Cehm

Question 1

bring together elements with similar properties.

Answer 1

The vertical groups

Question 2

are arranged in order of increasing atomic number from left to right. The groups are numbered at the top. and the periods at the extreme left in the periodic table on the inside front cover.

Answer 2

The horizontal periods

Question 3

The first two groups

Answer 3

the s block-

Question 4

the last six groups

Answer 4

the p block

Question 5

together constitute the main-group elements

Answer 5

S block and p block

Question 6

are known as the transition elements.

Answer 6

The d-block elements

Question 7

The f-block elements

Answer 7

sometimes called the innertransition elements

Question 8

sometimes called the innertransition elements

Answer 8

The f-block

Question 9

the f-block elements are extracted from the table and placed at the bottom

Question 10

The 14 elements following lanthanum (Z = 57) are called

Answer 10

lanthanides

Question 11

actinium

Answer 11

actinides.

Question 12

Have a shiny luster various colors,
although most are silvery
Solids are malleable and ductile
Good conductors of heat and electricity
Mest metal oxides ave ionic solids that
are basic
Tend to for cations in aqueous solution

Answer 12

Metals

Question 13

Donot have a luster: various colors
Salids are usuallvbrittle: some are hard.
and some are soft
Poor conductors of heat and electricity Mest nonmetallic oxides are molecular
substences that formacid ic solutions
Tend to form anions or orvanions in
aqueous solution

Answer 13

Nonmetals

Question 14

are elements that look like metals and in some ways behave like metals, but also have some nonmetallic properties.

Answer 14

Metalloids

Question 15

Atomic Properties and Periodic Table

Answer 15

●Atomic size
●Metallic and Non-metallic character
●Ionization (energy) potential
●Electron affinity
●Electronegativity

Question 16

As we increase the atomic number (or go down a group). . .
•each atom has another energy level,
•so the atoms get bigger.

Answer 16

Atomic Size

Question 17

Atomic Size - Period Trends

Answer 17

Going from left to right across a period, the size gets smaller.
•Electrons are in the same energy level.
•But, there is more nuclear charge.
•Outermost electrons are pulled closer

Question 18

is the tendency of an element to lose electrons and form positive ions (cations). For e.g., alkali metals are the most electropositive elements.
It is also known as electropositivity.

Answer 18

Metallic character

Question 19

The tendency of an element to accept electrons to form an anion is called

Answer 19

non-metallic or electronegative character

Question 20

For e.g., chlorine, oxygen and phosphorous show greater electronegative or.”

Answer 20

non-metallic character

Question 21

Trends across the period

Answer 21

Metallic character of elements decreases as we move to the right.
●Elements to the left have a pronounced metallic character while those to the right have a non-metallic character.
●Non-metallic character increases from left to right.

Question 22

Why does the metallic character decrease from left to right across the period?

Answer 22

The elements to the left of the periodic table have a tendency of losing electrons easily as compared to those to the right.
➢As we move from left to right of the period, the electrons of the outer shell experience greater pull of the nucleus.
➢This greater force of attraction is because the nuclear charge increases and the size of the atom decreases from left to right.

Question 23

Why does metallic character increase down the group?

Answer 23

●As we move down the group the number of shells increases.
●This causes the effective nuclear charge to decrease due to the outer shells being further away: in effect the atomic size increases.
●The electrons of the outermost shell experience less nuclear attraction and so can lose electrons easily thus showing increased metallic character.

Question 24

The amount of energy required to remove the most loosely bound electron from an isolated gaseous atom is called ionization energy (IE).

It is measured in the units of electron volts (eV) per atom or kilo joules per mole of atoms (kJ mol-1).

Answer 24

Ionization energy (IE).

Question 25

The ionization energy increases with increasing atomic number in a period.

Question 26

The ionization energy increases with increasing atomic number in a period. This is because

Answer 26

●The nuclear charge increases on moving across a period from left to right. ●The atomic size decreases along a period though the main energy level remains the same. ●Due to the increased nuclear charge and simultaneous decrease in atomic size, the valence electrons are more tightly held by the nucleus. Therefore more energy is needed to remove the electron and hence ionization energy keeps increasing.

