IB Sample Test
Transcript of IB Sample Test
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Question 1 Which of the following properties of the halogens increase
from F to I?I. Atomic radius
II. Melting pointIII. Electronegativity
Answer: B (atomic radius & melting point)Reason: atomic radius increases down a group since more
electron shells are being added each time; the meltingpoint increases since the atoms get larger & hence there aremore electrons to induce more temporary dipoles whichleads to stronger Van der Waals forces.
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Question 2 Which factors lead to an element having a low value of first
ionisation energy?I. Large atomic radius
II.High number of occupied energy levelsIII.High nuclear charge
Answer: A (large atomic radius & high number of occupiedenergy levels)
Reason: The larger the atom, the further away the outer(valence) electrons are from the nucleus; a large number ofoccupied energy levels gives rise to increased shielding;hence the outermost electron is more easily removed in theprocess: X(g) X+(g) + e-
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Question 3What is the correct number of each particle in a
fluoride ion, 19F?
protons neutrons electronsA. 9 10 8
B. 9 10 9
C. 9 10 10
D. 9 19 10
(Total 1 mark)
Answer: C
Reason: the fluoride anion means that there is 1 more electron than
proton to give the ion a single negative charge,; there are 10
neutrons as 10+9= 19 which is the mass number
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Question 4Which properties are typical of most non-metals in
period 3 (Na to Ar)?
I. They form ions by gaining one or more electrons. II. They are poor conductors of heat and electricity.
III. They have high melting points.
Answer: A (II & I only)
Reason: Non-metals usually form anions by gainingelectrons from metals; they are poor conductors due toonly Van der Waals forces being present which cannotcarry permanent charge
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Question 6 Rubidium is an element in the same group of the periodic table as
lithium and sodium.It is likely to be a metal which has a
A. high melting point and reacts slowly with water. B. high melting point and reacts vigorously with water.
C. low melting point and reacts vigorously with water.
D. low melting point and reacts slowly with water.
Answer: C
Reason: It is a group 1 alkali metal
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Question 7 When the following species are arranged in order ofincreasing radius,
what is the correct order?
A. Cl, Ar, K+
B. K+, Ar , Cl
C. Cl, K+, Ar
D. Ar, Cl, K+
Answer: D (ArK+
)Reason: Atomic radius increases down a group & from right to left across a
period. An increasing nuclear charge across a period causes the atom tocontract by pulling the outer (valence) electrons inwards. Ondescending a group more shells are being added each time so the
valence electrons become further away from the nucleus while at thesame time they are more shielded by the extra shells.
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Question 8 Which property decreases down group 7 in the periodic table?
A. atomic radius
B. electronegativity
C. ionic radius D. melting point
Answer: B (electronegativity)
Reason: On descending the group, more shells are being added so thevalence electrons are further away from the nucleus & are bettershielded by the extra inner shells. This outweighs the effect of theincreased nuclear charge & hence the attraction for the outer electronpair in a covalent bond is decreased.
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Question 9 What is the correct sequence for the processes occurring in
a mass spectrometer?
A. vaporization, ionization, acceleration, deflection
B. vaporization, acceleration, ionization, deflection
C. ionization, vaporization, acceleration, deflection
D. ionization, vaporization, deflection, acceleration
Answer: A (vaporization, ionization, acceleration, deflection)
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Question 10 Which of the reactions below occur as written?
I. Br2 + 2I 2Br + I2
II. Br2 + 2Cl 2Br + Cl2
A. I only B. II only
C. Both I and II
D. Neither I nor II
Answer: A
Reason: Bromine is more reactive (stronger oxidising agent) than iodine but lessreactive than chlorine & hence will only displace iodide ions from their salts &oxidise them to iodine. Reactivity decreases down group 7 due to the electronaffinity decreasing since more shells are being added, causing the valenceelectrons to be further away from the nucleus & better shielded.
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Question 11 Which general trends are correct for the oxides of the period 3 elements
(Na2O to Cl2O)?
I. Acid character decreases.
II. Electrical conductivity (in the molten state) decreases. III. Bonding changes from ionic to covalent.
Answer: C
Reason: Electricity conductivity decreases across period 3 for the oxidessince the bonding changes from ionic to covalent. Only ioniccompounds can conduct electricity in the molten state; covalentcompounds are uncharged species & hence are non-conductors. Acidcharacter increases across the period.
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Question 12 What increases in equal steps of one from left to right in the periodic table
for the elements lithium to neon?
A. the number of occupied electron energy levels
B. the number of neutrons in the most common isotope
C. the number of electrons in the atom
D. the atomic mass
Answer: C
Reason: The atomic number (no. of protons) increases in increments of 1
across period 2; the number of electrons increases in the same manner since
all atoms are neutral (the number of electrons must equal the number of
protons).
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Question 13 Which of the physical properties below decrease with increasing atomic
number for both the alkali metals and the halogens?
I. Atomic radius
II. Ionization energy
III. Melting point
A. I only
B. II only
C. III only
D. I and III only
Answer: B (II only)
Reason: The valence electrons are more easily removed ondescending a group since they are further away from thenucleus & better shielded.
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Question 14 Which is the correct description of polarity in F2 and HF molecules?
A. Both molecules contain a polar bond.
B. Neither molecule contains a polar bond.
C. Both molecules are polar. D. Only one of the molecules is polar.
Answer: D (Only one of the molecules is polar)
Reason: Only HF is polar since it is composed of 2 different atoms withvery different electronegativity values.
