The Valence Shell Electron Pair Repulsion: The VSEPR theory or the Valence Shell Electron Repulsion (VSEPR) theory is a model to predict the structure of molecules and polyatomic ions in which the central atom is a metal as well as a nonmetal. Lewis structures alone can not predict the structure of all molecules, but the Lewis structure, together with the VSEPR model, can predict the geometry of each atom in a molecule.

Postulates of VSEPR Theory

1. The number of valence shell electron pairs (bonded or non-bonded) around the central atom determines the shape of the molecule.
2. The electron pairs in the valence shell repel one another due to the negatively charged electron clouds.
3. In order to minimise repulsion, the electron pairs tend to occupy such positions in space that maximises the distance between them. 
4. The valence shell is taken as a sphere with the electron pairs localising on the spherical surface at a maximum distance from one another. 
5. A multiple bonds is treated as a single electron pair, and the two or three electron pairs of multiple bonds are treated as a single super pair. 
6. The VSEPR model is applicable to all resonating structures that represent a molecule.

The main idea of VSEPR theory is the repulsion between pairs of electrons (in bonds and lone pairs).

Lone pair (lp) – Lone pair (lp) > Lone pair (lp) – Bond pair (bp) > Bond pair (bp) – Bond pair (bp)

While the bonded pairs are shared between two atoms, the lone pairs are localised on the central atom. As a result, the lone pair electrons in a molecule occupy more space than the bonding pairs of electrons resulting in greater repulsion between lone pairs of electrons than the lone pair-bond pair and bond pair-bond pair repulsions. These repulsion effects result in deviations from idealised shapes and alterations in bond angles in molecules.

The molecule or polyatomic ion is given an \({\rm{A}}{{\rm{X}}_{\rm{m}}}{{\rm{E}}_{\rm{n}}}\) designation, where \({\rm{A}}\) is the central atom, \({\rm{X}}\) is a bonded atom, \({\rm{E}}\) is a non-bonding valence electron group (usually a lone pair of electrons), and \({\rm{m}}\) and \({\rm{n}}\) are integers. The number of groups is equal to the sum of \({\rm{m}}\) and \({\rm{n}}.\)

For the prediction of geometrical shapes of molecules with the help of VSEPR theory, the molecules are divided into two categories as-

1. Molecules in which the central atom has no lone pair, and 
2. Molecules in which the central atom has one or more lone pairs.

A. The geometries that are predicted from VSEPR when a central atom has no lone pair of electrons, but only bonded groups \(\left( {{\rm{n = 0}}} \right)\) are listed below

Groups around the central atom \(\left( {{\rm{m + 0}}} \right)\)	Geometry Name	Geometry Sketch	Predicted bond Angle	Example
\(2\)	linear		\({\rm{180}}^\circ \)
\(3\)	trigonal plane		\({\rm{120}}^\circ \)
\(4\)	tetrahedron		\({\rm{109.5}}^\circ \)
\(5\)	trigonal bipyramid		\({\rm{90}}^\circ \) and \({\rm{120}}^\circ \)
\(6\)	octahedron		\({\rm{90}}^\circ \)
\(7\)	pentagonal bipyramid		\({\rm{90}}^\circ \) and \({\rm{72}}^\circ \)
\(8\)	square antiprism		\(70.5^\circ ,\,99.6^\circ \) and \({\rm{109.5}}^\circ \)

B. The geometries that are predicted from VSEPR when a central atom has one or more lone pairs of electrons along with bond pairs are-

2 Electron Pairs

In \(2\) electron pair molecules, there are-
\(2\) bond pairs \({\rm{ + 0}}\) lone pair = linear
Hence, \({\rm{A}}{{\rm{X}}_{\rm{m}}}{{\rm{E}}_{\rm{n}}}\) for \(2\) electron pairs is \({\rm{A}}{{\rm{X}}_2}\) type.
\({\rm{A}}{{\rm{X}}_2}\) molecules: \({\rm{Be}}{{\rm{H}}_{\rm{2}}}\)
In \({\rm{Be}}{{\rm{H}}_{\rm{2}}},\)  the central atom, beryllium, contributes two valence electrons to the \({\rm{Be}}{{\rm{H}}_{\rm{2}}}\) structure. Each of the two hydrogen atoms contributes one valence electron to the \({\rm{Be}}{{\rm{H}}_{\rm{2}}}\) structure.

