worksheet (LKS) ini berisi materi ikatan kimia secara lengkap. mulai dari ikatan ion hingga ikatan kovalen. monggo dimanfaatkan

1. WORKSHEET OF CHEMISTRY
CHEMICAL BONDING
By: Dasianto, S.Pd
SMA BRAWIJAYA SMART SCHOOL MALANG
1

2. Jl. CIPAYUNG 10 MALANG
CHEMICAL BONDING
A. Differentiating Kinds of Chemical Bonding
To understanding differenciation kinds of chemical bonding, do the experiment:
1. Tittle: Differentiating kinds of chemical bonding
2. Purpose: Differentiating kinds of chemical bonding using boling point and electrical conductivity
3. Tools and equipment
Tools Equipment
Termometer Sugar crystal
Beaker glass Salt crystal
Electrolyte tools kit Water
Bunsen burner
Graduated cylinder
Digital scale
4. Procedure
a. Boiling point
- weigh 5 grams of sugar and 5 grams of salt
- dissolve into 50 ml of water in different beaker glass
- take 10 mL of solution and bring to the boil
- measuring the temperature required until boiling solution
b. Electrical conductivity
- Take 10 ml of solution
- measure the electrical conductivity of the electrolyte tools kit
5. Experiment data
Boiling point of Conduction of electricity
Sugar solution: Sugar solution: able/not
Salt solution: Salt solution: able/not
6. Data Analysis
No. Questions Answer
1 Which solution which has a higher boiling point?
2 Which solution that can conduct electric current?
(Electric current is marked with flashing lights or
bubbles formed around the carbon rod)
3 Sugar is one example of covalent compounds.
While salt is one example of ionic compounds. In
general how the tendency of the boiling point of
ionic compounds and covalent compounds? Why
its happen?
4 Sugar is one example of covalent compounds.
While salt is one example of ionic compounds. In
general how the tendency of electrical conductivity
of ionic compounds and covalent compounds?
Why its happen?
7. Conclussion
Based on the experimental results, the tendency of ionic compounds can/cannot conduct the electrical
current and the boiling point is higher/lower than covalent compounds.
Additional information: The following data are presented on the differences properties of ionic and
covalent compounds:
No Properties Ionic Compounds
Covalent Compounds
Non polar Polar
1 Melting/boiling point High Low High
2 Electrical conductivity Solid phase: not conducted
Liquid and solution phase:
Solid phase, liquid phase,
solution phase: not
Solid phase, liquid phase,:
not conducted

3. conducted conducted solution phase: conducted
3 Solubility in water Soluble Insoluble soluble
Based on the explanation above, issues that arise are: why the properties of ionic compounds and covalent
compounds are different? Why it is happen? The following will be explained about ionic and covalent
compounds in detail.
B. Ionic Compounds
Ionic compounds are compounds formed by the ionic bonding. Interaction or ionic bonding are attractive
forces between positive ions (cations) and negative ions (anions). Cation is an atom or group of atoms
positively charged, while the anion is an atom or group of atoms that are negatively charged.
There are two types of cations, namely simple cations (formed by removing an electron atoms) and
complex cation (group of atoms with a positive charge). As well as anions, simple anion formed by electron
accept by atoms, while anion complex is a group of atoms with negative charge.
The following examples of cations and anions is accompanied by its name:
Simple cations
Groups
IA
Name of cations Groups
IIA
Name of cations Groups
IIIA
Name of cations
Li+
Litium ion Mg2+
Magnesium ion Al3+
Aluminum ion
Na+
Sodium ion Ca2+
Calcium ion Ga3+
Galium ion
K+
Potassium ion Sr2+
Stronsium ion
Rb+
Rubidium ion Ba2+
Barium ion
Cs+
Ceasium ion
Another simple anion with severals chaerges:
Cation Name of cations Cation Name of cations
Cu+
Copper(I) ion Sn2+
Tin(II) ion
Cu2+
Copper(II) ion Sn4+
Tin(IV) ion
Fe2+
Iron(II) ion Co2+
Cobalt(II) ion
Fe3+
Iron(III) ion Co3+
Cobalt(III) ion
Complex Cation: NH4
+
(ammonium ion)
Simple anions
Groups
VA
Name of anions Groups
VIA
Name of anions Groups
VIIA
Name of anions
N3-
Nitride ion O2-
Oxide ion F-
Fluoride ion
S2-
Sulfide ion Cl-
Chloride ion
Se2-
Selenide ion Br-
Bromide ion
I-
Iodide ion
Complex anions
(-1) Name of anions (-2) Name of anions (-3) Name of anions
NO2
-
Nitrous ion SO3
2-
Sulphite ion PO3
3-
Phosphite ion
NO3
-
Nitrate ion SO4
2-
Sulphate ion PO4
3-
Phosphate ion
OH-
Hydroxide ion CO3
2-
Carbonate ion
C2O4
2-
Oxalate ion
CrO4
2-
Chromate ion
Cr2O7
2-
Dichromate ion
Based on data above, there are four types of ionic compound:
1. Combination between simple cation and simple anion. Example: Na+
+ Cl-
 NaCl (sodium chloride)
2. Combination between simple cation and complex anion. Example: Na+
+ NO3
-
 NaNO3 (sodium
nitrate)
3. Combination between complex cation and simple anion. Example: NH4
+
+ Cl-
 NH4Cl (ammonium
chloride)
4. Combination between complex cation and complex anion. Example: NH4
+
+ NO3
-
 NH4NO3
(ammonium nitrate)
Exercise: fullfill the folowing table:
Cation/anion Cl-
S2-
NO3
-
SO4
2-
PO3
3-

