Question

Which of the following has noble gas electronic configuration?
(A) $F{{e}^{2+}}$
(B) Cl$^{-}$
(C) K$^{+}$
(D) Cs$^{+}$
(E) Au$^{+}$

Answer 1

Hint: Noble gases are the most stable elements of the periodic table because they have complete octet. It means that their shells are complete. So they do not need to react to gain stability. All other elements react to achieve their electronic configuration state.

Complete step by step solution:
-We know that atoms can either gain / lose electrons to achieve the electronic configuration state of noble gases or they can share electrons. Ionic bonds are created by gaining/losing electrons and covalent bonds by sharing electrons.
-All the given options are ions and so they can achieve the noble gas configuration by losing or gaining electrons. Let us study all the given ions 1 by 1 to see if they reach noble gas configuration or not.
-The first ion is$F{{e}^{2+}}$. Generally, the configuration of Fe atoms is $\left[ Ar \right]3{{d}^{6}}4{{s}^{2}}$ . The outermost electron of Fe belongs to the 4s shell but the last electron of Fe belongs to the 3d shell. This is because the energy level of 3d is more than that of 4s. This is governed by (n+l) rule. So, when 1 electron is removed from the atom, the configuration will be$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{7}}$ .
Now, if one more electron is removed, there will be 6 electrons left in 3d shell and the configuration will be$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}$ . That is the electronic configuration of argon which is a noble gas.
-The next atom is Cl. It is a group 17 element and its configuration is $\left[ Ne \right]3{{s}^{2}}3{{p}^{5}}$ . When it gains electron, it appears in the outermost shell which is 3p and the configuration becomes that of argon gas which is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}$
-The next element is potassium. It is a group 1 element and its configuration is $\left[ Ar \right]4{{s}^{1}}$ . It loses one electron to become an ion which has the configuration of argon.
-The next element is Cs. It is a group 1 element and its configuration is $\left[ Xe \right]6{{s}^{1}}$ . It loses one electron to become an ion which has the configuration of xenon.
-The next atom is Au. It is group 11 element and its configuration is $\left[ Xe \right]4{{f}^{14}}5{{d}^{10}}6{{s}^{1}}$ . When it loses 1 electron from its outermost 6s shell, the configuration becomes $\left[ Xe \right]4{{f}^{14}}5{{d}^{10}}$. This is not the electronic configuration of any noble gas.
Therefore, except for gold, all other ions in the options will have the configuration of their nearest noble gases.

So the correct options are A, B, C, D.


Note: The filling of electrons in a shell is governed by many principles like the (n+l) rule, Aufbau’s principle, Hund’s rule of multiplicity and Pauli’s exclusion rule. But there are many sets of elements that do not follow these rules and are called exceptions of inorganic chemistry. For instance, elements like Nb, Cu, Cr, Mo, Tc, Ru, Rh, Pd, Ag, Au, Pt do not follow (n+l) rule.