MAIN GROUP ELEMENTS - all elements located in the s AND p sections. TRANSITION ELEMENTS INNER-TRANSITION ELEMENTS

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1 CHEM 105 Lect-02a Periodic Table of the Elements is a left-to-right arrangement of elements, sequential in terms of increasing atomic number. Whenever a noble gas is encountered then a new line is started under the previous line, sort of like the page-return / line-feed made by a typewriter. Vertical columns of elements are called groups. Groups of elements have similar chemical properties due to presence of similar numbers and types of outermost electrons. Outermost electrons are called valence shell electrons (vse) and are of paramount importance in chemistry. Several named P.T. sections based on the underlined criteria are identified below: A. Types of vse's - s, p, d and f sections as shown to the right s section - two elements across p six d ten f fourteen s d f p MAIN GROUP ELEMENTS - all elements located in the s AND p sections. TRANSITION ELEMENTS d INNER-TRANSITION ELEMENTS f B. Vertical Columns - Families of elements Alkali Metals Main Group I Li, Na, K, Rb, Cs have 1 s vse Alkaline Earths Main Group II Be, Mg, Ca, Sr, Ba, Ra 2 s Halogens Main Group VII F, Cl, Br, I 2 s + 5 p Noble Gases Main Group VIII He, Ne, Ar, Kr, Xe 2 s + 6 p C. lose or gain vse's? - metals or non-metals Metal elements tend to lose negative electrons forming positive ions (cations). Metal elements are located left of the diagonal line in the p section Non-metal elements tend to gain negative electrons forming negative ions (anions). Non-metal elements are located right of the diagonal line. The further an element is from the diagonal line the greater its metal or non-metal character. Elements that border the line are often called semi-metals. Applications of Periodic Table Information A. Determining CHARGES of common ions: (depends on location of element in P.T.) 1. MAIN GROUP elements: number of electrons lost (by metals) or gained (by non-metals) is predicted by number of "steps" to nearest noble gas. 2. TRANSITION elements: All share a common +2 charge, with other charges. 3. INNER-TRANSITION elements: All share a common +3 charge " " " " Determine expected ion charges for the following elements: Na, Cl, Ba, O, Ga, P B. Determining FORMULAS for simple BINARY compounds of MAIN GROUP elements 1. Cation of a metal element combined with an anion of a non-metal element forms an IONIC compound. The formula is the simplest ratio of ions that leads to charge neutralization. Determine expected formulas of binary compounds from ions of the following pairs of elements: Na & Cl, Mg & Br, K & P, Ba & As

2 2. Two or more non-metal elements form MOLECULAR compounds (or ions) Formulas for molecular compounds are not as easily predicted as for ionic compounds because no charges are present. In addition, the same set of elements often form more than one molecular compound (Law of Multiple Proportions). Suggest reasonable formulas for molecular compounds containing the pairs of elements shown: C & O, N & O, S & Cl, N & Cl C. Naming of BINARY Compounds: 1. BINARY IONIC compounds: name the metal first (as the element) and then name the non-metal - but change its ending to -IDE. Determine the names of the ionic compounds met above. 2. BINARY MOLECULAR compounds: name more electropositive nonmetal element first (it has lesser non-metal character) and then name the more electronegative non-metal element (it has greater non-metal character) - but change its ending to - IDE. Determine the names of the molecular compounds met above. D. Naming of other compounds / ions: 1. Polyatomic Molecular ions: (refer to text Table 2.2 on page 43) Molecular ions often involve oxyanions and corresponding oxyacids. Be familiar with the following: nitric acid and nitrate salts sulfuric acid and sulfates acetic acid and acetates nitrous acid and nitrite salts phosphoric acid and phosphates carbonic acid and carbonates Determine the names of the following examples: Ca ( C 3 H 5 O 2 ) 2, KSCN, Mg C 2 O 4 2. TRANSITION metal cations: (can have variable positive charges) The positive charge is written as a Roman numeral and spoken as a number. For example, Ti(III) and "titanium three cation" The charge on a transition metal cation can be assigned when the charge of the remaining component in the compound is known. Consider the compound Ti O 2 Determine the charge (i.e., oxidation state) of the titanium and then name the compound. This is a binary compound so oxygen has it lowest oxidation state of minus 2. There are TWO oxygen (with a net minus FOUR charge) for each titanium. In order to neutralize the charge titanium must have a plus four charge. The compound name is written as titanium(iv) oxide and spoken as "titanium four oxide". Determine the names of the following examples: Fe Cl 2, Cr Br 3, Ni ( C 2 H 3 O 2 ) 2, Ti Cl 4, V P O 4, Co 2 ( S O 4 ) 3

