Why are low spin tetrahedral complexes not formed?

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Q. Why are low spin tetrahedral complexes not formed?

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Concept understanding — Geometrical Isomerism Conditions

Geometrical Isomerism: The Intuition

Imagine you have two friends standing on opposite sides of a door. If the door is open, they can walk around and swap places easily — there's no real difference between who is on the left and who is on the right. But if the door is locked shut, they are stuck. One is permanently on the left side, the other on the right. That locked door creates two distinct arrangements: Friend A on the left, Friend B on the right versus Friend A on the right, Friend B on the left.

That locked door is the key idea behind geometrical isomerism.

In chemistry, molecules are three-dimensional. Atoms connected by a single bond can rotate freely — like an open door. But a double bond (or a ring structure) locks the atoms in place. If you have two different groups attached to each carbon of a double bond, you get two distinct spatial arrangements that cannot interconvert without breaking the bond. These are geometrical isomers (also called cis-trans or E-Z isomers).

The Precise Conditions

For a molecule to show geometrical isomerism, it must satisfy two conditions simultaneously:

C=C

Let's unpack each.

Condition 1: Restricted Rotation

A single bond ( $C-C$ ) allows free rotation — the atoms spin around the bond axis like a wheel. So no geometrical isomers exist there. A double bond ( $C=C$ ) has a pi ( $\pi$ ) bond that locks the molecule flat. Rotation would break the $\pi$ bond, which requires a lot of energy (about 250–270 kJ/mol). At room temperature, this rotation simply does not happen.

Rings (like cyclopropane, cyclobutane, etc.) also restrict rotation because the ring is a closed loop — atoms cannot rotate past each other without breaking the ring.

Condition 2: Two Different Substituents on Each End

This is the "different groups" rule. Look at each carbon of the double bond (or each ring carbon involved). If both carbons have two different groups attached, geometrical isomers exist. If even one carbon has two identical groups, there is only one possible arrangement.

Watch out

A common mistake: students check only one carbon. Both carbons must have two different substituents. If one carbon has two identical groups (like two hydrogens), the molecule is identical in both arrangements — no isomerism.

How to Check: A Step-by-Step Method

Take any molecule with a double bond. Follow these steps:

Identify the double bond (or ring). Mark the two carbon atoms involved.
List the two groups attached to the first carbon. Are they different from each other? If yes, proceed. If no → no geometrical isomerism.
List the two groups attached to the second carbon. Are they different from each other? If yes → geometrical isomerism exists. If no → no geometrical isomerism.

Tip

If the two groups on a carbon are identical, the molecule is symmetric about that carbon. Flipping the other side gives the same molecule — no isomers.

Examples to Cement the Idea

Example 1: But-2-ene ( $CH_3CH=CHCH_3$ )

Carbon 1 of the double bond: attached to $CH_3$ and $H$ → different ✓
Carbon 2 of the double bond: attached to $CH_3$ and $H$ → different ✓

Result: Two geometrical isomers exist — cis (both methyl groups on the same side) and trans (methyl groups on opposite sides).

cis-But-2-ene, with both methyl groups on the same side of the C=C double bond

trans-But-2-ene, with the two methyl groups on opposite sides of the C=C double bond

Example 2: 1,2-Dichloroethene ( $ClCH=CHCl$ )

Carbon 1: attached to $Cl$ and $H$ → different ✓
Carbon 2: attached to $Cl$ and $H$ → different ✓

Result: cis and trans isomers exist.

Example 3: 1,1-Dichloroethene ( $Cl_2C=CH_2$ )

Carbon 1: attached to $Cl$ and $Cl$ → identical ✗ …

Why this formula?

Geometrical Isomerism: Why the Conditions Hold

Geometrical isomerism (also called cis-trans or E-Z isomerism) arises when atoms or groups are arranged differently in space around a rigid part of a molecule — typically a double bond or a ring. The key is that rotation is restricted, so the spatial positions become fixed and distinct.