4. Chloride ion Sulfide ion Nitrate ion Sulphate ion Phosphate ion
Na+
Sodium ion
Mg2+
Magnesium ion
NH4
+
Ammonium ion
(NH4)2SO4
Ammonium sulphate
Fe2+
Iron(II) ion
FeS
Iron(II) sulfide
Fe3+
Iron(III) ion
The type of ionic compounds to be studied further is the first type. The main material is the process of
formation, and writing the symbols Lewis.
Formation of Ionic Compounds.
The first type of ionic compounds formed from simple cations and simple anions. Simple cation occurs if a
metal atom loses or releases a or amount of electrons, and simple anion formed when nonmetal atom
accept a or amounts of electrons. Release or accept electrons aims to achieve stability (reach the noble
gas configuration). Thus, the ionic compound can be formed by transfer of electrons between metal and
non metal atom. Interaction between cation and anion in ionic compound call as ionic bonding.
Example: Write formation of NaCl (sodium chloride) process.
11Na: [Ne] 3s1
(have a valence electron, to achieve stability by release an electron)
17Cl: [Ne] 3s2
3p5
(have 5 valence electrons, to achieve stability by accept an electron)
Lewis Symbol of: Na Cl NaCl
Exercise: write the formation process and Lewis Symbol of these ionic compounds:
No Ionic Compound No Ionic Compound
1 19K + 9F to form KF (potassium fluoride) 5 20Ca + 35Br to form CaBr2 (calcium bromide)
2 12Mg + 8O to form MgO (magnesium oxide) 6 13Al + 17Cl to form AlCl3 (aluminum chloride)
3 13Al + 7N to form AlN (aluminum nitride) 7 19K + 7N to form K3N (potassium nitride)
4 37Rb + 16S to form Rb2S (rubidium sulfide) 8 12Mg + 7N to form Mg3N2 (magnesium nitride)

5. Properties of ionc compounds:
1. Melting and boiling point
Alkaly Halide Melting Point (0
C)
NaF 993
NaCl 801
NaBr 747
NaI 651
From the table above, melting point of ionic compounds tend to have high, because the ionic bonding
in this compound is quite strong. In a group from top to bottom, melting point tend to decrease because
increasing bonding radii reduce bonding forces.
2. Physical state of ionic compound
Ionic compound commonly in solid phase, hard but brittle if struck by hammer. The slight movement of a
layer of ion within crystal suddenly places ions of the same charge next to each other, and for that
instant there are large repulsive forces that split the solid and crystal of ionic compound shatters.
3. Solubility
Ionic compound able to soluble in polar solvent.
4. Electrical conductivity
In the solid state, ionic compounds do not conduct electricity because the attractive force between
cations and anions revent the movement of ions through the crystal. When the solid melted, however,
the ions become free to move about and the liquid conducts electricity quite well. In solution, all of the
ions are become free and electrical conductivity more stronger.
NaCl(l)  Na+
(l) + Cl-
(l)
NaCl(s)  Na+
(aq) + Cl-
(aq)
C. Covalent Compounds
Covalent compounds are compounds formed by the covalent bonds. Covalent bonding occurs when two or
more atoms share electrons (use electrons together). Generally covalent bonds occur between nonmetal
atoms. There are four types of covalent bonding:
1. Single covalent bond
Example: formation H2 (hydrogen molecule)
2. Doble covalent bond
Example: formation O2 (oxigen molecule)
3. Triple covalent bond
Example:
4. Coordinate covalent bond

