Question

Element A and B with their respective electronic configurations $3{d^{10}}4{s^1}$ and $4{d^{10}}5{s^1}$ in their outermost shell are:
(A) Both are coinage metals
(B) Both are non-metals
(C) A is non-metal and B is coinage metal
(D) A is coinage metal and B is non-metal

Answer 1

Hint: Coinage metals are those metals used to make coins in early age. These metals belong to group $11$ hence coinage metals are d block elements. Coinage metals include copper, silver and gold. The last electron enters a particular sub-shell, then it is said to belong to that block. If the last electron enters the d subshell, then it belongs to the d block.

Complete Step by Step Solution:
Element A has electronic configuration $3{d^{10}}4{s^1}$. It appears that the last electron enters the sub-shell. Hence it should belong to s block. But it is not true.

According to the Aufbau principle, electrons are filled according to the increasing energy order of atomic orbitals. $4s$is of lower energy than $3d$ according to (n+l) rule. Hence $4s$ is filled first. So electronic configuration is $3{d^9}4{s^2}$. But $3{d^{10}}4{s^1}$is more stable than $3{d^9}4{s^2}$because of extra stability associated with an exactly half-filled orbital.
Also,$3{d^{10}}$is fully filled which is more stable than $3{d^9}$.

Copper has \[29\] atomic number. Thus, $3{d^{10}}4{s^1}$is the electronic configuration of copper.

Silver has $47$ atomic number of 47. $4{d^{10}}5{s^1}$is the electronic configuration of silver.
Both are coinage metals.
Hence, the correct option is A.

Note: Nowadays, coins are made from bronze, zinc and many other metals. But only silver, copper and gold are called coinage as these coins were found in the coinage period.
Thus, the general configuration of coinage metals is $n{d^{10}}(n + 1){s^1}$.