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The Notre Dame of Isulan, Inc.

LEARNING MODULE
General Chemistry 1
SECOND SEMESTER
QUARTER 2 Month 1-2

Name _____________________________________________________________
Section ________________________________________________________
Subject Teacher MR. STEPHEN CARL B. BEDANO Cell Phone No.09051346118
Class Adviser ___________________________________________________

Unit Title: All Compounds

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Introduction:
Curiosity on the finite of matter inspired the early scientists and even philosophers to propose laws and
theories about the atom and its properties. In trying to explain day-to-day transformation and behaviour of
matter, they were able to lay down the foundation for the principles that are now being studied in chemistry and
other fields of science. The same principles led to the advancement of technology and development of modern
materials and devices that improve the quality of life. New technologies in turn allow present-day scientists,
fuelled by the same curiosity, to study further matter and its properties.

Overview
This learning area is designed to provide a general background for the understanding composition,
structure, and properties of matter; quantitative principles, kinetics, and energetics of transformations of matter;
and fundamental concepts of organic chemistry.

Definition of Terms
Volume - is the space is occupies.
Mass - is the amount of matter.
Atomic Number - represents the number of protons in the nucleus of an element.
Mass number - indicates the total number of protons and neutrons.
Pressure gauge - measures the pressure in a closed system.
Barometer - measures the atmospheric pressure.
Lessons and Coverage: Month 1- 2
Lesson No. Topic/Title You’ll learn to...

Orbits, Orbitals  Use quantum numbers to describe an electron in an atom.

And Ionic Bonding  Determine the magnetic property of the atom based on its
Lesson 1 electronic configuration.
 Draw an orbital diagram to represent the electronic configuration
of atoms.
 Draw the Lewis structure of ions.
 Apply the octet rule in the formation of molecular covalent
Covalent Bonding compounds.
Lesson 2  Write the formula of molecular compounds formed by the
nonmetallic elements of the representative block.
 Draw Lewis structure of molecular covalent compounds.
 Describe the geometry of simple compounds.
 Determine the polarity of simple molecules.
 Describe the different functional groups.
 Describe some simple reactions of organic compounds:
Lesson 3 Organic compounds combustion of organic fuels, addition, condensation, and
saponification of fats.
 Describe the formation and structure of polymers.
 Describe the structure of proteins, nucleic acids, lipids, and
carbohydrates, and relate them to their function.

Expected Skills
To do well in this module, you need to remember and do the following:
1. Read and understand the lesson properly.
2. Analyse the questions properly before answering.
3. Complete all the given exercises properly. Read and follow directions carefully.
4. Take note of important details, processes, and methods in each lesson.
5. Go back to the lesson if you can’t answer the given exercises.
6. Be persevering and study hard.
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LESSON 1
ORBITS, ORBITALS AND IONIC BONDING

PRE-ASSESSMENT
Let’s find out how much you already know about the lesson. Let us evaluate your learning. Read and
understand the questions carefully and choose the letter that represents your answer and write them on the space
before the number.
Note: Formative Assessment are for your self-learning activities. Answer this activity as honestly as you can
before you look at the answer key at the last page of this module. The answers will be discussed during
ONLINE CLASSES.
Activity 1- Pre-Assessment: Formative Assessment 1 (FA 1)

Instruction: Multiple Choice: Read each item carefully. Choose the correct answer from the given choices.
Write the letter of your answer on the space provided before the number

____1. Which atomic orbital can accommodate a maximum of six electrons?

a. s b. p c. d d. f
____2. What are the possible values of the magnetic quantum number?
a. n b. 0 to n-1 c. - ℓ to + ℓ d. +½, -½
____3. What is the azimuthal quantum number for an electron in the f orbital?
a. s b. 1 c. 2 d. 3
____4. How many electrons can be placed in the third principal energy level (n=3)
a. 2 b. 8 c. 10 d. 18
____5. According to the Aufbau principle, which orbital should be occupied first?
a. 3s b. 3p c. 3d d. 4s
What do you think? Did you get the answers correctly? Now, I think we are ready to move further with
our lesson. Good start
EXPLORE
Let’s start our lesson as we recall our past learning in the previous grade level. As you go through the
different learning activities, have in mind on how do these activities can be of great help in real-life situations.

Activity 2 – READING 1
Note: Just read and understand the concepts presented so that you will know how to answer the
succeeding activities.
A set of quantum numbers gives and information about the atomic orbitals where an electron may be
found. Every electron in an atom is assigned a unique set of quantum numbers, of which there are four classes:
principal, azimuthal, magnetic, and spin. In this lesson let us learn more important ideas about quantum
numbers, electronic configuration and orbital diagram.
 Principal Quantum Number

The principal quantum number (n) indicates the

energy level or shell where an atomic orbital can be
found. It can have integral values (n = 1,2,3 and so on),
which correspond to the “orbits” in the Borh model.
This means that the bohr model was not exactly wrong.
 Azimuthal Quantum Number

The azimuthal quantum number (ℓ) specifies the

sublevel (or subshell) within a particular principal
energy level. It can have values of 0 to n-1. The
azimuthal quantum number represents the kind and
shape of the orbital (s, p, d, or f) that is being occupied
by an electron. The ℓ values corresponding to the type
of orbitals are shown in table 6-1.
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 Magnetic Quantum Number

The magnetic quantum number (mℓ) indicates the specific orbital within the sublevel where the electron
is found. It can have values of -ℓ to +ℓ. It also gives the number of orbitals in a sublevel and the spatial
orientation of these orbitals.

For example, in the first principal energy level (n=1), ℓ can only have a value of zero, which corresponds
to an s subshell. The magnetic quantum number can thus have only one value, mℓ =0, indicating that there is
only one s orbital. In the second principal energy level (n=2), ℓ can have values of 0 and 1. This means that in
this energy level, there are s and p subshells. Unlike the  Spin Quantum Number
s subshell, the p subshell can have magnetic quantum
number values of-1, 0, +1, which correspond to the According to the Pauli exclusion
three p orbitals. principle, only a maximum of two electrons can
occupy an orbital, and they must have opposite
spins to minimize repulsion between them. This
principle is observed through the spin quantum
number, which can only have values of +½ or -½
for each electron. This quantum number describes
the intrinsic spin of the electron in the orbital. As
a rule, no two electrons in an atom can have the
set of quantum numbers, for example in the 1s
orbitals, each electron has the same n, ℓ, and mℓ
but different ms.

Activity 3 – VIDEO VIEWING 1 (FOR THOSE WHO HAVE CONNECTIVITY. THIS IS OPTIONAL). But read the
discussion that follows.

