT because coordination number 6 in general excludes geometrical isomerism — many other CN-6 complexes (e.g. $MX_4L_2$, $MX_3L_3$) do show it.

[A] For an octahedral complex $MX_6$ (all six ligands identical) there is only one possible spatial arrangement, and for $MX_5L$ (five identical + one different) the single different ligand can occupy any of the six equivalent octahedral positions — again only one distinct structure results. So neither shows geometrical isomerism. [A] is TRUE.

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$[Co(NH_3)_4Cl_2]^+$ has two possible spatial arrangements of the two $Cl^-$ ligands on the octahedron (cis and trans) — geometrical isomerism — giving violet (cis) and green (trans) forms.

The complex $[Co(NH_3)_4Cl_2]^+$ is octahedral with formula type $[MA_4B_2]$ (4 $NH_3$ + 2 $Cl$ around $Co^{3+}$). Two distinct, non-interconvertible spatial arrangements are possible:

• cis-$[Co(NH_3)_4Cl_2]^+$: the two $Cl$ ligands occupy adjacent (90°) positions — this isomer is violet.
• trans-$[Co(NH_3)_4Cl_2]^+$: the two $Cl$ ligands occupy opposite (180°) positions — this isomer is green.

Both isomers have the same molecular formula, the same donor atoms, and the same metal oxidation state ($+3$) — only the spatial arrangement of ligands differs. Because the ligand geometry around the metal differs, the crystal-field splitting and hence the d–d transition energies (and so the colour absorbed/observed) differ between the two forms. This is the defining signature of geometrical (cis–trans) isomerism, not a difference in connectivity or ionisation.

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$[Co(NO_2)_3(NH_3)_3]$ is an octahedral complex of the type $MA_3B_3$, which exhibits two geometrical isomers — facial (fac) and meridional (mer) — depending on how the two sets of three identical ligands are arranged relative to each other.

In an octahedral complex $MA_3B_3$ (here $M=Co^{3+}$, $A = NO_2^-$, $B=NH_3$), simple cis–trans naming does not apply since there are three ligands of each type; instead the isomers are called facial (fac) and meridional (mer):

fac-$[Co(NO_2)_3(NH_3)_3]$: Picture an octahedron with vertices labelled 1–6 (1,2,3 forming the top triangular face; 4,5,6 the bottom face). All three $NO_2^-$ ligands occupy one triangular face (positions 1, 2, 3 — mutually cis, each at $90^\circ$ to the other two), while all three $NH_3$ ligands occupy the opposite triangular face (positions 4, 5, 6). The three like ligands thus form a triangular "face" of the octahedron on each side.

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The complex $[Co(NH_3)_5NO_2]Cl_2$ exhibits linkage isomerism because the $NO_2^-$ ligand can coordinate through either the nitrogen atom ($-NO_2$, nitro) or the oxygen atom ($-ONO$, nitrito), giving two distinct isomers.

Ionisation isomers have identical formulae but produce different ions in solution by interchanging a ligand inside the coordination sphere with the counter-ion; e.g. [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br.

Ionisation isomerism occurs when the counter-ion (the ion outside the coordination sphere) can itself act as a ligand and thus exchange places with a ligand inside the coordination sphere. The two isomers have the same overall formula but ionise to give different ions in solution.

Example:

• [Co(NH3)5Br]SO4 -> [Co(NH3)5Br]2+ + SO4^2- (gives sulphate ion; gives white ppt with BaCl2). …

Interchanging ligands between the two complex ions gives the coordination isomer $[Pt(NH_3)_4][CuCl_4]$.

Concept. Coordination isomerism (a CBSE Class-12 coordination-compounds topic) occurs in salts where both the cation and the anion are complex ions; the ligands can be distributed differently between the two metal centres.

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Geometrical (cis-trans) isomerism requires at least two different pairs of ligands arranged around a square planar or octahedral centre in more than one distinguishable way; only the MA2B2 square planar complex among the options qualifies.

Checking each option:

• [MnBr4]²⁻: tetrahedral geometry (Mn²⁺, d⁵, weak field with 4 identical Br⁻ ligands). Tetrahedral complexes of the type MA4 do not show geometrical isomerism, since all four positions are equivalent.
• [Pt(NH3)3Cl]⁺: square planar, type MA3B (3 identical NH3 + 1 Cl). With only one different ligand, there is only one possible spatial arrangement — no cis-trans isomerism possible. …