Question:

Among the following, which one is a wrong statement?

Updated On: Apr 20, 2025

$PH_5 $ and $BiCl_5$ do not exist
$p\pi -d \pi$ bonds are present in $SO_2$
$SeF_4$ and $CH_4$ have same shape
$I^+_3 $ has bent geometry

Hide Solution

Verified By Collegedunia

The Correct Option is C

Approach Solution - 1

Option 1: $PH_5$ and $BiCl_5$ do not exist.
✅ Correct Statement: $PH_5$ does not exist due to the absence of empty d-orbitals in Phosphorus. Similarly, $BiCl_5$ is unstable due to the inert pair effect in Bismuth.

Option 2: $p\pi - d\pi$ bonds are present in $SO_2$.
✅ Correct Statement: $SO_2$ has resonance structures involving $p\pi - d\pi$ bonding between sulfur and oxygen.

Option 3: $SeF_4$ and $CH_4$ have same shape.
❌ Wrong Statement:
- $CH_4$ has a tetrahedral shape (no lone pairs on central atom).
- $SeF_4$ has a see-saw shape due to the presence of one lone pair on selenium.

Option 4: $I_3^+$ has bent geometry.
✅ Correct Statement: The ion $I_3^+$ has a bent shape due to lone pair-lone pair and lone pair-bond pair repulsions.

Final Answer:
✅ $Option 3: SeF_4\ and\ CH_4\ have\ same\ shape$ is the wrong statement.

Was this answer helpful?

Hide Solution

Verified By Collegedunia

Approach Solution -2

Shape of Molecules

The shape of a molecule is determined by the VSEPR (Valence Shell Electron Pair Repulsion) theory, which takes into account the number of bonding and lone pairs around the central atom.

1. Shape of $ \text{SeF}_4 $ (Selenium Tetrafluoride)

The shape of $ \text{SeF}_4 $ is **see-saw**. This occurs because selenium (Se) in $ \text{SeF}_4 $ has five regions of electron density around it—four bonding pairs from fluorine and one lone pair. According to VSEPR theory, these five regions of electron density arrange themselves in a trigonal bipyramidal shape, but the lone pair occupies one of the equatorial positions, causing a **see-saw** shape for the molecule.

2. Shape of $ \text{CH}_4 $ (Methane)

The shape of $ \text{CH}_4 $ is **tetrahedral**. Carbon (C) in $ \text{CH}_4 $ has four bonding pairs of electrons from four hydrogen atoms and no lone pairs. According to VSEPR theory, these four bonding pairs arrange themselves in a tetrahedral geometry to minimize repulsion between them, resulting in a tetrahedral shape.

Conclusion:

Therefore, the shape of $ \text{SeF}_4 $ is **see-saw**, while the shape of $ \text{CH}_4 $ is **tetrahedral**.

Was this answer helpful?

Questions Asked in NEET exam

View More Questions

Concepts Used:

Chemical Bonding and Molecular Structure

Such a group of atoms is called a molecule. Obviously, there must be some force that holds these constituent atoms together in the molecules. The attractive force which holds various constituents (atoms, ions, etc.) together in different chemical species is called a chemical bond.

Types of Chemical Bonds:

There are 4 types of chemical bonds which are formed by atoms or molecules to yield compounds.

Ionic Bonds - Ionic bonding is a type of chemical bonding which involves a transfer of electrons from one atom or molecule to another.
Covalent Bonds - Compounds that contain carbon commonly exhibit this type of chemical bonding.
Hydrogen Bonds - It is a type of polar covalent bonding between oxygen and hydrogen wherein the hydrogen develops a partial positive charge
Polar Bonds - In Polar Covalent chemical bonding, electrons are shared unequally since the more electronegative atom pulls the electron pair closer to itself and away from the less electronegative atom.

Factors Affecting Bond Enthalpy in Chemical Bonding:

Size of the Atom
Multiplicity of Bonds
Number of Lone Pair of Electrons Present
Bond Angle