Let’s break down why the conditions are what they are.

1. The Core Requirement: Restricted Rotation

For two molecules to be geometrical isomers, they must have the same connectivity but different spatial arrangement due to a barrier to rotation.

Double bonds (C=C): The π-bond locks the two carbons in place — rotation requires breaking the π-bond (energy ~250 kJ/mol), so it doesn’t happen at room temperature.
Rings (e.g., cycloalkanes): The ring structure physically prevents free rotation about C–C single bonds within the ring.

Why this matters: Without restricted rotation, the molecule would freely interconvert between arrangements — no distinct isomers exist.

2. Condition 1: Two Different Groups on Each Carbon (for C=C)

Consider a general alkene:

\text{C} = \text{C}

Each carbon must have two different substituents (not counting the other carbon of the double bond).

Why?

If one carbon has two identical groups (e.g., both H), then swapping the groups on that carbon produces the same molecule — no isomerism.

Example:

1,2-dichloroethene ( $\text{ClHC=CHCl}$ ): Each carbon has H and Cl (different) → geometrical isomers exist.
1,1-dichloroethene ( $\text{Cl}_2\text{C=CH}_2$ ): One carbon has two Cl (identical) → no geometrical isomers.

Formal condition:

For a C=C bond, geometrical isomerism is possible iff each doubly bonded carbon bears two different substituents.

3. Condition 2: For Rings — Similar Logic

In a ring (e.g., cyclopropane, cyclohexane), the ring itself restricts rotation. Here, geometrical isomerism occurs when two substituents on different ring carbons can be on the same side (cis) or opposite sides (trans).

Why?

The ring is a closed loop — you cannot rotate one carbon relative to another without breaking bonds.
If the two substituents are on different carbons, their relative orientation (same side / opposite sides) is fixed.

Condition:

The ring must have at least two substituents (could be same or different) on different carbons.
If both substituents are on the same carbon, swapping them doesn’t change the molecule (no isomerism).

Example:

1,2-dimethylcyclopropane: Two methyl groups on adjacent carbons → cis and trans isomers exist.
1,1-dimethylcyclopropane: Both methyls on same carbon → no geometrical isomerism.

4. The E-Z Notation (Why It’s Needed)

When the four substituents on a C=C are all different, cis-trans naming fails. The Cahn-Ingold-Prelog priority rules assign E (opposite sides) or Z (same side).

Why this works:

Priority is based on atomic number (higher = higher priority). …

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Low spin tetrahedral complexes are not formed because the crystal field splitting energy ( $\Delta_t$ ) in a tetrahedral field is too small to overcome the pairing energy ( $P$ ) required to force electrons into the same orbital — the high spin configuration is always energetically favoured.

The Core Idea: Why Spin State Depends on Geometry

The spin state of a metal complex (whether it is high spin or low spin) is decided by a simple energy competition: is it cheaper to pair two electrons in the same orbital, or to keep them unpaired in separate orbitals? The answer depends on two numbers:

$\Delta$ — the crystal field splitting energy (the energy gap between the $t_{2g}$ and $e_g$ sets of d-orbitals).
$P$ — the pairing energy (the energy cost of putting two electrons in the same orbital, which includes Coulomb repulsion and exchange energy loss).

If $\Delta > P$ , the complex will be low spin — electrons prefer to pair up in the lower-energy orbitals rather than jump to the higher set. If $\Delta < P$ , the complex will be high spin — electrons stay unpaired because pairing is too expensive.

Now here is the critical point: the magnitude of $\Delta$ depends heavily on geometry.

The Tetrahedral Field: A Weaker Split

In a tetrahedral complex, the metal ion is at the centre of a tetrahedron with four ligands at the corners. The d-orbitals split into two sets:

The $e$ set ( $d_{x^2-y^2}$ , $d_{z^2}$ ) — lower in energy.
The $t_2$ set ( $d_{xy}$ , $d_{xz}$ , $d_{yz}$ ) — higher in energy.

