CHEM 212:CHEMISTSRY V 2
CHEM 212:CHEMISTSRY V 2
8th Edition
ISBN: 9781260304503
Author: SILBERBERG
Publisher: MCGRAW-HILL CUSTOM PUBLISHING
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Chapter 10, Problem 10.62P

(a)

Interpretation Introduction

Interpretation:

Lewis structure of PF5 is to be drawn.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

(a)

Expert Solution
Check Mark

Answer to Problem 10.62P

The Lewis structure of PF5 is,

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  1

Explanation of Solution

The total number of valence electrons of PF5 is calculated as,

  Total valence electrons(TVE)=[[(Total number of P atom)(Valence electrons of P)]+[(Total number of F atom)(Valence electrons of F)]]        (2)

Substitute 1 for the total number of P the atom, 5 for valence electrons of P, 7 for the total valence electrons of F, and 5 for the value of the total number of F atom in equation (1).

  Total valence electrons(TVE)=[(1)(5)+(5)(7)]=40.

PF5 has 40 valence electrons. 10 electrons are used to form 5 single bonds with the five fluorine atoms and the remaining 30 electrons are placed on the five fluorine atoms as 3 lone pairs on each.

Thus, the Lewis structure of PF5 is drawn as follows,

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  2

Conclusion

The molecule PF5 has a trigonal bipyramidal shape.

(b)

Interpretation Introduction

Interpretation:

Lewis structure of CCl4 is to be drawn.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

(b)

Expert Solution
Check Mark

Answer to Problem 10.62P

The Lewis structure of CCl4 would be:

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  3

Explanation of Solution

The total number of valence electrons of CCl4 is calculated as,

  Total valence electrons(TVE)=[[(Total number of C atom)(Valence electrons of C)]+[(Total number of Cl atom)(Valence electrons of Cl)]]        (2)

Substitute 1 for the total number of C the atom, 4 for valence electrons of C, 7 for the total valence electrons of Cl, 4 for the total number of Cl the atom in equation (2).

  Total valence electrons(TVE)=[(1)(4)+(4)(7)]=32.

The Lewis structure of CCl4 is,

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  4

Conclusion

The molecule CCl4 has a tetrahedral shape.

(c)

Interpretation Introduction

Interpretation:

Lewis structure of H3O+ is to be drawn.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

(c)

Expert Solution
Check Mark

Answer to Problem 10.62P

The Lewis structure H3O+ is as follows:

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  5

Explanation of Solution

The total number of valence electrons of H3O+ is calculated as,

  Total valence electrons(TVE)=[[(Total number of O atom)(Valence electrons of O)]+[(Total number of H atom)(Valence electrons of H)](positive charge)]        (3)

Substitute 1 for the total number of O the atom, 6 for valence electrons of O, 1 for the total valence electrons of H, 3 for the total number of H atom, and 1 for the positive charge in equation (3).

  Total valence electrons(TVE)=[(1)(6)+(3)(1)1]=8.

The total number of valence electrons in H3O+ is 8. The Lewis structure of H3O+ is given as,

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  6

Conclusion

The molecular shape of H3O+ is trigonal pyramidal.

(d)

Interpretation Introduction

Interpretation:

Lewis structure of ICl3 is to be drawn.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

(d)

Expert Solution
Check Mark

Answer to Problem 10.62P

Lewis structure of ICl3 is,

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  7

Explanation of Solution

The total number of valence electrons of ICl3 is calculated as,

  Total valence electrons(TVE)=[[(Total number of I atom)(Valence electrons of I)]+[(Total number of Cl atom)(Valence electrons of Cl)]]        (4)

Substitute 1 for the total number of I atom, 7 for valence electrons of I, 7 for the total valence electrons of Cl, and 3 for the total number of Cl atom in equation (4).

  Total valence electrons(TVE)=[(1)(7)+(3)(7)]=28.

With I as central atom, first three bond pairs are placed between each I-Cl bond then the remaining are placed as the lone pairs. Lewis structure of ICl3 is,

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  8

Conclusion

The electron group arrangement of ICl3 is trigonal bipyramidal.

