Identify which sets of quantum numbers are valid for an electron. each set is ordered (n, l,ml, ms).a. 3,1,1,-1/2b. 4,3,1,-1/2c. 2,0,0,-1/2d. 1,3.0,1/2e. 3,2,-1,-1/2f. 3,3,-1,1/2g. 3,2,1,-1h. 1,-1,-1,-1/2i. 3,3,1,1/2j. 3,0,0,1/2k. 4,3,4,-1/2l. 0,2,1,1/2

The valid set of quantum numbers are:

a. 3,1,1,-1/2

b. 4,3,1,-1/2

c. 2,0,0,-1/2

e. 3,2,-1,-1/2

j. 3,0,0,1/2

Explanation:

Quantum numbers (n, l, mℓ, ms) are the set of numbers that describe the state of an electron in an atom.

The four quantum numbers and their rules are:

Therefore,

a. 3,1,1,-1/2: Valid

b. 4,3,1,-1/2: Valid

c. 2,0,0,-1/2: Valid

d. 1,3,0,1/2: NOT Valid

Reason: ℓ ≤ (n-1). Therefore, ℓ can not be greater than n, .

e. 3,2,-1,-1/2: Valid

f. 3,3,-1,1/2: NOT Valid

Reason: ℓ ≤ (n-1). Therefore, ℓ can not be equal to n. (ℓ ≠n)

g. 3,2,1,-1: NOT Valid

Reason: the only allowed values of ms = (- s), (+ s) = (- 1/2), (+ 1/2)

h. 1,-1,-1,-1/2: NOT Valid

Reason: ℓ ≤ (n-1) and n ≥ 1. Therefore, ℓ can not be equal to n and also ℓ can't be negative.

i. 3,3,1,1/2: NOT Valid

Reason: ℓ ≤ (n-1). Therefore, ℓ can not be equal to n. (ℓ ≠n)

j. 3,0,0,1/2: Valid

k. 4,3,4,-1/2: NOT Valid

Reason: mℓ = (- ℓ) to (+ ℓ). Therefore, mℓ can't be greater than ℓ.

l. 0,2,1,1/2: NOT Valid

Reason: n ≥ 1. Therefore, value of n can't be 0 (n≠0)

substituting a hydrogen atom with a halogen in a hydrocarbon:

- the single bond remains single. a is wrong.

- the bond with hydrogen is as strong as with a halogen. c is wrong

- a hydrocarbon is saturated if there is no double bond. we do not know that so d is wrong.

- carbon will always have four valence electrons. e is wrong.

by elimination, the correct answer is b. the boiling point of the new compound increases.