Atom and Valence Electrons

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Arrange the following elements in order of increasing electronegativity:

germanium, bromine, selenium, arsenic Please answer this question according to the general rules you have learned regarding periodic trends.
DO NOT base your answer on tabulated values since exceptions may occur. |

germanium smallest arsenic selenium bromine largest

Feedback:
Electronegativity is the ability of an atom in a molecule to attract electrons to itself.

In general, electronegativity increases as the atomic radius decreases. Smaller atoms have higher electronegativities.

Notice that all of these elements are in row 4.

Across a row of the periodic table, atomic radius decreases with increasing atomic number. The electronegativity, therefore, increases.

The following table shows the actual radii and electronegativities for these elements. Please keep in mind that the actual values may sometimes deviate from the rule.

Element | Atomic Number | Radius (pm) | Electronegativity | germanium | 32 | 122 | 1.9 | arsenic | 33 | 120 | 2.1 | selenium | 34 | 119 | 2.4 | bromine | 35 | 114 | 2.8 |
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Arrange the following elements in order of increasing electronegativity:

germanium, silicon, tin, carbon Please answer this question according to the general rules you have learned regarding periodic trends.
DO NOT base your answer on tabulated values since exceptions may occur. |

tin smallest germanium silicon carbon largest

Feedback:
Electronegativity is the ability of an atom in a molecule to attract electrons to itself.

In general, electronegativity decreases as the atomic radius increases. Larger atoms have lower electronegativities.

Notice that all of these elements are in group 4A.

Within a group of the periodic table, atomic radius increases with atomic number. The electronegativity, therefore, decreases.

The following table shows the actual radii and electronegativities for these elements. Please keep in mind that the actual values may sometimes deviate from the rule.

Element | Atomic Number | Radius (pm) | Electronegativity | tin | 50 | 140 | 1.8 | germanium | 32 | 122 | 1.9 | silicon | 14 | 118 | 1.8 | carbon | 6 | 77 | 2.5 |
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3.Top of Form
Arrange the following elements in order of increasing electronegativity:

bismuth, polonium, lead, astatine Please answer this question according to the general rules you have learned regarding periodic trends.
DO NOT base your answer on tabulated values since exceptions may occur. |

lead smallest bismuth polonium astatine largest

Feedback:
Electronegativity is the ability of an atom in a molecule to attract electrons to itself.

In general, electronegativity increases as the atomic radius decreases. Smaller atoms have higher electronegativities.

Notice that all of these elements are in row 6.

Across a row of the periodic table, atomic radius decreases with increasing atomic number. The electronegativity, therefore, increases.

The following table shows the actual radii and electronegativities for these elements. Please keep in mind that the actual values may sometimes deviate from the rule.

Element | Atomic Number | Radius (pm) | Electronegativity | lead | 82 | 146 | 1.7 | bismuth | 83 | 150 | 1.8 | polonium | 84 | 168 | 1.9 | astatine | 85 | 140 | 2.1 |

1. For each bond, use + and - to show the direction of polarity. Indicate by letter which bond is expected to be the most polar.

(A) + Se-O -

(B) + Se-S -

(C) - O-S +

The most polar bond is (A,B,C) A.

Feedback:

When two atoms are joined by a covalent bond, the bond will have polar character when the electronegativities of the two atoms are different. The atom with the highest electronegativity will have a partial negative charge, while the atom with the lowest electronegativity will have a partial positive charge.

The most polar bond is expected to occur between atoms with the greatest electronegativity difference.

Note that O, S and Se are all in group 6A. The electronegativities of the atoms follow the expected trend to decrease as one moves down a group and are:

O = 3.5
S = 2.5
Se = 2.4
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2.For each bond, use + and - to show the direction of polarity. Indicate by letter which bond is expected to be the most polar.

(A) + Se-O -

(B) + Se-S -

(C) - O-S +

The most polar bond is (A,B,C) A.

Feedback:

When two atoms are joined by a covalent bond, the bond will have polar character when the electronegativities of the two atoms are different. The atom with the highest electronegativity will have a partial negative charge, while the atom with the lowest electronegativity will have a partial positive charge.

