Elektron Konfiguration, Orbitaler, og Kvantetal -converted PDF

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Electron Configuration, Orbitals, and Quantum Numbers

Orbital is simply the probable location to find electrons Quantum Numbers and their meanings:

n

Energy level, principal quantum number

n=3 n=2

n = energy level, principal quantum number that describes the size and energy of that orbital - as n increases, the distance from the nucleus increases as well - electrons that are further away from the nucleus exist at higher energy levels l

n=1

N

Shape, angular momentum quantum number -

When l = 0, s orbital

-

When l = 1, p orbital

-

When l = 2, d orbital

-

When l = 3, f orbital

n value always equal to the number of sublevels so, if n = 6, there are 6 sublevels

ml

Orientation, magnetic quantum number ml = describes the orientation of the orbital relative to other similar orbitals in that atom

ms

Magnetic electron spin Only two possibilities: either spin on clockwise direction or counterclockwise ms = +1/2, or ms = -1/2 When a spin is facing up, the electron spin is positive half When a spin is facing down, the electron spin is negative half

Example 1: Give the quantum numbers (n, l, ml, ms) that describes the 2p5 electron n: The principal quantum number corresponds to the coefficient number, so in this case, it is 2 -

indicating that this electron is in the second energy level

l: sublevel p orbital, in this case, it is 1 (Remember that every s orbital can hold max. 2 electrons, p can hold 6, d can hold 10, and f can hold 14) -

remember that s orbital has a sublevel of 0

-

remember that p orbital has a sublevel of 1

-

remember that d orbital has a sublevel of 2

-

remember that f orbital has a sublevel of 3

ml: since l=1, ml will vary between -1 and 1 (-1,0,1) -

focusing on the 5th electron on the p sublevel:

-

the fifth electron is found in the middle (0), so the m1 value is 0

-1

0

1

ms: since the arrow of the fifth electron is facing downwards in m l, the spin is going to be -1/2 Thus, the quantum numbers that describes the 2p5 electron are: (2,1,0, -1/2) Example 2: Give the quantum numbers (n, l, ml, ms) that describes the 3d5 electron n: The principal quantum number corresponds to the coefficient number, so in this case, it is 3 l: sublevel d orbital, in this case, it is 2 ml: since l=2, ml will vary between -2 and 2 (-2,-1,0,1,2) -

focusing on the 5th electron on the d sublevel:

-2

-

-1

0

1

2

the fifth electron is found in the right (2), so the m1 value is 2

ms: since the arrow of the fifth electron is facing upwards in ml, the spin is going to be +1/2 Thus, the quantum numbers that describes the 3d5 electron are: (3,2,2, +1/2)

Example 3: Give the quantum numbers (n, l, ml, ms) of the element Fluorine (F) Step 1: Focus on the last valence electron of the element, in this case 2p 5 electron n: The principal quantum number corresponds to the coefficient number, so in this case, it is 2 l: sublevel p orbital, in this case, it is 1 ml: since l=1, ml will vary between -1 and 1 (-1,0,1) -

focusing on the 5th electron on the p sublevel:

-

the fifth electron is found in the middle (0), so the m1 value is 0

-1

0

1

ms: since the arrow of the fifth electron is facing downwards in m l, the spin is going to be -1/2 Thus, the quantum numbers that describes the 2p5 electron are: (2,1,0, -1/2) Example 4: Give the quantum numbers (n, l, ml, ms) of the element Iron (Fe)

Step 1: Focus on the last valence electron of the element, in this case 3d 6 electron n: The principal quantum number corresponds to the coefficient number, so in this case, it is 3 l: sublevel d orbital, in this case, it is 2 ml: since l=2, ml will vary between -2 and 2 (-2,-1,0,1,2) focusing on the 6th electron on the d sublevel:

-

-2

-

-1

0

1

2

the sixth electron is found in the left (-2), so the m1 value is -2

ms: since the arrow of the sixth electron is facing downwards in ml, the spin is going to be -1/2 Thus, the quantum numbers that describes the 3d6 electron are: (3,2, -2, -1/2) Example 5: Give the quantum numbers (n, l, ml, ms) that describes the 4f5 electron n: The principal quantum number corresponds to the coefficient number, so in this case, it is 4 l: sublevel f orbital, in this case, it is 3 ml: since l=3, ml will vary between -3 and 3 (-3,-2,-1,0,1,2,3) -

focusing on the 5th electron on the f sublevel:

-

the fifth electron is found in the right (1), so the m1 value is 1

-3

-2

-1

0

1

2

3

ms: since the arrow of the fifth electron is facing upwards in ml, the spin is going to be +1/2 Thus, the quantum numbers that describes the 4f5 electron are: (4,3,1, +1/2) General Steps to follow to determine quantum number if given an electron or element: 1.

