Question

26. Identify the element whose last electron would have the following four quantum numbers:
a. 3, 1, -1, +1/2
b. 4, 2, +1, +1/2
c. 2, 1, +1, -1/2
d. 6, 1, 0, -1/2
e. 4, 3, +3, -1/2

Answer 1

we have to identify the element whose last electron would have the following four quantum numbers.

1. 3 , 1 , -1 , +1/2
2. 4 , 2 , +1, +1/2
3. 2 , 1 , +1, -1/2
4. 6 , 1 , 0, -1/2
5. 4, 3, +3, -1/2

solution : principle quantum number shows orbital size and energy level. it is denoted by n.

azimuthal quantum number shows orbital shapes or subshells. like s , p , d , f etc....

magnetic quantum number shows orbital orientation. it is denoted by m.

spin quantum number shows electron spin direction. there are two spins of each electron i.e., +1/2 and -1/2.

1. 3, 1 , - 1, +1/2

n = 3, l = 1 ( means p subshell)

so, 3p¹

now you can write electron configuration : 1s² 2s² 2p⁶ 3s² 3p¹

total number of electrons = 13

so it is Aluminium.

2. 4, 2, +1, +1/2,

n = 4, l = 2 (means d subshell )

so, 4d¹

now electron configuration = [Kr]4d¹5s²

atomic no = 39

so it is Yittrium.

3. 2, 1, +1, +1/2

n = 2, l = 1 (p subshell)

so 2p¹

i.e., 1s² 2s², 2p¹ => atomic no = 5 (boron)

it is Boron.

4. 6, 1, 0, -1/2

n = 6, l = 1 ( p subshell)

so 6p¹

configuration : [Xe]4f¹⁴5d¹⁰6s²6p¹ [atomic no 81]

it is Thallium.

5. 4, 3, +3, -1/2

n = 4, l = 3 (f subshell)

so, 4f¹

configuration : [Xe]4f¹5d¹ [ atomic no 58]

so it is Cerium.

Answer 2

Answer:

answer

Explanation:

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