Question 27

Variation down a group

Answer 27

●The ionization energy gradually decreases in moving from top to bottom in a group. This is due to the fact that: ●The nuclear charge increases in going from top to bottom in a group. ●An increase in the atomic size due to an additional energy shell (level) 'n'. As a result , the electron becomes less firmly held to the nucleus and so the ionization energy decreases as we move down the group.

Question 28

is the amount of energy released when an electron is added to an isolated gaseous atom.

Answer 28

Electron affinity

Question 29

This the relative tendency of an atom in a molecule to attract a shared pair of electrons towards itself

Answer 29

Electronegativity

Question 30

The value of electronegativity of an element describes the ability of its atom to compete for electrons with the other atom to which it is bonded.

Question 31

had the idea that all matter consisted of minute particles, which he the coined as “atomos.” However, his idea about atoms was not accepted by the community because he could not provide factual basis to his claims because it was purely based in philosophical reasoning.

Answer 31

Democritus

Question 32

(1803) an English school teacher, gave the first atomic Theory who provide precise and detailed description of the building blocks of matter

Answer 32

John Dalton

Question 33

is the smallest particle of an element that has the chemical characteristic of that element.

Answer 33

ATOM

Question 34

All matter is composed of elements. •All the atoms of a given elements are same or identical. •The atoms of different elements have different masses. •A compound is a specific combination of atoms of more than one element •In a chemical reaction, atoms are neither created nor destroyed but are merely rearranged to produce new substances.

Answer 34

John Dalton’s Atomic Theory

Question 35

Atoms may be disintegrated by transmutation process. (Transmutation is a process in which atom is converted either the complex or disintegrated atom.) •Not all atoms of a given element are identical or the same •Not all atoms of a given element have identical characteristics except mass. •Atoms of different elements have different

Answer 35

Modern Atomic Theory

Question 36

Electrons

Answer 36

Joseph Thomson

Question 37

In 1897, British physicist —— investigated the “cathode rays” that emitted rays when high voltage is applied between two electrons in a glass tube containing small amount of gas. His observation confirmed the earlier work, suggesting that a stream of particles move from one

Answer 37

Joseph Thomson

Question 38

emitted rays when high voltage is applied between two electrons in a glass tube containing small amount of gas

Answer 38

cathode rays

Question 39

Thomson discovered that the properties of rays were the same no matter what metal is used for the electrodes. Thomson showed that cathode rays are stream of negatively charged particles coming from the inside of the atoms that made up of the cathodes --- these particles are called electrons. He also measured the value of e/m, which is the ratio of the size of electron’s charge to its mass and found it to be ?????

Answer 39

1.76 x108 Coulomb/gram.

Question 40

an American, found that e = 1.6 x 10-19 C based on his oil drop experiment. Therefore, the mass of electrons (e) will be: Mass of electron = e x m/e = 1.6 x 10-19 C x 1/1.76 x108 Coulomb/gram

Answer 40

Robert Millikan

Question 41

is the positively charged subatomic particle with a mass of 1.6726 x 10-24g. It is located at the center of the atom called the nucleus. It was discovered by Ernest Rutherford, a New Zealander, using the gold foil experiment.

Answer 41

Proton

Question 42

Proton

Answer 42

Ernest Rutherford

Question 43

gold foil

Answer 43

Ernest Rutherford

Question 44

neutral subatomic particle with a mass of 1.6749 x 10-24 g

Answer 44

neutron

Question 45

neutron

Answer 45

James Chadwick

Question 46

Beryllium foil experiment.

Answer 46

James Chadwick

Question 47

He proposed that the electrons in an atom could only be in a certain orbits, or energy levels, around the nucleus; that is his theory was that the energy of the electrons is quantized and can only be lost or gained in discrete amount. This theory led to the modern theory of atomic

Answer 47

Niels Bohr

Question 48

British scientist, was the first to measure the atomic number accurately.

Answer 48

Henry Moseley

Question 49

equal to the number of protons in an atom

Answer 49

The atomic number (Z) is

Question 50

•To calculate for e-, p+ and n0. A – Z = neutron Z = proton Z = electron (if element is uncharged) Z – c = electron (if element is charged) n0 + p+ = mass number (A)

Question 51

One atomic mass unit is 1/12 the mass of an atom of Carbon-12. The mass of a single carbon-12 atom is 1.9926 x 10-23 g = 1.6605x10-24 g. 1 u = 1.9926 x 10-23 g = 1.6605x10-24 g 12

Question 52

are atoms that have the same atomic number but different atomic mass

Answer 52

Isotopes

Question 53

are isotopes that do not undergo radioactivity and do not disintegrate; thus they are abundant in nature

Answer 53

Stable isotopes

Question 54

are those that exhibit radioactivity and can be artificial or natural. Nevertheless, both emit radiation in the form of alpha, beta and gamma rays

Answer 54

Unstable isotopes

Question 55

consolidated average of the masses of the naturally existing isotopes relative to the mass of Carbon-12.

