ATOMIC STRUCTURE AND INTERATOMIC BONDING, Cheat Sheet of Materials science

Chromium has four naturally-occurring isotopes: 4.34% of 50Cr, with an atomic weight of 49.9460 amu, 83.79% of 52Cr, with an atomic weight of 51.9405 amu, 9.50% of 53Cr, with an atomic weight of 52.9407 amu, and 2.37% of 54Cr, with an atomic weight of 53.9389 amu. On the basis of these data, confirm that the average atomic weight of Cr is 51.9963 amu.

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HW #1
2.2 Chromium has four naturally-occurring isotopes: 4.34% of 50Cr, with an atomic weight of 49.9460 amu, 83.79%
of 52Cr, with an atomic weight of 51.9405 amu, 9.50% of 53Cr, with an atomic weight of 52.9407 amu, and 2.37% of
54Cr, with an atomic weight of 53.9389 amu. On the basis of these data, confirm that the average atomic weight of
Cr is 51.9963 amu.
Solution
The average atomic weight of silicon

(A
Cr )
is computed by adding fraction-of-occurrence/atomic weight
products for the three isotopes. Thus

A
Cr = f50Cr A50Cr + f52Cr A52Cr f53Cr A53Cr f54Cr A54Cr

(0.0434)(49.9460 amu) + (0.8379)(51.9405 amu) + (0.0950)(52.9407 amu) + (0.0237)(53.9389 amu) = 51.9963 amu
2.7 Give the electron configurations for the following ions: Fe2+, Al3+, Cu+, Ba2+, Br-, and O2-.
Solution
The electron configurations for the ions are determined using Table 2.2 (and Figure 2.6).
Fe2+: From Table 2.2, the electron configuration for an atom of iron is 1s22s22p63s23p63d64s2. In order to
become an ion with a plus two charge, it must lose two electronsin this case the two 4s. Thus, the electron
configuration for an Fe2+ ion is 1s22s22p63s23p63d6.
Al3+: From Table 2.2, the electron configuration for an atom of aluminum is 1s22s22p63s23p1. In order to
become an ion with a plus three charge, it must lose three electronsin this case two 3s and the one 3p. Thus, the
electron configuration for an Al3+ ion is 1s22s22p6.
Cu+: From Table 2.2, the electron configuration for an atom of copper is 1s22s22p63s23p63d104s1. In order
to become an ion with a plus one charge, it must lose one electronin this case the 4s. Thus, the electron
configuration for a Cu+ ion is 1s22s22p63s23p63d10.
Ba2+: The atomic number for barium is 56 (Figure 2.6), and inasmuch as it is not a transition element the
electron configuration for one of its atoms is 1s22s22p63s23p63d104s24p64d105s25p66s2. In order to become an ion
with a plus two charge, it must lose two electronsin this case two the 6s. Thus, the electron configuration for a
Ba2+ ion is 1s22s22p63s23p63d104s24p64d105s25p6.
Br-: From Table 2.2, the electron configuration for an atom of bromine is 1s22s22p63s23p63d104s24p5. In
order to become an ion with a minus one charge, it must acquire one electronin this case another 4p. Thus, the
electron configuration for a Br- ion is 1s22s22p63s23p63d104s24p6.
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HW

2.2 Chromium has four naturally-occurring isotopes: 4.34% of 50 Cr, with an atomic weight of 49.9460 amu, 83.79% of 52 Cr, with an atomic weight of 51.9405 amu, 9.50% of 53 Cr, with an atomic weight of 52.9407 amu, and 2.37% of (^54) Cr, with an atomic weight of 53.9389 amu. On the basis of these data, confirm that the average atomic weight of Cr is 51.9963 amu.

