QUESTIONS AND ANSWERED _ CHEMISTRY
PERIODIC CLASSIFICATION.
QUESTIONS
1.Multiple choice.
Choose the most correct answer.
(i) One of the following elements is not earth metal.
A: magnesium
B: calcium
C: aluminium
D : beryllium
(ii) Moving down the group ______ of metals decrease.
A. Electronegativity
B: reactivity
C: ionization energy
D: melting points.
(iii) Re-occurrence of similar properties of elements in the periodic table is called _____
A: periodicity
B: atomicity
C: ionization
D: electropositivity
(iv) They display both metallic and non-metallic characteristics.
A: alkali metals
B: alkali earth metals
C: halogens
D: metalloids
(v) Odourless, colourless and stable elements.
A: halogens
B: alkali metals
C: metalloids
D: noble gases
(vi) Always it represent the number of period in which an element found.
A: number of shells
B: electron in outer most shell
C: valency
D: valency shell
(vii) Element Q with electronic configuration 2:8:2 belongs to the group _____ and period ______
A: 2 and 2
B: 3 and 2
C: 2 and 1
D: 6 and 2
(viii) Hydrogen is placed in group I elements because______
A: it is a metal
B: it reacts by losing electron to form compounds
C: it reacts by gaining electron to form compounds
D: it is inert gas
(ix) When compared to metals, non-metals are ___________
A: good conductors of electricity
B: more brittle than metals
C: more ductile than metals
D: good conductor of heat.
(x) Which of the following statements does not describe the alkaline earth metals?
A: they burn in oxygen to form a solid white oxide.
B: they become less reactive as they move down the group.
C: they are good conductors of heat and electricity
D: they react with water
2. Matching items.
Match items in list A with responses in list B
List A
(i) Reactivity of metals in the periodic table.
(ii) Always it increases with increase of atomic number.
(iii) The distance from the centre of the atomic nucleus to the outer most shell.
(iv) Metallic characteristics
(v) Non-metallic characteristics
(vi) Elements with some metallic and non-metallic characteristics
(vii) Hallogens
(viii) Alkali metals
(ix) Alkaline earth metals
(x) Stable non-metallic elements.
List B
A: electropositivity
B: ionization energy
C: chlorine and fluorine
D: sodium and lithium
E: argon and neon
F: electronegativity
G: magnesium and calcium
H: boron and silicon
I : increase when you move down the group
J: atomic radius
K: atomic diameter
3. Why argon can not react with calcium?
4. Group the following elements into metals and non-metals.
-fluorine
-oxygen
-aluminium
-neon
-magnesium
-lithium
-nitrogen
-sodium
5. Identify an element which is more reactive between the following and state the reason for the element to be more reactive.
(a) sodium and magnesium
(b)chlorine and fluorine
(c) nitrogen and chlorine
(d) sodium and potassium
6. Define the following terms and state the general trends of elements along the periods and down the groups
(a) ionization energy
(b) electronegativity
(c) electropositivity
(d) atomic radius
(e) boiling point
(f) density
(g) melting point
ANSWERS:
1.(i) C vi) A
(ii) C vii) B
(iii) A viii) ,B
(iv) D ix) B
(v) D x) B
2.(i) I vi) H
(ii) B vii) C
(iii) J viii) D
(iv) A ix) G
(v) F x) E
3. Argon can not react with calcium because it is stable element with eight electrons in its outer shell such that it can not form bond with other elements to make compounds.
4. Metals Non- metals
-aluminium -fluorine
-magnesium -oxygen
-lithium -neon
-sodium -nitrogen
5. (a) Sodium is more reactive than magnesium because reactivity of metals increase from right to left along the periods of the periodic table and sodium found more left side than magnesium.
(b) Fluorine is more reactive than chlorine because the reactivity if of non-metals decrease with decrease of electronegativity. Down the group electronegativity of elements decrease.
(c) chlorine is more reactive than nitrogen because reactivity of non-metals increase from left to right along the periods of the periodic table and chlorine is found at right side of nitrogen.
(d) potassium is more reactive than sodium because the reactivity of metals increase with increase of electropositivity and down the group electropositivity of elements increases.
6. (a)Ionization energy
Is a minimum energy that required to remove bond forming electron(s) from an atom.
It increases from left to right along periods of the periodic table and decreases down the groups.
(b) electronegativity
Is a tendency of elements to attract electron(s) toward it self. It increases from left to right along periods and decreases down the groups.
(c) electropositivity
Is a tendency of electrons to lose electron(s). It increases when moving down the groups.
(d) atomic radius
Is a distance from the centre region of an atom to to its outer most shell. It remain constant in size at the period but increases down the group due to increase of number of shells.
