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Chapter 8: Problem 46

In the following pairs of binary compounds, determine which one is a molecular substance and which one is an ionic substance. Use the appropriate naming convention (for ionic or molecular substances) to assign a name to each compound: (a) \(\mathrm{TiCl}_{4}\) and \(\mathrm{CaF}_{2},(\mathbf{b}) \mathrm{ClF}_{3}\) and \(\mathrm{VF}_{3},(\mathbf{c}) \mathrm{SbCl}_{5}\) and \(\mathrm{AlF}_{3} .\)

Short Answer

Expert verified
(a) TiCl₄ is an ionic substance named titanium tetrachloride, and CaF₂ is an ionic substance named calcium fluoride. (b) ClF₃ is a molecular substance named chlorine trifluoride, and VF₃ is an ionic substance named vanadium trifluoride. (c) SbCl₅ is a molecular substance named antimony pentachloride, and AlF₃ is an ionic substance named aluminum fluoride.

Step by step solution

01

(a) Identify the type of substance for TiCl4 and CaF2

In TiCl₄, we have titanium (Ti), which is a metal, and chlorine (Cl), a non-metal. This indicates that TiCl₄ is an ionic substance. In CaF₂, we have calcium (Ca), a metal, and fluorine (F), a non-metal, making it an ionic substance as well.
02

(a) Name TiCl4 and CaF2

For ionic substances, we use the metal name followed by the non-metal name with its ending changed to "-ide." TiCl₄: Titanium tetrachloride CaF₂: Calcium fluoride
03

(b) Identify the type of substance for ClF3 and VF3

In ClF₃, we have chlorine (Cl) and fluorine (F), both non-metals, making it a molecular substance. In VF₃, we have vanadium (V), a metal, and fluorine (F), a non-metal, making it an ionic substance.
04

(b) Name ClF3 and VF3

For molecular substances, we use prefixes to indicate the number of atoms of each element. For ionic substances, we use the metal name followed by the non-metal name with its ending changed to "-ide." ClF₃: Chlorine trifluoride VF₃: Vanadium trifluoride
05

(c) Identify the type of substance for SbCl5 and AlF3

In SbCl₅, we have antimony (Sb), a metalloid, and chlorine (Cl), a non-metal. Considering metalloids have properties of both metals and non-metals but mostly behave as non-metals in compounds, SbCl₅ would be a molecular substance. In AlF₃, we have aluminum (Al), a metal, and fluorine (F), a non-metal, making it an ionic substance.
06

(c) Name SbCl5 and AlF3

For molecular substances, we use prefixes to indicate the number of atoms of each element. For ionic substances, we use the metal name followed by the non-metal name with its ending changed to "-ide." SbCl₅: Antimony pentachloride AlF₃: Aluminum fluoride

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Most popular questions from this chapter

Energy is required to remove two electrons from Ca to form \(\mathrm{Ca}^{2+},\) and energy is required to add two electrons to \(\mathrm{O}\) to form \(\mathrm{O}^{2-} .\) Yet \(\mathrm{CaO}\) is stable relative to the free elements. Which statement is the best explanation? (a) The lattice energy of CaO is large enough to overcome these processes. (b) CaO is a covalent compound, and these processes are irrelevant. (c) CaO has a higher molar mass than either Ca or O. (d) The enthalpy of formation of CaO is small. (e) CaO is stable to atmospheric conditions.

The compound chloral hydrate, known in detective stories as knockout drops, is composed of \(14.52 \% \mathrm{C}, 1.83 \% \mathrm{H}\) , \(64.30 \% \mathrm{Cl},\) and 13.35\(\% \mathrm{O}\) by mass, and has a molar mass of 165.4 \(\mathrm{g} / \mathrm{mol}\) . (a) What is the empirical formula of this substance? (b) What is the molecular formula of this substance? (c) Draw the Lewis structure of the molecule, assuming that the Cl atoms bond to a single \(C\) atom and that there are a \(C-C\) bond and two \(C-O\) bonds in the compound.

Although \(\mathrm{I}_{3}\) is a known ion, \(\mathrm{F}_{3}^{-}\) is not. (a) Draw the Lewis structure for \(\mathrm{I}_{3}^{-}\) (it is linear, not a triangle). (b) One of your classmates says that \(\mathrm{F}_{3}^{-}\) does not exist because \(\mathrm{Fis}\) too electronegative to make bonds with another atom. Give an example that proves your classmate is wrong. (c) Another classmate says \(\mathrm{F}_{3}^{-}\) does not exist because it would violate the octet rule. Is this classmate possibly correct? (d) Yet another classmate says \(\mathrm{F}_{3}^{-}\) does not exist because \(\mathrm{F}\) is too small to make bonds to more than one atom. Is this classmate possibly correct?

Arrange the bonds in each of the following sets in order of increasing polarity: (a) \(\mathrm{C}-\mathrm{F}, \mathrm{O}-\mathrm{F}, \mathrm{Be}-\mathrm{F}\) ; (b) \(\mathrm{O}-\mathrm{Cl}, \mathrm{S}-\mathrm{Br}, \mathrm{C}-\mathrm{P} ;(\mathbf{c}) \mathrm{C}-\mathrm{S}, \mathrm{B}-\mathrm{F}, \mathrm{N}-\mathrm{O}\)

Predict the ordering, from shortest to longest, of the bond lengths in \(\mathrm{CO}, \mathrm{CO}_{2},\) and \(\mathrm{CO}_{3}^{2-}\) .
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