The splitting energy is denoted $\Delta_t$ . There is a well-known relationship between $\Delta_t$ and the octahedral splitting $\Delta_o$ :

\Delta_t = \frac{4}{9} \Delta_o

This is not an arbitrary number — it comes from the fact that in a tetrahedral field, the ligands approach along axes that are not directly aligned with the d-orbitals, so the electrostatic interaction is weaker. Also, there are only four ligands instead of six, which further reduces the field strength.

The consequence is immediate: $\Delta_t$ is always much smaller than $\Delta_o$ — typically less than half.

The Pairing Energy: A Fixed Cost

The pairing energy $P$ is an intrinsic property of the metal ion and its oxidation state. It does not change with geometry. For a given $d^n$ configuration, $P$ is a fixed number.

So the competition becomes: is $\Delta_t$ ever larger than $P$ ?

Step-by-Step Reasoning

Consider the maximum possible $\Delta_t$ . Even with the strongest-field ligands (like CN $^-$ or CO), $\Delta_t$ is at most $\frac{4}{9}$ of the octahedral $\Delta_o$ for the same metal and ligands. For most metals, even the octahedral $\Delta_o$ is barely larger than $P$ for the $d^4$ , $d^5$ , $d^6$ , and $d^7$ configurations where spin-state ambiguity exists.
Compare magnitudes. For a typical first-row transition metal like Fe $^{2+}$ ( $d^6$ ), $\Delta_o$ for a strong-field ligand might be around 20,000–30,000 cm $^{-1}$ , while $P$ is roughly 15,000–20,000 cm $^{-1}$ . So $\Delta_o$ can exceed $P$ — low spin is possible in octahedral geometry. But $\Delta_t = \frac{4}{9} \Delta_o$ gives roughly 9,000–13,000 cm $^{-1}$ , which is always less than $P$ .
Check all $d^n$ configurations. The only configurations that can potentially show low-spin behaviour are $d^4$ , $d^5$ , $d^6$ , and $d^7$ (where there is a choice between pairing in the lower set or occupying the higher set). For each of these, the tetrahedral splitting is simply too small. For $d^8$ , $d^9$ , and $d^{10}$ , there is no spin-state ambiguity anyway — the ground state is fixed regardless of $\Delta$ . …

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Why Are Low Spin Tetrahedral Complexes Not Formed?

This question tests your understanding of crystal field theory (CFT) and how splitting energy ( $\Delta_t$ ) compares to pairing energy (P) in tetrahedral geometry.

Common Mistakes & How to Avoid Them

1. Confusing Tetrahedral and Octahedral Splitting

Mistake: Students often assume tetrahedral splitting ( $\Delta_t$ ) is large, like octahedral splitting ( $\Delta_o$ ).

Why it’s wrong:

In tetrahedral complexes, the crystal field splitting is much smaller:

\Delta_t = \frac{4}{9} \Delta_o

Since $\Delta_t$ is small, it is almost always less than the pairing energy (P) for any metal ion.

How to avoid:

Memorise the ratio: $\Delta_t \approx 0.44 \Delta_o$
Always compare $\Delta_t$ with $P$ — if $\Delta_t < P$ , electrons will not pair (high spin is favoured).

2. Forgetting That Pairing Energy Is Always Positive

Mistake: Thinking that pairing can happen “for free” in tetrahedral complexes.

Why it’s wrong:

Pairing two electrons in the same orbital costs energy (the pairing energy P). Since $\Delta_t$ is small, the energy gained by pairing (which would require $\Delta_t > P$ ) is never enough.

How to avoid:

Write the condition for low spin: $\Delta > P$
For tetrahedral: $\Delta_t \ll P$ always → low spin impossible.

