Humboldt State University ® Department of Chemistry

Richard A. Paselk
 

General Chemistry

Spring 2009
Exercise: Periodicity and Atomic Structure

© R. Paselk 2008
 
 

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Periodicity and Atomic Structure

1. Using Spectroscopic notation write the electronic configurations for the ground state atoms below:

Be: Counting across the Periodic Table as shown below starting with H = 1s1

Periodic Table - Outermost Electrons
 IA IIA IIIA IVA VA VIA VIIA VIIIA
   1s1  1s2
2s1 Be    B C N O F Ne

giving a configuration for Be = 1s2 2s2

Al: Counting across the Periodic Table as shown below starting with H = 1s1

Periodic Table - Outermost Electrons
 IA IIA IIIA IVA VA VIA VIIA VIIIA
   1s1   1s2
 2s1 2s2    2s2p1 2s2p2 2s2p3 2s2p4 2s2p5 2s2p6
3s1 3s2 IIIB IVB VB VI VIIB VIIIB IB IIB  Al Si P S Cl Ar

giving a configuration for Al = 1s2 2s2 2p6 3s2 3p1

Si: Counting across the Periodic Table as shown below starting with H = 1s1

Periodic Table - Outermost Electrons
 IA IIA IIIA IVA VA VIA VIIA VIIIA
   1s1   1s2
 2s1 2s2    2s2p1 2s2p2 2s2p3 2s2p4 2s2p5 2s2p6
3s1 3s2 IIIB IVB VB VI VIIB VIIIB IB IIB  3s2p1 Si P S Cl Ar

giving a configuration for Si = 1s2 2s2 2p6 3s2 3p2

S: Counting across the Periodic Table as shown below starting with H = 1s1

Periodic Table - Outermost Electrons
 IA IIA IIIA IVA VA VIA VIIA VIIIA
   1s1   1s2
 2s1 2s2    2s2p1 2s2p2 2s2p3 2s2p4 2s2p5 2s2p6
3s1 3s2 IIIB IVB VB VI VIIB VIIIB IB IIB  3s2p1 3s2p2 3s2p3 S Cl Ar

giving a configuration for S = 1s2 2s2 2p6 3s2 3p4

 

Ne: Counting across the Periodic Table as shown below starting with H = 1s1

Periodic Table - Outermost Electrons
 IA IIA IIIA IVA VA VIA VIIA VIIIA
   1s1   1s2
 2s1 2s2    2s2p1 2s2p2 2s2p3 2s2p4 2s2p5 Ne

giving a configuration for Ne = 1s2 2s2 2p6

2. Using Spectroscopic notation the Noble Gas Core convention write the electronic configurations for the ground state atoms below:

Hg: To use the Noble Gas Core convention we first find Hg on the Periodic Table in period six. The previous period ends with Xe, so we begin with [Xe]. If we then count across the table as we have done above,

Periodic Table of the Elements
   IA IIA IIIA IVA VA VIA VIIA VIIIA
   H  He
2 Li Be   B C N O F Ne
3       IIIB IVB VB VI VIIB VIIIB IB IIB            
4     3                                
5     4                                
6  s1  s2 5  d1  d2 d3 d4 d5 d6 d7 d8 d9  Hg Tl  Pb Bi Po At Rn

 

we start with 6s1 then s2. At this time note that we have to drop down one period for the d electrons giving 5d1 and count across to Hg = 6s2 5d10.

Ta: To use the Noble Gas Core convention we first find Hg on the Periodic Table in period six. The previous period ends with Xe, so we begin with [Xe]. If we then count across the table as we have done above,

Periodic Table of the Elements
   IA IIA IIIA IVA VA VIA VIIA VIIIA
   H  He
2 Li Be   B C N O F Ne
3       IIIB IVB VB VI VIIB VIIIB IB IIB            
4     3                                
5     4                                
6  s1  s2 5  d1  d2 Ta                        

we start with 6s1 then s2. At this time note that we have to drop down one period for the d electrons giving 5d1 and count across to Ta = 6s2 5d3.