The most polar bond is expected to occur between atoms with the greatest electronegativity difference.

Note that O, S and Se are all in group 6A. The electronegativities of the atoms follow the expected trend to decrease as one moves down a group and are:

O = 3.5
S = 2.5
Se = 2.4

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For each bond, use + and - to show the direction of polarity. Indicate by letter which bond is expected to be the most polar.

(A) + Si-P -

(B) + Si-Cl -

(C) - Cl-P +

The most polar bond is (A,B,C) B.

Feedback:

When two atoms are joined by a covalent bond, the bond will have polar character when the electronegativities of the two atoms are different. The atom with the highest electronegativity will have a partial negative charge, while the atom with the lowest electronegativity will have a partial positive charge.

The most polar bond is expected to occur between atoms with the greatest electronegativity difference.

Note that Si, P and Cl are all in period 3. The electronegativities of the atoms follow the expected trend to increase as one moves across a period and are:

Si = 1.8
P = 2.1
Cl = 3.0

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For each bond, use + and - to show the direction of polarity. Indicate by letter which bond is expected to be the most polar.

(A) + Br-F -

(B) + Br-Cl -

(C) - F-Cl +

The most polar bond is (A,B,C) A.

Feedback:

When two atoms are joined by a covalent bond, the bond will have polar character when the electronegativities of the two atoms are different. The atom with the highest electronegativity will have a partial negative charge, while the atom with the lowest electronegativity will have a partial positive charge.

The most polar bond is expected to occur between atoms with the greatest electronegativity difference.

Note that F, Cl and Br are all in group 7A. The electronegativities of the atoms follow the expected trend to decrease as one moves down a group and are:

F = 4.0
Cl = 3.0
Br = 2.8

For each bond, use + and - to show the direction of polarity. Indicate by letter which bond is expected to be the most polar.

(A) + Si-P -

(B) - Cl-P +

(C) - Cl-Si +

The most polar bond is (A,B,C) C.

Feedback:

When two atoms are joined by a covalent bond, the bond will have polar character when the electronegativities of the two atoms are different. The atom with the highest electronegativity will have a partial negative charge, while the atom with the lowest electronegativity will have a partial positive charge.

The most polar bond is expected to occur between atoms with the greatest electronegativity difference.

Note that Si, P and Cl are all in period 3. The electronegativities of the atoms follow the expected trend to increase as one moves across a period and are:

Si = 1.8
P = 2.1
Cl = 3.0

For each bond, use + and - to show the direction of polarity. Indicate by letter which bond is expected to be the most polar.

(A) + Si-P -

(B) + Si-S -

(C) - S-P +

The most polar bond is (A,B,C) B.

Feedback:

When two atoms are joined by a covalent bond, the bond will have polar character when the electronegativities of the two atoms are different. The atom with the highest electronegativity will have a partial negative charge, while the atom with the lowest electronegativity will have a partial positive charge.

The most polar bond is expected to occur between atoms with the greatest electronegativity difference.

Note that Si, P and S are all in period 3. The electronegativities of the atoms follow the expected trend to increase as one moves across a period and are:

Si = 1.8
P = 2.1
S = 2.5

Sorry, you only did 6 out of 7 parts correctly.

Answer(s):(Your response(s) are shown below, followed by the correct answer(s).) Chemical Formulas | Scientific Notation | Periodic Table | Tables |

For each bond, use + and - to show the direction of polarity. Indicate by letter which bond is expected to be the most polar.

(A) + I-F -

(B) + I-Cl -

(C) - F-Cl +

The most polar bond is (A,B,C) A.

Feedback:

When two atoms are joined by a covalent bond, the bond will have polar character when the electronegativities of the two atoms are different. The atom with the highest electronegativity will have a partial negative charge, while the atom with the lowest electronegativity will have a partial positive charge.

The most polar bond is expected to occur between atoms with the greatest electronegativity difference.

