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Chapter 2 Atomic Structure and Interatomic Bonding problems, Study notes of Chemistry

Nebraska Indian Community CollegeChemistry

Thus, the electron configuration for a P5+ ion is 1s22s22p6. P3–: From Table 2.2, the electron configuration for an atom of phosphorus is 1s22s22p63s23p3. In ...

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HW 2 solutions:
Chapter 2 Atomic Structure and Interatomic Bonding problems:
2.9, 2.10, 2.22a, 2.25, 2.27
2.9 Give the electron configurations for the following ions: P5+, P3–, Sn4+, Se2–, I–, and Ni2+.
Solution
The electron configurations for the ions are determined using Table 2.2 (and Figure 2.8).
P5+: From Table 2.2, the electron configuration for an atom of phosphorus is 1s22s22p63s23p3. In order to
become an ion with a plus five charge, it must lose five electrons—in this case the three 3p and the two 3s. Thus, the
electron configuration for a P5+ ion is 1s22s22p6.
P3–: From Table 2.2, the electron configuration for an atom of phosphorus is 1s22s22p63s23p3. In order to
become an ion with a minus three charge, it must acquire three electrons—in this case another three 3p. Thus, the
electron configuration for a P3– ion is 1s22s22p63s23p6.
Sn4+: From the periodic table, Figure 2.8, the atomic number for tin is 50, which means that it has fifty
electrons and an electron configuration of 1s22s22p63s23p63d104s24p64d105s25p2. In order to become an ion with a
plus four charge, it must lose four electrons—in this case the two 4s and two 5p. Thus, the electron configuration for
an Sn4+ ion is 1s22s22p63s23p63d104s24p64d10.
Se2–: From Table 2.2, the electron configuration for an atom of selenium is 1s22s22p63s23p63d104s24p4. In
order to become an ion with a minus two charge, it must acquire two electrons—in this case another two 4p. Thus,
the electron configuration for an Se2– ion is 1s22s22p63s23p63d104s24p6.
I–: From the periodic table, Figure 2.8, the atomic number for iodine is 53, which means that it has fifty three
electrons and an electron configuration of 1s22s22p63s23p63d104s24p64d105s25p5. In order to become an ion with a
minus one charge, it must acquire one electron—in this case another 5p. Thus, the electron configuration for an I– ion
is 1s22s22p63s23p63d104s24p64d105s25p6.
Ni2+: From Table 2.2, the electron configuration for an atom of nickel is 1s22s22p63s23p63d84s2. In order
to become an ion with a plus two charge, it must lose two electrons—in this case the two 4s. Thus, the electron
configuration for a Ni2+ ion is 1s22s22p63s23p63d8.
pf3
pf4
pf5

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HW 2 solutions:

Chapter 2 Atomic Structure and Interatomic Bonding problems:

2.9, 2.10, 2.22a, 2.25, 2.

2.9 Give the electron configurations for the following ions: P5+, P^3 – , Sn4+, Se^2 – , I–, and Ni2+.

Solution The electron configurations for the ions are determined using Table 2.2 (and Figure 2.8). P5+: From Table 2.2, the electron configuration for an atom of phosphorus is 1 s^22 s^22 p^63 s^23 p^3. In order to become an ion with a plus five charge, it must lose five electrons—in this case the three 3 p and the two 3 s. Thus, the electron configuration for a P5+^ ion is 1 s^22 s^22 p^6. P^3 – : From Table 2.2, the electron configuration for an atom of phosphorus is 1 s^22 s^22 p^63 s^23 p^3. In order to become an ion with a minus three charge, it must acquire three electrons—in this case another three 3 p. Thus, the electron configuration for a P^3 –^ ion is 1 s^22 s^22 p^63 s^23 p^6. Sn4+: From the periodic table, Figure 2.8, the atomic number for tin is 50, which means that it has fifty electrons and an electron configuration of 1 s^22 s^22 p^63 s^23 p^63 d^104 s^24 p^64 d^105 s^25 p^2. In order to become an ion with a plus four charge, it must lose four electrons—in this case the two 4 s and two 5 p. Thus, the electron configuration for an Sn4+^ ion is 1 s^22 s^22 p^63 s^23 p^63 d^104 s^24 p^64 d^10. Se^2 – : From Table 2.2, the electron configuration for an atom of selenium is 1 s^22 s^22 p^63 s^23 p^63 d^104 s^24 p^4. In order to become an ion with a minus two charge, it must acquire two electrons—in this case another two 4 p. Thus, the electron configuration for an Se^2 –^ ion is 1 s^22 s^22 p^63 s^23 p^63 d^104 s^24 p^6. I–: From the periodic table, Figure 2.8, the atomic number for iodine is 53, which means that it has fifty three electrons and an electron configuration of 1 s^22 s^22 p^63 s^23 p^63 d^104 s^24 p^64 d^105 s^25 p^5. In order to become an ion with a minus one charge, it must acquire one electron—in this case another 5 p. Thus, the electron configuration for an I–^ 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. Ni2+: From Table 2.2, the electron configuration for an atom of nickel is 1 s^22 s^22 p^63 s^23 p^63 d^84 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 a Ni2+^ ion is 1 s^22 s^22 p^63 s^23 p^63 d^8.