The Lewis electron structure of \({\rm{Be}}{{\rm{H}}_{\rm{2}}}\) is-

Both groups around the central atom are bonding pairs (BP). Thus, \({\rm{Be}}{{\rm{H}}_{\rm{2}}}\) is designated as \({\rm{A}}{{\rm{X}}_{\rm{2}}}\). Hence, according to VSEPR theory, the molecular geometry that minimises repulsions in \({\rm{Be}}{{\rm{H}}_{\rm{2}}}\) is linear.

Electron pairs	Bond pair (bp)	Lone pair (lp)	Bond angle	Geometry
\(2\)	\(2\)	\(0\)	\({\rm{180}}^\circ \)	Linear

3 Electron Pairs

In three electron pairs, there are-

\(3\) bond pairs \(+ \,0\) lone pair = trigonal planar
Hence, \({\rm{A}}{{\rm{X}}_{\rm{m}}}{{\rm{E}}_{\rm{n}}}\) for \(3\) electron pairs is \({\rm{A}}{{\rm{X}}_3}\) type.
\({\rm{A}}{{\rm{X}}_3}\) molecules: \({\rm{BC}}{{\rm{l}}_{\rm{3}}}\)

In \({\rm{BC}}{{\rm{l}}_{\rm{3}}}\) the central atom, boron, contributes three valence electrons to the \({\rm{BC}}{{\rm{l}}_{\rm{3}}}\) structure.
However, each of the chlorine atoms contributes seven valence electrons to the \({\rm{BC}}{{\rm{l}}_{\rm{3}}}\) structure.
The Lewis electron dot structure of \({\rm{BC}}{{\rm{l}}_{\rm{3}}}\) is-

All electron groups surrounding the Boron central atom are bonding pairs (BP), so the structure is designated as \({\rm{A}}{{\rm{X}}_3}.\)

Hence, according to VSEPR theory, the molecular geometry that minimises repulsions in \({\rm{BC}}{{\rm{l}}_{\rm{3}}}\) is trigonal planar.

\(2\) bond pairs \( + \,1\) lone pair = bent

A three electron pairs system consisting of \(2\) bond pairs and \(1\) lone pair consists of a bent structure
Hence, \({\rm{A}}{{\rm{X}}_{\rm{m}}}{{\rm{E}}_{\rm{n}}}\) for \(3\) electron pairs is \({\rm{A}}{{\rm{X}}_2}{\rm{E}}\) type.
\({\rm{A}}{{\rm{X}}_2}{\rm{E}}\) molecules: \({\rm{S}}{{\rm{O}}_{\rm{2}}}\)

In \({\rm{S}}{{\rm{O}}_{\rm{2}}},\)  the central atom, sulfur, has \(6\) valence electrons. So does each oxygen atom. Hence, there are \(18\) valence electrons in total in \({\rm{S}}{{\rm{O}}_{\rm{2}}}\) molecule. With \(18\) valence electrons, the Lewis dot electron structure is-

The \({\rm{S}}{{\rm{O}}_{\rm{2}}}\) the molecule consists of two double bonds and one lone pair, so the structure is designated as \({\rm{A}}{{\rm{X}}_2}{\rm{E}}.\) Hence, we have one BP–BP interaction and two LP–BP interactions.

The molecular geometry as described by VSEPR is bent or V-shaped. The O-S-O bond angle is expected to be less than \({\rm{120}}^\circ \)  because of the extra space taken up by the lone pair.

