In both water and dimethyl ether (CH3—O—CH3), oxygen atom is central atom, and has the same hybridisation, yet

NCERT Exemplar · Q41

Q. In both water and dimethyl ether (CH3—O—CH3), oxygen atom is central atom, and has the same hybridisation, yet they have different bond angles. Which one has greater bond angle? Give reason.

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Concept understanding — VSEPR Theory

VSEPR Theory: Why Molecules Have the Shapes They Do

Imagine you're in a crowded room. Everyone wants their personal space. If you're standing with a few friends, you'll naturally spread out so no one is too close to anyone else. That's exactly what happens inside a molecule.

The Core Intuition

Electron pairs are negatively charged. They repel each other. In a molecule, the electron pairs around a central atom will arrange themselves as far apart as possible — just like those people in the room. This simple idea is the entire foundation of VSEPR (pronounced "ves-per") Theory.

Note

VSEPR stands for Valence Shell Electron Pair Repulsion. The name tells you exactly what it's about: the repulsion between electron pairs in the valence shell.

The Precise Statement

VSEPR Theory states that the geometry around a central atom is determined by minimizing the repulsion between all electron pairs (both bonding and lone pairs) in its valence shell.

Two key points to hold onto:

All electron pairs repel — whether they are shared (bonding pairs) or unshared (lone pairs).
Lone pairs repel more strongly than bonding pairs. A lone pair is "fatter" — it's only attracted to one nucleus, so it spreads out more and pushes harder on its neighbours.

How to Predict Shape in 3 Steps

Step 1: Count the total electron pairs around the central atom.

Add the number of atoms bonded to the central atom plus the number of lone pairs on it. This gives you the steric number.

Step 2: Arrange those pairs as far apart as possible.

This gives you the electron-pair geometry — the shape if you pretend all pairs are identical.

Step 3: Replace lone pairs with "invisible" space.

The actual molecular geometry is the shape formed by the atoms alone, ignoring lone pairs.

The Common Geometries at a Glance

Steric Number	Electron-Pair Geometry	Lone Pairs	Molecular Geometry	Example	Bond Angle
2	Linear	0	Linear	$\text{CO}_2$	180°
3	Trigonal planar	0	Trigonal planar	$\text{BF}_3$	120°
3	Trigonal planar	1	Bent	$\text{SO}_2$	~119°
4	Tetrahedral	0	Tetrahedral	$\text{CH}_4$	109.5°
4	Tetrahedral	1	Trigonal pyramidal	$\text{NH}_3$	~107°
4	Tetrahedral	2	Bent	$\text{H}_2\text{O}$	~104.5°
5	Trigonal bipyramidal	0	Trigonal bipyramidal	$\text{PCl}_5$	90°, 120°
6	Octahedral	0	Octahedral	$\text{SF}_6$	90°

Watch out

A common mistake: thinking that $\text{NH}_3$ is tetrahedral. It has tetrahedral electron-pair geometry, but because one position is a lone pair, the molecular shape is trigonal pyramidal. The bond angle is 107°, not 109.5°.

Why Lone Pairs Squeeze Bond Angles

Take water ( $\text{H}_2\text{O}$ ). The central oxygen has 4 electron pairs: 2 bonding (to H atoms) and 2 lone pairs. The ideal tetrahedral angle is 109.5°. But the two lone pairs push harder on the bonding pairs, compressing the H–O–H angle to about 104.5°.

In ammonia ( $\text{NH}_3$ ), there's only one lone pair, so the compression is less — the H–N–H angle is about 107°. …

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Both molecules have $sp^3$ hybridised oxygen, but the bond angle in dimethyl ether ( $\approx 111.7^\circ$ ) is larger than in water ( $104.5^\circ$ ) because the two methyl groups are bulkier than hydrogen atoms, causing greater repulsion between bonded pairs and widening the angle.

The key to this question lies in VSEPR (Valence Shell Electron Pair Repulsion) Theory. This theory states that electron pairs around a central atom arrange themselves to minimise repulsion. But not all repulsions are equal — the strength of repulsion follows a clear hierarchy: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair.

In both water ( $H_2O$ ) and dimethyl ether ( $CH_3-O-CH_3$ ), the central oxygen atom has the same electronic geometry. Let's see why.

Count the electron pairs around oxygen.

