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UT CH 301 - Study Guide

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CH301 Fall 2009 Worksheet 3 Answer Key on Electronic Configuations and Periodic Table Trends 1. Write the electron configuration of gallium (Ga) in long notation and in short notation.1s22s22p63s23p64s23d104p1[Ar] 4s23d104p12. Write the electron configuration of a divalent vanadium cation (V2+) in long and in shortnotation.1s22s22p63s23p64s23d1[Ar] 4s23d13. Write the electron configuration of a monovalent iodine anion (I-) in long and in shortnotation.1s22s22p63s23p64s23d104p65s24d105p6[Kr] 5s24d105p64. State in your own words the Pauli exclusion principle. Which of the quantum numbers is itmost concerned with?In a given atom, all of the electrons must be described by a unique set of four quantumnumbers (n, l, mland ms). It is most concerned with ms, since for any value of n, l and ml,msmust be different, ensuring that the electrons have opposite spins. So, for our purposes,the Pauli exclusion principle states that any two electrons in the same orbital must haveopposite spins.5. Which ground state element corresponds to the following electron configurations?a. [Rn] 5f146d37s2Dubnium (Db)b. [Ne] 3s23p4Sulfur (S)c. [Xe] 4f145d66s2Osmium (Os)d. [Xe] 4f145d106s1Gold (Au)6. In your own words, what does Hund's rule state?Every orbital in a subshell must have one electron in it before any orbital can have a pair ofelectrons in it. In other words, we half fill all orbitals before completely filling any of them.(For the record, Hund had 3 rules, but we're only concerned with the 1st one.)7.In which region of the periodic table do we find most of the exceptions to Aufbau forneutral, ground state elements? What about cations (positively charged ionic species)?The d block elements, aka the transition metals, and their ions often will "borrow" anelectron or two from the nearest s subshell in order to fill or half-fill their d subshell.The p block metal ions, e.g. Pb3+, also have electron configurations that are exceptions toAufbau, particularly if they are isoelectronic with exception in the d block.8. Write the electron configuration for silver. Try to give at least one example of an ion thatwould have an identical electron configuration.[Kr] 5s14d10Cd+, In2+and Sn3+would all have the same electron configuration as Ag.9. Is the electron configuration for silver a violation of the Aufbau principle? Defend yourassertion.No, silver's electron configuration is not a violation of the Aufbau principle. Silver'sexperimentally determined electron configuration is [Kr] 5s14d10. The Aufbau principlestates that a ground state atom will have its electrons in their lowest energy possible, so if[Kr] 5s14d10is what is observed, it is by definition lower in energy than [Kr] 5s24d9andthus Aufbau has not been violated.10. What one adjective describes the following species: Ar, P3-, Cl-, Ca2+?Isoelectronic.11. Arrange the following ions in order of increasing ionic radiusSr2+ and Ca2+O-2 and S-2Sr2+ > Ca2+Ionic radius increases down a group because electrons are occupying shells with higherprincipal quantum numbers. All cations are smaller than their parent atoms because theylose electrons and their core is exposed.O-2 < S-2Again, ionic radius increases down a group. Anions are larger than their parent atoms. Moreelectrons are added to the valence shell and the repulsive effect of the electrons andrepel each other.12. Define electron affinity in your own words and explain why electron affinity decreasesfrom groups I to II.Electron affinity corresponds to the energy released when an electron is added to an atomin the gas phase. When a lot of energy is released the atom is said to have a high electronaffinity. These atoms are more likely to take in an electron. Group I atoms need 1 electronto complete their s orbital. When an electron is added to group II atoms the electronenters into a p orbital. The complete s orbital is more tightly bound to the nucleus thanthe p orbital.13. Explain why a p orbital experiences more shield than an s orbital.Remember that orbitals are found by squaring the angular wavefunction. An electron in thes orbital penetrates through inner shells and has no nodes. An electron in the p orbitalpenetrates much less because its angular wavefunction limits how close it can get to thenucleus (it also has a node through the nucleus unlike the s orbital). Also, electrons in thes orbital will shield those in the p orbital from the protons in the nucleus.14. Calculate the effective nuclear charge for all of the electrons in nitrogen.The actual nuclear charge of nitrogen is 7 (there are 7 protons in the nucleus).Electrons in the same shell will experience the same effective nuclear charge1s electrons: 7-0 = 7 No electrons shield the first row2s electrons: 7-2 = 52p electrons: 7-2 = 5Electrons in the same row will have the same effective nuclear charge15. Explain why the first ionization energy decreases from group V to group VI elements (anexception to the trend of increasing ionization energy from left to right).Filled and half filled shells have additional stability.16. Rank the following elements in increasing atomic radius and explain the trend: Mg, Cs,F, BrCs > Mg > Br > FAtomic radius decreases from left to right across a period and increases down a group.Down a group the electrons occupy shells that are farther from the nucleus and theseelectrons are more shielded. Across a period, the number of electrons and protons increasebut these electrons are not that shielded since they fill the same shell and experience anincreasing ENC.17. You travel to planet Darwin IV and discover that on this planet three electron spinsexist. Which element would have a higher electron affinity, an element with atomic number17 or an element with atomic number 26?On a 3 spin world.... (remember we live in a 2 spin world!)The electron configuration of element 17:1s3 2s3 2p9 3s2The electron configuration of element 26:1s3 2s3 2p9 3s3 3p8Element 26 only needs one more electron to complete the p subshell, therefore it has ahigher electron affinity.18. Where do you find metals in the periodic table? Which periodic trend dictates thelocation of metals on the periodic table?Metals are found in the lower left part of the periodic table. Their ionization energies arelow and readily lose electrons.19. Rank the following in order of increasing ionization energies: He, Ca, Sn, Sn+1Ca < Sn < Sn+1 < HeIn general, ionization energy increases across a period and decreases down a row. Thesecond ionization energy of an element is


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UT CH 301 - Study Guide

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