A central atom in a molecule has two lone pairs of electrons and forms three single bonds. The shape

Q: Why do lone pairs go to the equatorial positions instead of the axial ones?

Lone pairs exert more repulsion than bond pairs. In a Trigonal Bipyramidal geometry, equatorial positions are $120^\circ$ apart, while axial positions are only $90^\circ$ from the equatorial plane. Placing lone pairs in equatorial positions maximizes their distance from other electron pairs, minimizing overall repulsion.

Q: What is the difference between electron geometry and molecular shape?

Electron geometry considers the arrangement of ALL electron pairs (both bonds and lone pairs). Molecular shape, however, only describes the arrangement of the actual atoms. We use the electron geometry as a base, place the lone pairs, and then 'hide' them to see the final molecular shape.

Q: Why isn't the shape trigonal planar if there are three bonds?

Trigonal planar shape occurs when a central atom has exactly three bond pairs and ZERO lone pairs (steric number 3, $sp^2$ hybridisation). Here, the two lone pairs push the bonds out of a simple planar arrangement, fundamentally changing the base geometry to Trigonal Bipyramidal.

Q: How do I know the general formula is $AB_3L_2$?

The formula is just a shorthand. 'A' stands for the central atom. 'B' stands for the bonded atoms (we have 3, so $B_3$). 'L' stands for lone pairs (we have 2, so $L_2$). It helps quickly categorize the molecule type.

Q: What would the shape be if there was only one lone pair?

If there were 4 bond pairs and 1 lone pair ($AB_4L_1$), the steric number would still be 5. The single lone pair would occupy one equatorial position, pushing the remaining four atoms into a 'See-saw' shape.

Question

A central atom in a molecule has two lone pairs of electrons and forms three single bonds. The shape of this molecule is

Options:

(A) see-saw

(B) planar triangular

(C) T-shaped

(D) trigonal pyramidal

Sigma Percentile · Accepted Answer

### Analyzing the Setup
We are given a central atom that forms three single bonds and possesses two lone pairs of electrons. Our goal is to determine the final shape of this molecule.

To do this, we first need to find the **steric number**, which dictates the electron geometry around the central atom.

### The Master Equation: Steric Number
The steric number is the sum of the number of bond pairs ($BP$) and the number of lone pairs ($L$).

$$ 	ext{Steric Number} = BP + L $$

Substituting our given values:

$$ 	ext{Steric Number} = 3 + 2 = 5 $$

A steric number of 5 corresponds to an **$sp^3d$ hybridisation**.

### Electron Geometry vs. Molecular Shape
For an $sp^3d$ hybridized central atom, the base electron geometry is **Trigonal Bipyramidal (TBP)**.

In a TBP geometry, there are two types of positions:
1. **Equatorial positions:** Three positions lying in a plane, separated by $120^\circ$ angles.
2. **Axial positions:** Two positions perpendicular to the equatorial plane, separated by $90^\circ$ from the equatorial bonds.

### Minimizing Repulsion
According to **VSEPR (Valence Shell Electron Pair Repulsion) theory**, lone pairs exert greater repulsive forces than bond pairs. To minimize this repulsion, lone pairs will always occupy the positions that offer the maximum bond angle.

In the TBP geometry, the equatorial positions offer a $120^\circ$ angle, whereas axial positions only offer $90^\circ$. Therefore, the two lone pairs will occupy two of the three equatorial positions.

### Final Calculation: The Resulting Shape
With the two lone pairs safely tucked away in the equatorial plane, we are left with three bonded atoms: one in the remaining equatorial position, and two in the axial positions.

When we look at the molecule, we only "see" the atoms, not the lone pairs. The arrangement of these three atoms around the central atom forms a distinct **T-shape**.

Thus, a molecule with the general formula $AB_3L_2$ will always have a **T-shaped** molecular geometry.

Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: A central atom in a molecule has two lone pairs of electrons and forms three single bonds. The shape of this molecule is

Select Answer:

Visualized Solution

$A na l y z in g t h e G i v e n D a t a$

$S t er i c N u mb er an d H y b r i d i s a t i o n$

$V S E P R T h eor y an d G eo m e t r y$

$P os i t i o nin g t h e L o n e P ai r s$

$D e t er minin g t h e F ina l S ha p e$

$E x pl or in g R e l a t e d S t r u c t u r es$

The Sigma Insight: Hybridisation and VSEPR Theory

Analyzing the Setup We are given a central atom that forms three single bonds and possesses two lone pairs of electrons

The Master Equation

Electron Geometry vs

Minimizing Repulsion According to VSEPR (Valence Shell Electron Pair Repulsion) theory, lone pairs exert greater repulsive forces than bond pairs

Final Calculation

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List-I

List-II

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Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: A central atom in a molecule has two lone pairs of electrons and forms three single bonds. The shape of this molecule is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Given Data

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Steric Number and Hybridisation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VSEPR Theory and Geometry

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Positioning the Lone Pairs

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Determining the Final Shape

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Exploring Related Structures