
What is the shorthand electronic configuration of Pt?
Answer
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Hint : Pt is the symbol of the element Platinum of the periodic table. Its atomic number is \[78\]. Platinum belongs to the platinum group of elements, which is part of the periodic table's group $10$. It has six isotopes that exist naturally. It is the element of d-block.
Complete Step By Step Answer:
The electron configuration in shorthand begins with the symbol of the noble gas from the previous time, followed by the additional electron configuration for the given part.
For ex – Since, we know that the atomic number of sodium is $10$, so sodium has one more $3s$ electron than the noble gas neon, its shorthand notation is $[Ne]3s$.
Electronic configuration of any element is done on the basis of Aufbau principle. It states that the lowest energy orbitals are occupied first, followed by the higher energy orbitals.
Since, there are $78$ electrons in the platinum element, then all these electrons will be in the atomic orbital in the following order - $1s,2s,2p,3s,3p,4s,3d,4p,5s,4p,5d,6p,7s,.....$
Here, the maximum number of electrons which can be filled in the following orbitals are listed below
According to the rule, firstly $2$ electrons of platinum will be filled in $1s$ orbital, then next 2 electrons will be filled in $2s$ orbital and so on.
Like this, after filling the electrons in the orbitals, the electronic configuration of platinum is \[\left[ {{\mathbf{Xe}}} \right]{\mathbf{4}}{{\mathbf{f}}^{14}}{\mathbf{5}}{{\mathbf{d}}^9}{\mathbf{6}}{{\mathbf{s}}^1}\].
Note :
The real answer is $[Xe]4{f^{14}}5{d^8}6{s^2}$ but due to exception in platinum, the answer is given above. This is because the rule of thumb is applied here which means that half-filled orbitals are more stable.
Complete Step By Step Answer:
The electron configuration in shorthand begins with the symbol of the noble gas from the previous time, followed by the additional electron configuration for the given part.
For ex – Since, we know that the atomic number of sodium is $10$, so sodium has one more $3s$ electron than the noble gas neon, its shorthand notation is $[Ne]3s$.
Electronic configuration of any element is done on the basis of Aufbau principle. It states that the lowest energy orbitals are occupied first, followed by the higher energy orbitals.
Since, there are $78$ electrons in the platinum element, then all these electrons will be in the atomic orbital in the following order - $1s,2s,2p,3s,3p,4s,3d,4p,5s,4p,5d,6p,7s,.....$
Here, the maximum number of electrons which can be filled in the following orbitals are listed below
| Orbital | Maximum number of electrons |
| s | $2$ |
| p | $8$ |
| d | $18$ |
| f | $32$ |
According to the rule, firstly $2$ electrons of platinum will be filled in $1s$ orbital, then next 2 electrons will be filled in $2s$ orbital and so on.
Like this, after filling the electrons in the orbitals, the electronic configuration of platinum is \[\left[ {{\mathbf{Xe}}} \right]{\mathbf{4}}{{\mathbf{f}}^{14}}{\mathbf{5}}{{\mathbf{d}}^9}{\mathbf{6}}{{\mathbf{s}}^1}\].
Note :
The real answer is $[Xe]4{f^{14}}5{d^8}6{s^2}$ but due to exception in platinum, the answer is given above. This is because the rule of thumb is applied here which means that half-filled orbitals are more stable.
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