To summarise:

Electron pairs	Bond pair (bp)	Lone pair (lp)	Bond angle	Geometry
\(3\)	\(3\)	\(0\)	\({\rm{120}}^\circ \)	Trigonal planar
	\(2\)	\(1\)	\({\rm{118}}^\circ \)	Bent

4 Electron Pairs

In \(4\) electron pairs, there are:

\(4\) bond pairs \( + \,0\) lone pair = tetrahedral
Hence, \({\rm{A}}{{\rm{X}}_{\rm{m}}}{{\rm{E}}_{\rm{n}}}\) for \(4\) electron pair is \({\rm{A}}{{\rm{X}}_4}\) type.
\({\rm{A}}{{\rm{X}}_4}\) molecules: \({\rm{C}}{{\rm{H}}_4}\)

In \({\rm{C}}{{\rm{H}}_4},\) the central atom, carbon, contributes four valence electrons to the \({\rm{C}}{{\rm{H}}_4}\) structure.
Each of the four hydrogen atoms contributes one valence electron to the \({\rm{C}}{{\rm{H}}_4}\) structure.

The Lewis electron dot structure is-

In \({\rm{C}}{{\rm{H}}_4},\) there are four electron groups around the central carbon atom. As all the electron groups are bonding pairs, the structure is designated as \({\rm{A}}{{\rm{X}}_4}.\)

The repulsions between the groups are minimised by placing the groups in the corners of a tetrahedron with bond angles of \(109.5^\circ .\)
With four bonding pairs, according to VSEPR theory, the molecular geometry of methane is tetrahedral.

\(3\) bond pairs \( + \,1\) lone pair = trigonal pyramidal
Hence, \({\rm{A}}{{\rm{X}}_{\rm{m}}}{{\rm{E}}_{\rm{n}}}\) for \(4\) electron pair is \({\rm{A}}{{\rm{X}}_3}{\rm{E}}\) type.
\({\rm{A}}{{\rm{X}}_3}{\rm{E}}\) molecules: \({\rm{N}}{{\rm{H}}_3}\)

In ammonia, the central atom, nitrogen, has five valence electrons, and each of the three hydrogen atoms donates one valence electron to the \({\rm{N}}{{\rm{H}}_3}\) structure.
The Lewis electron dot structure of \({\rm{N}}{{\rm{H}}_3}\) is-

There are four electron groups (three bonding pairs and one lone pair) around the central nitrogen atom and is designated as \({\rm{A}}{{\rm{X}}_3}{\rm{E}}\) type. The repulsion between the groups is minimised by directing each hydrogen atom and the lone pair to the corners of a tetrahedron.
The LP–BP interactions cause H–N–H bond angles to deviate significantly from the angles of a perfect tetrahedron.

The molecular geometry according to the VSEPR theory of \({\rm{A}}{{\rm{X}}_3}{\rm{E}}\) type molecule is trigonal pyramidal.

\(2\) bond pairs \( + \,2\) lone pairs = bent

A four-electron pair system, consisting of \(2 \)bond pairs and \(2\) lone pairs results in a bent structure.
Hence, \({\rm{A}}{{\rm{X}}_{\rm{m}}}{{\rm{E}}_{\rm{n}}}\) for \(4\) electron pair is \({\rm{A}}{{\rm{X}}_2}{{\rm{E}}_2}\) type.
\({\rm{A}}{{\rm{X}}_2}{{\rm{E}}_2}\) molecules: \({{\rm{H}}_{\rm{2}}}{\rm{O}}\)

In a water molecule, the Oxygen atom contributes six valence electrons, and each of the two hydrogen atoms contributes one valence electron.
The Lewis electron dot structure of \({{\rm{H}}_{\rm{2}}}{\rm{O}}\) is-