3. Ignoring the d-Orbital Splitting Pattern

Mistake: Treating tetrahedral splitting like octahedral (e.g., thinking $t_{2g}$ is lower in energy).

Why it’s wrong:

In tetrahedral geometry, the splitting is inverted:

$e$ set ( $d_{x^2-y^2}, d_{z^2}$ ) → lower energy
$t_2$ set ( $d_{xy}, d_{yz}, d_{zx}$ ) → higher energy

This inversion does not change the fact that $\Delta_t$ is small — but students sometimes misapply the orbital labels and get confused.

How to avoid:

Draw the tetrahedral splitting diagram correctly:

Energy
  ↑
  |   t₂ (higher)
  |   ↑ Δ_t (small)
  |   e  (lower)

Remember: the magnitude matters more than the order for spin state.

4. Assuming Low Spin Exists for d⁴, d⁵, d⁶, d⁷ in Tetrahedral

Mistake: Applying octahedral rules directly to tetrahedral complexes.

Why it’s wrong:

In octahedral, low spin is possible for $d^4$ to $d^7$ (e.g., $[Co(NH_3)_6]^{3+}$ is low spin $d^6$ ).

In tetrahedral, no known example exists because $\Delta_t$ is too small.

How to avoid:

Memorise: All tetrahedral complexes are high spin (except for very rare cases with extremely strong field ligands — but these are not in the JEE/NEET syllabus).
For exam purposes: “Low spin tetrahedral complexes are not formed.”

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 mark
$Q. Draw the structure of geometrical isomers of [Co(NH3)4Cl2].$
›Reveal solutionSolution
[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.) …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Which complexes do not show geometrical isomerism?$
(a) Square planar complexes
(b) Tetrahedral complexes
(c) Octahedral complexes
(d) All of the above
›Reveal solutionSolution
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. …
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CBSE 2025Set ANNUAL1 mark
$[Fe(NH_3)_2(CN)_4]^-$
›Reveal solutionSolution

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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CBSE 2024Set ANNUAL1 markMCQ
$Q. Which kinds of isomerism are exhibited by octahedral Co(NH3)4Br2Cl ?$
(a) Geometrical and ionization
(b) Geometrical and Optical
(c) Optical and ionization
(d) Geometrical only
›Reveal solutionSolution

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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CBSE 2024Set ANNUAL1 markMCQ
$Q. Assertion [A] : Complexes of MX6 and MX5L type [X and L are unidentate] do not show geometrical isomerism. Reason [R] : Geometrical isomerism is not shown by the complexes of coordination number 6.$
(a) Both [A] and [R] are true and [R] is the correct explanation of [A].
(b) Both [A] and [R] are true, but [R] is not the correct explanation of [A].
(c) [A] is true, but [R] is false.
(d) [A] is false, but [R] is true.
›Reveal solutionSolution

$MX_6$ 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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CBSE 2024Set ANNUAL1 markMCQ
$[Co(NH_3)_4Cl_2]^+$
(a) linkage isomerism
(b) geometrical isomerism
(c) coordination isomerism
(d) ionization isomerism
›Reveal solutionSolution
$[Co(NH_3)_4Cl_2]^+$ 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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CBSE 2023Set ANNUAL1 mark
$[Co(NO_2)_3(NH_3)_3]$
›Reveal solutionSolution

$[Co(NO_2)_3(NH_3)_3]$ 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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CBSE 2020Set 56/2/11 mark
$[Co(NH_3)_5NO_2]Cl_2$
›Reveal solutionSolution
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.

Why This Question Tests a Key Concept

This problem isn't just about memorising a name — it's about recognising that a ligand can bind in more than one way. The $NO_2^-$ ion is an ambidentate ligand: it has two different donor atoms (N and O) that can form a coordinate bond with the central metal ion. That single fact is the entire foundation of linkage isomerism.

Important

Linkage isomerism arises only when an ambidentate ligand coordinates through different atoms. The complex must have the same molecular formula but differ in which atom of the ligand is bonded to the metal.