Oxygen has 6 valence electrons. In water, it forms two single bonds with hydrogen atoms, using 2 electrons. The remaining 4 electrons form two lone pairs. That gives 2 bond pairs + 2 lone pairs — a total of 4 electron domains.

In dimethyl ether, oxygen forms two single bonds with carbon atoms (from the methyl groups). Again, 2 electrons are used in bonding, leaving 4 as two lone pairs. So the electron domain count is identical: 2 bond pairs + 2 lone pairs.
Determine the hybridisation.

Four electron domains around a central atom always correspond to $sp^3$ hybridisation. The ideal bond angle for a perfect tetrahedron is $109.5^\circ$ . But the presence of lone pairs distorts this angle because lone pairs occupy more space and repel more strongly than bond pairs.
Compare the bond pair repulsions.

In water, the two bond pairs are attached to small hydrogen atoms. The repulsion between the two O–H bond pairs is relatively modest. The dominant repulsion comes from the two lone pairs pushing the bond pairs closer together, resulting in the well-known bond angle of $104.5^\circ$ .

In dimethyl ether, the bond pairs are attached to methyl groups ( $-CH_3$ ). These are much bulkier than hydrogen atoms. The bond pair–bond pair repulsion between the two O–C bonds is significantly stronger because the electron clouds of the methyl groups are larger and more polarisable. This increased repulsion pushes the two bond pairs further apart, opening the bond angle.
The result. …

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

COMEDK 2026Set 2026-A1 markMCQ
$I F_5$
›Reveal solutionSolution
The key is to count the total valence electrons, then assign them as bond pairs and lone pairs around the central iodine atom. In $\mathrm{IF_5}$ , iodine has 5 bond pairs and 1 lone pair, so the correct option is (A).

Concept & Intuition

This question tests your ability to apply the Lewis structure and VSEPR theory to a molecule. Instead of memorizing shapes, you can always work from first principles: count valence electrons, connect atoms with single bonds, then distribute remaining electrons as lone pairs to satisfy the octet (or expanded octet for period 5+ elements). Iodine is in period 5, so it can accommodate more than 8 electrons — here it forms five bonds and keeps one lone pair, giving a total of 12 electrons around it.

Step-by-step reasoning

Count total valence electrons

Iodine (I) is in group 17, so it has 7 valence electrons.

Each fluorine (F) is also group 17, so each has 7 valence electrons.

Total = $7 + 5 \times 7 = 7 + 35 = 42$ valence electrons.

Draw the skeletal structure

Iodine is the central atom (less electronegative, can expand octet).

Connect each fluorine to iodine with a single bond.

Each single bond uses 2 electrons, so 5 bonds use $5 \times 2 = 10$ electrons.

Remaining electrons = $42 - 10 = 32$ electrons.

Complete octets of fluorine atoms

Each fluorine needs 6 more electrons (3 lone pairs) to complete its octet.

For 5 fluorines: $5 \times 6 = 30$ electrons used.

Remaining electrons = $32 - 30 = 2$ electrons.

Place remaining electrons on central iodine

The 2 leftover electrons form one lone pair on iodine.

So iodine has: …
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COMEDK 2025Set 2025-A1 markMCQ
$\mathrm{XeF}_2 ~\&~ \mathrm{HgCl}_2$
›Reveal solutionSolution
The key is to apply VSEPR theory to determine the molecular geometry of each species. Both molecules in pair (A), XeF₂ and HgCl₂, are linear, so (A) is the correct answer.

Concept & Intuition

Molecular shape is determined by the number of bonding pairs and lone pairs around the central atom, following VSEPR (Valence Shell Electron Pair Repulsion) theory. Lone pairs repel more strongly than bonding pairs, so they occupy more space and can distort the geometry. To compare shapes, we must count the steric number (SN = number of atoms bonded + number of lone pairs) and then predict the arrangement.

Step-by-step reasoning

Analyze pair (A): XeF₂ and HgCl₂

XeF₂: Xe has 8 valence electrons. It forms two single bonds with F atoms (using 2 electrons), leaving 6 electrons = 3 lone pairs. Steric number = 2 (bonds) + 3 (lone pairs) = 5. The electron-pair geometry is trigonal bipyramidal, with lone pairs occupying the equatorial positions (to minimize repulsion). The two F atoms are axial, giving a linear shape.