Bi: First find Bi on the Periodic Table in period six. The previous period ends with Xe, so we begin with [Xe]. If we then count across the table as we have done above,

Periodic Table of the Elements
   IA IIA   IIIA IVA VA VIA VIIA VIIIA
   H  He
2 Li Be   2 B C N O F Ne
3       IIIB IVB VB VI VIIB VIIIB IB IIB 3            
4     3                     4            
5     4                     5            
6  s1  s2 5  d1  d2 d3  d4 d5 d6 d7 d8 d9  d10 6 p1 p2 Bi      

we start with 6s1 then s2. At this time note that we have to drop down one period for the d electrons giving 5d1 and count across the transition elements, then go back up to 6p1. Finally,

Bi = 6s2 5d106p3

Am: Americium is an the Inner Transition element, one of the Actinides, so it has 5f electrons. Looking at the Periodic Table the previous Noble Gas is radon, Rn. Starting with Fr, 7s1 we get:

Am = [Rn] 7s25f 7

Xe: Even though Xe is itself a Noble gas, we need to go to the previous Noble gas, Kr. The idea is that we need to show all of the outermost electrons, those which have the potential to do chemistry! Thus we get:

Xe = [Kr] 5s24d105p6

Mo: Molybdenum is in period 5, so you might expect it to be [Kr] 5s24d4, however, as we saw with Cr, moving an electron from the 5s orbital to the 4d orbital will a complete half-d-set resulting in two symmetrical orbital sets and greater stability, Thus:

Mo = [Kr] 5s14d5

Ag: Silver is also in period 5, so you might expect it to be [Kr] 5s24d9, however, as we saw with Cu, moving an electron from the 5s orbital to the 4d orbital will a complete d-set resulting in two symmetrical orbital sets and greater stability, Thus:

Ag = [Kr] 5s14d10

3. Using Spectroscopic notation write the electronic configurations for the ground state ions below:

Mg2+: Metal ions ALWAYS lose their outermost electrons first. That means electrons with the highest values of n are the first to be lost. For Mg that means we go from Mg = 1s2 2s2 2p6 3s2. Thus Mg will lose the 3s electrons giving:

Mg2+ = 1s2 2s2 2p6 3s0 = 1s2 2s2 2p6

P3-: Phosphorus will gain three electrons to fill its 3p orbitals, giving:

P3- = 1s2 2s2 2p6 3s2 3p6

Ni2+: Metal ions ALWAYS lose their outermost electrons first. That means electrons with the highest values of n are the first to be lost. For Ni that means we go from Ni = 1s22s22p63s23p64s23d8. Thus Ni will lose the 4s electrons giving:

Ni2+= 1s22s22p63s23p64s03d8 = 1s22s22p63s23p63d8

V3+: Note that you can write the orbitals in order as they fill across the Periodic Table as above, OR in numerical order. Thus V = 1s22s22p63s23p63d34s2, and for the ion we lose the two 4s and one 3d electron giving us:

V3+= 1s22s22p63s23p63d24s0 = 1s22s22p63s23p63d2

Ag+: Recall from above that due to symmetry silver has an electron configuration of Ag = 1s22s22p63s23p63d104s24p64d105s1. The ion thus loses the single 5s electron, giving:

Ag+= 1s22s22p63s23p63d104s24p64d105s0 = 1s22s22p63s23p63d104s24p64d10

Note how this structure explains the plus-one oxidation state of silver.

4. Using Orbital Filling Diagrams write the electronic configurations for the ground state atoms below:

B
1s
2s
2p
Note the the arrow for the 2p electron can be oriented either up or down.
Ti
1s
2s
2p
3s
3p
4s
3d

Note that the two 3d electrons in Ti go into separate orbitals AND they have the same spin (the arrows point the same direction).

Fe
1s
2s
2p
3s
3p
4s
3d

Note that the six 3d electrons in Fe are arranged to minimize pairing - only one pair is formed, the other four go into separate orbitals AND they have the same spin (the arrows point the same direction).

Cr
1s
2s
2p
3s
3p
4s
3d

As we have seen before, the electrons spread out in a single orbital set and all have the same spin. We also see in Cr the migration of one of the 4s electrons to give a half-filled d-subshell.

As
1s
2s
2p
3s
3p
4s
3d
4p

Note how the 4p electrons in As spread out over the three p orbitals AND they all have the same spin (the arrows point the same direction).

Se
1s
2s
2p
3s
3p
4s
3d
4p

In addition to these exercises you should familiarize yourself with the text materials.

Periodicity and Atomic Structure Module

© R A Paselk

Last modified 20 October 2009