Answer 55

Isotopic abundance or Relative Atomic Mass

Question 56

i’m Mass number = Σ (% Abundance) (mass of the isotope)

Question 57

represents a compound in terms of chemical symbols; it indicates the elements and numbers of atoms present in a compound. -In writing the formula of a compound, always consider the rule that the total positive charges must be equal to the sum of the total negative charges. Therefore the compound must have a net zero charge.

Answer 57

Chemical Formula

Question 58

should be written first followed by the negative ion.

Answer 58

The positive

Question 59

The net ionic charge of the positive ion and the negative ion is equal to zero.

Answer 59

The net ionic charge of the positive ion and the negative ion is equal to zero. Na+ and Cl- Na+1 + Cl-1 = NaCl (0) H+ and S-2 H+1+ S-2 = H2S (0) Mg+2 and Br- Mg+2 + Br-1 = MgBr2

Question 60

The shortcut method to get the correct formula of a binary compound is to write each ion with its charge. You should start with the positive ion then cross over the number (not the plus or minus sign) and write them as subscripts.

Question 61

consist only of two elements

Answer 61

Binary compounds

Question 62

Metal + non-metal

Question 63

the name of the positive element (metals) is written first followed by the name of negative element (non-metal) with the ending –ide.

Answer 63

the name of the positive element (metals) is written first followed by the name of negative element (non-metal) with the ending –ide. Al2S3 Aluminum sulfide Mg3N2 Magnesium nitride KCl Potassium chloride Na2O Sodium oxide

Question 64

VARIABLE OXIDATION NUMBER a) Stock System or Roman Numeral System •metallic elements written first followed by its oxidation number in Roman Numeral enclosed in parenthesis: Fe+3 + O-2 Fe2O3 Iron (III) oxide Hg+2 + Cl-1 HgCl2 Mercury (II) chloride Cu+2 + Br-1 CuBr2 Copper (II) bromide

Question 65

VARIABLE OXIDATION NUMBER b) Classical or Conventional System In this method the suffix “ous” is added to the Latin name of the electropositive metal with a lower oxidation state and the suffix “ic” applied with the higher oxidation state. FeCl2 Ferrous chloride SnF2 Stannous fluoride FeCl3 Ferric chloride SnF4 Stannic fluoride NAMING BINARY

Question 66

NAMING BINARY COMPOUNDS Common Cations with variable oxidation numbers Au+1 Aurous or gold (I) Au+3 Auric or gold (III) Co+2 Cobaltous or cobalt (II) Co+3 Cobaltic or cobalt (III) Cu+1 Cuprous or copper (I) Cu+2 Cupric or copper (II) Fe+2 Ferrous or Iron (II) Fe+3 Ferric or Iron (III)

Question 67

NAMING BINARY COMPOUNDS Common Cations with variable oxidation numbers Pb+2 Plumbous or Lead (II) Pb+4 Plumbic or Lead (IV) Mn+2 Manganous or Manganese (II) Mn+3 Manganic or manganese (III) Hg2+2 Mercurous or mercury (I) Hg+2 Mercuric or mercury (II) Sn+2 Stannous or Tin (II) Sn+4 Stannic or Tin (IV)

Question 68

NAMING BINARY COMPOUNDS 3. In naming binary compounds containing two non-metals, the rule is similar to naming ionic compounds except that prefixes are used.