Instructions: In order to have a better idea about quantum numbers, follow this link: (5007) Quantum
Numbers, Atomic Orbitals, and Electron Configurations - YouTube. Watch it carefully so that you are prepared
for the next task.
Process Questions:

1. What are the different quantum numbers?

FIRM-UP
In this section, you will be able to learn and understand the key concepts for you to answer the learning
activities. In this part, there are activities that will help you answer essential questions especially questions that
you have in your mind.
Try to deepen your understanding that it would look like taking a trip around the world for free.
Activity 4 –READING 2
1.
Note: Just read and understand the concepts presented so that you will know how to answer the succeeding
activities.
 Electronic Configuration and Orbital Diagram
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Each electron in an atom has a unique set of quantum numbers. How these electrons are
distributed among the orbitals in an atom is given by the electron configuration. In writing electron
configurations, each orbital and the electron(s) it holds are represented as

Take note that in a neutral atom, the number of electrons is equal to the number of protons. Thus, the
number of electrons in an atom is equal to its atomic number.
The Aufbau principle (also known as the building up principle) of filling up of orbitals states that
electrons must first occupy the orbitals with lower energies than those with higher energies. The first two
orbitals (ls and 2s) are each occupied first with two electrons.
The sequence of filling the orbitals is depicted in
figure 6-15. As the figure shows, the first orbital that is
assigned two electrons (of opposite spins) is the 1s
orbital, followed by two electrons for 2s, and another two
for each 2p orbital (a total of six electrons for Px, Py and
Pz). This sequence goes on until all the electrons in an
atom have been assigned to orbitals. Note that the
maximum number of electrons for the s orbital is two, six
for p, 10 for d, and 14 for f.
The orbital diagram is another representation of how electrons (arrows) occupy the orbitals (boxes)

The next element after boron is carbon

with six electrons. Will the sixth electron pair
with the fifth in the same orbital or occupy the
next p orbital? The answer is given by Hund's
rule of maximum multiplicity, which states that for
degenerate orbitals (orbitals with similar energies
such as the three p orbitals), the electrons will
singly occupy each orbital and with parallel spins
before they pair up. The same is true for the d and
f orbitals. This pairing up rule gives the atom the
most stable distribution of its electrons, just like
how the Aufbau principle predicts the tendency of
an atom to acquire its most stable electron
configuration (with minimum repulsion).
The outermost energy level of an atom is called its valence shell, and the electrons occupying these
shells are called valence electrons. The inner shells that are completely filled are called closed shells. Lithium
has only one valence electron, while oxygen has six. Notice that the valence shell of the noble gas neon is
completely filled, and is thus very stable as the shell of the other noble gases.
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Heavier atoms that have more electrons will have very long electron configurations. In this case, a noble
gas with a similar closed-shell configuration to the atom of interest is used as a shortcut and is referred to as a
core symbol written inside brackets. The resulting configuration is called noble gas electron configuration. The
two possible configurations for lead are given below.

Transition metals with d4 and d9 configurations tend to adopt a more stabilized half-filled or fully filled
configuration by using one electron from the nearby s orbital. For example, the electron configuration for
chromium is [Ar]4s13d5 instead of 4s23d4; that of copper is [Ar|4s13d10 instead of 4s23d9
 Paramagnetism and Diamagnetism
Paramagnetism and diamagnetism are properties of elements explained by electron distribution.
Paramagnetism refers to the characteristic of an element to be slightly attracted to a magnet. It results from the
presence of unpaired electrons in some of the atomic orbitals of an atom, which creates magnetic moment for
the atom. These electrons tend to align themselves in the direction of an external magnetic field. Arsenic
(A=33), which has three unpaired electrons in its 4p orbitals, is considered paramagnetic.

Elements without unpaired electrons in their orbitals exhibit diamagnetism. Diamagnetism is

characterized by non-attraction, or even a slight repulsion, of an element to a magnet. When the electrons in an
atom's orbitals are all paired, the atom does not have a net magnetic moment and is not attracted by an external
magnetic field. Krypton (A=36), which has fully filled 4p orbitals, is diamagnetic.
When paramagnetic species or atoms are close to each other (as in a metal rod or wire), their individual
magnetic moments could interact with each other and spontaneously align. Such is the case for a few metals
such as iron, cobalt, and nickel which are said to exhibit ferromagnetism. Ferromagnetism is a phenomenon that
greatly enhances the paramagnetism of a material in such a way that it becomes permanently attracted to a
magnet.

 Ion formation
Most metals have one to three valence electrons, which can be easily removed because of their relatively
low ionization energy. Group 1A elements, having the lowest ionization energies among the element groups,
can easily lose their one valence electron to achieve a noble gas configuration (helium configuration (duet) for
Period 2 metals, octet for Period 3 and above). Group 2A elements can lose two valence electrons using a
higher ionization energy than Group 1A. Group 3A elements have three valence electrons to lose with even
greater ionization energy than Group 2A elements. Note that the number of electrons that a neutral atom loses is
numerically equal to the charge of its ion.
To illustrate these concepts, consider lithium and magnesium atoms. A neutral lithium atom (Group 1A,
Period 2) has one valence electron that can easily be removed with low ionization energy to attain the helium
configuration, forming a +1 ion. A neutral magnesium atom (Group 2A, Period 3) can lose its two valence
electrons with relatively higher ionization energy to achieve the configuration of neon, forming a +2 ion.
 7

Nonmetals, having high electron affinity, can gain valence electrons to fill their
s and p orbitals and form an octet. Groups 5A, 6A, and 7A can gain three, two, and one
electron(s), respectively to form their anions. Similar to cation formation, the number
of electrons gained by an atom is numerically equal to the charge of its anion. For
example, chlorine (Group 7A) and nitrogen (Group 5A) gain one and three electrons to
form -1 and -3 ions, respectively.

 Lewis Structure of Ions

Ions may be represented using Lewis structures by simply removing (or adding)
the number of dots that corresponds to the electrons lost (or gained) by the neutral
atom. For instance, losing the single valence electron in the neutral atom of sodium
leaves a Lewis structure with no dot but with a +1 charge. The same principle applies
to magnesium.

For anions, gained electrons are represented as additional dots in the configuration of the neutral atom,
basically forming an octet enclosed in brackets. The charge of the ion is added at the upper right- hand corner of
the configuration.

Activity 5 – TEST YOUR UNDERSTANDING: FORMATIVE ASSESSMENT 1 (FA 2)

Formative Assessment is for your self-learning activity. Answer this activity as honestly as you can
before you look at the answer key at the last page of this module.

Instruction: Kindly give what is asked below. Write your answer on the activity sheet attached in your module

1. Compare and contrast paramagnetism and diamagnetism using Venn diagram.

2. Draw the lewis structures for the ions formed by the following:
a. Phosphorus
b. Bromine
c. Sulfur

DEEPEN
Your goal in this section is to take a closer look at some aspects of the topic. You will be going to learn
and understand chemical bonding for you to prepare for the final task.
You have also to remember that you are improving on the important skills that you will be needing as
you go through this part.
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Activity 6 – (Q2SA1)
Note: The SUMMATIVE ASSESSMENT (SA1) is a show case of your learnings and mastery about the
learning objectives. After doing the Formative Assessments and reading activities you are now ready to
answer the Summative Assessment. Entitle your work as Q2-Lesson 3- Activity 6 – (SA1)
Instruction: Kindly give what is asked below. Write your answer on the activity sheet attached in your module.
1. Write the electron configuration of the following element, draw its orbital diagram and determine the
magnetic property if it is paramagnetism or diamagnetism.
Element Electron Configuration Magnetic Property

1. rhodium (A=
45)

2. argon (A=18)

3. potassium
(A=19)

4. calcium (A=20)

5. copper (A= 29)