Step-by-Step Reasoning

Identify the coordination sphere and counter ions

The formula is $[Co(NH_3)_5NO_2]Cl_2$ . The square brackets enclose the coordination sphere: $Co^{3+}$ (cobalt in +3 oxidation state) surrounded by five $NH_3$ ligands and one $NO_2^-$ ligand. The two $Cl^-$ ions are outside the brackets — they are counter ions, not directly bonded to cobalt.

Recognise the ambidentate nature of $NO_2^-$

The nitrite ion can bind through:

Nitrogen atom: forming a nitro complex, $[Co(NH_3)_5(NO_2)]^{2+}$

Oxygen atom: forming a nitrito complex, $[Co(NH_3)_5(ONO)]^{2+}$

Both have the same overall formula $[Co(NH_3)_5NO_2]Cl_2$ , but the connectivity differs.

Check for other isomerism types

Geometrical isomerism requires different spatial arrangements of ligands (e.g., cis/trans in square planar or octahedral complexes). Here, all five $NH_3$ are identical, and the sixth position is occupied by $NO_2^-$ — there is no possibility of different geometric arrangements.

Optical isomerism requires chirality (non-superimposable mirror images). This complex has no chiral centre or plane of asymmetry. …
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CBSE 2018Set ANNUAL1 mark
$Q. Explain with an example the ionisation isomerism in complex compounds.$
›Reveal solutionSolution
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). …
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CBSE 2018Set 56/11 mark
$[Cu(NH_3)_4][PtCl_4]$
›Reveal solutionSolution

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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CBSE 2017Set ANNUAL1 markMCQ
$Q. Which complex exhibit geometrical isomerism?$
(a) [MnBr4]2+
(b) [Pt(NH3)3Cl]+
(c) [PtCl2(P(C2H5)3)2]
(d) [Fe(H2O)5NO]2+
›Reveal solutionSolution
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. …
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Q.Why are low spin tetrahedral complexes not formed?

Concept understanding — Geometrical Isomerism Conditions

Geometrical Isomerism: The Intuition

The Precise Conditions

Condition 1: Restricted Rotation

Condition 2: Two Different Substituents on Each End

How to Check: A Step-by-Step Method

Examples to Cement the Idea

Example 1: But-2-ene ( $CH_3CH=CHCH_3$ )

Example 2: 1,2-Dichloroethene ( $ClCH=CHCl$ )

Example 3: 1,1-Dichloroethene ( $Cl_2C=CH_2$ )

Why this formula?

Geometrical Isomerism: Why the Conditions Hold

1. The Core Requirement: Restricted Rotation

2. Condition 1: Two Different Groups on Each Carbon (for C=C)

Why?

3. Condition 2: For Rings — Similar Logic

Why?

4. The E-Z Notation (Why It’s Needed)

The Core Idea: Why Spin State Depends on Geometry

The Tetrahedral Field: A Weaker Split

The Pairing Energy: A Fixed Cost

Step-by-Step Reasoning

Concept: Crystal Field Theory (CFT) and Spin States in Tetrahedral Complexes

Method: Crystal Field Splitting Analysis

Why Are Low Spin Tetrahedral Complexes Not Formed?

Common Mistakes & How to Avoid Them

1. Confusing Tetrahedral and Octahedral Splitting

2. Forgetting That Pairing Energy Is Always Positive

3. Ignoring the d-Orbital Splitting Pattern

4. Assuming Low Spin Exists for d⁴, d⁵, d⁶, d⁷ in Tetrahedral

Why This Question Tests a Key Concept

Step-by-Step Reasoning

Q.Why are low spin tetrahedral complexes not formed?

The Core Idea: Why Spin State Depends on Geometry

The Tetrahedral Field: A Weaker Split

The Pairing Energy: A Fixed Cost

Step-by-Step Reasoning

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