With two bonding pairs and two lone pairs, the structure is designated as \({\rm{A}}{{\rm{X}}_2}{{\rm{E}}_2}\) type with a total of four electron pairs. The water molecule acquires a bent or V shape due to significant LP–LP, LP–BP, and BP–BP interactions. The H–O–H bond angle is even less than the H–N–H angles in \({\rm{N}}{{\rm{H}}_3}.\)

To summarise-

Electron pairs	Bond pair (bp)	Lone pair (lp)	Bond angle	Geometry
\(4\)	\(4\)	\(0\)	\(109.5^\circ \)	Tetrahedral
	\(3\)	\(1\)	\(107^\circ \)	Trigonal pyramidal
	\(2\)	\(2\)	\(105^\circ \)	Bent

5 Electron Pairs

\(5\) bond pairs \(+ \,0\) lone pair = trigonal bipyramidal

A five-electron pair system consisting of \(5\) bond pairs and \(0\) lone pairs comprises a trigonal bipyramidal structure.
Hence, \({\rm{A}}{{\rm{X}}_{\rm{m}}}{{\rm{E}}_{\rm{n}}}\) for \(5\) electron pair is \({\rm{A}}{{\rm{X}}_{\rm{5}}}\) type.
\({\rm{A}}{{\rm{X}}_{\rm{5}}}\) molecules: \({\rm{PC}}{{\rm{l}}_5}\)

In \({\rm{PC}}{{\rm{l}}_5},\)  the Phosphorus central atom contributes five valence electrons, and each of the five chlorine atoms contributes seven valence electrons. Hence, there are \(40\) valence electrons depicted by Lewis’s electron structure as shown-

All electron groups surrounding the central phosphorus atom are bonding pairs, so the structure is designated as \({\rm{A}}{{\rm{X}}_5}.\)

Hence, according to VSEPR theory, the molecular geometry of \({\rm{PC}}{{\rm{l}}_5},\) is trigonal bipyramidal.
\({\rm{PC}}{{\rm{l}}_5},\) molecule has three chlorine atoms in a plane in equatorial positions and two chlorine atoms above and below the plane in axial positions. The three equatorial chlorine atoms are separated by \(120^\circ \) from one another, and the two axial positions are at \(90^\circ \) to the equatorial plane. The axial and equatorial positions are not chemically equivalent.

\(4\) bond pairs \( + \,1\) lone pair = see-saw

A five electron pair system consisting of \(4\) bond pairs and \(1\) lone pair comprises a seesaw structure.
Hence, \({\rm{A}}{{\rm{X}}_{\rm{m}}}{{\rm{E}}_{\rm{n}}}\) for \(5\) electron pair is \({\rm{A}}{{\rm{X}}_{\rm{4}}}{\rm{E}}\) type.
\({\rm{A}}{{\rm{X}}_{\rm{4}}}{\rm{E}}\) molecules: \({\rm{S}}{{\rm{F}}_{\rm{4}}}\)

In \({\rm{S}}{{\rm{F}}_{\rm{4}}},\) the sulphur atom has six valence electrons, and each of the four fluorine atoms contributes seven valence electrons to the structure of \({\rm{S}}{{\rm{F}}_{\rm{4}}}.\)

The Lewis electron structure of \({\rm{S}}{{\rm{F}}_{\rm{4}}}\) is-

With an expanded valence, this species is an exception to the octet rule.

There are five groups around sulphur (four bonding pairs and one lone pair) atom; hence the lowest energy arrangement is trigonal bipyramid.
The lone pair of electrons are placed in the equatorial position and are more stable than the lone pair in the axial position.

The molecular structure is based on VSEPR theory is trigonal bipyramid described as a seesaw. The \({{\rm{F}}_{{\rm{axial}}}}{\rm{ – S – }}{{\rm{F}}_{{\rm{axial}}}}\) angle is \({\rm{173}}^\circ \) rather than \({\rm{180}}^\circ \) because of the lone pair of electrons in the equatorial plane.