VALUES INTEGRATION
In this section, you have learned how appreciates the gift of nature, preserve the uniqueness of
things, develop critical thinking and praise the work of others.
Now that you have a deeper understanding about the topic, you are ready to do the task in the next
section.
TRANSFER
Your goal in this section is to apply and transfer your learning to real-life situations. You will be
given a practical task to demonstrate your learning and understanding.
Activity 7 – (Q1PT1)

Illustrate the reactions at the molecular level in any of the following:

1. enzyme action 2. protein denaturation 3. separation of components in coconut milk
PERFORMANCE TASK
RUBRICS
Description Score Score obtain
Content provided are correct and appropriate 50
Sufficient elaboration of key ideas and examples are provided 30
Creativity 20
Total 100
EVALUATION
A. Multiple choice: Read each item carefully. Choose the letter of the correct answer from the given choices.
Write the letter your answer on activity sheet attached in the module.
1. Which element has electron in its d orbital?
a. Fe b. K c. Na d. s
2. Which Element has four electrons in its p orbital?
a. Be b. C c. Ne d. O
3. What noble gas can be used as the core symbol for the electron configuration of selenium (A=34)?
a. Ar b. Kr c. Ne d. Xe
4. How many valence electron are found in an atom with an electronic configuration 1s22s22p63s23p5
a. 3 b. 5 c. 7 d. 17
5. In terms of magnetic properties due to electrons, neon (A=10) may be described as _____________
a. diamagnetic b. electromagnetic c. ferromagnetic d. paramagnetic

B. Draw the lewis structures for the ions formed by the following:
 9

a. Phosphorus
b. Bromine
c. Sulfur

.
Phoenix Publishing House; General Chemistry, Exploring life Through Science series.

STUDENT’S MONITORING SHEET: Check the corresponding column for your accomplishments of different
learning tasks.
Learning Activity Accomplished Not Accomplished
Pre- assessment Formative Assessment 1 (FA1)
Act. 5 Formative Assessment 3 (FA3)
Act. 6 Summative Assessment 1 (SA1)
Act. 7 Performance Task 1 (PT1)
Evaluation Summative Assessment 2 (SA2)

LESSON 2
COVALENT BONDING
PRE-ASSESSMENT
Let’s find out how much you already know about the lesson. Let us evaluate your learning. Read and
understand the questions carefully and choose the letter that represents your answer and write them on the space
before the number.
Note: Formative Assessment are for your self-learning activities. Answer this activity as honestly as you can
before you look at the answer key at the last page of this module. The answers will be discussed during
ONLINE CLASSES.
Activity 1- Pre-Assessment: Formative Assessment 1 (FA 1)
Identification: Identify what is being described by the following statement. Write your answer on the space
provided
_______1. It is an arrangement of electron pairs reduces repulsions among four sets of electron pairs
_______2. It describes the three-dimensional arrangement of atoms within a molecule or polyatomic ion.
_______3. It is covalently bonded to more than one atom.
_______4. It is bonded to only one other atom.
_______5. It suggests that electron pairs around an atom assume an arrangement in space that reduces the
repulsions
between them.
What do you think? Did you get the answers correctly? Now, I think we are ready to move further with
our lesson. Good start
EXPLORE
Let’s start our lesson as we recall our past learning in the previous grade level. As you go through the
different learning activities, have in mind on how do these activities can be of great help in real-life situations.

Activity 2 – READING 1

Note: Just read and understand the concepts presented so that you will know how to answer the succeeding
activities.
 Formation of a covalent bond
Air contains oxygen; not in the form of ions (O2-) nor of neutral oxygen atoms (O), but as diatomic
molecules (O2). Molecules are a group of atoms bound together by covalent bonds. Covalent bonds exist
between nonmetal atoms, which have small electronegativity difference. Because nonmetal atoms have
relatively similar electronegativities, they tend to attract valence electrons equally (or almost equally) and just
share them achieve an octet (or duet). Some elements exist as diatomic or polyatomic molecules. These include
hydrogen (H2), oxygen (O2), the halogens (e.g., I2, Br2, F2), and sulfur (S8). The atoms in polyatomic ions (e.g.,
NH4+) are also bound by covalent bonds. Compounds that result from covalent bonding are called molecular
compounds such as butane (C2H8) in liquefied petroleum gas (LPG).
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 Types of Covalent Bonds

Two nonmetal atoms can form a maximum of three covalent bonds between each other depending on the
number of electron pairs that they need to share to attain a noble gas configuration. For example, two hydrogen
atoms-each with one valence electron-may share each other's electron to achieve the helium configuration. The
orbitals that contain each valence electron overlap to form an orbital common to both atoms. The sharing of one
electron pair forms a single covalent bond represented as (-). Other compounds that consist of single covalent
bonds are given below. The process of drawing these structures will be discussed in the succeeding lesson.

Two atoms, such as the diatomic oxygen, that share two pairs of electrons (i.e., four electrons in all)
form double bonds represented by (=). Double bonds are stronger than single bonds, and are thus depicted
shorter than single covalent bonds. They exist in many compounds such as carbon dioxide and ethene.

The two atoms sharing six valence electrons from triple covalent bond (≡). This is illustrated by
molecular nitrogen (N2). A neutral nitrogen atoms (Group 5A) has five valence electrons. Two nitrogen atoms
form three covalent bonds by sharing six electrons, three from each atom, for each to form an octet. Triple
bonds are the strongest among the covalent bonds and are, thus, represented as the shortest among the three.

Activity 3 – VIDEO VIEWING 1 (FOR THOSE WHO HAVE CONNECTIVITY. THIS IS OPTIONAL). But read the
discussion that follows.

Instructions: In order to have a better idea about covalent bonds, follow this link: (5057) Covalent
Bonding! (Definition and Examples) - YouTube. Watch it carefully so that you are prepared for the next task.

Process Questions:
1. What and how to identify a covalent bond?

FIRM-UP
In this section, you will be able to learn and understand the key concepts for you to answer the learning
activities. In this part, there are activities that will help you answer essential questions especially questions that
you have in your mind.
Try to deepen your understanding that it would look like taking a trip around the world for free.
Activity 4 –READING 2
Note: Just read and understand the concepts presented so that you will know how to answer the succeeding
 Molecular compounds and their properties
activities.
Ionic bonding discussed in the previous chapter differs from covalent bonding in many aspects.
An ionic bond forms between a metal and a nonmetal through an exchange of electrons; a covalent bond exist
between two nonmetals-different or identical elements-through sharing of electrons. In term of the compounds
they form, covalent compounds are electrically nonconducting because their constituent atoms are bound to the
molecules and do not dissociates into electrically charged particles when dissolved in water. Aqueous ionic
compound are conducting in water. The forces of attraction between covalent molecules are relatively weaker
than in ionic compounds; thus, they have low melting and boiling point and are soft. Note, however, that when
compounds undergo phase changes, only the attraction between molecules is altered and not the covalent bond
 11

between atoms. Other comparisons between ionic and covalent bonding are given in table 9-1

Many molecular compound naturally occur in the environment. Some are essential for biological
activities, while others contribute to environment pollution. Table 9-2 lists some of these compounds.

