An element with the electronic configuration $\left[ Xe \right]4{{f}^{14}}5{{d}^{7}}6{{s}^{2}}$ is:
(A) An alkaline earth element
(B) a transition element
(C) an inert gas
(D) a rare earth

Hint: Think about the elements which are known by the terms mentioned in the options. We have to identify the element from the given electronic configuration. Look at the electronic configuration and find out in which orbital the valence electrons are present. Find out how many electrons are present in the valence shell to get the answer.

Complete step by step answer:
- First, let’s get a brief idea about elements present in the given options.

(A) An alkaline earth element – Alkaline earth elements are Beryllium, Magnesium, Calcium, Strontium, Barium and Radium. The second group of s-block in the periodic table consists of alkaline earth metals. They have the valence shell electronic configuration, $n{{s}^{2}}$ where n is the valence shell.

(B) A transition element – Transition elements are the elements present in the d-block of periodic table. Some transition elements are Titanium, Chromium, Copper, Nickel, Manganese, etc. They have the valence shell electronic configuration, $(n-1){{d}^{1-10}}n{{s}^{2}}$ where n is the valence shell and (n-1) is the penultimate shell.

(C) An inert gas – Inert gases are noble gas elements like Helium, Argon, Neon, Krypton, Xenon and Radon. They belong to the last group in the periodic table. They have the valence shell electronic configuration, $n{{s}^{2}}n{{p}^{6}}$ where n is the valence shell.

(D) A rare earth – Rare earth elements are a group of 17 elements which are present in very low concentration levels on ground. Rare earth elements include all the lanthanides and Scandium and Yttrium.

- Now, let’s look at the given electronic configuration, $\left[ Xe \right]4{{f}^{14}}5{{d}^{7}}6{{s}^{2}}$
- We can see that the 5d orbital is incompletely filled. This indicates that the given element belongs to d-block and not s-block. So, this element is not an alkaline earth metal.
- It has the valence shell 6s which means the element belongs to the sixth period. We know that lanthanides are present in the sixth period. But the given element has completely filled 4f orbitals and 3 electrons left to complete 5d orbitals, which means, this element is not a lanthanide and thus, not a rare earth element.
- This element has a vacant 6p orbital, so it is not an inert gas element.
- So, three out of four options are eliminated. Therefore, our answer should be the transition element which is correct because d-block elements are transition elements.
- To find the name of an element present in d-block, let’s calculate its atomic number.
- Xenon is 54. So, $54+14+7+2=77$. So, the given electronic configuration is of Iridium.
- Therefore, an element with the electronic configuration $\left[ Xe \right]4{{f}^{14}}5{{d}^{7}}6{{s}^{2}}$ is a transition element.
So, the correct answer is “Option B”.

Note: Remember the names of some characteristic groups in the periodic table like alkali metals, transition metals, chalcogens, halogens, inert gases or noble gases, lanthanides, actinides and rare earth elements. For such questions, try breaking down the given electronic configuration and try the method of elimination to get the correct answer.