Question

1. What element is represented by the electron configuration 1s2 2s2 2p6 3s2 3p2 ? ____________

2. What element is represented by the orbital diagram: _____________

3. Draw the orbital diagrams (box/line notation) for the following species:

Mn2+ ____________________________________________

Cu ____________________________________________

4. Write the electron configuration for the following elements and ions using spdf notation (without noble gas abbreviations):

Cr ____________________________________________

Si ____________________________________________

Fe2+ ____________________________________________

P 3- ____________________________________________

5. Write the electron configuration for the following elements and ions using noble gas spdf notation: Sn2+ Br‾

6. What are the possible values of the angular momentum quantum number (l) when n = 4 ?

7. What are the possible values for the magnetic quantum number (ml ) when l = 2 ?

8. In the ground state of selenium, how many electrons share the quantum number: a) n = 4 _____________ b) l = 2 ______________ 1s 2s 2p 3s

9. List all the quantum numbers for each of the 9 electrons in fluorine. n l ml ms

a.

b.

c.

d.

e.

f.

g.

h.

i.

10. What are the possible quantum numbers for the last electron in zinc? n = ___________ l = ___________ ml = __________ ms = ___________

11. Which element has the largest atomic radius? a) Na or Li b) Zr or La or Sb or S

12. Which element has the smallest ionic radius? a) Na+ or Mg2+ b) Be2+ or O2- or F‾ or S2-

13. Which element has the highest ionization energy? a) Sr or Ba b) Br or I or Te

14. Which element has the greatest electron affinity (most negative ∆E)? a) Na or Cl b) Sr or Ba

15. Which alkaline earth metal has the smallest atomic radius? __________________

16. Which alkali metal has the highest ionization energy? __________________

17. Which halogen has the lowest electron affinity? __________________

Answer 1

1) 1s2 2s2 2p6 3s2 3p2: Add all number of electrons 2+2+6+2+2 =14

Atomic number = 14

Look at the periodic table to find the element with atomic number 14. It is silicon

2) Orbital diagram is missing in the question

3)

Mn²⁺:

Electronic configuration of Mn²⁺: atomic number of Mn is 25. but it has +2 charge. so Mn²⁺ will have 25-2 = 23 electrons and the electronic configuration is shown below

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

and Cu: atomic number of Cu is 29. So Cu will have 29 electrons and the electronic configuration is shown below

1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹

4

4)

Cr: atomic number is 24: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁴

Si: atomic number is 14: 1s² 2s² 2p⁶ 3s² 3p²

Fe²⁺ :atomic number is 26: It has +2 charge so 2 electrons are subtracted = 26-2 = 24

electronic configuration of Fe : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶

electronic configuration of Fe²⁺:1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶

P ^3- : atomic number is 15: It has -3 charge so 3 electrons are added = 15+ 3=18

electronic configuration of P: 1s² 2s² 2p⁶ 3s² 3p³

electronic configuration of P^3-:1s² 2s² 2p⁶ 3s² 3p⁶

5) Sn²⁺: atomic number of Sn is 50: It has +2 charge so 2 electrons are subtracted = 50 - 2=48

electronic configuration of Sn: [Kr] 5s² 4d¹⁰ 5p²

electronic configuration of Sn²⁺:[Kr] 5s² 4d¹⁰

Br‾:

atomic number of Br is 35: It has -1 charge so one electrons is added = 35+1=36

electronic configuration of Br: [Ar] 4s² 3d¹⁰ 4p⁵

electronic configuration of Br^-:[Ar]4s² 3d¹⁰ 4p⁶

6) When n = 4, l = n+1 values = 4+1 = 5 values they are 4, 3, 2, 1, 0

7) When l= 2,

ml values = 2l+1 = 2*2 +1 = 4+1 =5. They are -2,-1,0,+1,+2

8) Selenium atomic number is : 34

Electronic configuration of Se:[Ar] 4s² 3d¹⁰ 4p⁴

a) n= 4, (4s² and 4p⁴) so number of electrons sharing this orbit = 2+4 =6

b) l= 2 means it is the orbit d, (3d¹⁰). Number of electrons = 10

1s² 2s² 2p⁶ 3s²

9) Fluorine atomic number is 9. Electronic configuration : 1s² 2s² 2p⁵. For each electron we have to write quantum numbers

a) 1s¹: n=1 l=0 ml=1, ms =+1/2

b) 1s²: n=1 l=0 ml=1, ms =-1/2 (another electron in 1s)

c) 2s¹:n=2 l=0 ml=1, ms =+1/2

d) 2s²:n=2 l=0 ml=1, ms =-1/2

e) 2p¹:n=2 l=1 ml=-1, ms =+1/2

f) 2p²::n=2 l=1 ml=0, ms =+1/2

g) 2p³:n=2 l=1 ml=1, ms =+1/2

h)2p⁴:n=2 l=1 ml=-1, ms =-1/2

i) 2p⁴: n=2 l=1 ml=0, ms =-1/2

10) electronic configuration of Zn: [Ar] 4s² 3d¹⁰

n=4 l=0 ml=1 m_s=+1/2

11) According to periodic trends as we move down the group atomic radius increases.

So La is the answer

12) As we move from bottom to top ionic radius decreases and if any two ions are there in same period, then ion with highest charge will have lowest ionic radius.

So Be²⁺ is the answer

13) Periodic trend: Left to right in period, Ionization energy increases. Down the group ionization energy decreases.

So Br is the answer

14) Cl has the most electron affinity

15) Top to bottom atomic radius increases So Beriliyum has the smallest radius

16) Lesser the atomic radius higher the ionization energy. So Lithium will have highest ionization energy.

17) Down the group electron affinity decreases So Iodine has lowest electron affinity