\(3\) bond pairs \(+ \,2\) lone pairs = T-shape

A five-electron pair system consisting of \(3\) bond pairs and \(2\) lone pairs comprises a T-shape.
Hence, \({\rm{A}}{{\rm{X}}_{\rm{m}}}{{\rm{E}}_{\rm{n}}}\) for \(5\) electron pair is \({\rm{A}}{{\rm{X}}_3}{{\rm{E}}_2}\) type.
\({\rm{A}}{{\rm{X}}_3}{{\rm{E}}_2}\) molecules: \({\rm{Br}}{{\rm{F}}_{\rm{3}}}.\)

In \({\rm{Br}}{{\rm{F}}_{\rm{3}}},\)  the bromine atom contributes seven valence electrons, and each of the three fluorine atoms contributes seven valence electrons to the \({\rm{Br}}{{\rm{F}}_{\rm{3}}}\) molecule.
The Lewis electron dot structure of \({\rm{Br}}{{\rm{F}}_{\rm{3}}}\) is-

This molecule is an exception to the octet rule.

There are five groups around the central bromine atom, three bonding pairs and two lone pairs. Hence, the structural designation is \({\rm{A}}{{\rm{X}}_3}{{\rm{E}}_2}\) with a total of five electron pairs. Both the lone pairs are in the equatorial positions with minimum bp-lp repulsive energy. However, there is a deviation in bond angles because of the presence of the two lone pairs of electrons, and the molecule acquires a T-shape.

The \({{\rm{F}}_{{\rm{axial}}}}{\rm{ – Br – }}{{\rm{F}}_{{\rm{axial}}}}\) bond angle is \({\rm{172}}^\circ ,\) less than \({\rm{180}}^\circ \) because of LP–BP repulsions.

\(2\) bond pairs \( + \,3\) lone pairs = linear

A five-electron pair system consisting of \(2\) bond pairs and \(3\) lone pairs comprises a linear shape.
Hence, \({\rm{A}}{{\rm{X}}_{\rm{m}}}{{\rm{E}}_{\rm{n}}}\) for \(5\) electron pair is \({\rm{A}}{{\rm{X}}_2}{{\rm{E}}_3}\) type.
\({\rm{A}}{{\rm{X}}_2}{{\rm{E}}_3}\) molecules: \(I_3^ – \)

Each iodine atom in \(I_3^ – \) contributes seven electrons and the negative charge of one electron to the structure of \(I_3^ – .\)

The Lewis electron dot structure is-

There are five electron groups around the central iodine atom in \(I_3^ – ,\) two bonding pairs and three lone pairs. To minimise repulsions, the groups are directed to the corners of a trigonal bipyramid.

The lone pairs are placed in the axial positions that eliminate LP–LP repulsions.

There is no deviation in a bond angle because the three lone pairs of electrons have equivalent interactions with the three iodine atoms. The molecular geometry of \(I_3^ – \) as per VSEPR theory is linear. The ion has an I–I–I angle of \(180^\circ ,\) as expected.

Electron pairs	Bond pair (bp)	Lone pair (lp)	Bond angle	Geometry
\(5\)	\(5\)	\(0\)	\(120^\circ \) \(90^\circ \)	Trigonal bipyramidal
	\(4\)	\(1\)	\(118^\circ \) \(88^\circ \)	Seesaw
	\(3\)	\(2\)	\(86^\circ \)	T-shape
	\(2\)	\(3\)	\(180^\circ \)	Linear

6 Electron Pairs

\(6\) bond pairs \(+ \,0\) lone pair = octahedral

A six electron pair system consisting of \(6\) bond pairs and \(0\) lone pairs comprises an octahedral shape.
Hence, \({\rm{A}}{{\rm{X}}_{\rm{m}}}{{\rm{E}}_{\rm{n}}}\) for \(6\) electron pair is \({\rm{A}}{{\rm{X}}_{\rm{6}}}\) type.
\({\rm{A}}{{\rm{X}}_{\rm{6}}}\) molecules: \({\rm{S}}{{\rm{F}}_{\rm{6}}}\)

The central atom, sulfur, in \({\rm{S}}{{\rm{F}}_{\rm{6}}}\) contributes six valence electrons, and each fluorine atom has seven valence electrons, so the Lewis electron structure is-

With an expanded valence, this species is an exception to the octet rule.