6. Exercise: draw the formation of covalent compound below!
Chemical
Formula
Lewis Structure
Chemical
Formula
Lewis Structure
Cl2
(17Cl)
CO2
(6C, 8O)
PCl3
(15P, 17Cl)
XeF2
(54Xe, 9F)
NH3
(7N, 1H)
BrF5
(35Br, 9F)
SF4
(16S, 9F)
HCN
(1H, 6C, 7N)
H2O
(1H, 8O)
PCl3
(15P, 17Cl)
The above exercises are for simple molecules. To drawing structure of complex molecules such as H2SO4,
HNO3, H3PO3, H3PO4 required systematic ways. Step to drawing complex molecules are:
1. Determine the central atom
2. Determine the skeletal structure
3. Count the Total Valence Electron (TVE)
4. Place the sigma (single) covalent bond and making the substitute atom to obey octet rule
5. Calculate Total sigma electron (TSE) and Total Lone Pair Electron (TLPE)
6. Calculate the rest electron (RE). RE = TVE-TSE-TLPE
7. Place the rest electron on the central atom
8. Making the lowest formal charge for central atom by making phi (double) bonding
Formula to calculate formal charge is: Qf = NA – NLP – ½ NBP
NA = number of valence electron an atom
NLP = number of lone pair electron

7. N H
H
H
NBP = number of bonding pair electron
Example: determine formal charge of each atom in NH3 molecule
The molecule structure of NH3 molecule is
N H
H
H
Example: Draw the Lewis structure of NH3 molecule
1. The central atom is N
2. The skeletal structure
H N H
H
3. Count the Total Valence Electron (TVE)
TVE = 5 + 3.1 = 8
4. Place the sigma (single) covalent bond and making the substitute atom to obey octet rule
5. Calculate Total sigma electron (TSE) and Total Lone Pair Electron (TLPE)
TSE = 6
TLPE = 0
6. Calculate the rest electron (RE).
RE = TVE-TSE-TLPE
RE = 8-6-0 = 2
7. Place the rest electron on the central atom
N H
H
H
8. Calculate the formal charge of each atom
Qf N = NA – NLP – ½ NBP
= 5 – 2 – ½ 6
= 0
Qf H = NA – NLP – ½ NBP
= 1 – 0 – ½ 2
= 0
Exercise: Draw the Lewis symbol of these compound:
Chemical
Formula
Lewis Structure
Chemical
Formula
Lewis Structure
H2SO4
(1H, 16S, 8O)
HNO3
(1H, 7N, 8O)
Qf N = NA – NLP – ½ NBP
= 5 – 2 – ½ 6
= 0
Qf H = NA – NLP – ½ NBP
= 1 – 0 – ½ 2
= 0

8. H3PO3
(1H, 15P, 8O)
H3PO4
(1H, 15P, 8O)
D. Shape and Polarity of Molecules
Before discuss about shape and polarity of moleculs, do this experiment:
1. Tittle: Identify polarity of molecule
2. Purposes: identify polarity of molecule by experiment
3. Tools and equipment
Tools and equipments Quantities
Burette 6
Statif 6
Ruler/polythene 1
Beaker glass 1 (50 ml)
Wool 1 sheet
Aquadest 50 ml
Kerosene 50 ml
Palm oil 50 ml
Tab water 50 ml
4. Procedure
Figure Activity
Set the burette, statif look like the figure, and then
add:
1. Aquadest in burette 1
2. Kerosene in burette 2
3. Palm oil in burette 3
4. Tap water
Charge the polythene/ruler by scouring in wool
repeatedly