(e) boiling point
Is a point of temperature at which the substance change into vapour. It increases when moving down the group due to increase of metallic characteristics(electropositivity)
(f) density of element
Is a ratio of mass of an element to its volume.
It increases down the group as metallic of element increases.
(g) melting point.
Is the point at which solid substance turns to liquid.
It increases down the group as metallic characteristics increase.
BONDING, FORMULA AND NOMENCLATURE
(i)The bond between sulphur and oxygen in sulphur dioxide.
A: ionic bond
B: covalent bond
C: electrovalent bond
D: sulphur bond
(ii) The oxidation state of nitrogen in ammonium ions(NH4+) is___
A: -3 B: 5 C: 2 D: 4
(iii) Positively or negatively charged elements are called ____
A: valency
B: molecules
C: ions
D: atoms
(iv) An element with 19 electrons, its valency is____
A: 2. B: 3. C: 1. D: 4
(v) The following atoms can not exist freely as a single atom, except
A: sodium
B: chlorine
C: argon
D: magnesium
(vi) Which of the following elements is divalent element?
A: Lithium
B: calcium
C: aluminium
D: carbon
(vii) One of the following substances is radicle.
A: chlorine ions
B: potassium ions
C: nitrate ions
D: neon
(viii) Valency of carbonate ions
A: 2 B: 4 C: 1 D: 3
(ix) Ions formed when _________electrons
A: metals gain and non metals
B: elements share
C: metals lose and non-metals gain
D: elements tranfer
(x) Holds two or more substances together
A: radical
B: valency
C: chemical formula
D: bond
2. Matching items
Match items in list A with responses from list B
List A
(i) positively or negatively charged elements
(ii) covalent substances
(iii) electrovalent substances
(iv) formed when atoms lose electrons
(v)valency shell
(vi)react by either sharing or gaining electrons
(vii) the force of attraction between elements with opposite charges
(viii)monovalent elements
(ix) force of attraction that holds things together
(x) hydrogen ion
List B
A: inner most shell
B: bond
C: ionic bond
D: outermost shell
E: H+
F: H
G: H-
H: ionic bond
I: solid and crystalline subatances
K: liquid or gaseous substance
L: group i elements
M: group ii elements
N: non-metals
O: metals
P: ions
Q: cations
R: anions
S: chemical bond
3. Define the following terms
(i) valency
(ii)oxidation state
(iii) radicals
(iv)empirical formula
(v) molecular formular
(vi) covalent bond
(vii) electrovalent bond
(viii) binary compounds
(ix) chemical formula
(x) chemical bond
4. Why covalent compounds do not conduct electricity?
5. Find the oxidation number of the underlined atoms in the following compounds
(i) FeCl
(ii) KClO
(iii) NO3 -
(iv) SO3 2- (sulphur)
6. State the valence of each atom involved to form the compound below
(a) NaCl
(b) CO
(c) K2O
7. Classify the the ions below into cations and anions
(a) Cl-
(b) Zn2+
(c) SO4 2-
(d) NH4 +
(e) Li +
(f) I-
8.Name the following binary compounds.
(i)H2S
(ii) PCl5
(iii) CCl4
(iv) FeBr2
(v) H2Cl2
9. Write the chemical formula of the following compounds.
(i) iron ii chloride
(ii) carbon monoxide
(iii) Zinc chloride
(iv) potassium iodide
10. Calculate the empirical formula of hydrocarbon molecule made by 88.88% of carbon and 11.12% of hydrogen.
11. Show the difference between covalent compound and electrovalent compounds (four points).
12. A certain compound is made up of 69.58% of barium, 6.09% of carbon and 24.32% of oxygen. Calculate the empirical formula of the compound. R.A.M of barium is 137.3
13. What is the molecular formular of the compound made by 15.8% carbon and 84.2% sulphur? The molecular mass of the compound is 76
14. Calculate the molecular formula of an organic compound which consists 84.9% carbon and 15.1% hydrogen. Its molecular mass is 70
15. Show difference between oxidation state and valency.
ANSWERS:
1. (i) A (vi) B
(ii)A (vii) C
(iii)C (viii) A
(iv)C (ix) C
(v) C (x) D
2. (i) P vi) N
(ii) K vii) C
(iii) I viii) L
(iv) Q ix) S
(v) D x) E
3. (i) Valency:- is the combining power of an element.
This means ability of an element to combine with other elements depends on valency.
(ii) Oxidation state:- is a number of electron(s) that element can lose, gain or share when forming a compound.
(iii) Radical:- is a group of atoms with unpaired electrons and acts as a single atom.
(iv) Empirical formula:- is a simplest chemical formula which express its composition by mass.
