Question

Which ion has the same electronic configuration as $C{{l}^{-}}$?
(a) ${{F}^{-}}$
(b) ${{P}^{+}}$
(c) $S{{c}^{3+}}$
(d) $S{{i}^{4+}}$

Answer 1

Hint: To solve this question you need to know the periodic table. Find the element placed nearest to Chlorine in the periodic table or look at the atomic number of elements and find the element with atomic number as that of $C{{l}^{-}}$.

Complete step by step answer:
If the number of electrons is the same in two elements, its electronic configuration will be the same.
So, let us first find the number of electrons in Chlorine.
Chlorine is a p-block element, belonging to group 17 and period 3.
From this, we can say that it has 7 electrons in its outermost orbital and that it has 3 orbitals.
Atomic number of Chlorine is 17 and its electronic configuration is –
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{5}}$
When Chlorine gains one electron, it becomes $C{{l}^{-}}$. Therefore, its electronic configuration becomes –
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}$
Now, let us look at the options and find the atomic number of each given element.
F – Fluorine
Atomic number = 9
${{F}^{-}}$ has 9 + 1 = 10 electrons
P – Phosphorus
Atomic number = 15
${{P}^{+}}$has 15 – 1 = 14 electrons
Sc – Scandium
Atomic number = 21
$S{{c}^{3+}}$has 21 – 3 = 18 electrons
Si – Silicon
Atomic number = 14
$S{{i}^{4+}}$has 14 – 4 = 10 electrons
As we can see, the number of electrons in $S{{c}^{3+}}$ is the same as that of $C{{l}^{-}}$.
Therefore, the answer is – option (c)

Additional information:
Metals lose electrons in order to have a stable electronic configuration. Therefore, their valency is represented by ‘+’. Whereas, non-metals gain electrons to achieve stable configuration. Therefore, their valency is represented by ‘-’.

Note: The configuration $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}$ is the most stable electronic configuration of Chlorine since its octet is complete. It is also the electronic configuration of the noble gas Argon.