Solution The average atomic weight of silicon

( A Cr )is computed by adding fraction-of-occurrence/atomic weight products for the three isotopes. Thus

A Cr = f (^50) Cr A (^50) Cr + f (^52) Cr A (^52) Cr  f (^53) Cr A (^53) Cr  f (^54) Cr A (^54) Cr

 (0.0434)(49.9460 amu) + (0.8379)(51.9405 amu) + (0.0950)(52.9407 amu) + (0.0237)(53.9389 amu) = 51.9963 amu

2.7 Give the electron configurations for the following ions: Fe2+, Al3+, Cu+, Ba2+, Br-, and O2-.

Solution The electron configurations for the ions are determined using Table 2.2 (and Figure 2.6).

Fe2+: From Table 2.2, the electron configuration for an atom of iron is 1 s^22 s^22 p^63 s^23 p^63 d^64 s^2. In order to become an ion with a plus two charge, it must lose two electrons—in this case the two 4 s. Thus, the electron configuration for an Fe2+^ ion is 1 s^22 s^22 p^63 s^23 p^63 d^6. Al3+: From Table 2.2, the electron configuration for an atom of aluminum is 1 s^22 s^22 p^63 s^23 p^1. In order to become an ion with a plus three charge, it must lose three electrons—in this case two 3 s and the one 3 p. Thus, the electron configuration for an Al3+^ ion is 1 s^22 s^22 p^6. Cu+: From Table 2.2, the electron configuration for an atom of copper is 1 s^22 s^22 p^63 s^23 p^63 d^104 s^1. In order to become an ion with a plus one charge, it must lose one electron—in this case the 4 s. Thus, the electron configuration for a Cu+^ ion is 1 s^22 s^22 p^63 s^23 p^63 d^10. Ba2+: The atomic number for barium is 56 (Figure 2.6), and inasmuch as it is not a transition element the electron configuration for one of its atoms is 1 s^22 s^22 p^63 s^23 p^63 d^104 s^24 p^64 d^105 s^25 p^66 s^2. In order to become an ion with a plus two charge, it must lose two electrons—in this case two the 6 s. Thus, the electron configuration for a Ba2+^ ion is 1 s^22 s^22 p^63 s^23 p^63 d^104 s^24 p^64 d^105 s^25 p^6. Br-: From Table 2.2, the electron configuration for an atom of bromine is 1 s^22 s^22 p^63 s^23 p^63 d^104 s^24 p^5. In order to become an ion with a minus one charge, it must acquire one electron—in this case another 4 p. Thus, the electron configuration for a Br-^ ion is 1 s^22 s^22 p^63 s^23 p^63 d^104 s^24 p^6.

O2-: From Table 2.2, the electron configuration for an atom of oxygen is 1 s^22 s^22 p^4. In order to become an ion with a minus two charge, it must acquire two electrons—in this case another two 2 p. Thus, the electron configuration for an O2-^ ion is 1 s^22 s^22 p^6.

2.9 With regard to electron configuration, what do all the elements in Group VIIA of the periodic table have in common?

Solution Each of the elements in Group VIIA has five p electrons.

2.21 Using Table 2.2, determine the number of covalent bonds that are possible for atoms of the following elements: germanium, phosphorus, selenium, and chlorine.

Solution

For germanium, having the valence electron structure 4 s^24 p^2 , N' = 4; thus, there are 8 – N' = 4 covalent bonds per atom. For phosphorus, having the valence electron structure 3 s^23 p^3 , N' = 5; thus, there is 8 – N' = 3 covalent bonds per atom. For selenium, having the valence electron structure 4 s^24 p^4 , N' = 6; thus, there are 8 – N' = 2 covalent bonds per atom. For chlorine, having the valence electron structure 3 s^23 p^5 , N' = 7; thus, there are 8 – N' = 1 covalent bond per atom.

2.23 Explain why hydrogen fluoride (HF) has a higher boiling temperature than hydrogen chloride (HCl) (19.4 vs.

- 85°C), even though HF has a lower molecular weight.

Solution

The intermolecular bonding for HF is hydrogen, whereas for HCl, the intermolecular

bonding is van der Waals. Since the hydrogen bond is stronger than van der Waals, HF will

have a higher melting temperature