9. Polythene placed near the drop of liquid, and
observes the current of drop of liquid.
5. Experiment data
Liquid
The current of drop of liquid
Bent Not bent
Aquadest
Kerosene
Palm oil
Tap water
6. Data analysis
No Questions Answer
1 Which liquid make the current of drop of liquid
bent? And explain why it’s happen!
2 Which liquid make the current of drop of liquid
not bent? And explain why it’s happen!
3 Explain, how do these molecules have this
properties?
7. Conclussion
Based on this experiment, the polar molecules are ....................................... because able to change
of current drop liquid. The non polar molecules are ........................................... because not able to
change of current drop liquid
Concepts previously learned is to symbolization compounds in two-dimensional media. The fact that there
are in nature, each molecule has a different shape, depending on the type of atoms involved. In this section
will be learned about the shape and polarity of a molecule. 3D molecular shape describe the shape of a
molecule, polarity of molecule can be determine based on the shape of the molecule. Polarity of molecule is
a concept to predict the distribution of electrons in a molecule. If the electrons tend to localize, said non-
polar molecules. Whereas in case of the spread of the electrons will be polar molecules.
Table below describe shape and polarity of molecules:
Coordinati
on number
General
formula
Shape of
molecule
Geometri Angels Example Polarity
2 AX2 Linear 1800
CO2, BeCl2 Non polar
3 AX3 Trigonal planar 1200
BF3, BCl3 Non polar
3 AX2E “V” shape/
bent
<1200
SO2 Polar

10. 4 AX4 Tetrahedral 109,50
CH4, CCl4 Non polar
4 AX3E Trigonal
pyramidal
<109,50
NH3, PCL3 Polar
4 AX2E2 “V” shape/
bent
<109,50
H2O, SCl2 Polar
4 AXE3 Linear 1800
HF, OH-
Polar
5 AX5 Trigonal
bipyramidal
Axial:900
Equatorial:
1200
PF5, PCl5 Non polar
5 AX4E Seesaw Axial:<900
Equatorial:
<1200
SF4, SeF4 Polar
5 AX3E2 “T” shape
Axial:<900
ClF3, BrF Polar
5 AX2E3 Linear Axial:1800
XeF2, I3-
Non Polar
6 AX6 Octahedral Axial:900
Equatorial:
900
SF6, SeF6 Non polar
6 AX5E Square
pyramidal
Axial:900
Equatorial:
900
BrF5, IF5 Polar
6 AX4E2 Square planar Equatorial:
900
BrF4-
, XeF4 Non polar
Step to determine shape of molecule and predict polarity of molecule are:
1. Draw the Lewis structure of molecule
2. Calculate coordination number (CN) and predict shape of molecule
CN = (BEP + LEP)
CN = coordination number
BEP = bonding electron pairs around central atom
LEP = lone electron pairs around central atom
3. Calculate moment dipole (µ). Moment dipole is vector resultant of bond in molecule.
Example: Predict shape and polarity of ammonia (NH3)
1. Lewis structure of ammonia is

11. N H
H
H
2. CN of ammonia
CN = BEP + LEP
CN = 3 + 1
CN = 4
Shape of molecule:
(trigonal pyramide)
3. Moment dipole (µ) of molecule
this is polar molecule
Exercise: predict shape and polarity oh these molecules:
Chemical
Formula
Shape and polarity
Chemical
Formula
Shape and polarity
Cl2
(17Cl)
CO2
(6C, 8O)
PCl3
(15P, 17Cl)
XeF2
(54Xe, 9F)
NH3
(7N, 1H)
BrF5
(35Br, 9F)
SF4
(16S, 9F)
HCN
(1H, 6C, 7N)