Question 69

compounds which is composed of hydrogen and a more electronegative element is named in the same as binary compound of nonmetal. -The word hydrogen is named first followed by the second element which is made by affixing the suffix –ide to the root word of the nonmetal. HCl Hydrogen chloride HF Hydrogen fluoride HBr Hydrogen bromide

Question 70

When the above substances are dissolved in water, they become aqueous acids. In these acids the prefix “hydro” is attached to the root word of the nonmetal and the suffix -ic is added followed by the word acid. HCl(aq) Hydrochloric acid HF(aq) Hydrofluoric acid HBr(aq) Hydrobromic acid HI(aq) Hydroiodic acid H2S(aq) Hydrosulfuric acid

Question 71

Polyatomic compounds are formed in the same way as binary compounds. A polyatomic ion is a stable group of atoms that carries an overall electrical charge.

Question 72

1. In naming compounds containing polyatomic ions the positive ion should be written first followed by the negative ion. However, parenthesis is placed around the polyatomic ion and the subscript is written just after the close parenthesis whenever a multiple of the polyatomic ion is necessary. K2CrO4 Potassium chromate Zn(NO2)2 Zinc nitrite Fe2(SO4)3 Ferric sulfate Mg3(PO4)2 Magnesium phosphate NaHCO3 Sodium bicarbonate

Question 73

another polyatomic compounds are compounds made up of hydrogen, oxygen and another elements. Suffix is used to differentiate the non-metals between two ternary oxyacids. The name of the acid containing the non-metal with the lower oxidation number is “ous”. The name of the acid containing a nonmetal with the higher oxidation number carries the suffix “ic.”

Answer 73

TERNARY OXIDE

Question 74

TERNARY OXIDE HNO3 Nitric acid HNO2 Nitrous acid H3PO4 Phosphoric acid H3PO3 Phosphorous acid H2SO4 Sulfuric acid H2SO3 Sulfurous acid

Question 75

TERNARY OXIDE HClO4 Perchloric acid HClO3 Chloric acid HClO2 Chlorous acid HClO Hypochlorous acid HIO4 Periodic acid HIO3 Iodic acid HIO2 Iodous acid HIO Hypoiodous acid

Question 76

with more than one positive ion can be formed from an acid containing two or more hydrogen atoms by placing one hydrogen or both with different metals

Answer 76

SALTS

Question 77

is formula of the compound gives the smallest and simplest number ratio of atoms that make up the compound.

Answer 77

Empirical Formula (EF)

Question 78

STEPS IN DETERMINING THE EF: 1.Change the percent to gram, using 100 g of compound as basis 2.Calculate the number of moles of each element 3. Divide the number of moles of each element by the smallest number of moles. If it does not result in a whole number, multiply the mole ratio by a common factor to convert these to whole number.

Answer 78

Example 1 Determine the empirical formula of a compound containing 74.19% Na and 25.81% O.

Question 79

Calculate the empirical Formula of a compound containing 52.9% Al and 47.1% O.

Answer 79

Example 2 ANSWER: EF = Al2O3

Question 80

MOLECULAR FORMULA

Question 81

Molecular Formula (MF)

Answer 81

is the true formula of the compound and shows the actual number of atoms of each element present in one molecule of a compound.

Question 82

STEPS IN DETERMINING THE MF: 1.Determine the Empirical Weight (EW) of the Empirical Formula (EF). 2.Divide the given Molecular weight (MW) by the Empirical Weight to solve for the Molecular Ratio (MR). MR = MW/ EW 3. Multiply the value of MR to the EF to get the Molecular Formula (MF). MF = MR x EF

Question 83

Vanillin is the active flavoring agent in vanilla bean, that contains 63.14% Carbon, 5.31% Hydrogen and 31.55% Oxygen. What is the MF of vanillin with Molecular weight of 152 g/mole?

Answer 83

ANSWER: MF = C8H8O3

Question 84

What is the Empirical Formula of C4H10?

Answer 84

ANSWER: EF = C2H5

Question 85

Which of the following is an Empirical Formula? A. H2O2 B. P4O10 C. NO2 D. C2H4

Answer 85

ANSWER: C. NO2

Question 86

TO MEASURE ELEMENT

Answer 86

STOICHIOMETRY

Question 87

Greek) meaning ELEMENT

Answer 87

stoichelon

Question 88

Greek) meaning TO MEASURE

Answer 88

metron

Question 89

is the sum of the atomic weights of all the atoms present in a molecule.

Answer 89

MOLECULAR WEIGHT

Question 90

in SI system is the amount of a substance that contains as many elementary entities (atoms, molecules or other particles) as there are atoms in exactly 12 grams of the C-12 isotope.

Answer 90

MOLE