(v) Molecular formula:- is a simplest chemical formula which shows the actual number of atoms present in the compound.
(vi) Covalent bond:- is a bond which formed by sharing electrons between non-metals.
(vii) Electrovalent bond:- is a bond which formed when electron(s) transfered from metal to non-metal.
(viii) Binary compound:- is a compound which made up of two two ions.
(ix) Chemical formula:- refers to representation of chemical compound by using symbols of elements in correct proportions.
(x) Chemical bond:-is a force of attraction that holds two or more elements together.
4. Covalent compounds do not conduct electricity because they have no free ions which are necessary for carrying electric current in molten or solution form.
5. (i) Solution.
Data:
Oxidation states
Fe=needed
Cl= -1
Fe + (-1x3)=0
Fe -3=0
Fe=0+3
Oxidation state of iron=3
(ii) Data:
Oxidation states
K=1
Cl=needed
O=-2
1 + Cl + (-2 x 3)=0
1 + Cl + (-6) =0
Cl - 5 =0
Cl=5
Oxidation state of Cl =5
(iii) Data:
Oxidation states.
N=needed
O=-2
N + (-2 x 3)= -1
N + (-6) =-1)
N= -1 + 6
N=5
Oxidation state of N = 5
(iv) Data:
Oxidation states.
S=needed
O=-2
S + (-2 x 3) = -2
S + (-6)=-2
S=-2 + 6
S=4
Oxidation state of sulphur = 4
6. We write chemical formula by interchanging valences of reacting species.
In order to determine the valency of an atom in a chemical formula, firstry the valence of individual atom must be identified.
(i) NaCl
The falency of Na=1
Cl=1
(ii) CO
The valency of C=4
O=2
During exchange of the valences, 4 placed at Oxygen(O4) and 2 placed at carbon (C2) then C2O4 obtaned and simplified by dividing by 2 to get CO2
(iii) K2O
The valency of K=1
O=2
7. Cations. Anions.
b, d, e a, c, f
8. (i) dihidrogen sulphide.
(ii) phosphorous pentachloride
(iii) carbon tetrachloride
(iv) iron(ii) bromide
(v) hydrogen chloride
9. (i) FeCl
(ii) CO
(iii)ZnCl
(iv) KI
10. Solution.
Data given.
Masses C=12g
H=1g
%composition C= 88.88
H=11.12
Elements symbols C H
Mass 12 1
%comp./mass 88.88/12 11.12/1
=7.407 =11.12
Divide by
Smallest no.
7.407/7.404 11.12/7.404
=1 1.5~2
C1. H2
Epirical formula is CH2
11. (i) covalent compound are either liquids ir gaseous While electrovalent are solid crystalline.
(ii) covalent compounds do not conduct electricity in molten state While electrovalent compounds conduct electricity in molten state.
(iii) covalent compound are generally insoluble in water While electrovalent compounds are generally soluble in water.
(iv) covalent compounds have low melting points while electrovalent compounds have high melting points
(v)covalent compounds formed when non-metals react While electrovalent compounds formed when metals react with non-metals.
12. Solution.
Data given.
Relative atomic masses (R.A.M )
Barium(Br)=137.3
Carbon=12
Oxygen=16
%composition Br=59.58
C=6.09
O=24.32
Elements symbols Br C O
%comp./R.AM =0.508 =0.508 =1.52
Divide by
Lowest number =1. =1. =3
Br1. C1. O3
Empirical formuls is BrCO
13. Solution.
Data given.
R.A.M C=12
S=32
%comp. C=15.8
S=84.2
Elements symbols. C S
%comp./R.A.M. 15.8/12 84.2/16
=1.32 =2.63
Divide by
Smallest No. 1.32/1.32 2.63/1.32
=1 =2
C1 S2
Empirical formula is CS
Molecular formula= (Ef)n
(Empirical formula)n=76
(CS2)n = 76
(12 + (16x2))n=76
(44)n=76
n=76/44
n~1
Molecular formula= C1x1 S2x1
=CS2
14. Solution.
Data given.
R.A.M C=12
H=1
%composition.
C= 84.9
H= 15.5
Elements symbols C H
%comp./R.A.M 84.9/12 15.1/1
=7.08 =15.1
Divide by
Smallest No. 7.08/7.08 15.1/7.08
=1 =2
C1 H2
Empirical formular is CH2
M.F = (E.F)n
M.F = (CH2)n
(CH2)n = 70
(12+(1x2))n=70
14n=70
n=70/14
=5
M.F=(CH2)5
=C5H
Molecular formular is C5H
15. √Oxidation state is arbitrary value While Oxidation state is a fixed value
√Oxidation state is either negative or positive While valence has no charge.
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