12. H2O
(1H, 8O)
PCl3
(15P, 17Cl)
Properties of covalent compounds
No Properties
Covalent Compounds
Non polar Polar
1 Melting/boiling point Low High
2 Electrical conductivity
Solid phase, liquid phase,
solution phase: not conducted
Solid phase, liquid phase,: not
conducted
solution phase: conducted
3 Solubility in water
(polar solvent)
Insoluble
Soluble
4 Solubility in CCl4 (non
polar solvent)
Soluble
Insoluble
E. Metallic Bonding
Metalic bonding is a bond that occurs due to the interaction between the metal atoms. In the electron-sea
model, crystal of metal is viewed as 3-dimensional array of metal cations immersed in a sea of delocalized
electrons that are free to move throughout the crystal as shown below:
Atom in metallic solid pack closely together in regular structure, if the atoms in metal are represented as
spheres, they can only touch the others spheres, and leaving gaps or called as interstitial sites. Amounts of
atoms that surrounding an atom called coordination number. Give attention to these pictures:
(a) (b) (c)
Packing of atoms in metals, can be describe in three kinds, they are: hexagonal close packed (HCP), cubic
close packed (CCP) and body centered cubic (BCC). Filled this table below!
Type packing
atom in metals
Picture Explanation
Hexagonal close
packed (HCP)
Coordination number:
+ + + + + +
+ + + + + +
+ + + + + +
+ + + + + +
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_
_ _
_
_
_
_ _
__
_
_ _
_
__
_
_ _
_
__
_
_
_
_
_
_
_
_
_
_
_
_ _
_ _
__
_ _
_
_
_
_
_
_
_
_
_
_
_
_
positive ions from
the metallic atom
with regular
arangement
delocalized
electrons
Layer A
Layer B
c c c
c c c
c
a a
a
a
a a
a
c
Interstitial site Interstitial site
b
b b
b
b
b
b
c c
c c
c c
Layer A

13. The unit cell of HCP
How to determine coordination number
based on this picture?
A, b, c describe the unit cell. What is
the unit cell?
Cubic close packed
(CCP)
Coordination number
Layer A
Layer A
Layer B
Layer B
Interstitial sites b
Layer B
Layer B
Layer A
Layer A
Layer A
Interstitial sites a
Layer A
Layer B
a a
a
a
a a
a
Layer C
b b b
Layer A
LayerA
Layer C
Layer B
Layer B
LayerB
Layer A
Layer C
Layer A
LayerA
Layer C
Layer B Layer C
LayerA

14. The unit cell of CCP
Body centered
cubic (BCC)
Primitive cubic
Coordination number:
Coordination number:

15. Properties of metallic bonding
In generall, properties of metals are: dense, malleable and ductile, conductor of heat and electricity, have
high melting and boiling point, and lustrous.
Give attention the this table:
No Metals Melting Point (0
C) Boiling Point (0
C)
1 Li 180 1330
2 Na 97.8 890
3 K 63.7 774
4 Na 97.81 892
5 Mg 651 1107
6 Al 660 2467
1. Based on the table, melting/boiling point metals in a group (data no. 1-3) tend to ...............................
because .........................................................
2. Based on the table, melting/boiling point metals in a period (data no. 4-6) tend to ...............................
because .........................................................
F. Intermolecular Forcess
Ionic bonding, covalent bonding and metallic bonding are types of intermolecular force. Intermolecular force
are interaction between atoms in a molecul. If there are two or more molecule interact each others, called
as intermolecular force. So, intermolecular force present on covalent bonding. There are four kinds of
intermolecular force:
1. London Force
London force occurs when two or more non-polar molecules interact with each other. Example:
interaction between H2 and H2, CH4 and H2 etc. Explanation about London force are:
2. Dipole-induced dipole force
Dipole-induced dipole force occurs when polar molecule and non polar molecules interact with each
other. Example: interaction between HCl and H2, HCl and CH4 etc. Explanation about dipole-induced
dipole force are:

16. 3. Dipole-dipole force
Dipole-dipole force occurs when two or more polar molecules interact with each other. Example:
interaction between HCl and HCl, HCl and CH3Cl etc. Explanation about dipole- dipole force are:
4. Hydrogen bonding
Hydrogen bonding occurs when an atom have hydrogen atom and interact with F, O, N atom from
another molecule. Example: interaction between H2O and H2O, H2O and HF, H2O and NH3 etc.
Explanation about hydrogen bonding are:
The following is the strengthes of intermolecular forces: London < dipole-induced dipole < dipole-dipole <
hydrogen bonding.
Effect intermolecular force to physical properties of compounds
1. Melting or boiling point
a. the stronger the intermolecular forces the higher boiling point
b. the stronger the London force in non-polar molecule, higher boiling point