HgCl₂: Hg is in group 12, with 2 valence electrons. It forms two single bonds with Cl atoms, using both electrons — no lone pairs. Steric number = 2 + 0 = 2. Electron-pair geometry and molecular shape are both linear.

Both are linear → same shape. This pair is a candidate.

Analyze pair (B): BF₃ and NH₃

BF₃: B has 3 valence electrons, forms three single bonds with F, no lone pairs. SN = 3, shape = trigonal planar.

NH₃: N has 5 valence electrons, forms three single bonds with H (using 3 electrons), leaving 2 electrons = 1 lone pair. SN = 3 + 1 = 4. Electron-pair geometry is tetrahedral, but the lone pair pushes the H atoms down, giving a trigonal pyramidal shape.

Trigonal planar ≠ trigonal pyramidal → not the same shape.

Analyze pair (C): XeF₄ and SF₄

XeF₄: Xe has 8 valence electrons, forms four single bonds with F (using 4 electrons), leaving 4 electrons = 2 lone pairs. SN = 4 + 2 = 6. Electron-pair geometry is octahedral; lone pairs occupy opposite positions (trans), giving a square planar shape.

SF₄: S has 6 valence electrons, forms four single bonds with F (using 4 electrons), leaving 2 electrons = 1 lone pair. SN = 4 + 1 = 5. Electron-pair geometry is trigonal bipyramidal; the lone pair occupies an equatorial position, giving a see-saw shape. …
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COMEDK 2025Set 2025-E1 markMCQ
$\mathrm{BF}_3, \mathrm{NH}_3, \mathrm{PF}_3$
›Reveal solutionSolution
The bond angles fall as $\text{XeF}_2\,(180^\circ) > \text{BF}_3\,(120^\circ) > \text{NH}_3\,(107^\circ) > \text{PF}_3\,(\sim97^\circ)$ — option (B).

Analyse each molecule's geometry and bond angle.

$\text{XeF}_2$ : $\text{Xe}$ is $sp^3d$ hybridised with 3 lone pairs in the equatorial plane; the shape is linear, so the $\text{F–Xe–F}$ angle is $\mathbf{180^\circ}$ .

$\text{BF}_3$ : $\text{B}$ is $sp^2$ hybridised with no lone pair; trigonal planar, angle $\mathbf{120^\circ}$ .

$\text{NH}_3$ : $\text{N}$ is $sp^3$ with one lone pair; pyramidal, lone-pair repulsion compresses the angle to $\mathbf{\sim107^\circ}$ . …
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COMEDK 2025Set 2025-M1 markMCQ
$\mathrm{CH}_4$
›Reveal solutionSolution
The question tests molecular geometry, dipole moments, and bond length concepts. The correct statement is (D): an AB₅E molecule (5 bond pairs + 1 lone pair) adopts a square pyramidal geometry.

Concept & Intuition

Each option probes a different chemical principle: hybridisation in methane vs. sulfur tetrafluoride, dipole moment trends in nitrogen trihalides, bond length reasoning in carbon dioxide, and VSEPR geometry for a molecule with five bonding pairs and one lone pair. The key is to recall that lone pairs occupy more space than bonding pairs, distorting ideal geometries, and that inductive effects are not the primary factor in CO₂ bond length.

Step-by-Step Reasoning

Option (A): Hybridisation in CH₄ and SF₄

CH₄: Carbon has 4 bond pairs, no lone pairs → tetrahedral geometry → sp³ hybridisation. Correct.

SF₄: Sulfur has 5 valence electrons, forms 4 bonds and has 1 lone pair → 5 electron domains → sp³d hybridisation (trigonal bipyramidal electron geometry, seesaw molecular shape).

Since SF₄ is sp³d, not sp³, statement (A) is false.

Option (B): Dipole moments of NF₃ vs. NH₃

Both have trigonal pyramidal geometry with a lone pair on nitrogen.

In NH₃, N–H bonds are less polar than N–F bonds (F is more electronegative), but the lone pair’s dipole moment direction adds to the bond dipoles in NH₃, while in NF₃ the bond dipoles point away from the lone pair, partially cancelling.

Result: NH₃ has a larger net dipole moment (1.47 D) than NF₃ (0.24 D).

Thus, statement (B) is false.