If given an element, focus on the last valence electron from the electron configuration

2.

n is always the coefficient number, that you see in front of the last valence electron configuration

3.

l is always s orbital = 0, p orbital = 1, d orbital = 2, and f orbital = 3

4.

m1 always varies between -l and l, so fx. if l=2, ml will vary from -2 to 2 (-2,-1,0,1,2)

5.

ms is always given by the direction of the arrow the last electron lands on. Remember when adding arrows (from the superscript of the electron configuration), always fill 1 upper arrow until all boxes are checked then begin with downward arrow. Note where the last arrow is. If arrow is facing up, the spin is positive (+1/2), if arrow is facing down, the spin is negative (-1/2).

Example 6: Given the unique quantum numbers (n = 3, l = 2, ml = 1, ms = -1/2), determine the electron -

Since n = 3 and l = 2, we know that it is a 3d orbital which has 5 shells

To determine how many electrons there are we need to check the ml and ms value -

Since ml = 1, ms = -1/2

-

We know that the last electron landed in 1 shell among -2, -1, 0, 1, 2 and since the spin is negative, we also know that it is a downward facing arrow

-

We set up the shells varying from -2 to 2 (l value) and start by adding arrows facing upwards. When the shells are full, start with downward facing arrows and note where the last electron lands

-2

-1

0

1

2

-

In this case, the last electron lands on 1 so now we count how many electrons there are in total (9)

-

Now we can write out electron configuration for the given unique quantum numbers as 3d9

How to write electron configurations, and fill up orbital diagram: Energy levels and sublevels: 1s 2s 3s 4s

2p 3p 4p

3d 4d

4f

Note: s orbital can hold up to max. 2 electrons p orbital can hold up to max. 6 electrons d orbital can hold up to max. 10 electrons f orbital can hold up to max. 14 electrons

Example: Write down the electron configuration of the element Phosphorus (P): We know that phosphorus holds a total of 15 electrons, so we start by filling up the shells

Note: PE = Potential Energy = increases as you go up the shells Note: When you already write down the electron configuration and then begin filling up the shells, do not fill all arrows up and then all arrows down like you would when determining quantum numbers. Instead, for example, if you have 1s2, it means that in 1s orbital there is 1 arrow facing up, one arrow facing down. Only when you reach the final orbital, start by adding the arrows up, then arrows down, count the number of total electrons of the element.

-

Remember that when writing electron configurations, refer to the before-group noble gas and then write the remaining.

-

For Example, the electron configuration of Chlorine (Cl) can be given as [Ne]3s23p5 or as 1s22s22p63s23p5

How to write electron configurations for ions: Step 1: Write the electron configuration of the element Step 2: Determine whether the ion loses or gains electrons -

If it is a cation, it has a plus charge, and it means that it loses electrons

-

If it is an anion, it has a negative charge, and it means that it gains electrons

Step 3: Always remove the electrons from the highest energy level in the electron configuration Step 4: Always add the electrons to the highest energy level in the electron configuration

Determining between Paramagnetic and Diamagnetic: Paramagnetic substance contains unpaired electrons therefore they are attracted to an external magnetic field Diamagnetic substance contains no unpaired electrons and is weakly repelled by a magnetic field Example 1: Write the electron configuration for Mg2+ Mg = 1s22s22p63s2 Mg2+ = 1s22s22p6 -

Is a cation, loses 2 electrons, the highest energy level is in third shell, 3s2, so we remove 2 electrons from there

Example 2: Write the electron configuration for P3P = 1s22s22p63s23p3 (Paramagnetic) P3- = 1s22s22p63s23p6 -

Is an anion, gains 3 electrons, highest energy level is in third shell, 3p3, so we add 3 more electrons to it

Example 3: Write the electron configuration for Fe2+ and Fe3+ Fe = 1s22s22p63s23p64s23d6 Fe2+ = 1s22s22p63s23p63d6 -

Loses 2 electrons from 4s shell because it has the highest energy level

3+

Fe = 1s22s22p63s23p63d5 -

Loses 3 electrons so 2 electrons from the 4s shell because it has the highest energy level, then 1 electron from the 3d shell because after 4s shell is removed, the 3d shell has the highest energy level

Example 4: Write the electron configuration for S2S = 1s22s22p63s23p4 S2- = 1s22s22p63s23p6

Example 5: Write the electron configuration for N3N = 1s22s22p3 N3- = 1s22s22p6

When writing electron configurations (the normal method), there are exceptions to these following elements: -

Cr = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵ = [Ar] 4s¹ 3d⁵ (most paramagnetic)

-

Cu = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d10 = [Ar] 4s¹ 3d10

-

Mo = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹ 4d⁵ = [Kr] 5s¹ 4d⁵

-

Ag = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹ 4d¹⁰ = [Kr] 5s¹ 4d10

-

Au = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s¹ 4f¹⁴ 5d¹⁰ = [Xe] 6s¹ 4f¹⁴ 5d¹⁰...