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Electron Configurations for the Ions of Transition Metals & Main Group Elements Chemistry 11
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The Order of Orbital “Filling” 1.Lowest energy to higher energy. 2.Adding electrons can change the energy of the orbital. 3.Half filled orbitals have a lower energy. 4.Makes them more stable. 5.Changes the filling order
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Practice: Writing Electron Configurations Titanium - 22 electrons Titanium - 22 electrons 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 2 Vanadium - 23 electrons Vanadium - 23 electrons 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 3 Chromium - 24 electrons Chromium - 24 electrons 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 4 is expected But this is wrong!! Why? But this is wrong!! Why?
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True Configuration of Chromium 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 5 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 5 Why? Why? This gives us two half filled orbitals. This gives us two half filled orbitals. Slightly lower in energy. Slightly lower in energy. The same principal applies to copper. The same principal applies to copper.
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Copper’s Electron Configuration Copper has 29 electrons so we expect Copper has 29 electrons so we expect 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 9 But the actual configuration is But the actual configuration is 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 10 This gives one filled orbital and one half filled orbital. This gives one filled orbital and one half filled orbital. Remember these exceptions Remember these exceptions
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Valence Electrons Are the electrons in the highest occupied energy level of an element’s atom. Are the electrons in the highest occupied energy level of an element’s atom. Are usually the only electrons used in the formation of chemical bonds. Are usually the only electrons used in the formation of chemical bonds.Note: Many of the similar chemical properties of elements in the same group (vertical column) are related to the number of “ s ” and “ p ” electrons in the highest occupied energy level. These electrons are valence electrons.
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Example 1: Sodium 11 Na 11 Na 1s 2 2s 2 2p 6 3s 1 1s 2 2s 2 2p 6 3s 1 Highest energy level = 3 Highest energy level = 3 Single “s” electron Single “s” electron No “p” electrons for this element No “p” electrons for this element Na has 1 valence electron Na has 1 valence electron Ion: Ion:
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Example 2: Phosphorus 15 P 15 P 1s 2 2s 2 2p 6 3s 2 3p 3 1s 2 2s 2 2p 6 3s 2 3p 3 Highest energy level = 3 Highest energy level = 3 Two “s” electrons Two “s” electrons Three “p” electrons Three “p” electrons 5 valence electrons 5 valence electrons Ion: Ion:
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Example 3: Potassium 19 K 19 K Electron Configurations Electron Configurations 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 Highest energy level = 4 Highest energy level = 4 1 valence electron 1 valence electron Ion: Ion:
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Example 4: Carbon 6 C 6 C Electron Configuration Electron Configuration 1s 2 2s 2 2p 2 1s 2 2s 2 2p 2 Highest energy level = 2 Highest energy level = 2 4 valence electrons 4 valence electrons Ion: Ion:
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Example 5: Magnesium 12 Mg 12 Mg Electron Configuration Electron Configuration 1s 2 2s 2 2p 6 3s 2 1s 2 2s 2 2p 6 3s 2 Highest energy level = 3 Highest energy level = 3 2 valence electrons 2 valence electrons Ion: Ion:
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Example 6: Oxygen 8 O 8 O Electron Configuration Electron Configuration 1s 2 2s 2 2p 4 1s 2 2s 2 2p 4 Highest energy level = 2 Highest energy level = 2 6 valence electrons 6 valence electrons Ion: Ion:
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Electron Configurations & Periodicity Of the 3 subatomic particles, the electron plays the greatest role in determining the physical and chemical properties of an element. Of the 3 subatomic particles, the electron plays the greatest role in determining the physical and chemical properties of an element. There is a relationship between the electron configuration of elements and their arrangement in the table. There is a relationship between the electron configuration of elements and their arrangement in the table.
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Transition Metals Do not form ions with noble gas configurations Do not form ions with noble gas configurations If they did, they would have to lose 4 or more electrons, and the energy needed to do this would be too high. If they did, they would have to lose 4 or more electrons, and the energy needed to do this would be too high. These metals do form cations (“+” charged ion) with more than one charge (multi-valent) These metals do form cations (“+” charged ion) with more than one charge (multi-valent) Electrons are first moved from the outer “s” and then inner “d” sublevels. Electrons are first moved from the outer “s” and then inner “d” sublevels.
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Example 1: 25 Mn ( 18 Ar)4s 2 3d 5 25 Mn ( 18 Ar)4s 2 3d 5 Mn 2+ loses 2 electrons from the “s” sublevel Mn 2+ loses 2 electrons from the “s” sublevel Mn 2+ ( 18 Ar)4s 0 3d 5 Mn 2+ ( 18 Ar)4s 0 3d 5
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Example 2: 26 Fe( 18 Ar)4s 2 3d 6 26 Fe( 18 Ar)4s 2 3d 6 Fe 2+ loses 2 electrons from the “s” sublevel Fe 2+ loses 2 electrons from the “s” sublevel Fe 2+ ( 18 Ar)4s 0 3d 6 Fe 2+ ( 18 Ar)4s 0 3d 6 Fe 3+ ( 18 Ar)4s 0 3d 5 Fe 3+ ( 18 Ar)4s 0 3d 5
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Question 1 Read the following electron configurations to determine the identities of the following elements: a) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3 d10 4p 6 5s 2 b) 1s 2 2s 2 2p 6 3s 2 3p 3 c) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 5d 10 6p 6 d) 1s 2 2s 2 2p 2 e) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 5
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Question 2 Explain what is meant by the term “ isoelectronic ”. Answer : The prefix “ iso ” means “the same ”, so isoelectronic means that two atoms or ions have the same number of electrons or the same electron configuration.
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Question 3 What are four ions, with their charges, that are isoelectronic with each of the following: a) neon : N 3– O 2– F 1– Na 1+ Mg 2+ Al 3+ b) argon : P 3– S 2– Cl 1– K 1+ Ca 2+ Sc 3+ c) krypton : As 3– Se 2– Br 1– Rb 1+ Sr 2+ Y 3+ d) an S 2– ion : P 3– Cl 1– K 1+ Ca 2+ Sc 3+
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Question 4 A calcium ion, Ca2+, is isoelectronic with argon. Does this mean that calcium has “turned into” argon? Explain. Answer : No, calcium has not turned into Argon. Remember, it is the atomic number (number of protons) that determines what type of element an atom is. The calcium ion has the same number of electrons as an Argon atom. This just means that they are Both stable octet with 18 electrons.
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Question 5 Write the electron configuration for the element with atomic number 117. In which chemical group does it belong? Predict the charge on the ion that it will form.
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Question 5 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 6 7s 2 5f 14 6d 10 7p 5 This element belongs to the Halogen family or Group 17 (VIIB). It will gain one electron to have the same configuration as a Noble Gas, so it will form ions with a charge of 1-.
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