Question

Please help explain both of these sections I’m not understanding

2. Indicate the maximum number of electrons in an atom that can have the following quantum number(s): **these quantum numbers may not apply to any electron, if so, indicate zero** a) n 3 1s b) n = 4, l = 3, mi :-1, m. +1/2 d) n: 3,1:1. mi : 2 3. Write the quantum number(s) associated with each of the following a) the fifth principal energy level b) the 6s subshell c) the last orbital of the 3d subshell d) the first electron added to the 4f subshell

Answer 1

Q2.

Recall Pauli Exclusion principle, which states that no two electrons can have the same quantum numbers. That is, each electron has a specific set of unique quantum numbers.

Now, let us define the quantum numbers:

n = principal quantum number, states the energy level of the electron. This is the principal electron shell. As n increases, the electron gets further and further away. "n" can only have positive integer numbers, such as 1,2,3,4,5,... Avoid negative integers, fractions, decimals and zero.

l =  Orbital Angular Momentum Quantum Number. This determines the "shape" of the orbital. This then makes the angular distribution. Typical values depend directly on "n" value. then l = n-1 always. Note that these must be then positive integers, avoid fractions, decimals. Since n can be 1, then l = 1-1 = 0 can have a zero value.

ml = Magnetic Quantum Number. States the orientation of the electron within the subshell. Therefore, it also depends directly on the "l" value. Note that orientation can be negative as well, the formula:

ml = +/- l values, therefore, 0,+/-1,+/- 2,+/-3 ... Avoid fractions and decimals

ms = the electron spin, note that each set can hold up to two electrons, therefore, we must state each spin (downwards/upwards). It can only have two values and does not depends on other values,

ms can cave only +1/2 or -1/2 spins. avoid all other numbers. also, avoid 0.5 or -0.5

a)

if n = 3, then

2*n^2 = 2*(3^2) = 18 electrons max.

b)

max no. of e- is 1, since only 1 electron can have this configuration

c)

n = 4, l = 2, implies (0,1,2, ... s,p,d)

Max amount for "d" orbital --> 10 electrons

d)

if ml = 2, then, only

ms can change

ms possible values = +1/2 and -1/(2

therefore, total of 2 electrons can be present

e)

n = 2, l = 2

recall that if l = 2, then

s,p,d --> d value can have up to 10 electorns

f)

similar as in d),

ms possible values = +1/2 and -1/(2

therefore, total of 2 electrons can be present

Q3.

a)

5th "n" level --> 5n

b)

6s subshell --> n = 6, l = 0

c)

last orbital of 3d --> n = 3, (s,p,d,f --> l = 2) and last subshell -->ml = (0,1,2)

d)

1st electron in the 4f subshell

n = 4,

l = 3

ml = 0

ms= +1/2