Target: 10 Questions in 10 minutes

An IB Chemistry data booklet is helpful

1. Which of the following statements is FALSE?
  • A. The nucleus of an atom contains protons and neutrons
  • B. Negatively charged electrons occupy the space outside the nucleus
  • C. Atoms contain a positively charged dense nucleus
  • D. The number of electrons in an atom equals the number of neutrons

2. What is the atomic number of an atom which has 30 electrons and 35 neutrons?

  • A.   30
  • B.   35
  • C.   65
  • D.   105

3. How many protons, neutrons and electrons are present in each atom of 39K?

  Protons Neutrons Electrons
A 19 20 19
B 19 39 19
C 20 39 20
D 20 19 20

4. What is the correct symbol for the species which contains 13 protons, 14 neutrons and 10 electrons?

  • A. 27 14 Si3+
  • B. 27 13 Si
  • C. 27 13 Al3+
  • D. 27 13 Al+

5. Which of the following species are isoelectronic (contain the same number of electrons)?

I. 16 8 O2-
II. 24 12Mg2+
III. 29 14Si
  • A.   I and II only
  • B.   I and III only
  • C.   II and III only
  • D.   I, II and III
6. Which statement about two isotopes of an element is TRUE?
  • A.   They have the same number of neutrons
  • B.   They have different numbers of electrons
  • C.   They have the same chemical properties
  • D.   The have different atomic numbers

7. Which of the species below have the smallest number of neutrons?

  • A. 198 78 Pt
  • B. 197 79 Au
  • C. 205 81 Tl
  • D. 204 84 Po

8. A sample of an unknown element “Z” consists of 3 isotopes as shown below:

Isotope Percentage Abundance /%
64Z 89.43
65Z 7.15
66Z 3.42
 

What is the relative atomic mass of element “Z” to 2dp?

  • A.   64.11
  • B.   64.12
  • C.   64.13
  • D.   64.14

9. The relative atomic mass of Indium is 114.816. Indium consists of two isotopes, 113In and 115In. What is the percentage abundance of the isotope 113In?

  • A.   8.8%
  • B.   9.0%
  • C.   9.2%
  • D.   9.4%

10. What is the relative atomic mass of zirconium from the information shown in the table below?

Isotope Percentage Abundance /%
90Zr 51.5
91Zr 11.2
92Zr 17.1
94Zr 17.4
96Zr 2.8
 
  • A.   90.9
  • B.   91.3
  • C.   91.7
  • D.   92.1
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Question 1:

Let's evaluate each statement to determine which one is false:

A. The nucleus of an atom contains protons and neutrons.
This is true. The nucleus is composed of protons (positively charged) and neutrons (neutral).

B. Negatively charged electrons occupy the space outside the nucleus.
This is true. Electrons are negatively charged and are found in orbitals surrounding the nucleus.

C. Atoms contain a positively charged dense nucleus.
This is true. The nucleus is small and dense, with a positive charge due to the protons.

D. The number of electrons in an atom equals the number of neutrons.
This is false. In a neutral atom, the number of electrons equals the number of protons (not neutrons). The number of neutrons can vary (isotopes) and is not necessarily equal to the number of electrons. For example, carbon-12 has 6 protons, 6 electrons, and 6 neutrons (equal in this case), but carbon-14 has 6 protons, 6 electrons, and 8 neutrons (not equal).

Thus, statement D is false.


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Question 2:

The atomic number of an atom is defined as the number of protons in the nucleus.

Given:

In a neutral atom (with no charge), the number of electrons equals the number of protons. Therefore, the number of protons is 30.

Thus, the atomic number is 30.

The options are:

Answer: A. 30


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Question 3:

To determine the number of protons, neutrons, and electrons in an atom of 39K (potassium-39):

  1. Atomic number (Z): The symbol K represents potassium, which has an atomic number of 19. This means every potassium atom has 19 protons.

  2. Electrons: In a neutral atom, the number of electrons equals the number of protons. So, there are 19 electrons.

  3. Mass number (A): The superscript 39 is the mass number, which is the sum of protons and neutrons. Therefore:

    Number of neutrons=A−Z=39−19=20

    So, there are 20 neutrons.

Now, match these values to the options:

Answer: A. 19 protons, 20 neutrons, 19 electrons


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Question 4:

To determine the correct symbol for the species with 13 protons, 14 neutrons, and 10 electrons:

Number of protons (atomic number, Z) = 13
The element with atomic number 13 is Aluminum (Al).