3 ANSWERS to exercises present in the Handout finding charges of common ions: for MAIN GROUP elements, Na (sodium) Na (at.no. 11) has Ne (neon, at.no.10) as its nearest noble gas. Na has ONE MORE electron than Ne. It attains the same number of electrons as the noble gas by LOSING ONE electron. At this point sodium has: 11 protons, but only 10 electrons, so it has a charge of PLUS ONE. The ion is written as: Na 1+ Cl (chlorine) Cl (at.no. 17) has Ar (argon, at.no.18) as its nearest noble gas. Cl has ONE electron less than Ar. It attains the same number of electrons as the noble gas by GAINING ONE electron. At this point chlorine has: 17 protons, but 18 electrons, so it has a charge of MINUS ONE. The ion is written as: Cl 1- Ba (barium) O (oxygen) Ga (gallium) Ba (at.no. 56) has Xe (xenon, at.no.54) as its nearest noble gas. Ba has TWO electrons more than Xe. It attains the same number of electrons as the noble gas by LOSING TWO electrons. At this point barium has: 56 protons, but 54 electrons, so it has a charge of PLUS TWO. The ion is written as: Ba 2+ O (at.no. 8) has Ne (neon, at.no.10) as its nearest noble gas. O has TWO electrons less than Ne. It attains the same number of electrons as the noble gas by GAINING TWO electrons. At this point oxygen has: 8 protons, but 10 electrons, so it has a charge of MINUS TWO. The ion is written as: O 2- Ga (at.no. 31) has Kr (krypton, at.no.36) as its nearest noble gas -OR DOES IT? For purposes of finding ion charge it does not! Ga has THREE VSE more than the noble gas Ar (argon, at.no.18), rather than five less then the nobel gas krypton. This happens b/c the first-rowof-transition-metals the ten elements from at.no.21 through 30), is encountered before gallium appears in the table. However, VSE (d) electrons of transition metals do not constitute VSE for main group elements, so they are ignored. Ga attains the same number of VSE as argon by LOSING THREE electrons. At this point gallium has: 31 protons, but 28 electrons, so it has a charge of PLUS THREE. The ion is written as: Ga 3+ P (phosphorus) P (at.no. 15) has Ar (neon, at.no.18) as its nearest noble gas. P has THREE electrons less than Ar. It attains the same number of electrons as the noble gas by GAINING THREE electrons. At this point phosphorus has: 15 protons, but 18 electrons, so it has a charge of MINUS THREE. The ion is written as: P 3- writing formulas for simple BINARY compounds: for MAIN GROUP elements metal elements combining with non-metal elements, form ionic compounds Na forms a +1 cation, and chlorine forms a -1 anion. The ratio of ions necessary to achieve balance of charges is 1:1, so the formula is written as: NaCl

4 Mg forms a +2 cation, and bromine forms a -1 anion. The ratio of ions necessary to achieve balance of charges is 1:2, so the formula is written as: MgBr 2 K forms a +1 cation, and phosphorus forms a -3 anion. The ratio of ions necessary to achieve balance of charges is 3:1, so the formula is written as: K 3 P Ba forms a +2 cation, and arsenic forms a -3 anion (just like phosphorus - note that arsenic is in the same periodic table group as phosphorus). The ratio of ions necessary to achieve balance of charges is 2:3, so the formula is written as: Ba 3 As 2 non-metal elements combining with non-metal elements form a molecular compound or ion. Examples of multiple proportions are the rule here, and predicting a simplest, or "best" formula is not the goal here. Examples would be: C and O form: CO CO 2 C 2 O 3 N and O form: N 2 O NO N 2 O 3 N 2 O 4 N 2 O 5 S and Cl form: S 2 Cl 2 SCl 2 SCl 4 naming of simple compounds N and S form: S 2 N 2 S 4 N 2 S 11 N 2 binary ionic compounds NaCl sodium chloride MgBr 2 magnesium bromide K 3 P potassium phosphide Ba 3 As 2 barium arsenide binary molecular compounds CO carbon monoxide CO 2 carbon dioxide C 2 O 3 dicarbon trioxide N 2 O dinitrogen oxide (common name is nitrous oxide) NO nitrogen oxide (common name is nitric oxide) N 2 O 3 dinitrogen trioxide N 2 O 4 dinitrogen tetroxide N 2 O 5 dinitrogen pentoxide polyatomic molecules and ions, usually involves oxyacids and oxyanions: HNO 3 nitric acid, HNO 2 nitrous acid, H 2 SO 4 sulfuric acid, H 3 PO 4 phosphoric acid, HC 2 H 3 O 2 acetic acid, H 2 CO 3 carbonic acid, 1- NO 3 nitrate anion 1- NO 2 nitrite anion 2- SO 4 sulfate anion 3- PO 4 phosphate anion 1- C 2 H 3 O 2 acetate anion 2- CO 3 carbonate anion