2.9 Give the electron configurations for the following ions: P5+, P^3 – , Sn4+, Se^2 – , I–, and Ni2+.

P5+: P5+^ ion is 1 s^22 s^22 p^6. P^3 – :P^3 –^ ion is 1 s^22 s^22 p^63 s^23 p^6. Sn4+: 1 s^22 s^22 p^63 s^23 p^63 d^104 s^24 p^64 d^10. Se^2 – : 1 s^22 s^22 p^63 s^23 p^63 d^104 s^24 p^6. I–: 1 s^22 s^22 p^63 s^23 p^63 d^104 s^24 p^64 d^105 s^25 p^6. Ni2+: 1 s^22 s^22 p^63 s^23 p^63 d^8. 2.10 Potassium iodide (KI) exhibits predominantly ionic bonding. The K+^ and I–^ ions have electron structures that are identical to which two inert gases? Solution The K+^ ion is just a potassium atom that has lost one electron; therefore, it has an electron configuration the same as argon (Figure 2.8). The I–^ ion is a iodine atom that has acquired one extra electron; therefore, it has an electron configuration the same as xenon.

Mixed Bonding

2.25 Compute the %IC of the interatomic bond for each of the following compounds: MgO, GaP, CsF, CdS, and FeO. Solution The percent ionic character is a function of the electron negativities of the ions X A and X B according to Equation 2.16. The electronegativities of the elements are found in Figure 2.9. For MgO, X Mg = 1.3 and X O = 3.5, and therefore, For GaP, X Ga = 1.8 and X P = 2.1, and therefore, For CsF, X Cs = 0.9 and X F = 4.1, and therefore, For CdS, X Cd = 1.5 and X S = 2.4, and therefore, For FeO, X Fe = 1.7 and X O = 3.5, and therefore,

MgO=70.1%

GaP = 2.2%

CsF = 92.3%

CdS = 18.3%

FeO= 55.5%

%IC = ⎡ 1 − exp(− 0.25)(3.5−1.3)^2 ⎣

× 100 = 70.1%

%IC = ⎡ ⎣ 1 − exp(−0.25)(2.1− 1.8)^2 ⎤ ⎦ × 100 = 2.2% %IC = ⎡ ⎣ 1 − exp(− 0.25)(4.1− 0.9)^2 ⎤ ⎦ × 100 = 92.3% %IC = ⎡ ⎣ 1 − exp(−0.25)(2.4 − 1.5)^2 ⎤ ⎦ × 100 = 18.3% %IC = ⎡ 1 − exp(− 0.25)(3.5−1.7)^2 ⎣

× 100 = 55.5%

Bonding Type-Material Classification Correlations

2.27 What type(s) of bonding would be expected for each of the following materials: solid xenon, calcium fluoride (CaF 2 ), bronze, cadmium telluride (CdTe), rubber, and tungsten? Solution For solid xenon, the bonding is van der Waals since xenon is an inert gas. For CaF 2 , the bonding is predominantly ionic (but with some slight covalent character) on the basis of the relative positions of Ca and F in the periodic table. For bronze, the bonding is metallic since it is a metal alloy (composed of copper and tin). For CdTe, the bonding is predominantly covalent (with some slight ionic character) on the basis of the relative positions of Cd and Te in the periodic table. For rubber, the bonding is covalent with some van der Waals. (Rubber is composed primarily of carbon and hydrogen atoms.) For tungsten, the bonding is metallic since it is a metallic element from the periodic table. Solid Xenon: van der Waals (inert gas) CaF 2 : ionic Bronze: metallic CdTe: covalent Rubber: covalent + van der Waals Tungsten: metallic