1. Identify the following three elements. (Capitalization counts!) 2. Give the e

Question

1. Identify the following three elements.
(Capitalization counts!)

2. Give the electron configurations for the following atoms. Do not use the noble gas notation. Write out the complete electron configuration. Be sure to input your notation as the example shows and to leave spaces between each orbital designation.

Example: Mg 1s2 2s2 2p6 3s2

3. How many electrons in an atom can have the following sets of quantum numbers? Enter the maximum number of electrons into the table.

Element Symbol The ground-state electron configuration is [Ar]4s23d104p2. __________ The ground-state electron configuration is [Kr]5s24d105p4. __________ An excited state of this element has the electron configuration 1s22s22p53s13p1. __________

Explanation / Answer

Question 1

Part (a) -- This is Ge. You can count the electrons across the periodic table. Two electrons into the p block is Ge.

Part (b) -- This is Te. Four electrons into the p block in period five is Te.

Part (c) -- This must be Mg. You can promote an electron from the 3s orbital to the 3p orbital; you are simply changing the quantum number l by one, which is allowed.

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Question 2

This is mostly simple electron counting. You go across the periodic table for these elements. Feel free to check the answers online:

He | 1s2

P |  1s22s22p63s23p3

Be | 1s22s2

B | 1s22s22p1

Na | 1s22s22p63s1

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Question 3

Recall the rules for the quantum numbers:

> n can be any positive integer greater than 0. (e.g. 1, 2, 3, 4...)

> l can be any positive integer which fits the criteria n - 1. It can be zero. It represents the type of orbital as well (s, p, d, f, etc).

> ml can be any number from -l to +l.

> msis the spin quantum number and can only be +1/2 or -1/2.

Thus we look to see which of these quantum numbers is possible:

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Part (a)

If n = 4 and l = 3 (a f orbital), then:

ml can be -3, -2, -1, 0, 1, 2, 3.

Each of the ml numbers can hold two electrons: one with a +1/2 spin and the other with -1/2 spin. That gives us 7*2 = 14 electrons. The seven is the number of ml values we have; 2 electrons per ml value.

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Part (b)

If an electron has a quantum number of n =1, then the only l value is 0.

If l = 0, then the only ml value is 0.

ms can be +1/2 and -1/2 for an ml value of 0. That means two electrons are possible.

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Part (c)

Only one electron can have this set of quantum numbers. The give-away is the specifity of ms, which can only have two values.

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Part (d)

Only one electron can have this set of quantum numbers for the same reason as (c).

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