Number of neutrons = 14
The mass number (A) = protons + neutrons = 13 + 14 = 27.

Number of electrons = 10
Since a neutral atom of Al would have 13 electrons, having only 10 electrons indicates a charge of +3 (because it has lost 3 electrons).

Therefore, the species is Al³⁺ with mass number 27.

Now, evaluate the options:

A.
Silicon (Si) has atomic number 14 (not 13), so incorrect.

B.
This is neutral silicon with atomic number 14 and mass 27, but we need aluminum with charge, so incorrect.

C.
This has atomic number 13 (Al), mass number 27, and charge +3, which matches exactly.

D.
This has atomic number 13 (Al) but mass number 24 (should be 27) and charge +1 (should be +3), so incorrect.

The answer is C.


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Question 5:

To determine which species are isoelectronic (have the same number of electrons), we need to calculate the number of electrons in each species based on their atomic numbers and charges.

Recall:

  • Atomic number (Z) = number of protons

  • For an ion: number of electrons = Z - charge (if positive) or Z + |charge| (if negative)

Let's analyze each species:

I. 

  • Atomic number (Z) = 8

  • Charge = -2, so it has gained 2 electrons.

  • Number of electrons = 8 + 2 = 10 electrons

II. 

  • Atomic number (Z) = 12

  • Charge = +2, so it has lost 2 electrons.

  • Number of electrons = 12 - 2 = 10 electrons

III. (This is written as neutral silicon? But note: the mass number 29 is unusual for Si, but we care about electrons.)

  • Atomic number (Z) = 14

  • It is written without a charge, so we assume it is neutral.

  • Number of electrons = 14 (since neutral atom has same number of electrons as protons)

So:

  • I has 10 electrons

  • II has 10 electrons

  • III has 14 electrons

Therefore, only I and II are isoelectronic (both have 10 electrons).

Answer: A. I and II only

Note: Option III has 14 electrons, so it is not isoelectronic with I and II.


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Question 6:

Let's evaluate each option for two isotopes of the same element:

Isotopes are atoms of the same element (same atomic number) that have different mass numbers due to different numbers of neutrons.

Thus, the true statement is C.


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Question 7:

To determine which species has the smallest number of neutrons, we need to calculate the number of neutrons for each option. The number of neutrons is given by:

Number of neutrons=Mass number−Atomic number

Let's calculate for each:

The smallest number of neutrons is 118 from option B.


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Question 8:

To calculate the relative atomic mass of element "Z" based on the given isotopic abundances, use the formula:

Relative atomic mass=[(% abundance of isotope 1)/100 × mass of isotope 1] + [(% abundance of isotope 2)/100 × mass of isotope 2] + [(% abundance of isotope 3)/100 × mass of isotope 3]

Given:

Now, compute step by step:

  1. Contribution from 64Z:

    89.43100/100 ×64=0.8943×64=57.2352
  2. Contribution from 65Z:

    7.15/100×65=0.0715×65=4.6475
  3. Contribution from 66Z:

    3.42/100×66=0.0342×66=2.2572
  4. Sum the contributions:

    57.2352+4.6475+2.2572=64.1399
  5. Round to two decimal places:
    64.1399≈64.14

Therefore, the relative atomic mass is 64.14.

Answer: D. 64.14


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Question 9:

Let the percentage abundance of 113In be x%. Then the percentage abundance of 115In is (100−x)%

The relative atomic mass is given by the weighted average:

[x⋅113+(100−x)⋅115] / 100 =114.816

Now, solve for x :

  1. Multiply both sides by 100:

    x⋅113+(100−x)⋅115=11481.6
  2. Expand the left side:

    113x+11500−115x=11481.6
  3. Combine like terms:

    −2x+11500=11481.6
  4. Subtract 11500 from both sides:

    −2x=11481.6−11500
  5. Divide both sides by -2:

    x =

Therefore, the percentage abundance of 113In is 9.2%.

Answer: C. 9.2%


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Question 10:

To calculate the relative atomic mass of zirconium, use the weighted average based on the isotopic abundances and masses:

Given:

Now compute each contribution:

  1. 90Zr
    0.515×90=46.35

  2. :
    0.112×91=10.192

  3. :
    0.171×92=15.732

  4. :
    0.174×94=16.356

  5. :
    0.028×96=2.688

Now sum these contributions:

46.35+10.192=56.542

56.542+15.732=72.274

72.274+16.356=88.63

88.63+2.688=91.318

So, the relative atomic mass is approximately 91.3.

Answer: B. 91.3


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