Phương pháp dạy học hóa học bằng tiếng Anh

3. Base your answers to questions 16–20 on the following electron configurations of neutral atoms.

a. 1s²2s²2p⁶s¹ c. 1s² e. 1s²2s²2p⁵

b. 1s²2s²2p² d. 1s²2s¹2p³ f. 1s²2s²2p⁴

16. Which electron configuration represents a noble gas?

17. Which electron configuration shows an atom in the excited state?

18. Which electron configuration shows an alkali metal?

19. Which electron configuration represents a halogen?

20. Which ground state configuration shows a total of 4 valence electrons?

4. Give the electron configuration, and short electron configuration of the ground state of S (Z = 16), Ne (Z = 10)

5. Give the electron configuration, and short electron configuration of the ground state of Fe (Z = 56), Cu (Z = 29), Cr (Z = 24)

6. Two elements in Period 5 are adjacent to one another in the periodic table. The ground state atom of one element has only s electrons in its valence shell; the other has at least one d electron in an unfilled shell. Identify the elements.

7. Two elements are in the same column of the periodic table, one above the other. The ground state atom of one element has two s electrons in its outer shell, and no d electrons any where in its configuration. The other element has d electrons in its configuration. Identify the elements.

8. Naturally occurring silver exists as two isotopes , has mass of 106.9u and relative abundance 51.8%; has mass of 108.9u and 48.2%. Calculate the average atomic mass of silver.

9. The two stable isotopes of boron exist in the following proportions: 19.78% (10.01 u) and 80.22% (11.01 u). Calculate the average atomic mass of boron

10. Boron exists as two naturally occurring isotopes: (10.01 u) and (11.01 u). Calculate the relative abundance of each isotope of boron.

Answer

1.

1. [c. electron]

2. [f. kernel]—Think of a kernel of popcorn. Valence electrons are not included in the kernel.

3. [d. mass number]—Because the electrons have essentially no mass, the protons and the neutrons make up the vast majority of the mass of an atom.

4. [l. orbital]—Orbitals are not always full, but when they are, they can hold a maximum of two electrons.

5. [i. positive]—Remember, losing negative charges (electrons) leaves an atom with “extra” positive charges.

6. [j. quantum numbers]

7. [e. atomic number]—The atomic number is also equal to the nuclear charge of an atom.

8. [g. negative]—When a neutral atom gains extra negative charges, it will

have a net negative charge.

9. [a. proton]

2.

10. [hydrogen]—H is the elemental symbol for hydrogen.

11. [1]—The atomic number is shown in the lower left-hand corner of the notation.

12. [3]—The mass number is shown in the upper left-hand corner of the notation.

13. [1]—The number of protons is equal to the atomic number.

14. [2]—The number of neutrons is equal to (the mass number – the atomic number), or 3 – 1 = 2.

15. [0]—The number of electrons is equal to (the atomic number – the charge of atom), or 1 – (+1) = 0.

3. 16. [c. 1s² ]—Although it only contains 2 electrons, helium is considered a noble gas because its valence shell is full.

17. [d. 1s²2s¹2p³]- The 2s sublevel is not full, yet there are electrons in the 2p sublevel. An electron must have “jumped up” from the 2s sublevel to the 2p sublevel.

18. [a. 1s²2s²2p⁶3s¹]—With a total of 11 electrons, this neutral configuration must represent sodium.

19. [e. 1s²2s²2p⁵]—With seven valence electrons, this atom would be located incolumn 17. With a total of 9 electrons, it must be fluorine.

20. [b. 1s²2s²2p²]—The question clearly asked for a ground state configuration, so that ruled out answer d. The only other configuration that shows 4 valence electrons is b.

4. Give the electron configuration, and short electron configuration of the ground state of S (Z = 16): 1s²2s²2p⁶3s²3p⁴; [Ne]3s²3p⁴

Ne (Z = 10): 1s²2s²2p⁶; [He]2s²2p⁶

5. Give the electron configuration, and short electron configuration of the ground state of Fe (Z = 56): 1s22s22p63s23p63d64s²; [Ar]3d⁶4s²

Cu (Z = 29): 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹; [Ar]3d¹⁰4s¹

Cr (Z = 24): 1s²2s²2p⁶3s²3p⁶3d⁵4s¹; [Ar]3d⁵4s¹

6. Ag and Cd

7. Ca and Sr

8. Average atomic mass of Ag = 106.9 u (0.518)+108.9 u (0.482)=107.9 u

9. Average atomic mass of B = 10.01 u (19.78)+11.01 u (80.22)=10.81 u

10. Average atomic mass = x(atomic mass B-10)+(1−x)(atomic mass B-11)

10.81 = x(10.01)+(1−x)(11.01)

10.81 = 10.01x+11.01−11.01x11.01x−10.01

x = 11.01−10.81

x = 0.2000

The abundance of boron-10 is 0.2000.

The abundance of boron-11 is 1−x, or 1−0.2000=0.8000.

The abundance of is therefore 20.00%. The abundance of is 80.00%

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