(e)

Interpretation Introduction

Interpretation:

Lewis structure of BeH2 is to be drawn.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

(e)

Expert Solution
Check Mark

Answer to Problem 10.62P

Lewis structure of BeH2 is depicted as,

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  9

Explanation of Solution

The total number of valence electrons of BeH2 is calculated as,

  Total valence electrons(TVE)=[[(Total number of  Be atom)(Valence electrons of Be )]+[(Total number of H atom)(Valence electrons of H )]]        (5)

Substitute 1 for the total number of Be atom, 2 for valence electrons of Be, 1 for the total valence electrons of H, and 2 for the total number of H atom in equation (5).

  Total valence electrons(TVE)=[(1)(2)+(2)(1)]=4.

With Be as central atom, first two bond pairs are placed between each Be-H bond then the remaining are placed as the lone pairs. Lewis structure of BeH2 is depicted as,

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  10

Conclusion

The electron group arrangement of BeH2 is linear.

(f)

Interpretation Introduction

Interpretation:

Lewis structure of PH2 is to be drawn.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

(f)

Expert Solution
Check Mark

Answer to Problem 10.62P

Lewis structure of PH2 is,

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  11

Explanation of Solution

The total number of valence electrons of PH2 is calculated as,

  Total valence electrons(TVE)=[[(Total number of P atom)(Valence electrons of P)]+[(Total number of H atom)(Valence electrons of H  )]+(negative charge)]        (6)

Substitute 1 for the total number of P atom, 5 for valence electrons of P, 1 for the total valence electrons of H, 2 for the total number of H atom, and  1 for the negative charge in equation (6).

  Total valence electrons(TVE)=[(1)(5)+(2)(1)+1]=8.

The Lewis structure of PH2 is,

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  12

Conclusion

The electron group arrangement of PH2 is tetrahedral.

(g)

Interpretation Introduction

Interpretation:

Lewis structure of GeBr4 is to be drawn.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

(g)

Expert Solution
Check Mark

Answer to Problem 10.62P

Lewis structure of GeBr4 is drawn as,

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  13

Explanation of Solution

The total number of valence electrons of GeBr4 is calculated as,

  Total valence electrons(TVE)=[[(Total number of Ge atom)(Valence electrons of Ge)]+[(Total number of Br atom)(Valence electrons of Br)]]        (7)

Substitute 1 for the total number of Ge the atom, 4 for valence electrons of Ge, 7 for the total valence electrons of Br, 4 for the total number of Br the atom, in equation (7).

  Total valence electrons(TVE)=[(1)(4)+(4)(7)]=32.

The total number of valence electrons in GeBr4 is 32. The Lewis structure of GeBr4 is,

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  14

Conclusion

The electron group arrangement of GeBr4 is tetrahedral.

(h)

Interpretation Introduction

Interpretation:

Lewis structure of CH3 is to be determined.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

(h)

Expert Solution
Check Mark

Answer to Problem 10.62P

The Lewis structure for CH3 anion is as follows:

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  15

Explanation of Solution

The total number of valence electrons of CH3 is calculated as,

  Total valence electrons(TVE)=[[(Total number of C atom)(Valence electrons of C)]+[(Total number of H atom)(Valence electrons of H)]+(negative charge)]        (8)

Substitute 1 for the total number of C atom, 4 for valence electrons of C, 1 for the total valence electrons of H, 3 for the total number of H atom, and 1 for the negative charge in equation (8).

  Total valence electrons(TVE)=[(1)(4)+(3)(1)+1]=8.

The total number of valence electrons in CH3 is 8. These eight electrons are placed such that three electrons form bonding pair and the last resides on carbon as shown below:

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  16

Conclusion

The electron group arrangement of CH3 is tetrahedral.

(i)

Interpretation Introduction

Interpretation:

Lewis structure of BCl3 is to be determined.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

(i)

Expert Solution
Check Mark

Answer to Problem 10.62P

The Lewis structure for BCl3 is as follows:

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  17

Explanation of Solution

The total number of valence electrons of BCl3 is calculated as,

  Total valence electrons(TVE)=[[(Total number of B atom)(Valence electrons of B)]+[(Total number of Cl atom)(Valence electrons of Cl  )]]        (9)

Substitute 1 for the total number of B atom, 3 for valence electrons of B, 7 for the total valence electrons of Cl, and 3 for the total number of Cl atom in equation (9).

  Total valence electrons(TVE)=[(1)(3)+(3)(7)]=24.

These 24 electrons are placed such that three of these form bonding pairs and the remaining ones reside as lone pairs as shown below:

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  18

Conclusion

The electron group arrangement of BCl3 is trigonal planar.

