# Exercises for Topic Electronic Structures of Atoms

Exercises for Topic Electronic Structures of Atoms

1. Calculate the *energy of the light emitted* with a frequency of 2.16 x 10^14 Hz.

Solution

E = hv = 6.63 x 10-34 x 2.16 x 10^-14

= 1.43 x 10^-19 J

2. A photon of light has a wavelength of 410 nm. Calculate the *frequency* of this photon.

v=c/λ

c = 3 x 10^8

Solution

= 3 x 10^8 / ( 410 x 10^-9 ) = 7.32 x 10^14 Hz

3. How many lines would be formed in the *Lyman series* from the *first five* energy levels? Sketch the energy level diagram.

Solution

Lyman series is formed from electronic transitions from higher energy levels to n= 1. For five energy levels, there are four electronic transitions. Therefore, four lines will be formed.

4. The second line in the Lyman series has a frequency of 2.92 x 10^15 Hz.

(a) State the electronic transition that produces this line.

(b) Calculate the energy of this line.

Solution

(a) The second line is caused by the electronic transition from n = 3 to n = 1.

(b) Energy, E = hv = 6.63 x 10^-34 x 2.92 x 10^15 = 1.94 x 10^-18 J

5. The diagram below shows some electronic transitions in atomic hydrogen.

Which electronic transition will produce the line with the

(a) lowest frequency?

(b) shortest wavelength?

Solution

(a) The line with the lowest frequency is produced by the electronic transition of the lowest energy, that is T4.

(b) The line with the shortest wavelength will have the highest frequency. Thus, it is produced by the electronic transition of the highest energy, that is T1.

a. T4 itu walaupun ia berada di n=4, tetapi ia hanyalah paschen siries, sebab ia emit daripada n=4 kepada n=3, sebab asal dia dari n=3. dan gari T4 paling pendek, sebab tu kita kata ia lowest frequency.

b. T1 walaupun ia kembali kepada n =1, tetapi ia mempunyai gari panjang dan menunjukkan frequency tinggi, vice versa wavelenght pendek, guna rumus ini v=c/λ atau v=1/λ

6. The wavelength of red light is 656 nm. What is the frequency of red light?

V = C/λ = 3.00 x 10^8 / 656 x 10^-9 = 4.57 x 10^14 Hz

7. Calculate the energy of the radiation that has a frequency of 4.57 x 10^14 Hz.

8. A photon has an energy of 2.09 x 10-18 kJ. What is its wavelength?

9. How many lines would be formed in the Balmer series from the first six energy levels?

4 lines caused by n6 to n2, n5 to n2, n4 to n2, and n3 to n2

10. The diagram below shows some electronic transitions in atomic hydrogen.

?????

(a) State the electronic transitions that will produce lines in the

(i) Lyman series

(ii) Balmer series

(b) Which electronic transition will produce the line with the lowest frequency?

11. The Lyman series of atomic hydrogen is shown below.

(a) State the electronic transition that will produce line L3.

(b) Which of the labelled lines has the longest wavelength?

(a) n = 4 to n = 1

(b) L, because the longest wavelength corresponds to the lowest frequency.

12. What is the maximum number of electrons that can fill the first three energy levels?

n = 1 can fill a maximum of 2 electrons.

n = 2 can fill a maximum of 8 electrons.

n = 3 can fill a maximum of 18 electrons.

Total number of electrons = 28

13. (a) Define an orbital.

(b) An electron is in the energy level n = 2. What type of orbitals could be occupied by the electron?

(a) An orbital is a region of space around the nucleus where the probability of finding an electron is maximum/ high

(b) 2s or 2p orbitals.

14. How many orbitals are found in the energy level n = 3? State the orbitals in the level.

9 orbitals. One 3s-orbital, three 3p-orbitals and five 3d-orbitals

15. The electronic configuration of an ion X2+ is 1s2 2s2 2p6 3s2 3p6.

What is the electronic configuration of atom X?

Solution

X2+ ion has 18 electrons. Therefore, atom X should have 20 electrons. Electronic configuration of atom X is 1s2 2s2 2p6 3s2 3p6 4s2.

16. The electronic configuration of an ion R3+ is 1s2 2s2 2p6 3s2 3p6 3d2.

What is the electronic configuration of atom R?

Solution

R3+ ada 20 electron, dia kurang 3 e, sepatutnya R ada 23 e, cuba guna gambarajah hujan utk susun semula 23 e tu.

R3+ ion has 20 electrons. Therefore, atom R should have 23 electrons. Electronic configuration of atom R is 1s2 2s2 2p6 3s2 3p6 3d3 4s2.

1s2 2s2 2p6 3s2 3p6 4s2 3d3 jawapan ini diterima juga, cuma ia perlukan pindah 4s2 ke hujung kanan, sebab 4s2 akan dikeluarkan terlebih dahulu (remove first) sebelum 3d3 diambil kerana 3d3 ialah sangat stabil, reasons half-filled. tetapi jawapan 1s2 2s2 2p6 3s2 3p6 3d5 tidak boleh diterima, ia salah.