Option (C): C–O bond length in CO₂

CO₂ has two C=O double bonds (linear, sp hybridised carbon). The bond length is about 116 pm (not 110 pm). …
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COMEDK 2025Set 2025-M1 markMCQ
$Q. From the given covalent compounds (A to F) identify the pair of molecules which have:$
(i) Two lone pairs of electrons on the central atom.
(ii) One lone pair of electrons on the central atom. A. $\mathrm{SO}_2$ B. $\mathrm{ClF}_3$ C. $\mathrm{BF}_3$ D. $\mathrm{BrF}_5$ E. XeF 4 F. $\mathrm{SF}_6$ (A)
(i) $A \& F\qquad$
(ii) $A \& E$ (B)
(i) $D \& F\qquad$
(ii) $\mathrm{C} \& \mathrm{D}$ (C)
(i) $B$ & $E\qquad$
(ii) $\mathrm{A} \& \mathrm{D}$ (D)
(i) $\mathrm{C} \& \mathrm{~A}\qquad$
(ii) $\mathrm{B} \& \mathrm{C}$
›Reveal solutionSolution
The key is to count the total valence electrons around the central atom, then use VSEPR theory to determine lone pairs. For the given compounds: (i) two lone pairs → ClF₃ (B) and XeF₄ (E);

(ii) one lone pair → SO₂ (A) and BrF₅ (D). The correct option is (C).

Concept & Intuition

The number of lone pairs on the central atom in a covalent molecule is found by:

Counting the total valence electrons of the central atom.

Adding one electron for each bond (shared pair) to the central atom.

Subtracting the number of bonding electrons (each bond = 2 electrons) to get the remaining non-bonding electrons.

Dividing by 2 gives the number of lone pairs.

Alternatively, use the VSEPR formula: Steric number = number of atoms bonded + number of lone pairs. The steric number equals the number of electron domains around the central atom. By knowing the molecular geometry and the number of bonded atoms, we can deduce lone pairs.

Let’s work through each compound step by step.

SO₂ (A)

Central atom: S (6 valence electrons).

Forms two double bonds to two O atoms (each double bond = 2 bonding pairs, but counts as 1 domain).

Total electron domains = 3 (two double bonds + one lone pair).

Bonded atoms = 2 → lone pairs = 1.

Result: One lone pair.

ClF₃ (B)

Central atom: Cl (7 valence electrons).

Forms three single bonds to F atoms (each uses 1 electron from Cl).

Remaining electrons = 7 − 3 = 4 → two lone pairs.

Steric number = 5 (3 bonds + 2 lone pairs), T-shaped geometry.

Result: Two lone pairs.

BF₃ (C)

Central atom: B (3 valence electrons).

Forms three single bonds to F atoms (uses all 3 electrons).

No remaining electrons → zero lone pairs.

Result: Zero lone pairs.

BrF₅ (D)

Central atom: Br (7 valence electrons).

Forms five single bonds to F atoms (uses 5 electrons).

Remaining electrons = 7 − 5 = 2 → one lone pair.

Steric number = 6 (5 bonds + 1 lone pair), square pyramidal geometry.

Result: One lone pair.

XeF₄ (E)

Central atom: Xe (8 valence electrons).

Forms four single bonds to F atoms (uses 4 electrons).

Remaining electrons = 8 − 4 = 4 → two lone pairs.

Steric number = 6 (4 bonds + 2 lone pairs), square planar geometry.

Result: Two lone pairs.

SF₆ (F)

Central atom: S (6 valence electrons).

Forms six single bonds to F atoms (uses 6 electrons). …
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COMEDK 2024Set 2024-A1 markMCQ
$\mathrm{SF}_4$
›Reveal solutionSolution
The key idea is that molecular shape is determined by the total number of electron domains (bonding and lone pairs) around the central atom, not just the lone pair count. SF₄ has one lone pair (not two) and is see-saw shaped; ClF₃ has two lone pairs (not one) and is T-shaped. Thus the assertion is correct but the reason is wrong.

Concept and Intuition

To predict molecular geometry, we use VSEPR (Valence Shell Electron Pair Repulsion) theory. The central atom’s electron domains (bonds and lone pairs) repel each other and arrange themselves as far apart as possible. Lone pairs occupy more space than bonding pairs, so they distort the ideal geometry. The key is to count total electron domains and then note how many are lone pairs. A common mistake is to mis-count lone pairs by forgetting to account for the central atom’s valence electrons and the bonding pattern.