17. Molecules
formula
Name of
compouds
Melting
point (o
C)
Mass
molecule reltif
Boiling
point (o
C)
Physical state at
room temperature
CH4 metana -181,9 16 -163,9 Gas
C2H6 Etana -183,2 30 -88,5 Gas
C3H8 Propana -189,6 44 -42,0 Gas
C4H10 Butana -138,3 58 -0,4 Gas
C5H12 Pentana -129,9 72 36,2 Cair
2. Density
The stronger the intermolecular forces, the higher the viscosity of the compound
G. Evaluation
1. Berikut ini yang merupakan kumpulan senyawa
ionic adalah…
a. HCl, NaCl, HNO3 d. O3, NO2, BeCl2
b. KCl, NaCl, NH4NO3 e. KNO3, BeCl2, KF
c. NH3, SO2, CH4
2. Diketahui beberapa jenis ion sebagai berikut:
NH4
+
, SO4
2-
, Ca2+
, PO4
3-
, NO3
-
dan Al3+
, maka
rumus kimia senyawa berikut yang tidak benar
adalah....
a. (NH4)2SO4 c. (NH4)3NO e. AlPO4
b. CaSO4 d . Al2(SO4)3
3. Di antara kelompok senyawa di bawah ini yang
semuanya mempunyai ikatan kovalen
adalah ....
a. KCl, NaCl dan HCl d. NH3, CaO dan K2O
b. HCl, SO2 dan NH3 e. KBr, NaCl dan
CaBr2
c. H2O, Na2O dan N2O5
4. Jika unsur Q (Ar = 24) dengan jumlah netron
12 berikatan dengan unsur R (Ar = 35) yang
mempunyai jumlah netron 18, maka senyawa
yang terbentuk adalah….
a. QR2, ikatan ionic d. Q2R, ikatan ionik
b. QR2, ikatan kovalen e. QR, ikatan
kovalen
c. Q2R, ikatan kovalen
5. Berikut ini adalah rumus elektron dari ion nitrat
N
O
x O O
P
Q
RS
T
Ikatan kovalen koordinat pada gambar
tersebut ditunjukkan oleh anak panah huruf…
a. P c. Q e. R
b. S d. T
6. Sebuah atom netral X mempunyai konfigurasi:
[Ne] 3s2
3p5
. Jika unsure tersebut membentuk
hidrida, maka senyawa yang mungkin
adalah…
a. XH2, ionic c. XH, kovalen
e. XH2, kovalen
b. XH, ionic d. XH3, kovalen
7. Unsur A mempunyai massa atom 27 dan
memiliki 14 neutron. Unsure B mempunyai 16
neutron dan 16 proton. Jika keduanya
berikatan, maka senyawa tersebut memiliki
massa molekul relative sebesar…
a. 59 c. 76 e. 150
b. 74 d. 86
8. Di dalam senyawa NH4Cl terdapat…
a. Ikatan ionik
b. Ikatan kovalen
c. Ikatan kovalen dan ionic
d. Ikatan kovalen dan kovalen koordinasi
e. Ikatan ionic, kovalen dan kovalen
koordinasi