Note that F, Cl and I are all in group 7A. The electronegativities of the atoms follow the expected trend to decrease as one moves down a group and are:

F = 4.0
Cl = 3.0
I = 2.5

Valence Electron Configurations for Main Group Elements and Lewis Dot Structures
The outermost electons in an atom are the valence electrons. For the main group elements, the valence electrons are those with the highest value of the shell number n. The valence electron configurations for the main group elements in the first three periods are given below.
Note that except for He * the number of valence electrons is equal to the group number. * elements in the same group have the same valence electron configuration, when the period number is replace by the symbol n
In the Lewis representation of the valence electrons for an atom, each of the valence electrons is represented by a dot.
Note: number of dots = number of valence electrons = group number | Group | 1A | 2A | 3A | 4A | 5A | 6A | 7A | 8A | Valence Configuration | ns1 | ns2 | ns2np1 | ns2np2 | ns2np3 | ns2np4 | ns2np5 | ns2np6 | Lewis Representation | | | | | | | | | 1st Period | H
1s1 | | | | | | | | 2nd Period | Li
2s1 | Be
2s2 | B
2s22p1 | C
2s22p2 | N
2s22p3 | O
2s22p4 | F
2s22p5 | Ne
2s22p6 | 3rd Period | Na
3s1 | Mg
3s2 | Al
3s23p1 | Si
3s23p2 | P
3s23p3 | S
3s23p4 | Cl
3s23p5 | Ar
3s23p6 |

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If this element is in period 5, its valence electron configuration is Correct answer: | 5s25p4 | | 5s^2^5p^4^ | Your response: | ns2np4 | | | | |
.

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The Lewis diagram shows 6 valence electrons, and represents a valence electron configuration of the type ns2np4. In period 5, n = 5 and the valence electron configuration is 5s25p4.

See the information pages for a summary of Lewis representations of valence electron configurations.

The following Lewis diagram represents the valence electron configuration of a main-group element.

This element is in group 3A .

Feedback:

The Lewis diagram shows 3 valence electrons, and represents a valence electron configuration of the type ns2np1.

Elements with 3 valence electrons are in group 3A.

See the information pages for a summary of Lewis representations of valence electron configurations.

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The electron configurations for four main-group elements are given. Match the electron configuration on the left with the Lewis structure on the right.

One of these is a noble gas. Can you tell which one?

2 1s22s22p63s23p64s23d104p4

4 1s22s22p63s23p64s23d104p65s24d105p6

1 1s22s22p63s23p5

3 1s22s1 | | 1) | | | | 2) | | | | 3) | | | | 4) | |

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For the main group elements, the valence electrons are the ones in the outermost shell. These are the ones with the highest value of n. The number of valence electrons is equal to the group number. Each dot in the Lewis structure represents a valence electron. electron configuration | valence electrons | group number | Lewis structure | element | 1s22s22p63s23p5 | 3s23p5 | 7A | | Cl | 1s22s22p63s23p64s23d104p4 | 4s24p4 | 6A | | Se | 1s22s1 | 2s1 | 1A | | Li | 1s22s22p63s23p64s23d104p65s24d105p6 | 5s25p6 | 8A | | Xe |

The noble gas is xenon. Note that Xe has 8 electrons in the outermost shell.

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The Lewis representation above depicts a reaction between hydrogen (blue) and a main-group element from group 6A (red).

In this representation, each Y atom needs 2 electron(s) to complete its octet, and gains these electrons by forming 2 bond(s) with atoms of H.

There are 2 unshared electron pair(s) and 2 bonding electron pair(s) in the product molecule.

The bonds in the product are covalent .

Feedback:

Since Y has 6 valence electrons, and shares electrons with other atoms to complete its octet, it is a nonmetal in group 6A. To complete its octet, it must gain 2 electrons. It does this by forming bonds with 2 hydrogen atoms.

Each shared electron pair is a two-electron bond and is called a bonding electron pair.
Electron pairs that are in the valence shell of only one atom are called unshared pairs.

Bonds that are represented as a shared pair of electrons are called covalent bonds.

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The element iodine would be expected to form 1 covalent bond(s) in order to obey the octet rule.