5 ions of transition metals, can display several different charges: The charge of the cation is specified by Roman numeral as in: TiO 2 named as titanium(iv) oxide, showing that titanium is present as a +4 cation in this compound. How to find charge on a transition metal cation, or polyatomic anion, and name the compound - when formula is given. when a MAIN GROUP metal is present: Ca(C 3 H 5 O 2 ) 2 Note the ratio of ions present in the formula is 1:2. Apply some simple algebra so all charges present add up to zero. (# Ca ions)*(charge Ca ion) + (# anion)*(charge anion) = 0 the formula shows there are TWO anions, and b/c Ca is a main group element, we can use nearest noble gas to find its charge. substitute: ( 1 )*( +2 ) + ( 2 )*(charge anion) = 0 solve: charge on [C 3 H 5 O 2 ] anion is MINUS ONE. KSCN Presume SCN is a polyatomic anion and leave it intact. Do the algebra: (# K ions)*(charge K ion) + (# anion)*(charge anion) = 0 the formula shows only one ion of each type. Potassium is a main group element, we can use nearest noble gas to find its charge. substitute: ( 1 )( +1 ) = ( 1 )(charge anion) = 0 solve: charge on [ SCN ] anion is MINUS ONE. MgC 2 O 4 Presume C 2 O 4 is a polyatomic anion and leave it intact. Do the algebra: (# Mg ions)*(charge Mg ion) + (# anion)*(charge anion) = 0 the formula shows there is one anion and one cation. b/c Mg is a main group element, we can use nearest noble gas to find its charge. substitute: ( 1 )*( +2 ) + ( 1 )*(charge anion) = 0 solve: charge on [ C 2 O 4 ] anion is MINUS TWO. C 2 O 4 2- is named oxalate anion. when a main group non-metal is present: FeCl 2 as before... (# Fe ions)*(charge Fe ion) + (# anion)*(charge anion) = 0 the formula shows there is one cation and two anions. b/c Cl is a main group element, we can use nearest noble gas to find its charge. substitute: ( 1 )*(charge cation) + ( 2 )*( -1 ) = 0 solve: charge on [ Fe ] cation is PLUS TWO, and this cmpd would be named: iron(ii) chloride

6 CrBr 3 as before... (# Cr ions)*(charge Cr ion) + (# anion)*(charge anion) = 0 the formula shows there is one cation and three anions. b/c Br is a main group element, we can use nearest noble gas to find its charge. substitute: ( 1 )*(charge cation) + ( 3 )*( -1 ) = 0 solve: charge on [ Cr ] cation is PLUS THREE, and this cmpd would be named: chromium(iii) bromide Ni(C 2 H 3 O 2 ) 2 need to know charge of acetate anion (-1) (# Ni ions)*(charge Ni ion) + (# anion)*(charge anion) = 0 the formula shows there is one cation and two anions. substitute: ( 1 )*(charge cation) + ( 2 )*( -1 ) = 0 solve: charge on [ Ni ] cation is PLUS TWO, and this cmpd would be named: nickel(ii) acetate TiCl 4 as before... (# Ti ions)*(charge Ti ion) + (# anion)*(charge anion) = 0 the formula shows there is one cation and four anions. b/c Cl is a main group element, we can use nearest noble gas to find its charge. substitute: ( 1 )*(charge cation) + ( 4 )*( -1 ) = 0 solve: charge on [ Ti ] cation is PLUS FOUR, and this cmpd would be named: titanium(iv) chloride VPO 4 recognize that [ PO 4 ] is an oxyanion, and leave it intact. Need to know the charge of phosphate anion (-3) (# V ions)*(charge V ion) + (# anion)*(charge anion) = 0 the formula shows there is one cation and one anion. substitute: ( 1 )*(charge cation) + ( 1 )*( -3 ) = 0 solve: charge on [ V ] cation is PLUS THREE, and this cmpd would be named: vanadium(iii) phosphate Co 2 (SO 4 ) 3 Would need to know charge on sulfate anion in this case. Note, from above, it has a minus 2 charge. Do the algebra: (# Co ions)*(charge Co ion) + (# anion)*(charge anion) = 0 substitute: ( 2 )(charge Co ion) + ( 3 )( -2 ) = 0 solve: charge on cobalt is PLUS THREE, and this compound would be named as: cobalt(iii) sulfate