Question

What is the correct order of size of Pd , Pt , Ni ?

Answer 1

When we will move down the group in the periodic table, the atomic size will increase as one more shell will be added to the atom. $Ni$ , $Pd$ and $Pt$ are all transition elements and all of them belongs to the same group i.e. Group $10$ . By writing the electronic configuration of each atom using the Aufbau principle, we can easily determine the correct order of size of Pd, $Pt$ , and $Ni$ .

Complete Step by step answer: We know that $Ni$ lies in the ${4^{th}}$ period, $Pd$ lies in the ${5^{th}}$ period and $Pt$ lies in the ${6^{th}}$ period.
Now we will write the atomic number of each element: -
Atomic number of $Ni = 28$
Atomic number of $Pd = 46$
Atomic number of $Pt = 78$
By using Aufbau principle which states that in the ground state of an atom, atomic orbitals are filled by electrons in the order of their increasing energies, we will write the electronic configuration: -
$Ni = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}4{p^6}$
$Pd = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}4{p^6}5{s^0}4{d^{10}}$
$Pt = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^{10}}4{s^2}4{p^6}4{d^{10}}5{s^2}5{p^6}4{f^{14}}5{d^9}6{s^1}$
Now here we see that one shell is being added in the d-orbital of the element as we are moving from $Ni \to Pd \to Pt$ . $Ni$ is $3d$ , $Pd$ is $4d$ while $Pt$ is $5d$ which means that the atomic size is increasing on moving down the group.

So, the correct order of size will be: - $Ni < Pd < Pt$

Additional information: Nickel is silvery white, tough, and harder than iron, is widely familiar because of its use in coinage but is more important either as the pure metal or in the form of alloys for its many domestic and industrial applications. Palladium is a grey-white metal, which is extremely ductile and used especially as a catalyst. Platinum is a very heavy, silver-white metal which is soft and ductile and has a high melting point and good resistance to corrosion and chemical attack.

Note: We can also write the configuration of $Pd$ and $Pt$ like: -
$Pd = [Kr]4{d^{10}} \\\ Pt = [Xe]4{f^{14}}5{d^{10}}6{s^2} \\\$
We do not write the configuration of $Pd$ as $[Kr]5{s^2}4{d^8}$ but as $[Kr]5{s^0}4{d^{10}}$ since we know that fully filled d-orbitals are more stable than partially filled orbitals, so the two electrons from $5{s^2}$ move to $4{d^8}$ to make it $5{s^0}$ and $4{d^{10}}$ .