Use the octet rule to predict the formula of the compound that would form between iodine and fluorine, if the molecule contains only one iodine atom and only single bonds are formed. Formula: | Correct answer: | IF | | IF | Your response: | | | | | | |

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Remember that electrons in covalent bonds are shared and therefore "belong" to both of the bonded atoms.

I is a nonmetal in group 7A, and therefore has 7 valence electrons. In order to obey the octet rule, it needs to gain 1 electron. It can do this by forming 1 single covalent bond.

F is a nonmetal in group 7A. It has 7 valence electrons and therefore each F atom will form one single covalent bond to complete its octet.

The reaction is depicted below, where X = F and Y = I

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The Lewis representation above depicts a reaction between hydrogen (blue) and a main-group element from group 5A (red).

In this representation, each Y atom needs 3 electron(s) to complete its octet, and gains these electrons by forming 3 bond(s) with atoms of H.

There are 1 unshared electron pair(s) and 3 bonding electron pair(s) in the product molecule.

The bonds in the product are covalent .

Feedback:

Since Y has 5 valence electrons, and shares electrons with other atoms to complete its octet, it is a nonmetal in group 5A. To complete its octet, it must gain 3 electrons. It does this by forming bonds with 3 hydrogen atoms.

Each shared electron pair is a two-electron bond and is called a bonding electron pair.
Electron pairs that are in the valence shell of only one atom are called unshared pairs.

Bonds that are represented as a shared pair of electrons are called covalent bonds.

The element oxygen would be expected to form 2 covalent bond(s) in order to obey the octet rule.

Use the octet rule to predict the formula of the compound that would form between oxygen and hydrogen, if the molecule contains only one oxygen atom and only single bonds are formed. Formula: | Correct answer: | OH2 | | OH_2_ | Your response: | | | | | | |

Feedback:

Remember that electrons in covalent bonds are shared and therefore "belong" to both of the bonded atoms.

O is a nonmetal in group 6A, and therefore has 6 valence electrons. In order to obey the octet rule, it needs to gain 2 electrons. It can do this by forming 2 single covalent bonds.

Hydrogen has 1 valence electron, and needs 1 more to have the same 2-electron valence configuration as the noble gas helium. Each hydrogen atom will form 1 single covalent bond to gain that 1 electron.

The reaction is depicted below, where Y = O

Each line in a Lewis structure represents a two-electron bond. Each pair of dots represents an unshaired pair of electrons.

Therefore, each bond to the central atom counts as two electrons and each unshared pair on the central atom counts as two electrons. If the total adds up to 8, then the cental atom obeys the octet rule. Lewis structure | cental atom | total electrons | octet rule | | B | 6 | no | | N | 8 | yes | | C | 8 | yes | | Br | 10 | no |

Each line in a Lewis structure represents a two-electron bond. Each pair of dots represents an unshaired pair of electrons.

Therefore, each bond to the central atom counts as two electrons and each unshared pair on the central atom counts as two electrons. If the total adds up to 8, then the cental atom obeys the octet rule. Lewis structure | cental atom | total electrons | octet rule | | Al | 6 | no | | C | 8 | yes | | N | 8 | yes | | Br | 12 | no |

Feedback:

Each line in a Lewis structure represents a two-electron bond. Each pair of dots represents an unshaired pair of electrons.

Therefore, each bond to the central atom counts as two electrons and each unshared pair on the central atom counts as two electrons. If the total adds up to 8, then the cental atom obeys the octet rule. Lewis structure | cental atom | total electrons | octet rule | | Be | 4 | no | | C | 8 | yes | | P | 8 | yes | | As | 12 | no |

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To answer the questions, interpret the following Lewis diagram for SF2. |

1. For the central sulfur atom: ... | The number of lone pair electrons | = 4 | | The number of bonding electrons | = 4 | | The total number of electrons | = 8 |

2. The central sulfur atom A ... | A. obeys the octet rule. | | B. has more than an octet. | | C. has less than an octet. |

Feedback: |

A Lewis diagram shows all of the valence electrons in a species.

Each dot in a Lewis diagram represents a single electron. A pair of dots on one atom symbol is called a lone pair. Lone pair electrons are non-bonding and "belong" to only one atom.

