Question

Atomic number of platinum is?
A) 68
B) 78
C) 74
D) 64

Answer 1

Atomic number of an element is equal to the number of protons in the nucleus. The platinum is 5d transition metal and it is a group VII element. The 5d transition series has a lanthanide series in it. Thus the general electronic configuration for the 5d transition element is, $\text{ }{{\left[ \text{Xe} \right]}^{54}}\text{ 4}{{\text{f}}^{\text{14}}}5{{\text{d}}^{1-10}}6{{\text{s}}^{1-2}}\text{ }$ .

Complete step by step answer:
The atomic number of an element is equal to the number of protons in the nucleus. In other words, the atomic number is a unique property of an element.
Platinum is found in the middle of the periodic table. It is a group VII element. It is a transition metal. It is a neighbour to the other elements like iridium, osmium, palladium, and ruthenium. Platinum is present in the sixth period.
The sixth-period elements are the transition metal atom. This is called the transition metal as the valence electron or last shell electrons enter the d-subshell. The 5d transition series starts with the lanthanum elements and ends on the mercury. Platinum is present in the 5d transition series.
Since it is a 5d transition series, the valence or the last shell electron enters into the 5d orbital. We are interested in determining the atomic number of the platinum. Platinum is present in group VII, thus its last shell configuration is equal to $\text{ n}{{\text{d}}^{\text{1}-10}}\text{ }\left( \text{n+1} \right){{\text{s}}^{\text{1}-2}}\text{ }$ .
Platinum is a 5d transition series element the general electronic configuration for the platinum is$\text{ 5}{{\text{d}}^{8}}\text{ 6}{{\text{s}}^{2}}\text{ }$. However, one should not forget that the sixth period contains a lanthanide series .The lanthanide series starts with the lanthanum $\text{ La }$to Lutetium $\text{ Lu }$ . These are a total of 15 elements. The .lanthanide series is in group III. Thus, the general configuration for the sixth-period transition metal is modified as $\text{ }\left( \text{n}-\text{1} \right){{\text{f}}^{\text{14}}}\text{ n}{{\text{d}}^{\text{1}-10}}\text{ }\left( \text{n+1} \right){{\text{s}}^{\text{1}-2}}\text{ }$

Therefore, the last shell electronic configuration of the platinum will be modified as $\text{ 4}{{\text{f}}^{\text{14}}}\text{5}{{\text{d}}^{8}}\text{ 6}{{\text{s}}^{2}}\text{ }$Now, the nearest Noble gas is xenon $\text{ Xe }$ , which has the atomic number equal to 54.
Thus, the final electronic configuration of platinum is,
$\text{ Pt = }{{\left[ \text{Xe} \right]}^{54}}\text{4}{{\text{f}}^{\text{14}}}\text{5}{{\text{d}}^{\text{8}}}\text{6}{{\text{s}}^{\text{2}}}\text{ }$
On calculating the number of electrons, we have
$\text{ 54 + 14 + 8 + 2 = 78 }$
In a neutral state, the number of electrons on the elements is equal to the atomic number. Thus the atomic number of platinum is equal to 78.

Hence, (B) is the correct option.

Note: Note that, if we follow the Aufbau principle then the electronic configuration of platinum is$\text{ Pt = }{{\left[ \text{Xe} \right]}^{54}}\text{4}{{\text{f}}^{\text{14}}}\text{5}{{\text{d}}^{\text{8}}}\text{6}{{\text{s}}^{\text{2}}}\text{ }$. But platinum is an exception. A thumb rule states that half-filled orbitals are stabilized thus one electron is shifted from s orbital to d orbitals. These are a relativistic effect. The exceptional electronic configuration is $\text{ Pt = }{{\left[ \text{Xe} \right]}^{54}}\text{4}{{\text{f}}^{\text{14}}}\text{5}{{\text{d}}^{9}}\text{6}{{\text{s}}^{1}}\text{ }$.