There are six electron groups around the central sulphur atom, each a bonding pair. Hence, according to VSEPR theory, the geometry that minimises repulsions is octahedral.

\(5\) bond pairs \( + \,1\) lone pair = square pyramidal

A six-electron pair system consisting of \(5\) bond pairs and \(1\) lone pair comprises a square pyramidal shape.
Hence, \({\rm{A}}{{\rm{X}}_{\rm{m}}}{{\rm{E}}_{\rm{n}}}\) for \(6\) electron pair is \({\rm{A}}{{\rm{X}}_{\rm{5}}}{\rm{E}}\) type.
\({\rm{A}}{{\rm{X}}_{\rm{5}}}{\rm{E}}\) molecules: \({\rm{Br}}{{\rm{F}}_5}\)

The central atom, bromine, in \({\rm{Br}}{{\rm{F}}_5}\) contributes seven valence electrons to \({\rm{Br}}{{\rm{F}}_5}\)  molecule.
So does each fluorine atom. Hence, the Lewis electron structure is-

With its expanded valence, this species is an exception to the octet rule.

In \({\rm{Br}}{{\rm{F}}_5},\) there are six electron groups around the \({\rm{Br}}\) central atom with five bonding pairs and one lone pair. Placing five \({\rm{F}}\) atoms around \({\rm{Br}}\) while minimising BP–BP, and LP–BP repulsions gives the following structure:

The \({\rm{Br}}{{\rm{F}}_5}\) structure has four fluorine atoms: coplanar in an equatorial position, one fluorine atom, and a lone pair of electrons in the axial positions.

This molecular structure is square pyramidal. The \({{\rm{F}}_{{\rm{axial}}}}{\rm{ – B – }}{{\rm{F}}_{{\rm{equatorial}}}}\) angles are \({\rm{85}}{\rm{.1}}^\circ ,\) less than \(90^\circ \) because of LP–BP repulsions.

\(4\) bond pairs \(+ \,2\) lone pairs = square planar

A six electron pair system consisting of \(4\) bond pairs and \(2\) lone pairs comprises a square planar shape.
Hence, \({\rm{A}}{{\rm{X}}_{\rm{m}}}{{\rm{E}}_{\rm{n}}}\) for \(6\) electron pair is \({\rm{A}}{{\rm{X}}_{\rm{4}}}{{\rm{E}}_2}\) type.
\({\rm{A}}{{\rm{X}}_{\rm{4}}}{{\rm{E}}_2}\) molecules: \({\rm{ICl}}_{\rm{4}}^{\rm{ – }}\)

The central atom, iodine, contributes seven electrons to the structure of \({\rm{ICl}}_{\rm{4}}^{\rm{ – }}.\) Also, each chlorine contributes seven electrons, and there is a single negative charge. The Lewis electron structure is-

There are six electron groups around the central atom, four bonding pairs, and two lone pairs. The molecular geometry that minimises LP–LP, LP–BP, and BP–BP repulsions is square planar.

\(3\) bond pairs \({\rm{ + }}\,{\rm{3}}\) lone pairs = T-shape

A six electron pair system consisting of \(3\) bond pairs and \(3\) lone pairs comprises a T-shape.
Hence, \({\rm{A}}{{\rm{X}}_{\rm{m}}}{{\rm{E}}_{\rm{n}}}\) for \(6\) electron pair is \({\rm{A}}{{\rm{X}}_{\rm{3}}}{{\rm{E}}_3}\) type.

