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TOPIC 7 : BONDING, FORMULA AND NOMENCLATURE QUESTIONS WITH ANSWERS ~ CHEMISTRY FORM 2

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TOPIC 3 : WATER QUESTIONS WITH ANSWERS ~ CHEMISTRY FORM 2

TOPIC 7 : BONDING, FORMULA AND NOMENCLATURE QUESTIONS WITH ANSWERS ~ CHEMISTRY FORM 2

(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

TOPIC 7 : BONDING, FORMULA AND NOMENCLATURE QUESTIONS WITH ANSWERS ~ CHEMISTRY FORM 2

(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

  1. 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

TOPIC 7 : BONDING, FORMULA AND NOMENCLATURE QUESTIONS WITH ANSWERS ~ CHEMISTRY FORM 2

  1. 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

  1. Why covalent compounds do not conduct electricity?
  1. Find the oxidation number of the underlined atoms in the following compounds

(i) FeCl3

(ii) KClO3

(iii) NO3 –

(iv) SO3 2- (sulphur)

  1. State the valence of each atom involved to form the compound below

  (a) NaCl

  (b) CO2

  (c) K2O

  1. 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

  1. Write the chemical formula of the following compounds.

  (i) iron ii chloride

  (ii) carbon monoxide

  (iii) Zinc chloride

  (iv) potassium iodide

TOPIC 7 : BONDING, FORMULA AND NOMENCLATURE QUESTIONS WITH ANSWERS ~ CHEMISTRY FORM 2

  1. Calculate the empirical formula of hydrocarbon molecule made by 88.88% of carbon and 11.12% of hydrogen.
  1. Show the difference between covalent compound and electrovalent compounds (four points).
  1. 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
  1. 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
  1. Calculate the molecular formula of an organic compound which consists 84.9% carbon and 15.1% hydrogen. Its molecular mass is 70
  1. 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

  1. (i) Pvi) N

   (ii) K           vii) C

  (iii) I            viii) L

  (iv) Q            ix) S

   (v) D             x) E

TOPIC 7 : BONDING, FORMULA AND NOMENCLATURE QUESTIONS WITH ANSWERS ~ CHEMISTRY FORM 2

  1. (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.

  1. Covalent compounds do not conduct electricity because they have no free ions which are necessary for carrying electric current in molten or solution form.
  1. (i) Solution.

     Data:

     Oxidation states

           Fe=needed

           Cl= -1

     Fe + (-1×3)=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

TOPIC 7 : BONDING, FORMULA AND NOMENCLATURE QUESTIONS WITH ANSWERS ~ CHEMISTRY FORM 2

  (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

  1. 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) CO2

       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

TOPIC 7 : BONDING, FORMULA AND NOMENCLATURE QUESTIONS WITH ANSWERS ~ CHEMISTRY FORM 2

(iii) K2O

       The valency of K=1

                                 O=2

  1. Cations.Anions.

     b, d, e            a, c, f

  1. (i) dihidrogen sulphide.

    (ii) phosphorous pentachloride

    (iii) carbon tetrachloride

    (iv) iron(ii) bromide

    (v) hydrogen chloride

  1. (i) FeCl2

   (ii) CO

    (iii)ZnCl2

    (iv) KI

  1. 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

  1. (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.

  1. Solution.

      Data given.

Relative atomic massea (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  BrCO3

  1. 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

TOPIC 7 : BONDING, FORMULA AND NOMENCLATURE QUESTIONS WITH ANSWERS ~ CHEMISTRY FORM 2

Empirical formula is  CS2

Molecular formula= (Ef)n

       (Empirical formula)n=76

       (CS2)n = 76

        (12 + (16×2))n=76

        (44)n=76

        n=76/44

        n~1

Molecular formula= C1x1 S2x1

                                =CS2

  1. 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+(1×2))n=70

                   14n=70

               n=70/14

                 =5

       M.F=(CH2)5

              =C5H10

Molecular formular is C5H10

  1. √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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