Each line connecting two atom symbols represents a two-electron bond (a shared pair of electrons).

Electrons in bonds are called bonding electrons and "belong" to both of the bonded atoms.

An atom that has 8 electrons in its valence shell is said to obey the octet rule. The central sulfur atom has 4 bonding electrons and 4 lone pair electrons for a total of 4 + 4 = 8 electrons in its valence shell. Since the total number of electrons around S is equal to 8, the octet rule is obeyed.

To answer the questions, interpret the following Lewis diagram for SF4. |

1. For the central sulfur atom: ... | The number of lone pair electrons | = 2 | | The number of bonding electrons | = 8 | | The total number of electrons | = 10 |

2. The central sulfur atom C ... | A. obeys the octet rule | | B. has less than an octet. | | C. has more than an octet. |

Feedback: |

A Lewis diagram shows all of the valence electrons in a species.

Each dot in a Lewis diagram represents a single electron. A pair of dots on one atom symbol is called a lone pair. Lone pair electrons are non-bonding and "belong" to only one atom.

Each line connecting two atom symbols represents a two-electron bond (a shared pair of electrons).

Electrons in bonds are called bonding electrons and "belong" to both of the bonded atoms.

An atom that has 8 electrons in its valence shell is said to obey the octet rule. The central sulfur atom has 8 bonding electrons and 2 lone pair electrons for a total of 8 + 2 = 10 electrons in its valence shell. This is an example of an expanded octet.

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To answer the questions, interpret the following Lewis diagram for NO2-. |

1. For the central nitrogen atom: ... | The number of lone pair electrons | = 2 | | The number of bonding electrons | = 6 | | The total number of electrons | = 8 |

2. The central nitrogen atom A ... | A. obeys the octet rule. | | B. has less than an octet. | | C. has more than an octet. |

Feedback: |

A Lewis diagram shows all of the valence electrons in a species.

Each dot in a Lewis diagram represents a single electron. A pair of dots on one atom symbol is called a lone pair. Lone pair electrons are non-bonding and "belong" to only one atom.

Each line connecting two atom symbols represents a two-electron bond (a shared pair of electrons).
Two lines connecting two atom symbols represents two bonds, and is called a double bond.
Electrons in bonds are called bonding electrons and "belong" to both of the bonded atoms.

An atom that has 8 electrons in its valence shell is said to obey the octet rule. The central nitrogen atom has 6 bonding electrons and 2 lone pair electrons for a total of 6 + 2 = 8 electrons in its valence shell. Since the total number of electrons around N is equal to 8, the octet rule is obeyed.

Drawing Lewis Structures
For molecules and polyatomic ions composed of nonmetals 1.Write the skeletal structure (the arrangement of atoms within the molecule): | Central atom=the atom with the lowest electronegativity (usually). | Hydrogen is always a terminal atom (on the end). | | 2.Count the total number of valence electrons: | Group number for each element = # valence electrons. | Add electrons for negatively charged ions. | Subtract electrons for positively charged ions. | | 3.Draw a bond between the central atom and each surrounding atom. | Single bond = 1 pair of electrons | | 4.Place lone (unshared) pairs of electrons about each terminal atom to complete their octets. | Octet= 4 electrons pairs around an atom (eight electrons) | Hydrogen can only have 2 electrons. | | 5.If there are more electrons left, place them as lone pairs on the central atom. | This will sometimes lead to an "expanded octet" around the central atom. | Expanded octet=five or six electron pairs around an atom. | Only central atoms from the third period and above can have expanded octets.
You will not be asked about species with expanded octets in this assignment. | | 6.If the central atom has an incomplete octet, use the electrons from surrounding atoms to make double or triple bonds. | Do not add electrons."Borrow" them from surrounding atoms. | Double bond = 2 pairs of electrons | Triple bond = 3 pairs of electrons | Only C, N, O, P, and S form multiple bonds. | F and Cl do not form multiple bonds. | | 7.Sometimes you just can't complete the octet for a central atom. | If there is an odd number of electrons, give the central atom 7 electrons instead of 8. | Boron and Beryllium just don't have enough electrons to go around and often have incomplete octets. |

1.Draw a Lewis structure for SiCl4 in which the central Si atom obeys the octet rule, and answer the following questions based on your drawing.