In \({\rm{A}}{{\rm{X}}_{\rm{3}}}{{\rm{E}}_3}\) type, there are \(6\) groups around the central atom, three bonding pairs, and \(3\) lone pairs. The lone pairs are in the equatorial positions with minimum bp-lp repulsive energy. However, there is a deviation in bond angles because of the presence of the \(3\) lone pairs of electrons, and the molecule acquires a T-shape

The \({{\rm{X}}_{{\rm{axial}}}}{\rm{ – A – }}{{\rm{X}}_{{\rm{axial}}}}\) bond angle is less than \({\rm{180}}^\circ \) because of LP–BP repulsions.

\(2\) bond pairs \( + \,4\) lone pairs = linear

A six electron pair system consisting of \(2\) bond pairs and \(4\) lone pairs comprises a linear shape.
Hence, \({\rm{A}}{{\rm{X}}_{\rm{m}}}{{\rm{E}}_{\rm{n}}}\) for \(6\) electron pair is \({\rm{A}}{{\rm{X}}_{\rm{2}}}{{\rm{E}}_4}\) type.

In \({\rm{A}}{{\rm{X}}_{\rm{2}}}{{\rm{E}}_4}\) type, there are six electron groups around the central atom, two bonding pairs, and four lone pairs.

The four lone pairs of electrons have equivalent interactions with the central atom; hence there is no deviation in bonding angles. The ion has an X–A–X angle of \(180^\circ ,\) as expected.

Electron pairs	Bond pair (bp)	Lone pair (lp)	Bond angle	Geometry
\(6\)	\(6\)	\(0\)	\(90^\circ \)	Octahedral
	\(5\)	\(1\)	\(88^\circ \)	Square pyramidal
	\(4\)	\(2\)	\(90^\circ \)	Square planar
	\(3\)	\(3\)	\(88^\circ \)	T-shape
	\(2\)	\(4\)	\(180^\circ \)	Linear

Limitations of VSEPR Theory

VSEPR fails for isoelectronic species: Isoelectronic species are elements, ions, and molecules that consist of an equal number of electrons. The VSEPR shapes of molecules are based on the number of valence electrons (i.e., bond pairs and lone pairs). Hence, two isoelectronic species, in spite of having the same number of valence electrons, can differ in geometry.

For example, both \({\rm{I}}{{\rm{F}}_{\rm{7}}}\) and \({\left[ {{\rm{Te}}{{\rm{F}}_{\rm{7}}}} \right]^ – }\) have \(56\) valence electrons; hence, their structure through VSEPR theory predicts should be pentagonal bipyramidal. But, experimental data indicate that the equatorial F atoms of \({\left[ {{\rm{Te}}{{\rm{F}}_{\rm{7}}}} \right]^ – }\) are not coplanar, the bond lengths of equatorial \({\rm{I – F}}\) and \({\rm{Te – F}}\) bonds are also different.

VSEPR fails for transition metal compounds: Due to the inert pair effect of d-block elements, the valence shell s-electrons in these elements tend to adopt a non-bonding role, and the VSEPR theory fails to estimate the correct geometries for these transitional metal complex compounds. For example, in \({\left[ {{\rm{SeC}}{{\rm{l}}_{\rm{6}}}} \right]^{{\rm{2 – }}}}{\rm{,}}{\left[ {{\rm{TeC}}{{\rm{l}}_{\rm{6}}}} \right]^{{\rm{2 – }}}}\) and \({\left[ {{\rm{Br}}{{\rm{F}}_{\rm{6}}}} \right]^{\rm{ – }}}\) the VSEPR structure is predicted to be pentagonal bipyramidal geometry.

However, due to the stereochemical inert pair effect, these molecules are found to be regular octahedral. This is because one of the electron pairs in these compounds is stereochemically inactive.