(j)

Interpretation Introduction

Interpretation:

Lewis structure of BrF4+ is to be drawn.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

(j)

Expert Solution
Check Mark

Answer to Problem 10.62P

The Lewis structure for BrF4+ is represented as follows:

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  19

Explanation of Solution

The total number of valence electrons of BrF4+ is calculated as,

  Total valence electrons(TVE)=[[(Total number of Br atom)(Valence electrons of Br )]+[(Total number of F atom)(Valence electrons of F )](positive charge)]        (10)

Substitute 1 for the total number of Br atom, 7 for valence electrons of Br, 7 for the total valence electrons of F, 4 for the total number of F atom, and 1 for the positive charge in equation (10).

  Total valence electrons(TVE)=[(1)(7)+(4)(7)1]=34.

The Lewis structure for BrF4+ is represented as follows:

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  20

Conclusion

The electron group arrangement of BrF4+ is trigonal bipyramidal.

(k)

Interpretation Introduction

Interpretation:

Lewis structure of XeO3 is to be drawn.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

(k)

Expert Solution
Check Mark

Answer to Problem 10.62P

The Lewis structure of XeO3 is as follows:

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  21

Explanation of Solution

The total number of valence electrons of XeO3 is calculated as,

  Total valence electrons(TVE)=[[(Total number of Xe atom)(Valence electrons of Xe)]+[(Total number of O atom)(Valence electrons of O)]]        (11)

Substitute 1 for the total number of Xe the atom, 8 for valence electrons of Xe, 6 for the total valence electrons of O, 3 for the total number of O the atom in equation (11).

  Total valence electrons(TVE)=[(1)(8)+(3)(6)]=26.

The total number of valence electrons in XeO3 is 26. The Lewis structure of XeO3 is thus drawn as:

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  22

Conclusion

The electron group arrangement of XeO3 is tetrahedral.

(l)

Interpretation Introduction

Interpretation:

Lewis structure of TeF4 is to be drawn.

Concept introduction:

The steps to draw the Lewis structure of the given molecule are as follows:

Step 1: Choose the least electronegative central metal atom and place the atoms relative to each other.

Step 2: Determine the total number of valence electron.

Step 3: Place a single electron pair between each atom and subtract 2 electrons corresponding to each of these bonds from the total number of valence electrons.

Step 4: Distribute the remaining electrons in pairs around each atom as non bonding electrons such that each atom gets a complete share of eight electrons.

(l)

Expert Solution
Check Mark

Answer to Problem 10.62P

The Lewis structure for TeF4 is drawn as follows:

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  23

Explanation of Solution

The total number of valence electrons of TeF4 is calculated as,

  Total valence electrons(TVE)=[[(Total number of Te atom)(Valence electrons of Te )]+[(Total number of F atom)(Valence electrons of F )]]        (12)

Substitute 1 for the total number of Te atom, 6 for valence electrons of Te, 7 for the total valence electrons of F, and 4 for the total number of F atom in equation (12).

  Total valence electrons(TVE)=[(1)(6)+(4)(7)]=34.

Analogous to GeBr4 Lewis structure for TeF4 may be drawn as follows:

CHEM 212:CHEMISTSRY V 2, Chapter 10, Problem 10.62P , additional homework tip  24

Conclusion

The electron group arrangement of TeF4 is trigonal bipyramidal.

