IB Chemistry HL Structure 1: Models of the particulate nature of matter questions

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Question 1

Multiple choiceMedium1 mark

A copper ion is represented by 2965Cu2+^{65}_{29}\mathrm{Cu}^{2+} What are its numbers of protons, neutrons and electrons, respectively?

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Answer: A. For a nuclide, the proton number is ZZ and the neutron number is A−ZA-Z a 2+2+ charge means two fewer electrons than protons. Thus p=29p=29 n=65−29=36n=65-29=36 and e=29−2=27e=29-2=27

B: adds 2 to the proton number when finding electrons, reversing the charge effect.
C: uses the electron count instead of the proton number in A−ZA-Z giving 38 neutrons.
D: assumes the ion has the same number of electrons as a neutral copper atom, ignoring its charge.

Remember: ion formation changes the number of electrons, not the nucleus.

Question 2

Multiple choiceMedium1 mark

Three groups of lines in hydrogen’s emission spectrum converge at different photon energies. For the group whose lines converge at the highest photon energy, which energy level do the transitions end at, and what is the maximum number of electrons that level can hold?

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Answer: A. The highest-energy convergence limit corresponds to transitions ending at the lowest energy level, n=1n=1 Its maximum capacity is 2n2=2(1)2=22n^2=2(1)^2=2 electrons.
B: assigns the limit to n=2n=2 then correctly applies 2n22n^2
C: assigns the limit to n=3n=3 then correctly applies 2n22n^2
D: identifies n=1n=1 but omits the factor of 2 in the capacity relation.
Remember: the maximum number of electrons in level nn is 2n22n^2

Question 3

Multiple choiceMedium1 mark

A fixed amount of an ideal gas has pressure 8.0×104 Pa8.0 \times 10^4\ \mathrm{Pa} volume 6.0×10−3 m36.0 \times 10^{-3}\ \mathrm{m^3} and temperature 300 K300\ \mathrm{K} It is changed to volume 3.0×10−3 m33.0 \times 10^{-3}\ \mathrm{m^3} and temperature 450 K450\ \mathrm{K} What is its final pressure?

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Answer: D. For a fixed amount of gas, P∝T/VP \propto T/V Thus, P2=P1(V1/V2)(T2/T1)=(8.0×104)(2.0)(1.5)=2.4×105 PaP_2=P_1(V_1/V_2)(T_2/T_1)=(8.0\times10^4)(2.0)(1.5)=2.4\times10^5\ \mathrm{Pa}

A: reverses the temperature ratio, using T1/T2T_1/T_2 instead of T2/T1T_2/T_1
B: accounts for the temperature increase but omits the volume decrease.
C: accounts for the volume decrease but omits the temperature increase.
Remember: at fixed amount, pressure increases with temperature and decreases with volume.

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