Solved Examples

Example 1: Using the VSEPR model, predict the molecular geometry of the given ion.
\({{\rm{H}}_{\rm{3}}}{{\rm{O}}^{\rm{ + }}}\) (hydronium ion)
Steps to follow:

Step 1: Draw the Lewis dot electron structure of the molecule or polyatomic ion.
The central atom, \({\rm{O}},\) has six valence electrons, and each \({\rm{H}}\) atom contributes one valence electron to the \({{\rm{H}}_{\rm{3}}}{{\rm{O}}^{\rm{ + }}}\) structure. The Lewis dot electron structure is-

Step 2: Determination of the electron group arrangement around the central atom that minimises repulsions.
There are four electron groups (three bonding pairs and one lone pair) around the central oxygen atom. Like \({\rm{N}}{{\rm{H}}_{\rm{3}}}{\rm{r}}\) epulsions are minimised by directing each hydrogen atom and the lone pair to the corners of a tetrahedron.

Step 3: An \({\rm{A}}{{\rm{X}}_{\rm{m}}}{{\rm{E}}_{\rm{n}}}\) designation is assigned; then LP–LP, LP–BP, or BP–BP interactions are identified to predict deviations in bond angles.
With three bonding pairs and one lone pair, the structure is designated as \({\rm{A}}{{\rm{X}}_{\rm{3}}}{\rm{E}}.\)  Due to LP–BP interactions, the bonding pair of angles deviate significantly from the angles of a perfect tetrahedron.

Step 4: Describe the molecular geometry.
The molecular geometry, as per VSEPR theory, is trigonal pyramidal. The H–O–H bond angles are less than the ideal angle of \({\rm{109}}{\rm{.5}}^\circ \) due to LP–BP repulsions:

Summary

The Lewis electron-pair approach provides an insight into the number and types of bonds between the atoms and the lone pairs of electrons on the constituting atoms. However, it fails to explain the actual arrangement of atoms in space. Therefore, the valence-shell electron-pair repulsion (VSEPR) model and Lewis electron dot structure are used to predict the shapes of many molecules and polyatomic ions. However, this model provides no information about bond lengths or the presence of multiple bonds.

In this article, we learned the different postulates of VSEPR theory along with its limitations. We learned the various steps essential in predicting the shape and geometry of a polyatomic ion or molecule. We also learned how the number of bonded and non bonded electrons play a vital role in determining the shape of molecules.

Frequently Asked Questions

Q.1. What are the two main problems with VSEPR theory?
Ans: Although the VSEPR model is useful in predicting the molecular geometry of polyatomic ions or molecules, it fails to predict the shapes of isoelectronic species and transition metal compounds. The VSEPR model does not take into account the stereochemically inactive lone pairs and relative sizes of substituents.

Q.2. What drives the VSEPR theory?
Ans: The VSEPR theory is based on the fact that the bonded and non-bonded electron pairs repel each other and will therefore adopt a geometry that places these electron pairs as far as possible to minimise repulsion.

Q.3.What is the importance of VSEPR theory?
Ans: VSEPR theory predicts the shape of nearly all polyatomic ions or molecules with a central atom, as long as the central atom is not a metal. Each shape has a name and an idealised bond angle associated with it.

Q.4. What are the 5 Vsepr shapes?
Ans: The VSEPR theory describes five main shapes of simple molecules: linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral.

Q.5. How to determine the VSEPR shape of a molecule or an ion?
Ans: The following steps are followed in determining the VSEPR shape of a molecule or an ion:
1. Count the number of valence electrons.
2. Find the number of valence electrons of bonded atoms.
3. Find the number of lone pairs on the central atom by subtracting the number of valence electrons on bonded atoms (Step 2) from the total number of valence electrons (Step 1).
Divide the number of VEs not in bonds (from Step 3) by \(2\) to find the number of LPs. For example,
If the number is \(0,\) there are no lone pairs on the central atom.
If the number is \(2,\) there is one lone pair on the central atom.
If the number is \(4,\) there are two lone pairs on the central atom.
If the number is \(6,\) there are three lone pairs on the central atom.
Use the Table of VSEPR electron and molecular geometries (Table 1) to determine the VSEPR geometry.

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