Question

Please help me understand how to do this? Im very confused on how to do this assignment.

The Problem: Table of Isotopes Mass number Isotopic mass (u) Natural abundance (%) 54 53.9396 5.845 56 55.9349 91.754 57 56.9354 2.119 58 57.9333 0.282 1. Use the table of isotopes, masses, and abundances to find the atomic weight of the element, and then use atomic weight to identify the element. 2. Write an elemental symbol for each isotope in 2x format. Indicate the number of protons, neutrons, and electrons in each isotope. You may assume you have atoms and not ions when identifying the number of electrons. 3. Write the complete electron configuration for the element. Why don't you write an electron configuration for each isotope? 4. Fill in the orbital diagram for the element. 5. Write an allowed set (just 4 numbers) of quantum numbers (principle (n), orbital angular momentum (1), magnetic quantum number (mi), and spin quantum number (ms) for the 23rd electron in the element

Answer 1

1) The atomic mass of the element is the weighted average of all the isotopes

               = {$\Sigma$ (isotope mass x percentage abundance) } / 100

               = [ (53.9396 x 5.845 ) + (55.9349 x 91.754) + ( 56.9354 x 2.119) + ( 57.933 x 0.282) ] / 100

                = [315.276962 + 5132.2508146 + 120. 6461126 + 16.337106 ] / 100

                = 55.8451

                 = 55.85

Thus the atomic weight of the unknown element is 55.85 g / mol

2) The atomic weight of 55.85 is corresponds to the element Iron.

    Thus the unknown element is Iron ( Fe )

3)

All the isotopes has the same atomic number. The mass number varies.

Number of protons = number of electrons = Atomic number = 26 for all the isotopes.

The number of neutrons = mass number - number of protons

                                   = mass number - atomic number

The results are tablulated.

Symbal At.no/MASS NO # Z n A 54 26 54 26/28/26 fe 26 56fe 26 56 26 30 26 26 575e 26 57 26 31 26 26 s8fe 26 58 26 3226 26

4)

       The complete electronic configuration is given by

               

Fei 26 Z - 6 2 2 2 635 ²3 p 6 45 3d IS 25

since the number of electrons are the same for all the isotopes, their electronic configurations are identical. Hence there is no need of writing electronic configuration for each isotope.

5)

The orbital diagram is as follows:

за 11 |1 11|1 4S or ЗР И» (Ф(ty 38 у ар (му о 10 2S (1ъ 13 ИИ

6) The position of the 23rd electron is in the 3d orbital.because 20 electrons occupy upto 4S orbital. The next 6 electrons are in the 3d orbital.

Fei Z S 26 ܝ2_6 6 2 633? 23°2p IS 3 249 3d P 20 67

                    Hence n = 3

              since l = n-1 = 3-1 = 2

The detailed positions of the 6 electrons in the 3d orbital are given below:

21 2223 2日 25 Lab 十 十%。 که [2%。 VIT 1|1|1 Conces 20 -2 +2 4)

               m(l) has the values   +2, +1, 0 , -1,-2. The 23rd electron occupies the middle of the 3d orbital and hence its m(l) value is 0

                It is an unpaired electron with positive spin. Hence m(s) = +1/2

Thus the four quantum numbers of the 23rd electron are

              n = 3, l = 2 , m(l) = 0 and m(s) = +1/2

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