The number of unshared pairs (lone pairs) on the central Si atom is: 0

The central Si atom forms 4 single bonds.

The central Si atom forms 0 double bonds.

Feedback: The Lewis structure for SiCl4 is | |

There are no lone pairs on the central Si atom.

The central Si atom forms 4 single bonds.

The central Si atom forms no double bonds.

See the content pages for help in drawing Lewis structures.

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Draw a Lewis structure for SO2 in which the central S atom obeys the octet rule, and answer the following questions based on your drawing.

The number of unshared pairs (lone pairs) on the central S atom is: 1

The central S atom forms 1 single bonds.

The central S atom forms 1 double bonds.

Feedback: The Lewis structure for SO2 is | |

There is 1 lone pair on the central S atom.

The central S atom forms 1 single bond.

The central S atom forms 1 double bond.

See the information page for help in drawing Lewis structures.

3.Draw a Lewis structure for PH3 in which the central P atom obeys the octet rule, and answer the following questions based on your drawing.

The number of unshared pairs (lone pairs) on the central P atom is: 1

The central P atom forms 3 single bonds.

The central P atom forms 0 double bonds.

Feedback: The Lewis structure for PH3 is | |

There is 1 lone pair on the central P atom.

The central P atom forms 3 single bonds.

The central P atom forms no double bonds.

See the information page for help in drawing Lewis structures.

Molecular Geometry
Experiments show that molecules and molecular ions have definite shapes. The technical term for 'molecular shape' is 'molecular geometry'. If you have not already done so, you will soon learn about a model that will enable you to predict the geometries of molecules from a knowledge of their Lewis structures.
The purpose of this unit is to let you explore some of the geometries adopted by molecules, and to introduce you to the names for these geometries.
Before you get started answering questions, take a minute to review the Jmol interface using the model for sulfuric acid shown in the window below. You will notice that you have a much better feeling for the three dimensional shape of the molecule while it is rotating. | H2SO4

Jmol troubleshooting |
Some Jmol operations: * Rotate molecule: Click and drag in window. * Element symbols: Click on 'Jmol' (lower right) for menu. Choose: Labels > Element Symbol * Distance: Double-click on an atom (look for +) and then move cursor over another atom (click off of the molecule to end) * Angle: Double-click on an atom, click on a second and move cursor over a third (click off of the molecule to end) * Space-filling: Click on 'Jmol' (lower right) for menu. Choose: Render>Scheme>CPK Spacefill

The geometry choices are shown below: | | | | | | | A model for NO2- is shown in the Jmol window. NO2- has bent geometry. This geometry is sometimes called angular or V-shaped geometry. | * Rotate molecule: Click and drag in window * Element symbols: Click on 'Jmol' (lower right) for menu. Choose: Labels > Element Symbol * Distance: Double-click on an atom (look for +) and then move cursor over another atom (click off of the molecule to end) * Angle: Double-click on an atom, click on a second and move cursor over a third (click off of the molecule to end)Jmol troubleshooting |

Rotate the molecule until you have a feeling for its three-dimensional shape. How many atoms are bonded to the central atom? 2

Measure the bond angle at the central atom. What is the approximate numerical value of this angle? 120 degrees.
To measure an angle, see the instructions to the right of the model.

You could view this geometry as being derived from a trigonal planar molecule by removing one atom, or as being derived from a tetrahedral molecule by removing two atoms. Which does the bond angle suggest? a trigonal planar molecule a tetrahedral molecule a trigonal planar molecule

For practice, type in the name of the geometry of the molecule: | |

* | Status : | 1 1 2 3 4 5 6 7 | 1:10 PM |
Top of FormScore:
Sorry, you only did 6 out of 7 parts correctly.

