Give the symbol of the element of lowest atomic number whose ground state has (a) a p electron. (b) four felectrons. (c) a completed d subshell. (d) six s electrons.

Short Answer

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Question: Identify the symbols of the elements with the following electron configurations in their ground state: a) The first element with a p electron Answer: Boron (B) b) The element with four f electrons in the least possible atomic number Answer: Cerium (Ce) c) The element with a completed d subshell in the least possible atomic number Answer: Nickel (Ni) d) The element with six s electrons Answer: Not feasible. There is no element with six s electrons in one subshell.

Step by step solution

01

Identify the first p subshell

We know that the p subshell corresponds to l = 1. The first p subshell is found when the principal quantum number n is 2. Therefore, we're looking for the 2p subshell.
02

Find the element of least atomic number with a p electron

In the 2p subshell, there can be up to 6 electrons (since ml can take the values -1, 0, 1). The atomic number of an element with the first p electron is the number when we start filling electrons in the 2p subshell. For n = 2, we have already filled the 2s subshell (l = 0) with 2 electrons (since there is only ml = 0 for l = 0). Therefore, the element we're looking for has 2 protons (from the 1s subshell) and 1 more proton (from the 2s subshell), which makes it atomic number 5. Thus, the symbol of the element with the first p electron is Boron (B). #b) four f electrons#
03

Identify the first f subshell

We know that the f subshell corresponds to l = 3. The first f subshell is found when the principal quantum number n is 4. Therefore, we're looking for the 4f subshell.
04

Find the element of least atomic number with four f electrons

In the 4f subshell, there can be up to 14 electrons (since ml can take the values -3, -2, -1, 0, 1, 2, 3). The atomic number of an element with four f electrons is the number when we start filling electrons in the 4f subshell and have already filled 4 of the possible 14 spaces. However, before reaching the 4f subshell, we have already filled the 1s, 2s, 2p, 3s, 3p, 4s, 3d, and 4p subshells. Counting the electrons (2 in s, 6 in p, 10 in d subshells), we have 2(1s) + 2(2s) + 6(2p) + 2(3s) + 6(3p) + 2(4s) + 10(3d) + 6(4p) + 4(4f) = 58 electrons. Thus, the element with four 4f electrons has atomic number 58. The symbol of the element is Cerium (Ce). #c) completed d subshell#
05

Identify a completed d subshell

A completed d subshell means that it has all 10 possible electrons (since ml can take the values -2, -1, 0, 1, 2). The lowest possible value of n for a d subshell is 3 (l = 2), so we're looking for the 3d subshell.
06

Find the element with a completed 3d subshell

The atomic number of an element with a completed 3d subshell is the number when we have filled all 10 electrons in the 3d subshell. Before reaching the 3d subshell, we've already filled the 1s, 2s, 2p, 3s, and 3p subshells. Counting the electrons (2 in s, 6 in p subshells), we have 2(1s) + 2(2s) + 6(2p) + 2(3s) + 6(3p) = 18 electrons. Adding 10 more electrons from the filled 3d subshell, we get a total of 28 electrons. Thus, the element with a completed 3d subshell has atomic number 28. The symbol of the element is Nickel (Ni). #d) six s electrons#
07

Analyze the request

Since there are only two electrons possible in any s subshell (l = 0, only 1 value of ml), having six s electrons in one element is not feasible. So, this question does not have an answer.

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