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Chapter 7: Problem 34

Predict the geometry of the following species: (a) NNO (b) \(\mathrm{ONCl}\) (c) \(\mathrm{NH}_{4}{ }^{+}\) (d) \(\mathrm{O}_{3}\)

Short Answer

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Question: Based on the VSEPR theory, predict the geometries of the following species: (a) NNO, (b) ONCl, (c) NH4+, and (d) O3. Answer: (a) NNO has a linear geometry, (b) ONCl has a bent geometry, (c) NH4+ has a tetrahedral geometry, and (d) O3 has a bent geometry.

Step by step solution

01

Identify the central atoms

For each given species, determine the central atom, which usually has the lowest electronegativity. (a) NNO: The central atom is N. (b) ONCl: The central atom is O. (c) NH4+: The central atom is N. (d) O3: The central atom is O.
02

Determine electron pairs and lone pairs around the central atoms

Using Lewis structures, count the number of bonding electron pairs (single, double, or triple bonds) and lone pairs of electrons around each central atom. (a) NNO: NN triple bond and NO double bond. Total of 2 bonding pairs and 0 lone pairs. (b) ONCl: O single bond with N and N single bond with Cl. Total of 2 bonding pairs and 1 lone pair. (c) NH4+: N has 4 single bonds with H. Total of 4 bonding pairs and 0 lone pairs. (d) O3: central O double bond with one O and single bond with another O. Total of 2 bonding pairs and 1 lone pair.
03

Apply VSEPR theory to predict geometries

Use the VSEPR theory to predict the geometries of the given species by minimizing repulsion between electron pairs. The geometries are determined by the arrangement of electron pairs around the central atom. (a) NNO: linear geometry. (b) ONCl: bent geometry. (c) NH4+: tetrahedral geometry. (d) O3: bent geometry.

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Most popular questions from this chapter

Oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), is a poisonous compound found in rhubarb leaves. Draw the Lewis structure for oxalic acid. There is a single bond between the two carbon atoms, each hydrogen atom is bonded to an oxygen atom, and each carbon is bonded to two oxygen atoms.

In which of the following molecules does the sulfur have an expanded octet? For those that do, write the Lewis structure. (a) \(\mathrm{SO}_{2}\) (b) \(\mathrm{SF}_{4}\) (c) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (d) \(\mathrm{SF}_{6}\)

Two different molecules have the formula \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}_{2}\). Draw a Lewis structure for each molecule. (All the \(\mathrm{H}\) and \(\mathrm{Cl}\) atoms are bonded to carbon. The two carbon atoms are bonded to each other.)

Consider the dichromate ion. It has no metal-metal nor oxygenoxygen bonds. Write a Lewis structure for the dichromate ion. Consider chromium to have six valence electrons.

Give the formula of a polyatomic ion that you would expect to have the same Lewis structure as (a) \(\mathrm{Cl}_{2}\) (b) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (c) \(\mathrm{CH}_{4}\) (d) \(\mathrm{GeCl}_{4}\)
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