 Covalent Molecules
Covalent molecules and polyatomic ions may be represented using Lewis structures and molecular
shape, which both depend on the constituent atoms and their valence electron.
 Lewis Structure of molecules and Polyatomic Ions
Constituent atoms of covalent molecules and polyatomic ions also satisfy the octet rule (or duet rule) as
observed in their Lewis structures. In writing the Lewis structure of a molecular compound, the atoms are
arranged in such a way that there are eight (or two) electrons in each atom, including those that form the
covalent bonds. For simple molecules with more than two atoms, the single atom in the formula (except H and
F) is normally assigned as the central atom surrounded by the other terminal atoms. A central atom is covalently
bonded to more than one atom; a terminal atom is bonded to only one other atom. A molecule may have more
than one central atom in its Lewis structure. Drawing the Lewis structure of molecules and polyatomic ions
observes these general guidelines, which will be applied to specific molecules in Sample Problems 9.1.
- 1. Finding the number of electrons available for distribution is done by adding the valence electrons of
all the atoms in the molecule or polyatomic ion. For positively charged polyatomic ion, the numerical
charge is deducted from the total valence electrons. For negatively charged ions, the numerical charge is
added.
Examples:
CO2 : (valence e of C) + (valence e- of Ox 2 atoms) = 4 + (6×2)= 16 e-
OH- : (valence e of H-) + (valence e- of H) + charge = 6 + 1+1=8 e-
NH4+: (valence e- of N) + (valence e- of H x 4 atoms) – charge = 5 + (1×4) -1=8 e-
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- 2. In arranging the atoms, the least electronegative atom is assigned as the central atom. Hydrogen can
form only one covalent bond so it is automatically designated as the terminal atom. Halogens are also
usually assigned as terminal atoms; but in some cases, they serve as central atoms. When atoms are
arranged, they are each connected by a single covalent bond first.

Each bond accounts for two electrons. To find the remaining electrons available for distribution, subract those
used for bonding from the total electrons computed in item 1.
CO2 : remaining e- = 16 - 4=12
OH- : remaining e- =8 2 = 6
NH4+ : remaining e- = 8 - 8 = 0
- 3. The remaining electrons (from item 2) are distributed to the atoms such that there are two electrons
for hydrogen and eight electrons for the other atoms. There are three possible ways to distribute the
remaining electrons in CO2

- 4. If there are not enough electrons for the atoms to have eight electrons each (e.g., CO), the single
bonds are changed to double or triple bonds by shifting nonbonding pairs of electrons as needed.
Consider structure (1) for CO2.

Note that carbon can form single, double, or triple covalent bonds. Nitrogen can form single or triple
bonds.
- 5. If there are more than one Lewis structure that result in item 4, the one that is symmetrical is usually
chosen. The concept of formal charge, which will be discussed later, also helps determine the acceptable
Lewis structure. For CO2, the preferred structure is the one at the left.

- 6. The structure of a polyatomic ion is enclosed in brackets and the charge is placed at the upper right
corner of the bracket.
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 Shape of molecules

Molecular geometry describes the three-dimensional arrangement of atoms within a molecule or polyatomic
ion. The molecular geometry of molecules or ions that contain only a few atoms can be predicted by using the
molecule's Lewis structure and the valence shell electron pair repulsion (VSEPR) theory. The VSEPR theory
suggests that electron pairs around an atom assume an arrangement in space that reduces the repulsions between
them. This arrangement depends on the number and type of electron pairs (whether bonding or nonbonding)
present in the molecule or ion. The theory as applied to a central atom with two, three, and four electron pairs
surrounding it are as follows:

1. For two electron pairs in an atom to be as far apart as possible, they should be
at 180° angle from each other (figure 9-5). This makes the atoms assume a linear
arrangement.

2. Three electron pairs in an atom are widely separated at 120° angle to one
another, forming an equilateral triangle (figure 9-6). This arrangement is
described as trigonal planar.

3. A tetrahedral arrangement of electron pairs reduces repulsions among four

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sets of electron pairs (figure 9-7). A tetrahedron is a four-sided polygon, with all
four sides as identical equilateral triangles. The angle between any two electron
pairs is 109.5°.
Note the three types of lines used in figure 9-7. All lines represent a
covalent bond. The single line (-) represents a bond on the plane of the paper.
The solid wedge ( ) suggests a bond protruding out of the paper and towards
the reader. The dashed wedge means the bond protrudes to the back of the paper.
In the valence shell of an atom in a molecule, each electron pair occupies
its own domain. Each domain is attracted and gets as close as possible to the
central atom, but keeps other domains as far away as possible. This follows that
two electron domains are separated by 180°, three domains by 120°, and four
domains by 109.5°. The electron domain (ED) geometry is not necessarily the
molecular geometry. The molecular geometry only looks at the arrangement of
the atoms; the ED geometry considers the effect of the nonbonding domains to
the shape of the molecule or ion. For example, the ED geometry of water (H 2O)
is tetrahedral, while its molecular geometry is bent.

Since electron domains tend to repel each other, the ideal arrangement of atoms in a molecule or
polyatomic ion is that which minimizes this repulsion. Therefore, a nonbonding domain tends to spread out and
occupy a larger space than a bonding domain does. But all domains-bonding and nonbonding-are all attracted to
the central atom.
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Answer:
General Formula No. of Bonding No. of Bonding ED geometry Molecular
Domains Domains Geometry
AB2N3 2 3 Trigonal bipyramid Linear
AB2N 2 1 Trigona planar Bent
AB4N 4 1 Trigonal seesaw
bipyramidal
 Electronegativity and bond polarity
Electronegativities can be used to predict the nature of the chemical bond that can exist between atoms.
The larger the electronegativity difference ( EN) between the two atoms, the more polar the bond will be. Ionic
bonds has the highest EN and are therefore the most polar of the chemical bonds. Covalent bonds may be
polar or nonpolar depending on the electronegativities of the atoms involved.

Sample Problems 9.4

Us the electronegativity value above to determine the type of bond that will be formed between the
given atoms. Rank the pairs according to increasing polarity. For these exercise, the ionic bond will also be
presented as (–) for ease of illustrative comparison.
Ca – Cl C- S Se- P C-O K-F
Solution:
Electronegativity difference Types of Bond
Ca – Cl 3.0 – 1.0 = 2.0 ionic
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C–S 4.0 – 0.8 = 3.2 ionic

Se – P 2.5 – 2.5 = 0 nonpolar covalent
Se – P 2.5 – 2.1 = 0.3 nonpolar covalent
K–F 3.5 – 2.5 = 1.0 polar covalent
Increasing polarity: C – S < Se – P < C – O < Ca – Cl < K – F

Bond polarity results from the electron distribution (electron density) between atoms as a consequence
of their electronegativities. Two atoms with equal electronegativities equally share the electrons, which results
in an even electron density. This produces a nonpolar bond. Unequal electronegativities result in an uneven
electron distribution in a bond, with one partially positive end (δ+) and one partially negative end (δ−). The
resulting bond is described as polar. The charge separation resulting from the unequal sharing of the electrons is
known as bond dipole represented as () where the arrowhead points to the more electronegative atom. The
greater the electronegativity difference between the bonded atoms, the greater is the bond dipole.
Consider the HCl and H2 molecules. Chlorine has an electronegativity value of 3.0 and hydrogen has
2.1. The result is a polar bond where the electron pair is displaced toward the more electronegative atom (Cl),
which acquires a partial negative charge (δ−). The less electronegative atom (H) has a partial positive charge
(δ+). In an H2 molecule, both atoms have the same electronegativity value of 2.1. There is equal sharing of
electrons; therefore, the covalent bond between them is considered nonpolar.