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Question 15 For which element are the group number and the period
number the same?
A. Li
B. Be
C. B
D. Mg
Answer: B (Be)
Reason: Beryllium is in both group 2 & period 2
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Question 22i Crocetin consists of the elements carbon, hydrogen and oxygen. Determine the
empirical formula of crocetin, if 1.00 g of crocetin forms 2.68 g of carbon dioxideand 0.657 g of water when it undergoes complete combustion.
Moles C = Moles CO2 = 2.68 g/44.01 g mol-1 = 0.0609 mol
Moles H = 2 x moles H2O = 2 x (0.657 g/ 18.02 g mol-1) = 0.0729 mol
Mass of C = 0.0609 mol x 12.01 g mol-1 = 0.731 g
Mass of H = 0.0729 mol x 1.01 g mol-1 = 0.0736 g
The remainder must be the mass of oxygen in crocetin
Mass of O = 1.00 - 0.731 - 0.0736 = 0.195 g
Moles O = 0.195/16 = 0.0122
Carbon Hydrogen Oxygen
Moles 0.0609 0.0729 0.0122
Mole ratio 0.0609/0.0122
0.0730/0.0122
0.0122/0.0122
Empericalformula
4.99 5.98 1
Empirical formula: C5H6O
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Question 22ii Determine the molecular formula of crocetin given
that 0.300 mole of crocetin has a mass of 98.5 g
Molar mass of crocetin = 98.5 g / 0.300 mol = 328 g mol-1
Total molecular mass of EF, (C5H6O) = 82.11
Molar mass of MF/ Molar mass of EF =328/82.11 = 4
Molecular formula: C20H24O4
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Question 23i Nitrogen is found in period 2 and group 5 of the
periodic table.
(i) Distinguish between the termsperiodandgroup.
Period is a horizontal row in the periodic table & a groupis a vertical column
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Question 23ii State the electron arrangement of nitrogen and explain
why it is found in period 2 and group 5 of the periodictable.
Electron arrangement: 2,5
Found in period 2 as electrons are in 2 energy
levels/shellsFound in group 5 as 5 outer/valence electrons
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Question 24ai A sample of germanium is analysed in a mass spectrometer. The first and
last processes in mass spectrometry are vaporization and detection.
State the names of the other three processes in the order in which they
occur in a mass spectrometer.
Ionization, acceleration, deflection
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Question 24aii For each of the processes named in (a) (i), outline how
the process occurs
Ionization: sample bombarded with high energy/high-speed electrons
Acceleration: use electric field/oppositely charged plates
Deflection: magnetic field
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Question 24bi The sample of germanium is found to have the
following composition:
Isotope70
Ge72
Ge74
Ge76
GeRelative abundance /% 22.60 25.45 36.73 15.22
Define the term relative atomic mass.
Average/weighted mean of masses of all isotopes of an
element;
Relative to 1 atom of12C
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Question 24bii Calculate the relative atomic mass of this sample of
germanium, giving your answer to two decimal places.
Ar = (70 x 0.226) + (72 x 0.2545) + (74 x 0.3673) + (76 x0.1522)
= 72.89
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Question 25a State the meaning of the term electronegativity.
the ability of an element/atom/nucleus to attract abonding pair of electrons in a covalent bond
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Question 25b State and explain the trend in electronegativity across
period 3 from Na to Cl.
Electronegativity increases across the period;
The number of protons/nuclear charge increases
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Question 28ai Explain the following statements.
(a) The first ionization energy of sodium is less than
that of magnesium.
Na has a lower nuclear charge/number of protons;
electrons being removed from the same energy
level/shell
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Question 28aii Explain the following statements.
(ii)The first ionization energy of sodium greater than
that of potassium
The electrons in sodium are closer to the nucleus/in a
lower energy level/ Na has less shielding effect
hence harder to remove the valence electrons
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Question 28b Explain the following statements.
The electronegativity of chlorine is higher than that ofsulfur.
Chlorine has a higher nuclear charge;
Attracts the electron pair/electrons in the bond morestrongly
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Question 29i Explain why the ionic radius of chlorine is less than that
of sulfur.
Chlorine has an extra proton/greater nuclear charge;
Same amount of shielding (2 inner shells);
Hence outer electrons attracted more strongly
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Question 29ii Explain what is meant by the term electronegativity
and explain why the electronegativity of chlorine isgreater than that of bromine.
Electronegativiy is the ability of an atom to attract abonding pair of electrons in a covalent bond
chlorine has a smaller radius/electrons are closer tothe nucleus/;
hence repelled by fewer inner electrons/decreased
shielding effect
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Question 30 A toxic gas, A, consists of 53.8% nitrogen and 46.2% carbon by mass. At
273 K and1.01105 Pa, 1.048 g of A occupies 462 cm3. Determine the empiricalformula of A.
Calculate the molar mass of the compound and determine itsmolecular structure.
Empirical formula of A: CN
Carbon Nitrogen
Moles 46.2/12.01 = 3.85 53.8/14.01 = 3.85
Mole ratio 1 1
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Question 30 (contd) Moles A = volume of A/ molar volume =0.462dm3 /
22.4dm mol-3 = 0.0206
Molar mass A = Mass/moles = 1.048g / 0.0206 mol =50.8 g mol-1
Mr of empirical formula (CN) = 12.01 + 14.01 = 26.02
Molar mass of A/Mr of empirical formula = 50.8/26.02
= 2.00
Hence Molecular formula of A: C2N2
Possible structure: NCCN (Cyanogen)