Question

Electronic configuration of silver atoms in ground state.
(A) $\left[ Kr \right]3{{d}^{10}}4{{s}^{1}}$
(B) $\left[ Xe \right]4{{f}^{14}}5{{d}^{10}}6{{s}^{1}}$
(C) $\left[ Kr \right]4{{d}^{10}}5{{s}^{1}}$
(D) $\left[ Kr \right]4{{d}^{9}}5{{s}^{2}}$

Answer 1

The concept of determining the electronic configuration for an atom in its ground state will help us solve the given illustration. Some exceptions can also interfere in the general electronic configuration as exceptions play an important role in determining the actual facts.
The ground state is the lowest energy state in which electrons occupy the normal energy level assigned to them.

Complete answer:
Let us study about the given element and its electronic configuration;
Silver (Ag):
It is an element with the atomic number 47. It is a d-block element of group 11 and period 5. It is a transition metal.
As, it is d-block element the general electronic configuration is given as $\left( n-1 \right){{d}^{\left( 1-10 \right)}}n{{s}^{\left( 0-2 \right)}}$ .
Ground state-
As we know, ground state is the lowest energy state in which electrons occupy the normal energy level assigned to them.
Thus,
The specific electronic configuration of Ag in its ground state is given as;
The Ag element belongs to the ninth column in d-block thus, the electronic configuration must consist of $4{{d}^{9}}$ .
Electronic configuration- $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}4{{s}^{2}}4{{p}^{6}}4{{d}^{9}}5{{s}^{2}}$
But in this case, there is an exception, silver’s 4d orbital will be completely filled which implies that only one electron will be present in 5s orbital. Thus,
Electronic configuration- $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}4{{s}^{2}}4{{p}^{6}}4{{d}^{10}}5{{s}^{1}}$
Now, the electronic configuration of Kr is given as- $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}4{{s}^{2}}4{{p}^{6}}$
Therefore,
The shorthand electron configuration of the elements can be given by considering the corresponding noble gases as,
Noble gas shorthand electron configuration- $\left[ Kr \right]4{{d}^{10}}5{{s}^{1}}$

Hence, option (C) is correct.

Note:
Do note that students may get confused in option (C) and (D), but remember silver has an exception and hence an electron is shifted to fulfil the d-orbital.
Also, option (B) can never be the answer as silver is a d-block element and that option consists of f-orbital.