Step-by-step reasoning

Determine the electron domains for SF₄

Sulfur (S) has 6 valence electrons.

In SF₄, sulfur forms four single bonds with four fluorine atoms, using 4 electrons.

Remaining electrons: $6 - 4 = 2$ electrons → one lone pair.

Total electron domains around S = 4 bonds + 1 lone pair = 5 domains.

The ideal geometry for 5 domains is trigonal bipyramidal. With one lone pair, the shape becomes see-saw (the lone pair occupies an equatorial position to minimize repulsion).

So Assertion (A) is correct: SF₄ is see-saw shaped.

Determine the electron domains for ClF₃

Chlorine (Cl) has 7 valence electrons.

In ClF₃, chlorine forms three single bonds with three fluorine atoms, using 3 electrons.

Remaining electrons: $7 - 3 = 4$ electrons → two lone pairs.

Total electron domains around Cl = 3 bonds + 2 lone pairs = 5 domains.

Again, ideal geometry is trigonal bipyramidal. With two lone pairs, both occupy equatorial positions (to minimize 90° repulsions), leaving the three fluorine atoms in a T-shape (one axial, two equatorial).

So Assertion (A) is also correct: ClF₃ is T-shaped.

Evaluate the Reason (R)

Reason (R) states: “SF₄ has two lone pair of electrons. But ClF₃ has one pair of electrons.”

From steps 1 and 2, we found SF₄ has one lone pair, and ClF₃ has two lone pairs.

Therefore, Reason (R) is factually incorrect. …
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COMEDK 2024Set 2024-M1 markMCQ
$\mathrm{ClF_3}$
›Reveal solutionSolution
VSEPR theory predicts molecular shape from the total number of electron domains (bonding + lone pairs) around the central atom. The correct matches are: ClF₃ → T‑shaped, BrF₅ → square pyramidal, SF₄ → see‑saw, IF₇ → pentagonal bipyramidal, which corresponds to option (C).

Concept & Intuition

VSEPR (Valence Shell Electron Pair Repulsion) theory says that electron pairs—whether in bonds or as lone pairs—arrange themselves as far apart as possible around a central atom. The total number of electron domains (each bond = 1 domain, each lone pair = 1 domain) determines the electron‑pair geometry. The actual molecular shape then depends on how many of those domains are lone pairs (which are “invisible” in the shape name). So the trick is:

Count the total domains → get the base geometry.

Subtract lone pairs → get the molecular shape.

Let’s apply this step‑by‑step to each molecule.

1. ClF₃ (Chlorine trifluoride)

Central atom: Cl (Group 17, 7 valence electrons).

Three F atoms each form a single bond → 3 bonding domains.

Remaining electrons: $7 - 3 = 4$ electrons → 2 lone pairs.

Total domains = 3 bonds + 2 lone pairs = 5 → electron‑pair geometry = trigonal bipyramidal.

With 2 lone pairs, they occupy equatorial positions (to minimise repulsion), leaving the three F atoms in a T‑shape.

Shape: T‑shaped → matches S in Column II.

Tip

In a trigonal bipyramid, lone pairs always go equatorial first because they have more space (90° to two axial positions, 120° to the other equatorial). Two lone pairs force a T‑shape.

2. BrF₅ (Bromine pentafluoride)

Central atom: Br (Group 17, 7 valence electrons).

Five F atoms → 5 bonding domains.

Remaining electrons: $7 - 5 = 2$ electrons → 1 lone pair.

Total domains = 5 bonds + 1 lone pair = 6 → electron‑pair geometry = octahedral.

One lone pair can occupy any vertex; the five F atoms then form a square pyramidal shape (like a pyramid with a square base).

Shape: Square pyramidal → matches R in Column II.

Watch out

A common mistake is to think 6 domains always give octahedral shape. But with one lone pair, the shape is square pyramidal, not octahedral. The lone pair “pushes” the other atoms slightly, but the name changes.

3. SF₄ (Sulfur tetrafluoride)

Central atom: S (Group 16, 6 valence electrons).

Four F atoms → 4 bonding domains.

Remaining electrons: $6 - 4 = 2$ electrons → 1 lone pair.

Total domains = 4 bonds + 1 lone pair = 5 → electron‑pair geometry = trigonal bipyramidal.