Answer(s):(Your response(s) are shown below, followed by the correct answer(s).) Chemical Formulas | Scientific Notation | Periodic Table | Tables | | | When one hears the word pyramid, one usually thinks of the Great Pyramids of Egypt shown in the photo to the left. These pyramids have a square base and would be called square pyramids.

A trigonal pyramid, as the name implies, has a triangular base instead of a square base. |

A model for NF3 is shown in the Jmol window. NF3 has trigonal pyramidal geometry. | * Rotate molecule: Click and drag in window * Element symbols: Click on 'Jmol' (lower right) for menu. Choose: Labels > Element Symbol * Distance: Double-click on an atom (look for +) and then move cursor over another atom (click off of the molecule to end) * Angle: Double-click on an atom, click on a second and move cursor over a third (click off of the molecule to end)Jmol troubleshooting |

Rotate the molecule until you have a feeling for its three-dimensional shape. How many atoms are bonded to the central atom? 3

If you consider only the three outer atoms, what shape do they define? triangle

Measure the bond angles at the central atom. Do they all have approximately the same numerical value? yes no yes
To measure an angle, see the instructions to the right of the model.

What is the approximate numerical value of this angle? 109 degrees.

Does the central atom lie in the plane defined by the three other atoms? yes no no

Are all three positions about the central atom equivalent, or is one of them different from the other two. all the same one is different all the same

For practice, type in the name of the geometry of the molecule: trigonal pyramidal

Feedback: |

If one of the outer atoms is removed from a molecule that has tetrahedral geometry, the resulting species is a trigonal pyramid or has trigonal pyramidal geometry.The central atom is at the apex of the pyramid, and it is bonded to three other atoms that form the triangular base of the pyramid. The ideal bond angles at the central atom are tetrahedral or 109.5 degrees, although in reality they can vary from this. All three of the positions around the central atom are equivalent. | | |

The molecule shown in the question is NF3. Some other species that have this geometry are CH3- and SbF3: CH3- | | SbF3 | | | |
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A model for CH4 is shown in the Jmol window. CH4 has tetrahedral geometry. | * Rotate molecule: Click and drag in window * Element symbols: Click on 'Jmol' (lower right) for menu. Choose: Labels > Element Symbol * Distance: Double-click on an atom (look for +) and then move cursor over another atom (click off of the molecule to end) * Angle: Double-click on an atom, click on a second and move cursor over a third (click off of the molecule to end)Jmol troubleshooting |

Rotate the molecule until you have a feeling for its three-dimensional shape. How many atoms are bonded to the central atom? 4

If you take any three of the outer atoms, what shape do they define? triangle

Measure the bond angles at the central atom. Do they all have the approximately the same numerical value? yes no yes
To measure an angle, see the instructions to the right of the model.

What is the approximate numerical value of this angle? 109 degrees.

Are all four positions about the central atom equivalent, or is one of them different from the other three. all the same one is different all the same

For practice, type in the name of the geometry of the molecule: tetrahedral

Feedback:

In tetrahedral geometry the central atom is bonded to four other atoms that are arranged in such a way that all of the ideal bond angles at the central atom are 109.5 degrees. When the distances of the four atoms from the central atom are the same, any three of the outer atoms lie at the vertices of an equilateral triangle. All four of the positions around the central atom are equivalent.

The molecule shown in the question is CH4. Some other species that have this geometry are XeO4 and SiCl4: XeO4 | |

A model for SO3 is shown in the Jmol window. SO3 has trigonal planar geometry. This is sometimes called triangular planar geometry. | * Rotate molecule: Click and drag in window * Element symbols: Click on 'Jmol' (lower right) for menu. Choose: Labels > Element Symbol * Distance: Double-click on an atom (look for +) and then move cursor over another atom (click off of the molecule to end) * Angle: Double-click on an atom, click on a second and move cursor over a third (click off of the molecule to end)Jmol troubleshooting |

Rotate the molecule until you have a feeling for its three-dimensional shape. How many atoms are bonded to the central atom? 3

Rotate the molecule and look at it edge on. Does the central atom lie in the plane defined by the other three atoms? yes no yes

Measure the bond angles at the central atom. Do they all have approximately the same numerical value? yes no yes
To measure an angle, see the instructions to the right of the model.

