Question

Give the symbol of the element of lowest atomic number that has (a) an \(\mathrm{f}\) subshell with 7 electrons. (b) twelve d electrons. (c) three 3 p electrons. (d) a completed p subshell.

Short answer

Question: Identify the element with the lowest atomic number that has (a) an f subshell with 7 electrons, (b) 12 d electrons, (c) three 3p electrons, and (d) a completed p subshell. Answer: (a) Gadolinium (Gd), (b) Invalid as no element can have 12 d electrons in a single d subshell, (c) Phosphorus (P), and (d) Neon (Ne).

Step-by-step solution

(a) Element with 7 f electrons

Begin with the element that introduces the f subshell, which is Cerium (Ce) with an atomic number of 58. It starts the fourth row of the f block (also known as the lanthanides), with an electron configuration of $$[Xe] 4f^1 5d^1 6s^2$$. Counting up to the element with 7 f electrons, we arrive at Gadolinium (Gd) with an electron configuration of $$[Xe] 4f^7 5d^1 6s^2$$. So the element is Gadolinium (Gd).

(b) Element with 12 d electrons

As mentioned in the analysis, no element can have 12 d electrons in a single d subshell as it can only hold up to 10 electrons. Therefore, this option is invalid.

(c) Element with three 3p electrons

To find the element with three 3p electrons, we first look at the element that fills the 2s subshell, which is Neon (Ne) with an electron configuration of $$1s^2 2s^2 2p^6$$. The next element is Sodium (Na) which starts the 3s subshell, followed by Magnesium (Mg) with $$3s^2$$. The next element, Aluminum (Al), has one 3p electron with a configuration of $$[Ne] 3s^2 3p^1$$. Phosphorus (P) is the element after Aluminum in the p block, and has the electron configuration $$[Ne] 3s^2 3p^3$$. Thus, Phosphorus (P) has three 3p electrons.

(d) Element with completed p subshell

We will look for a completed p subshell in the lowest energy level, which is the second energy level. Starting with Helium (He), it has a completed 1s subshell with a configuration of $$1s^2$$. The next element, Lithium (Li), has one 2s electron with a configuration of $$1s^2 2s^1$$ and Beryllium (Be) fills the 2s subshell with $$1s^2 2s^2$$. Then, Boron (B) starts the 2p subshell with one electron, giving it a configuration of $$[He] 2s^2 2p^1$$. Moving along the p block, Neon (Ne) has a completed 2p subshell and an electron configuration of $$[He] 2s^2 2p^6$$. Hence, Neon (Ne) has a completed p subshell.