One lone pair goes equatorial; the four F atoms then occupy the two axial and two equatorial positions, giving a see‑saw shape (also called distorted tetrahedron).

Shape: See‑saw → matches P in Column II. …
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COMEDK 2023Set 2023-M1 markMCQ
$\mathrm{I}: \mathrm{SF}_4 : \mathrm{II}: \mathrm{ICl}_3 : \mathrm{III}: \mathrm{SO}_2$
›Reveal solutionSolution
Applying VSEPR, the central atoms of $\mathrm{SF_4}$ , $\mathrm{ICl_3}$ and $\mathrm{SO_2}$ each carry lone (non-bonding) electron pairs, so all three qualify.

Count lone pairs on the central atom:

I: $\mathrm{SF_4}$ — S has 6 valence electrons; 4 are used in S–F bonds, leaving 1 lone pair (see-saw shape).

II: $\mathrm{ICl_3}$ — I has 7 valence electrons; 3 in I–Cl bonds, leaving 2 lone pairs (T-shape). …
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KCET 2022Set B-31 markMCQ
$Q. Alkali halides do not show dislocation defect because (A) Cations and anions have almost equal size (B) There is large difference in size of cations and anions (C) Cations and anions have low co-ordination number. (D) Anions cannot be accommodated in vacant spaces.$
›Reveal solutionSolution

The Frenkel (dislocation) defect requires a large cation–anion size difference so the smaller ion can squeeze into an interstitial void; alkali halides have ions of nearly similar size, so they show Schottky — not Frenkel — defects.

Step 1 — Identify the defect being named.

The Frenkel defect is also called the dislocation defect: an ion (usually the smaller cation) leaves its regular lattice site and lodges in an interstitial void, creating a vacancy + an interstitial. Since nothing leaves the crystal, the density is unchanged.

(Contrast with the Schottky defect: equal numbers of cations and anions are simply missing → density decreases.)

Step 2 — What structural condition does Frenkel need?

An interstitial void in a close-packed lattice is small. For an ion to fit into it, that ion must be much smaller than the ions forming the lattice. Hence:

Frenkel defect is shown by ionic solids with a large difference in the sizes of cation and anion, and a low coordination number (a loosely packed lattice with room in the voids).

The classic examples bear this out — in each, the cation is far smaller than the anion:
$\text{ZnS},\quad \text{AgCl},\quad \text{AgBr},\quad \text{AgI}$
(e.g. in AgBr, $Ag^+$ = 115 pm vs $Br^-$ = 195 pm.)

Step 3 — Apply this to alkali halides. …
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COMEDK 2021Set 2021-B1 markMCQ
$Q. Which one of the following Oxoacids of Chlorine has tetrahedral geometry? HOClO3, HOClO, HOClO2, HOCl (A) HOClO3 (B) HOCl (C) HOClO (D) HOClO2$
›Reveal solutionSolution
HOClO $_3$ = HClO $_4$ ; its central Cl has four O around it (sp $^3$ , no lone pair) $\Rightarrow$ tetrahedral.

Geometry of the oxoacids (based on groups around Cl):

HOCl (HClO): Cl has 1 O + 3 lone pairs — bent/linear-ish.

HOClO (HClO $_2$ ): 2 O + 2 lone pairs — bent (pyramidal-ish). …
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←Previous questionQ40 — Explain why PCl5 is trigonal bipyramidal whereas IF5 is square pyramid…→Next questionQ42 — Write Lewis structure of the following compounds and show formal charg…

Q.In both water and dimethyl ether (CH3—O—CH3), oxygen atom is central atom, and has the same hybridisation, yet they have different bond angles. Which one has greater bond angle? Give reason.

Concept understanding — VSEPR Theory

VSEPR Theory: Why Molecules Have the Shapes They Do

The Core Intuition

The Precise Statement

How to Predict Shape in 3 Steps

The Common Geometries at a Glance

Why Lone Pairs Squeeze Bond Angles

1. ClF₃ (Chlorine trifluoride)

2. BrF₅ (Bromine pentafluoride)

3. SF₄ (Sulfur tetrafluoride)

Q.In both water and dimethyl ether (CH3—O—CH3), oxygen atom is central atom, and has the same hybridisation, yet they have different bond angles. Which one has greater bond angle? Give reason.

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