18. 9. Tabel berikut menunjukkan sifat senyawa Q
dan R
Senyawa
Titik leleh
(o
C)
Daya Hantar Listrik
Lelehan Larutan
Q -115 Tidak Mampu
R 810 mampu Mampu
Dari data tersebut, ikatan yang terdapat
dalam senyawa Q dan R berturut-turut
adalah…
a. Kovalen polar dan ion
b. Kovalen non polar dan ion
c. Kovalen non polar dan kovalen polar
d. Kovalen koordinasi dan ion
e. Kovalen non polar dan hydrogen
10.Diberikan beberapa rumus molekul dan nama
masing-masing molekul. Pasangan rumus
molekul dan nama yang benar adalah….
a. PBr3 = fosfor bromida
b. NCl3 = nikel triklorida
c. Cl2O7 = klorida pentoksida
d. P2O5 = difosfor pentoksida
e. SF6 = sulfida heksaflurida
11.Pasangan rumus kimia senyawa yang
terbentuk dari ion-ion serta namanya yang
benar adalah ....
a. Cu2(PO4)3: Tembaga(II) pospat
b. Cu(SO4)2 : Tembaga(II) sulfat
c. Fe3(PO4)2 : Besi(III) pospat
d. Fe2(SO4)3 : Besi(III) sulfat
e. K2SO4 : Kalsium sulfat
12.Bentuk molekul yang mungkin untuk molekul
NH3 adalah (7N, 1H)…
13.Senyawa berikut yang dapat membentuk
ikatan hydrogen adalah…
1. H2O (nomor atom O = 6, H = 1)
2. CH3OH (nomor atom O = 6, H = 1, C = 6)
3. HF (nomor atom F = 9, H = 1)
4. HBr (nomor atom Br = 35, H = 1)
14.Senyawa berikut yang memiliki gaya dipole-
dipole sebagai gaya antar molekul terkuatnya
adalah…
1. H2O (nomor atom O = 6, H = 1)
2. CH3Cl (nomor atom Cl = 17, H = 1, C = 6)
3. HF (nomor atom F = 9, H = 1)
4. HBr (nomor atom Br = 35, H = 1)
15.Tabel berikut menunjukkan sifat senyawa Q
dan R
Senyawa
Titik leleh
(o
C)
Daya Hantar Listrik
Lelehan Larutan
Q -115 Tidak Mampu
R 810 Mampu Mampu
Dari data tersebut, ikatan yang terdapat
dalam senyawa Q dan R berturut-turut
adalah…
a. Kovalen polar dan ion
b. Kovalen non polar dan ion
c. Kovalen non polar dan kovalen polar
d. Kovalen koordinasi dan ion
e. Kovalen non polar dan hydrogen
16.Perhatikan gambar berikut:
Ikatan/interaksi antar molekul yang
ditunjukkan nomor 3 adalah....
a. Gaya dipol-dipol induksi
b. Gaya dipol-dipol
c. Ikatan hidrogen
d. Gaya London
e. Gaya ion-dipol
17.Apabila unsur 16X berikatan dengan unsur 35Y,
maka rumus kimia dan jenis ikatan yang
terbentuk adalah...
a. XY ionik c. XY2 ionik e. X2Y
kovalen
b. X2Y ionik d. XY2 kovalen
18.Rumus kimia dan jenis ikatan yang terjadi jika
20Ca berikatan dengan unsure Z adalah…
a. CaZ2, kovalen d. CaZ2, ionik
b. Ca2Z, kovalen e. Ca2Z, ionik
c. CaZ, ionik
19.Senyawa yang tersusun dari buah unsure 6X
dan 17Y, bila berikatan akan memiliki bentuk
molekul dan dan kepolaran berturut-turut
adalah…
a. Tetrahedral, polar d. bentuk V,
non polar
b. Tetrahedral, non polar e. Bentuk V,
polar
c. Trigonal bipiramida, polar
20.Berikut ini adalah rumus elektron dari ion nitrat
N
O
x O O
P
Q
RS
T
Ikatan kovalen koordinat pada gambar
tersebut ditunjukkan oleh anak panah huruf…

19. c. P c. Q e. R
d. S d. T
21.Perhatikan grafik titik didih beberapa senyawa
hidrida golongan IV-A, V-A, dan VI-A berikut
ini! (UN 2013)
Senyawa yang mempunyai ikatan hidrogen
antar molekulnya adalah nomor...
a. 1 dan 2 c. 1 dan 3 e. 4 dan 5
b. 4 dan 6 d. 5 dan 6
22.Perhatikan data hasil percobaan berikut.
No Sifat fisik Zat A Zat B
1 Wujud zat Padat Padat
2 Kelarutan dalam air Larut Tidak larut
3 Daya hantar listrik
larutan
Konduktor Isolator
4 Titik leleh dan titik didih Tinggi Rendah
Berdasarkan data tersebut, dapat disimpulkan
bahwa jenis ikatan yang terdapat pada zat A
dan zat B berturut-turut adalah….
A. ionik dan kovalen nonpolar
B. kovalen polar dan ionik
C. kovalen nonpolar dan ionik
D. kovalen koordinasi dan logam
E. hidrogen dan kovalen