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Chapter 10 Solutions

CHEM 212:CHEMISTSRY V 2

Ch. 10.2 - Prob. 10.6AFPCh. 10.2 - Prob. 10.6BFPCh. 10.2 - Prob. 10.7AFPCh. 10.2 - Prob. 10.7BFPCh. 10.2 - Prob. 10.8AFPCh. 10.2 - Prob. 10.8BFPCh. 10.3 - Prob. 10.9AFPCh. 10.3 - Prob. 10.9BFPCh. 10.3 - Prob. B10.1PCh. 10.3 - Prob. B10.2PCh. 10 - Prob. 10.1PCh. 10 - When is a resonance hybrid needed to adequately...Ch. 10 - Prob. 10.3PCh. 10 - Prob. 10.4PCh. 10 - Draw a Lewis structure for (a) SiF4; (b) SeCl2;...Ch. 10 - Draw a Lewis structure for (a) ; (b) C2F4; (c)...Ch. 10 - Prob. 10.7PCh. 10 - Prob. 10.8PCh. 10 - Prob. 10.9PCh. 10 - Draw Lewis structures of all the important...Ch. 10 - Prob. 10.11PCh. 10 - Draw Lewis structures of all the important...Ch. 10 - Prob. 10.13PCh. 10 - Prob. 10.14PCh. 10 - Draw the Lewis structure with lowest formal...Ch. 10 - Draw the Lewis structure with lowest formal...Ch. 10 - Prob. 10.17PCh. 10 - Prob. 10.18PCh. 10 - Prob. 10.19PCh. 10 - Prob. 10.20PCh. 10 - These species do not obey the octet rule. Draw a...Ch. 10 - These species do not obey the octet rule. Draw a...Ch. 10 - Molten beryllium chloride reacts with chloride ion...Ch. 10 - Prob. 10.24PCh. 10 - Prob. 10.25PCh. 10 - Phosgene is a colorless, highly toxic gas that was...Ch. 10 - If you know the formula of a molecule or ion, what...Ch. 10 - In what situation is the name of the molecular...Ch. 10 - Prob. 10.29PCh. 10 - Prob. 10.30PCh. 10 - Consider the following molecular shapes. (a) Which...Ch. 10 - Use wedge-bond perspective drawings (if necessary)...Ch. 10 - Prob. 10.33PCh. 10 - Determine the electron-group arrangement,...Ch. 10 - Determine the electron-group arrangement,...Ch. 10 - Prob. 10.36PCh. 10 - Prob. 10.37PCh. 10 - Prob. 10.38PCh. 10 - Prob. 10.39PCh. 10 - Determine the shape, ideal bond angle(s), and the...Ch. 10 - Prob. 10.41PCh. 10 - Determine the shape around each central atom in...Ch. 10 - Prob. 10.43PCh. 10 - Prob. 10.44PCh. 10 - Prob. 10.45PCh. 10 - Prob. 10.46PCh. 10 - Arrange the following ACln species in order of...Ch. 10 - State an ideal value for each of the bond angles...Ch. 10 - Prob. 10.49PCh. 10 - Prob. 10.50PCh. 10 - Prob. 10.51PCh. 10 - Prob. 10.52PCh. 10 - How can a molecule with polar covalent bonds not...Ch. 10 - Prob. 10.54PCh. 10 - Consider the molecules SCl2, F2, CS2, CF4, and...Ch. 10 - Consider the molecules BF3, PF3, BrF3, SF4, and...Ch. 10 - Prob. 10.57PCh. 10 - Prob. 10.58PCh. 10 - Prob. 10.59PCh. 10 - Prob. 10.60PCh. 10 - Prob. 10.61PCh. 10 - Prob. 10.62PCh. 10 - Prob. 10.63PCh. 10 - Prob. 10.64PCh. 10 - Prob. 10.65PCh. 10 - Prob. 10.66PCh. 10 - When SO3 gains two electrons, forms. (a) Which...Ch. 10 - The actual bond angle in NO2 is 134.3°, and in it...Ch. 10 - Prob. 10.69PCh. 10 - Propylene oxide is used to make many products,...Ch. 10 - Prob. 10.71PCh. 10 - Prob. 10.72PCh. 10 - Prob. 10.73PCh. 10 - Prob. 10.74PCh. 10 - Prob. 10.75PCh. 10 - Prob. 10.76PCh. 10 - Prob. 10.77PCh. 10 - A gaseous compound has a composition by mass of...Ch. 10 - Prob. 10.79PCh. 10 - Prob. 10.80PCh. 10 - Prob. 10.81PCh. 10 - Prob. 10.82PCh. 10 - Pure HN3 (atom sequence HNNN) is explosive. In...Ch. 10 - Prob. 10.84PCh. 10 - Prob. 10.85PCh. 10 - Oxalic acid (H2C2O4) is found in toxic...Ch. 10 - Prob. 10.87PCh. 10 - Hydrazine (N2H4) is used as a rocket fuel because...Ch. 10 - Prob. 10.89PCh. 10 - Prob. 10.90PCh. 10 - Prob. 10.91PCh. 10 - Consider the following molecular shapes: Match...Ch. 10 - Prob. 10.93PCh. 10 - Prob. 10.94PCh. 10 - Prob. 10.95PCh. 10 - Phosphorus pentachloride, a key industrial...
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