 Polarity of molecules
A polar molecule always contains one or more polar bonds; but some molecules with polar bonds can be
nonpolar overall. For example, water contains polar O-H covalent bonds and is also a polar molecule. Carbon
dioxide, however, has polar C= O covalent bonds but is a nonpolar molecule. This difference can be explained
by the net dipole moment (µ, molecular dipole), measured in debyes and which is affected by the shape of the
molecule. The higher the value of µ, the more polar is the molecule. Depending on the symmetry in the
molecule, dipole moments may cancel out of add up.
Consider the bent shape of the water molecule in figure 9-9. Two dipoles from each side, both directed
towards the central atom O, results to an upward net dipole moment, making the water molecule polar.
In the case of carbon dioxide, the linear shape allows the two equal but oppositely directed dipoles to cancel
out, giving a net dipole moment of zero. The CO2, molecule is, therefore, nonpolar.

The polarity of substances determines their solubility in different solvents. Solubility follows the general
rule "like dissolves like," which pertains to the similarity of the polarity of the solute and the solvent. A polar
solute always dissolves in a polar solvent. A nonpolar solute dissolves in a nonpolar solvent. A nonpolar (or
polar) solute will not dissolve in a polar (or nonpolar) solvent.

Activity 5 – TEST YOUR UNDERSTANDING: FORMATIVE ASSESSMENT 1 (FA 2)

Formative Assessment is for your self-learning activity. Answer this activity as honestly as you can
before you look at the answer key at the last page of this module.
 17

A. Instruction: Kindly answer the following questions below. Write your answer on the activity sheet attached
in your module.

1. How does polarity affect the way we mix common substances?

2. How is covalent bonding evidenced in everyday compounds?

DEEPEN
Your goal in this section is to take a closer look at some aspects of the topic. You will be going to learn
and understand chemical bonding for you to prepare for the final task.
You have also to remember that you are improving on the important skills that you will be needing as
you go through this part.
Activity 6 – (Q2SA1)
Note: The SUMMATIVE ASSESSMENT (SA1) is a show case of your learnings and mastery about the
learning objectives. After doing the Formative Assessments and reading activities you are now ready to
answer the Summative Assessment. Entitle your work as Q2-Lesson 3- Activity 6 – (SA1)

Instruction: Kindly complete the table below. Write your answer on the activity sheet attached in
A. By understanding octet rule, draw the Lewis structure of the following compound and give its molecular
formula
Compound Lewis structure Molecular formula
dihydrogen sulphide
carbon tetraflouride
nitrite ion

VALUES INTEGRATION
In this section, you have learned how to demonstrate the value of sharing and giving, uphold sense
of unity, respect individual differences, and communicate the importance of acceptance and respect ideas of
others. Now that you have a deeper understanding about the topic, you are ready to do the task in the
next section.

EVALUATION
Instruction: Kindly give what is being asked below. Write your answer on the activity sheet attached in the
module. Complete the table by putting the correct number.

A. Complete the given table. To answer this, determine first the Lewis structure of molecules.
Compound No. of Bonding No. of Bonding ED Geometry Molecular
Domains Domains Geometry
O3
CF4
OF2

B. Determine the types of bond in each pair of atoms. Which is the most polar bond? Least polar? Support your
answer by calculating the electronegativity difference in each pair. You can refer for the electronegativity
figure from sample 9.4 in the instruction delivery of this module.
Electronegativity difference Types of Bond
Cl – Ca
O–S
P–N
Na – Cl

References:
 18

Phoenix Publishing House; General Chemistry 1, Exploring life Through Science series.
STUDENT’S MONITORING SHEET: Check the corresponding column for your accomplishments of different
learning tasks.
Learning Activity Accomplished Not Accomplished
Pre- assessment Formative Assessment 1 (FA1)
Act. 5 Formative Assessment 2 (FA2)
Act. 6 Summative Assessment 1 (SA1)
Evaluation Summative Assessment 2 (SA2)

LESSON 3
ORGANIC COMPOUNDS
PRE-ASSESSMENT
Let’s find out how much you already know about this lesson. It’s now time to evaluate your learning.
Read and understand the questions carefully and choose the letter of the answer that you think best answers the
questions. Note: This is a self-checked activity and it will be recorded.

Activity 1- Pre-Assessment: Formative Assessment 1 (FA 1)

Note: Formative Assessment are for your self-learning activities. Answer this activity as honestly as
you can before you look at the answer key at the last page of this module.

Direction: Identify what is being describe by the following statement. Write your answer on the space provided
after the question.
__________1. These are large molecules composed of repeating smaller molecules called monomers.
__________2. It consist of repeating monosaccharide units, usually glucose.
__________3. It contain a glycerol and three fatty acids, each bound by an ester linkage.
__________4. These are biomolecules essential to the transmission of hereditary information and to the
manufacture of proteins in cells.
__________5. These are also unsaturated hydrocarbons having the formula CnH2n-2.

EXPLORE
In this lesson, you will understand how the lessons will serve as a means of enhancing yourself by
understanding various concepts that will give you a chance to discover more about yourself.
Let’s start the module by recalling our lessons in the previous grade levels. As you go through the
different learning activities in this lesson, always have in mind on how do these activities can be of great help in
real-life situations
Activity 2 – READING 1
Note: Just read and understand the concepts presented so that you will know how to answer the succeeding
activities.

In chemistry, organic compounds are generally any chemical compounds that contain carbon-
hydrogen bonds. Due to carbon's ability to catenate (form chains with other carbon atoms), millions of organic
compounds are known. The study of the properties, reactions, and syntheses of organic compounds comprise
the discipline known as organic chemistry. For historical reasons, a few classes of carbon-containing
compounds (e.g., carbonate salts and cyanide salts), along with a few other exceptions (e.g., carbon dioxide),
are not classified as organic compounds and are considered inorganic. Other than those just named, little
consensus exists among chemists on precisely which carbon-containing compounds are excluded, making any
rigorous definition of an organic compound elusive.
Although organic compounds make up only a small percentage of Earth's crust, they are of central
importance because all known life is based on organic compounds. Living things incorporate inorganic carbon
compounds into organic compounds through a network of processes (the carbon cycle) that begins with the
conversion of carbon dioxide and a hydrogen source like water into simple sugars and other
organic molecules by autotrophic organisms using light (photosynthesis) or other sources of energy. Most
synthetically-produced organic compounds are ultimately derived from petrochemicals consisting mainly
of hydrocarbons, which are themselves formed from the high pressure and temperature degradation of organic
matter underground over geological timescales. This ultimate derivation notwithstanding, organic compounds
are no longer defined as compounds originating in living things, as they were historically.
 19

Activity 3 – VIDEO VIEWING 1 (FOR THOSE WHO HAVE CONNECTIVITY. THIS IS OPTIONAL). But read the
discussion that follows.

