(C) In $SeCl_4$,the central atom $Se$ has $6$ valence electrons. It forms $4$ bonds with $Cl$ atoms and has $1$ lone pair of electrons. Thus,the total

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

If the Assertion is correct but Reason is incorrect.

(C) In $SeCl_4$,the central atom $Se$ has $6$ valence electrons. It forms $4$ bonds with $Cl$ atoms and has $1$ lone pair of electrons. Thus,the total number of electron pairs is $4 + 1 = 5$,which corresponds to $sp^3d$ hybridization. Due to the presence of one lone pair,the geometry is a see-saw shape,not tetrahedral. Therefore,the Assertion is correct,but the Reason is incorrect because $Se$ has only one lone pair,not two.

Class 11 Chemistry Chemical Bonding and Molecular Structure VSEPR Theory

Difficult

Assertion : $SeCl_4$ does not have a tetrahedral structure.
Reason : $Se$ in $SeCl_4$ has two lone pairs.

AIIMS 2005 All AIIMS

A
If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
B
If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
C
If the Assertion is correct but Reason is incorrect.
D
If both the Assertion and Reason are incorrect.

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Chemistry Questions

Chapter

Chemical Bonding and Molecular Structure

Subtopic

VSEPR Theory

Previous Year

AIIMS 2005 Paper All AIIMS years →

Match the molecules in List-$I$ with their respective shapes in List-$II$.
List-$I$ (Molecule) List-$II$ (Shape)
$A. SF_4$ $I. T\text{-shaped}$
$B. ClF_3$ $II. \text{Square planar}$
$C. BrF_5$ $III. \text{See-saw}$
$D. XeF_4$ $IV. \text{Square pyramidal}$

List-$I$ (Molecule)	List-$II$ (Shape)
$A. SF_4$	$I. T\text{-shaped}$
$B. ClF_3$	$II. \text{Square planar}$
$C. BrF_5$	$III. \text{See-saw}$
$D. XeF_4$	$IV. \text{Square pyramidal}$

EasyTS EAMCET 2024

View Solution

The five $P-F$ bond lengths in $PF_5$ are equal whereas the $P-Cl$ bond lengths in $PCl_5$ are not equal because

Medium

View Solution

In which of the following molecules are the bond lengths not equal?

Easy

View Solution

How will you account for the $104.5^{\circ}$ bond angle in water?

Easy

View Solution

Which of the following molecules has a regular geometry as expected?

MediumMHT CET 2025

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Assertion : $SeCl_4$ does not have a tetrahedral structure.Reason : $Se$ in $SeCl_4$ has two lone pairs.

Similar Questions

Match the molecules in List-$I$ with their respective shapes in List-$II$.List-$I$ (Molecule)List-$II$ (Shape)$A. SF_4$$I. T\text{-shaped}$$B. ClF_3$$II. \text{Square planar}$$C. BrF_5$$III. \text{See-saw}$$D. XeF_4$$IV. \text{Square pyramidal}$

The five $P-F$ bond lengths in $PF_5$ are equal whereas the $P-Cl$ bond lengths in $PCl_5$ are not equal because

In which of the following molecules are the bond lengths not equal?

How will you account for the $104.5^{\circ}$ bond angle in water?

Which of the following molecules has a regular geometry as expected?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Assertion : $SeCl_4$ does not have a tetrahedral structure.
Reason : $Se$ in $SeCl_4$ has two lone pairs.

Match the molecules in List-$I$ with their respective shapes in List-$II$.
List-$I$ (Molecule) List-$II$ (Shape)
$A. SF_4$ $I. T\text{-shaped}$
$B. ClF_3$ $II. \text{Square planar}$
$C. BrF_5$ $III. \text{See-saw}$
$D. XeF_4$ $IV. \text{Square pyramidal}$