What is the approximate numerical value of this angle? 120 degrees.

For practice, type in the name of the geometry of the molecule: trigonal planar

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Trigonal means three angles. Planar means lying on a plane. When the distances of the three atoms bonded to the central atom are the same, the three outer atoms lie at the vertices of an equilateral triangle, but this is not always the case.

The ideal angles at the central atom are 120°

The molecule shown in the question is SO3. Some other species that have this geometry are NO3- and AlBr3: NO3- | | |

A. The Lewis diagram for CH4 is: | | | The electron-pair geometry around the C atom in CH4 is tetrahedral. | There are 0 lone pair(s) around the central atom, so the geometry of CH4 is tetrahedral. |

B. The Lewis diagram for NH3 is: | | | The electron-pair geometry around the N atom in NH3 is tetrahedral. | There are 1 lone pair(s) around the central atom, so the geometry of NH3 is trigonal pyramid. |

A. The Lewis diagram for SeOF2 is: | | | The electron-pair geometry around the Se atom in SeOF2 is tetrahedral. | There are 1 lone pair(s) around the central atom, so the geometry of SeOF2 is trigonal pyramid. |

B. The Lewis diagram for AlBr3 is: | | | The electron-pair geometry around the Al atom in AlBr3 is trigonal planar. | There are 0 lone pair(s) around the central atom, so the geometry of AlBr3 is trigonal planar. | A. The Lewis diagram for BeBr2 is: | | | The electron-pair geometry around the Be atom in BeBr2 is linear. | There are 0 lone pair(s) around the central atom, so the geometry of BeBr2 is linear. |

B. The Lewis diagram for SO3 is: | | Recall that for predicting geometry, double and triple bonds count as only one electron pair. | The electron-pair geometry around the S atom in SO3 is trigonal planar. | There are 0 lone pair(s) around the central atom, so the geometry of SO3 is trigonal planar. |

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A.BeBr2 | There are 2 electron pairs around Be, so the electron-pair geometry is linear. | | | | Of the 2 electron pairs, 2 are bond pairs and 0 are lone pairs. The geometry around the Be atom, therefore, is linear. | | | A. The Lewis diagram for NH2Cl is: | | | The electron-pair geometry around the N atom in NH2Cl is tetrahedral. | There are 1 lone pair(s) around the central atom, so the geometry of NH2Cl is trigonal pyramid. |

B. The Lewis diagram for BrO4- is: | | | The electron-pair geometry around the Br atom in BrO4- is tetrahedral. | There are 0 lone pair(s) around the central atom, so the geometry of BrO4- is tetrahedral. | B. The Lewis diagram for NO2Cl is: | | Recall that for predicting geometry, double and triple bonds count as only one electron pair. | The electron-pair geometry around the N atom in NO2Cl is trigonal planar. | There are 0 lone pair(s) around the central atom, so the geometry of NO2Cl is trigonal planar. | Valence
Electron Pairs | Electron Pair
Geometry | Bond
Pairs | Lone
Pairs | Ideal
Geometry | | Example
Click and drag to rotate molecule. | 2 | | 2 | 0 | | | BeCl2 | | | 3 | | 3 | 0 | | | BF3 | | | 2 | 1 | | | SO2 | | | 4 | | 4 | 0 | | | CH4 | | | 3 | 1 | | | NH3 | | | 2 | 2 | | | H2O | The Lewis diagram for CS2 is: | | Recall that for predicting geometry, double and triple bonds count as only one electron pair. | The electron-pair geometry around the C atom in CS2 is linear. | There are 0 lone pair(s) around the central atom, so the geometry of CS2 is linear. |

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A.NO2Cl
Recall that for predicting geometry, double and triple bonds count as only one electron pair. | There are 3 electron pairs around N, so the electron-pair geometry is trigonal planar. | | | B. The Lewis diagram for BH2- is: | | | The electron-pair geometry around the B atom in BH2- is trigonal planar. | There are 1 lone pair(s) around the central atom, so the geometry of BH2- is bent. |

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