Instructions: In order to have a better idea about organic compound, follow this link: The Functional
Group Concept Explained | Organic Chemistry | FuseSchool - YouTube the properties of minerals. Watch it
carefully so that you are prepared for the next task.
Process Questions:
1. What is stoichiometry?
FIRM-UP
In this section, you will be able to learn and understand the key concepts for you to answer the learning
activities. In this part, there are activities that will help you answer essential questions especially questions that
you have in your mind.
Try to deepen your understanding that it would look like taking a trip around the world for free.
Activity 4 –READING 2
2.
Note: Just read and understand the concepts presented so that you will know how to answer the succeeding
3.
activities.
Carbon can covalently bond to hydrogen, oxygen, nitrogen, sulphur, phosphorus, and the halogens to
form the bulk of the compound that are essential to life. These compounds, referred to as organic compound
because of their carbon composition, are either structural components of the body or utilized by the body for
specific functions.
 Classification of organic compounds
Most organic compounds are classified based on their specific groups and the bonding of atoms that
render their general properties. These specific group are known as functional groups.
 Hydrocarbons
Hydrocarbons are compounds that contain only carbon and hydrogen atoms. They may be described as
saturated or unsaturated. Saturated hydrocarbons are those in which all carbon-carbon bonds are single bonds;
unsaturated hydrocarbons contain one or more carbon-carbon multiple bonds (double bonds, triple bonds, or
both). The three types of hydrocarbons are alkanes, alkenes, and alkynes.

1. Alkanes
Alkanes are saturated hydrocarbons with the general formula C nH2n+2. They can exist as straight- chain or
as branched compounds.
 Straight-Chain Alkanes
Straight-chain alkanes are named according to the number of carbon atoms they contain. With the
exception of the first four compounds (methane, ethane, propane, butane) whose names have historical origins,
the alkanes have Greek prefixes to reflect the number of their carbon atoms. The suffix -ane identifies the
molecule as an alkane. Alkanes are commonly used as fuels. Methane, ethane, propane, and butane are all
components of natural gas. Propane and butane are the major gases in liquefied petroleum gas (LPG).

Table 10-1
IUPAC Names for the First 10 Alkanes
Number of Carbon Prefix IUPAC Names Condensed Structural Formula
Atoms
1 Meth- Methane CH4
2 Eth- Ethane CH3CH3
3 Prop- Propane CH3CH2CH3
4 But- Butane CH3CH2CH2CH3
5 Pent- Pentane CH3CH2CH2CH2CH3
6 Hex- Hexane CH3CH2CH2CH2CH2CH3
7 Hept- Heptane CH3CH2CH2CH2CH2CH2CH3
8 Oct- Octane CH3CH2CH2CH2CH2CH2CH2CH3
9 Non- Nonane CH3CH2CH2CH2CH2CH2CH2CH2CH3
 20

10 Dec- Decane CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3

2. Alkenes
Alkenes, having double bonds in their structures, are unsaturated hydrocarbons with the general formula C nH2n
Straight-chain alkenes use the same prefixes as alkanes, but end in -ene. Examples are ethene (ethylene) and 2-
pentene. Ethene is responsible for the ripening of fruits; 2-pentene is an organic solvent.

As nonpolar compounds, alkenes are insoluble in water and soluble in nonpolar solvents. They show
trends in boiling points similar to alkanes. Alkenes may exhibit cis-trans or geometric isomerism, in which the
atoms are joined in the same order but have different arrangement in space. The cis isomer has the bulky (non-
H) substituents of the double-bonded carbon on the same side of the molecule. In the trans isomer, these bulky
substituents are oriented oppositely.

3. Alkynes

Alkynes are also unsaturated hydrocarbons having the formula C nH2n-2. They contain triple bonds in their
structure. Straight-chain alkynes also use the Greek prefixes, but a suffix of -yne. Alkynes are nonpolar and
exhibit the same trends in boiling point and physical state as the other hydrocarbons. The simplest and smallest
alkyne is ethyne, more commonly known as acetylene, which burns with pure oxygen, producing the intense
heat in welding torches.

4. Cycloalkanes
Cycloalkanes are saturated hydrocarbons in which the carbon atoms are connected in a cyclic
arrangement. They are usually represented as polygons; each corner corresponds to a carbon atom attached with
two hydrogen atoms. Cycloalkanes follow the general formula C 2H2n They are named by adding the prefix
cyclo- before the name of the open-chain compound having the same number of carbon atoms as there are in the
ring. The simplest cycloalkane is cyclopropane, which is used as inhalation anesthetic.

5. Aromatic Hydrocarbons
Aromatic hydrocarbons or arenes are compounds that contain a benzene ring in their structure. Most aromatic
hydrocarbons are nonpolar and are generally used as solvents for other nonpolar substances. Benzene, a six-
carbon ring with the formula C6H6, is the simplest arene.
 21

The term "aromatic" is derived from the characteristically strong and pungent smell emitted by most arenes.
Among these substances are naphthalene (commonly known as mothballs) and toluene. Naphthalene is used as
insect repellent. Toluene is an ingredient of petroleum products, lacquers, paints, and adhesives.

 Halogen – Containing Compounds

Organic compounds may also contain other elements aside from carbon and hydrogen. Some of these
are alkyl halides, where carbon atoms are bonded with halogens.
 Alkyl Halides
Alkyl halides are hydrocarbons in which one or more hydrogen atoms around a carbon atom have been
replaced by halogen atoms. They have the general formula R-X, where R is any alkyl group and X is a halogen.
The successive substitution of chlorine atoms to each hydrogen of methane forms different alkyl halides.

The melting and boiling points of alkyl halides increase as their molecular mass rises. Methyl chloride has
the least number of chlorine atoms and, thus, has the lowest molecular mass and boiling point among the given
alkyl halides. Carbon tetrachloride, with the greatest molecular mass, has the highest boiling point. In general,
alkyl halides have higher melting and boiling points compared to hydrocarbons with comparable molecular
masses.
Alkanes where hydrogen atoms are substituted with chlorine and fluorine are collectively called
chloroflourocarbons (CFCS). These compounds, widely used before as refrigerants and as foaming agents for
plastics, are known ozone-depleting substances whose uses are now being controlled and reduced.

 Oxygen – Containing Compounds

Some hydrocarbon derivatives have oxygen-containing functional groups, in which the oxygen either (1) participates
in two single bonds (alcohols, phenols, and ethers); (2) participates in a double bond (aldehydes and ketones); or (3)
participates in single bonds and another in a double bond (carboxylic acids and its derivatives).
 Alcohols and Phenols
Alcohols are compounds that have a hydroxyl group (-OH) bonded to a carbon atom and which makes them polar.
They have the general formula R-OH. They may be classified as primary (1 o), secondary (2°), or tertiary (3 o) depending
on the number of substituents (R) bonded to the carbon atom with the -OH group.

Alcohols are named by appending the suffix -ol to the hydrocarbon name from which it is derived. For example,
methane becomes methanol, which is an alternative engine fuel. Similarly, ethanol is derived from ethane and is used as a
disinfectant, wine ingredient, and fuel. Both methanol and ethanol are primary alcohols.
 22

Phenols are similar to alcohols except that the hydroxyl group is attached to a carbon of a benzene ring. They are
also named with the same suffix as alcohols. Phenols are more soluble in water and have higher boiling points than
alcohols of similar molecular mass. The simplest example of a phenol is monohydroxybenzene (also known as phenol),
which is used as an antiseptic and a raw material for making plastics and aspirin. Another example is hydroquinone, a skin
bleaching agent in personal care products.

