Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 133

ClCCI ClC(Cl)(Cl)Cl The \(\mathrm{Cl}^{\text { }} \mathrm{Cl}\) bond angle in \(1,1,2,2\) -tetrachloroethene and tetrachloromethane respectively are about (a) \(90^{\circ}\) and \(109.5^{\circ}\) (b) \(109.5^{\circ}\) and \(120^{\circ}\) (c) \(120^{\circ}\) and \(109.5^{\circ}\) (d) \(120^{\circ}\) and \(120^{\circ}\)

Short Answer

Expert verified
Answer: The bond angle in 1,1,2,2-tetrachloroethene is about 120° and in tetrachloromethane is about 109.5°.

Step by step solution

01

Write the Lewis structure for 1,1,2,2-tetrachloroethene.

From the given SMILES ClCCI, we can write the molecular structure of 1,1,2,2-tetrachloroethene as: ClC=CCl with Cl atoms attached to the carbons.
02

Identify VSEPR structure for 1,1,2,2-tetrachloroethene.

The central atoms, Carbon, are sp2 hybridized having double bond with another carbon. Therefore, the structure of 1,1,2,2-tetrachloroethene is trigonal planar around each carbon.
03

Write the Lewis structure for tetrachloromethane.

From the given SMILES ClC(Cl)(Cl)Cl, we can write the molecular structure of tetrachloromethane as: C(Cl)(Cl)(Cl)(Cl) with four chloride atoms attached to the central carbon atom.
04

Identify VSEPR structure for tetrachloromethane.

The central atom, Carbon, is sp3 hybridized having single bonds with four chlorine atoms. Therefore, the structure of tetrachloromethane is tetrahedral.
05

Determine bond angles for both molecules

For 1,1,2,2-tetrachloroethene (trigonal planar), Cl-C-Cl bond angle is around 120°. For tetrachloromethane (tetrahedral), Cl-C-Cl bond angle is around 109.5°. Now, we compare our calculated bond angles to the choices provided: (a) \(90^{\circ}\) and \(109.5^{\circ}\) (b) \(109.5^{\circ}\) and \(120^{\circ}\) (c) \(120^{\circ}\) and \(109.5^{\circ}\) (d) \(120^{\circ}\) and \(120^{\circ}\)
06

Choose the correct option

Our calculated bond angles match the option (c): \(120^{\circ}\) and \(109.5^{\circ}\). Therefore, the bond angle in 1,1,2,2-tetrachloroethene and tetrachloromethane are about \(120^{\circ}\) and \(109.5^{\circ}\), respectively.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The compound which has only one isopropyl group is (a) \(2,2,2,3\) -tetramethyl pentane (b) 2,4 -dimethyl pentane (c) \(2,2,3\) -tri methyl pentane (d) 2 -methyl pentane

The correct bondline structure of 4 -hydroxy \(-4\) -methyl pentan-2-one is

The dipole moment of vinyl chloride is lower than that of ethyl chloride. This is because of (a) resonance (b) inductive effect (c) Hyperconjugation (d) unstable nature of vinyl chloride

4-Nitroaniline is a weaker base compared to 4 -cyanoaniline. This is because (a) both nitro and cyano are strong electron withdrawing groups. (b) the lone pair of nitrogen of the \(\mathrm{NH}_{2}\) group is delocalised in 4 -nitro aniline and not in 4 -cyanoaniline. (c) the negative charge resides on the more electronegative oxygen in nitro aniline in the resonance delocalization. (d) cyano group is aligned planar with the ring.

Statement 1 Alkyl carbanion such as \(\mathrm{CH}_{3}^{-}\) has pyramidal shape. and Statement 2 The carbon atom carrying the negative charge has an octet of electrons.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free