Soalan: Nak susun 3d dulu ke 4s?

Jawapan:

Guna teori:
1. kalau nak susun, ikut gambarajah hujan tu,
2. kalau nak remove electron, remove orbital paling tinggi dulu.

kalau nak susun, ikut gambarajah hujan tu, maka 4s dahulu selepas itu 3d, selepas itu baru nilaikan, sama ada boleh tak dapat sama ada half-filled, atau fully-filled. sebab ia lebih stabil. half-filled dan fully filled hanya untuk orbital pdf sahaja, orbital s tak termasuk.

kemudian, susun semula (rearrange) maka 4s orbital mesti diletakkan di sebelah kanan sekali, sebab kalau kita nak remove electron, remove orbital paling tinggi dulu. maka 4s yang kena berada di sebelah kanan sekali, sebab elektron pada 4s akan keluar dulu. 4s orbital energy level lebih tinggi berbanding 3d. 4 lebih tinggi daripada 3.

sudah pasti yang half atau fully filled tak boleh keluar dulu sebab ia sangat stabil,

contoh,

Zn = 30. 1s2 2s2 2p6 3s2 3p6 3d10 4s2
(kita letak 4s di hujung kerana elektron 4s akan dikeluarkan terlebih dahulu).

Cu = 29. 1s2 2s2 2p6 3s2 3p6 3d9 4s2.
rearrange untuk dapat extra stability ialah 1s2 2s2 2p6 3s2 3p6 3d10 4s1.
(kita letak 4s di hujung kerana elektron 4s akan dikeluarkan terlebih dahulu).

*3 ni je yg penting,*

1. nak susun, guna gambarajah hujan.
2. susun semula, kalau boleh dapat half filled atau fully filled.
3. nak remove electron mesti daripada 4 dahlu berbanding 3 orbital

17. (a) State the orbitals that are found in the first three energy levels.

(b) Arrange these orbitals in ascending order of their energies.

18. Write the electronic configuration of the following atoms.

(a) Phosphorus (b) Calcium (c) Iron (d) Copper (e) Rubidium (f) Barium

19. Write the electronic configuration of the following cations.

(a) Mg2+

(b) K+

(c) O+

(d) V3+

(e) Mn2+

(a) Mg2+: 1s2 2s2 2p

(b) K+: 1s2 2s2 2p6 3s2 3p6

(c) O+: 1s2 2s2 2p3

(d) V3+: 1s2 2s2 2p6 3s2 3p6 3d2

(e) Mn2+: 1s2 2s2 2p6 3s2 3p6 3d5

20. Write the electronic configuration of the following anions.

(a) F-

(b) N3-

(c) S2-

(a) F-: 1s2 2s2 2p6

(b) N3-: 1s2 2s2 2p6

(c) S2-: 1s2 2s2 2p6 3s2 3p6

21. An ion Q+ has 18 electrons. What is the electronic configuration of atom Q?

22. The electronic configuration of an ion X3+ is 1s2 2s2 2p6 3s2 3p6 3d3. What is the electronic configuration of atom X?

23. An ion Y2+ has 26 electrons and 31 neutrons.

(a) What is the nucleon number of Y?

(b) Write the electronic configuration of atom Y.

(a) lon Y2+ has 26 electrons,

thus atom Y has 28 electrons and 28 protons.

Nucleon number of Y = 28 + 31 = 59

(b) 1s2 2s2 2p6 3s6 3p6 3d8 4s2

24. The proton number of iron is 26.

(a) Write the electronic configuration of the Fe2+ and Fe3+ ions.

(b) Based on the electronic configurations above, which ion is more stable? Explain your answer.

25. An ion Y has 18 electrons and is attracted to the anode in an electrolytic cell. It can form an ionic compound Na Y. What is the electronic configuration of the Y atom?

26. An element has a proton number of 26

(a) What is its valence electronic configuration?

(b) In which block in the Periodic Table does it belong to?

Solution

(a) Electronic configuration is 3d6 4s2.

(b) The d-orbitals are being filled. Thus, it is in d-block.

27. *Hund’s rule* is used to determine the ground state electronic configuration of an atom. State Hund’s rule.

*Hund’s rule* states that when filling a set of degenerate orbitals, the electrons fill each orbital singly with parallel spins before pairing up.

28. Use Hund’s rule to explain the filling of electrons in the p-orbitals of a *nitrogen atom*.

Electronic configuration of nitrogen is 1s2 2s2 2p3.

In the filling of the p-orbitals, three electrons are filled into each of the p-orbitals so that this gives the maximum number of parallel spins. The p-orbitals are half filled of electrons. This is a stable electronic configuration.