 Ethers
Ethers are compounds with the general formula R-O-R' where R and R'are hydrocarbon groups, which may
be the same or different. An example of ether with similar R and R' is diethyl ether, which is used as an organic
solvent and sometimes as anesthetic. Its name "diethyl" indicates the two ethyl groups attached to the oxygen
atom. The last name "ether" is a standard term for all ethers.

Methyl phenyl ether is an example of asymmetrical ether, or ether with dissimilar hydrocarbon groups.
The boiling points of ethers are generally higher than those of hydrocarbons, but lower than those of alcohols of
comparable molecular mass. Ethers are more soluble in water than hydrocarbons but less soluble than alcohols
because of the absence of polar groups.

 Aldehydes and Ketones

Both aldehydes and ketones contain the carbonyl (C=O) group. The difference lies on the substituents
attached to the carbon atom of the carbonyl; aldehydes contain R and H, while ketones have R and R' groups.

Aldehydes and ketones are named by replacing the terminal -e of the corresponding alkane name with -
al and -one, respectively. For example, methanal is the corresponding aldehyde of methane, and propanone is
the corresponding ketone of propane. Methanal is the simplest aldehyde and is usually available as 40%
aqueous solution known as formalin used to preserve biological specimens. Propanone, commonly known as
acetone, is a solvent for plastics and a nail polish remover.

Compared to their corresponding alkanes and alcohols, aldehydes and ketones have higher boiling points
because of the presence of carbonyl groups, but lower than those of alcohols due to the absence of the hydroxyl
groups.
 23

All aldehydes and ketones with more than three carbon atoms are soluble in nonpolar solvents while
those with less than three are soluble in water in all proportions.
 Carboxylic Acid
Carboxylic acid contains the carboxyl functional group have the general formula shown below.

Compared to other compounds of similar molecular mass, carboxylic acids have higher melting and
boiling points because of their polar nature. They are soluble in water, but their solubility decreases as the
number of carbon atoms increases.
Simple open-chain carboxylic acids are named by replacing the terminal -e of the corresponding alkane
name with -oic and adding the word acid. Examples are methanoic acid and ethanoic acid, which are derivatives
of the alkanes methane and ethane, respectively. Ethanoic acid, commonly called acetic acid, is an ingredient of
vinegar. Methanoic acid is secreted by ants. This acid causes the sting in their bites.

 Ester
Esters are almost similar in form to carboxylic acids except that the H in the carboxylic acid (-COOH) is
replaced with another alkyl group (R’). The general formula for esters is given below.

The names of esters are derived from the name of the alkyl group attached to oxygen (R') followed by
the carboxylic acid name whose -ic suffix is replaced with -ate. Ethyl acetate, used as an artificial food
flavoring because of its fruity smell, contains an ethyl and a carboxylic acid groups.

Esters have lower boiling points compared to carboxylic acids. Those with relatively lower molecular
mass are soluble in water, while those containing more than four carbons have limited solubility.
 Nitrogen Containing Compounds
 Amines
Amines are derived from ammonia (NH3) whose hydrogen atoms are replaced by hydrocarbon groups. They
may be classified either as primary (RNH,), secondary (R,NH), or tertiary (R,N) depending on the number of
substituents attached to the nitrogen atom.
 24

Amines are named using those of the substituent groups bonded to nitrogen, then adding the suffix -amine. The
prefixes di- and tri- are used for identical alkyl groups. The simplest primary amine is methylamine, used in making
pharmaceuticals, insecticides, and surfactants. It has one methyl group. Dimethylamine and trimethylamine has two and
three methyl groups, respectively. Dimethylamine is used as a solvent, while trimethylamine is responsible for the
unpleasant smell of decomposing plants and animals.

 Amides

Amides are another derivative of carboxylic acids and have the general formula given below.

Simple amides are named after their corresponding carboxylic acids; changing the -ic ending (common
names) or the -oic ending (IUPAC names) of the acid to -amide. Examples are methanamide and ethanamide.
Methanamide is a raw material in manufacturing herbicides and pesticides. Ethanamide is used as a solvent and
plasticizer.

 Reactions Involving Organic Compounds

An organic compound with a certain functional group can be transformed into another through chemical
reactions. Five of the most common reactions involving the different organic compounds are addition,
substitution, elimination, oxidation, and reduction.
 Addition
Addition reactions are characterized by the addition of an atom or a group of atoms to an unsaturated bond of a
compound. This type of reaction converts a triple or a double covalent bond into a double or a single bond,
respectively. Examples include the addition of Br 2 and an aqueous acid (H+) to ethene to form 1, 2-
dibromoethane and ethanol, respectively.
 Substitution
Substitution reactions involve the replacement of an atom or a group of atoms in an organic compound with
another atom or group of another compound. An example of this type is the reaction between chloromethane
and potassium hydroxide that produces methanol and potassium chloride.
 Elimination
Elimination reactions are those that involve the removal of an atom or a group from an organic compound.
These reactions form unsaturated bonds in the organic compound. For example, the H and Br atoms in
bromoethane are eliminated in the presence of potassium hydroxide and heat to produce ethene.
 Oxidation
Oxidation in organic chemistry is primarily characterized by either the removal of H from or the addition of
O to an organic compound. Below are some possible general oxidation reactions of organic compounds.
 Reduction
Reduction happens when an organic compound either gains H or loses O atoms. For example, acetone is
reduced to 2- propanol by sodium borohydride, with ethanol as the solvent.
 Combustion
Combustion is a special type of reduction- oxidation reaction in organic compounds used as
fuels. In general, combustion occurs between a hydrocarbons or its derivatives and oxygen gas,
producing mainly carbon dioxide and water. As an exothermic process, combustion releases and
provides energy to power engines. The combustion of methane is an example.

 Polymers and biomolecules

 25

Polymers and biomolecules are two of the complex types of organic compounds with wide range of
applications.
 Polymers
Polymers are large molecules composed of repeating smaller molecules called monomers. These
monomers are covalently bonded to each other through a process called polimirazation. A polymer may
consist of one or more types of monomers.
 Biomolecules
 Carbohydrates
Carbohydrates contain carbon, hydrogen, and oxygen in the ratio C (H,O) They include simple sugars,
starches, and cellulose, among many others. They serve as energy storage and as structural framework in
organisms.
Carbohydrates may be polyhydroxy (with multiple-OH) aldehydes, polyhydroxy ketones, or compounds
that can be broken down into them. The simplest form of carbohydrates are called simple sugars or
monosaccharides. The combination of two simple sugars produces a disaccharide. Carbohydrates composed of
three to 10 monosaccharide molecules are called oligosaccharides, while those containing more than 10 are
called polysaccharides.

Monosaccharides
Monosaccharides such as glucose, fructose, and galactose are the basic subunits of a carbohydrate. They
have many hydroxyl groups (-OH) in their molecular structure; thus, they are polar and therefore very soluble in
water. They can exist as a chain or ring. Glucose is a six-carbon pentahydroxy (with 5 -OH groups) aldehyde,
while fructose is a six-carbon pentahydroxy ketone. Glucose and galactose are structural isomers.

Dissacharides
Two monosaccharide units can be combined through the process called condensation reaction to produce a
disaccharide molecule. During such process, a glycosidic bond is formed between the two FOH groups of the
monosaccharide units. In a reverse process called hydrolysis, the glycosidic bond is broken.
The most common disaccharides are sucrose, lactose, and maltose. Sucrose, or common table sugar, is a
disaccharide of glucose and fructose. Lactose, the sugar present in milk, forms from the condensation of glucose
and galactose. Maltose is made from two glucose units. Disaccharide molecules are also sources of cell energy,
but they must be hydrolyzed first into monosaccharide units before they are used by the cells.

Polysaccharides
Polysaccharides consist of repeating monosaccharide units, usually glucose. Among these complex sugars
are starch, glycogen, and cellulose, which serve as energy storage or as cell framework. Starch exists as either
linear chain (amylose) or branched chain (amylopectin) of glucose units. This polysaccharide obtained from
plants should be hydrolyzed first into its glucose units before it can be used as energy source.

 Lipids
Lipids are macromolecules composed of several simpler compound groups. They have mainly hydrogen and
carbon atoms and a few oxygen atoms that make them generally nonpolar. As nonpolar compounds, they are
soluble in nonpolar solvents and insoluble in water. Triglycerides, phospholipids, and steroids are common
types of lipids.

Triglycerides
Triglycerides, commonly termed fats, contain a glycerol and three fatty acids, each bound by an ester
linkage. These compounds are the most abundant energy storage lipids in organisms. They can supply more
energy than carbohydrates. Fatty acids, which are carboxylic acids, as saturated or unsaturated. Saturated fatty
acids contain only may be described single bonds. They tend to solidify at room temperature. This explains why
fats with more saturated fatty acids are considered less healthful and are associated with heart diseases.
Unsaturated fatty acids contain double bonds and tend to make fats liquid at room temperature.

Phospolipids
Macromolecules composed of a glycerol, a phosphate group, two fatty acids, and an organic group like
choline are called phospholipids. The fatty acids form the "tail" of a phospholipid molecule; the three other
make its "head." The tail end is hydrophobic (water-fearing, water-insoluble), and components up. the head is
hydrophobic (water-loving, water-soluble). These two opposing properties make the whole phospholipid
molecule amphiphatic, which explains the roles of the lipid bilayer in cell membranes-to hold the cells together
and control the substances that pass into and out of the cells.
 26

Steroids: Cholesterol
Steroids are compounds whose backbone contains three six-carbon rings and one five-carbon ring. One
of the common types of steroids is cholesterol, which is a structural component of animal cell membranes.
 Protein
Proteins are polymers composed of amino acid monomers. An amino acid is an organic compound that
contains an amino (-NH,) group and a carboxyl (-COOH) group bonded to a carbon atom called alpha carbon.
Essential amino acids such as arginine, lysine, phenylalanine, and tryptophan, are not synthesized by the body
and should be obtained from diet. Nonessential amino acids are produced by the body and include alanine,
glycine, and proline. Amino acids also combine through condensation reactions and are linked by peptide
bonds. Two amino acids form a dipeptide; more units form a polypeptide. One or more polypeptides make up
protein.

 Nucleic acids
Nucleic acids are biomolecules essential to the transmission of hereditary information and to the manufacture of
proteins in cells. The two kinds of nucleic acids in cells are ribonucleic acid (RNA) and deoxyribonucleic acid
(DNA). The RNA participates in the formation of polypeptides, while the DNA constitutes the hereditary
material. The building blocks of nucleic acids are nucleotides, which are composed of a base, a sugar, and a
phosphate group.
The bases found in DNA and RNA are heterocyclic amines, which may either be purine or pyrimidine bases.
Adenine, guanine, cytosine, and thymine constitute the DNA. The first three bases (A, G, C) and uracil make up
the RNA. The sugar component of RNA is D-ribose, while that of DNA is 2-deoxy-D-ribose. Each sugar group
in a nucleic acid is attached to a phosphate group through phosphodiester linkage.

Activity 5 – TEST YOUR UNDERSTANDING: FORMATIVE ASSESSMENT 2 (FA 2)

Formative Assessment is for your self-learning activity. Answer this activity as honestly as you can.

Instruction: Now that you have read our lesson about animal’s specialized organ. Instruction: Kindly give what
is being asked below. Write your answer on the activity sheet attached in your module. Indicate the following
Q1-lesson 5 – Activity 5- (FA2)

1. Describe the different functional groups.

2. Describe some simple reactions of organic compounds

Now that you know the important ideas about this topic, let’s go deeper by moving on to the next
section.
DEEPEN
Your goal in this section is to take a closer look at some aspects of the topic. You will be going to learn
and understand about variables, constants and exponents for you to prepare for the final task.
You have also to remember that you are improving on the important skills that you will be needing as
you go through this part.
Activity 6 – (Q2SA1)
Note: The SUMMATIVE ASSESSMENT (SA1) is a show case of your learnings and mastery about the
learning objectives. After doing the Formative Assessments and reading activities you are now ready to
answer the Summative Assessment. Entitle your work as Q1-Lesson 3- Activity 6 – (SA1)

Kindly give what is being asked below. Write your answer on the activity sheet attached in your module.

Naming of compounds.

Molecular Formula Name of Compounds

C5H10 1.
C9H16 2.
C10H22 3.
C7H16 4.
C8H18 5.
 27

Formula Writing
Name Formula
1. Pentane
2. Hexane
3. heptane
4. butane
5. octane
VALUES INTEGRATION
In this section, you have learned to promote the advancement of science and technology to improve the
way of life, develop awareness respect for human body, exhibit careful judgment in choosing which food to eat
and practice sense of awareness and responsibility in using substances.

Now that you have a deeper understanding about the topic, you are ready to do the task in the next
section.
EVALUATION

Instruction: MULTIPLE CHOICE: Read each item carefully. Choose the correct answer from the given
choices. Write your answer on the activity sheet attached in your module.
1. These reactions form unsaturated bonds in the organic compound.
a. elimination reactions b. oxidation c. reduction d. combustion
2. It happens when an organic compound either gains H or loses O atoms.
a. elimination reactions b. oxidation c. reduction d. combustion
3. It contains the carboxyl functional group.
a. amines b. alkyns c. carboxylic acid d. polyssacharides
4. A macromolecules composed of a glycerol, a phosphate group, two fatty acids, and an organic group like
choline are called phospholipids.
a. Lipids b. carbohydrates c. protein d. phospolipids
5. These are compounds whose backbone contains three six-carbon rings and one five-carbon ring. One of the
common types of steroids is cholesterol, which is a structural component of animal cell membranes.
a. enzymes b. proteins c. monosaccharide d. steroids

B. Instruction: Kindly give what is being asked below. Write your answer on the activity sheet attached in your
module.

1. Describe the formation and structure of polymers.

2. Describe the structure of proteins, nucleic acids, lipids, and carbohydrates, and relate them to their function.

Reference:
Phoenix Publishing House; General Chemistry 1, Exploring life Through Science series.
STUDENT’S MONITORING SHEET: Check the corresponding column for your accomplishments of different
learning tasks.
Learning Activity Accomplished Not Accomplished
Pre- assessment Formative Assessment 1 (FA1)
Act. 5 Formative Assessment 2 (FA2)
Act. 6 Summative Assessment 1 (SA1)
Evaluation Summative Assessment 2 (SA2)