Numford
Electronic configuration

Platinum (Pt)

[Xe] 4f¹⁴ 5d⁹ 6s¹, an exception. 2 unpaired electrons, period 6, group 10, d block.

Element or ion

Type a symbol, a name or an atomic number, add a charge such as Fe3+ or Cl-, or tap an element in the table. Orbitals fill from m = −l up, spin +½ first.

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H1He2Li3Be4B5C6N7O8F9Ne10Na11Mg12Al13Si14P15S16Cl17Ar18K19Ca20Sc21Ti22V23Cr24Mn25Fe26Co27Ni28Cu29Zn30Ga31Ge32As33Se34Br35Kr36Rb37Sr38Y39Zr40Nb41Mo42Tc43Ru44Rh45Pd46Ag47Cd48In49Sn50Sb51Te52I53Xe54Cs55Ba56Hf72Ta73W74Re75Os76Ir77Pt78Au79Hg80Tl81Pb82Bi83Po84At85Rn86

PtPlatinum, Z = 78, 195.084 u

[Xe] 4f14 5d9 6s1

In full: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 4f¹⁴ 5s² 5p⁶ 5d⁹ 6s¹

[Xe]4f5d6s
  • Blockd block
  • Period6
  • Group10
  • Valence10
  • Unpaired2
  • MagneticPara

Last electron 5d

  • n5shell
  • l2d subshell
  • m+1orbital
  • s−½spin down
Why Pt is an exception

Plain filling would give [Xe] 4f¹⁴ 5d⁸ 6s². For heavier atoms the (n − 1)d, (n − 2)f and ns levels lie very close together, so one electron sits in d. These are the configurations found by experiment and printed in the NCERT tables.

Working

Step by step

  1. Platinum has 78 electrons.
  2. Fill in (n + l) order: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d⁸.
  3. That gives [Xe] 4f¹⁴ 5d⁸ 6s², but the accepted configuration is [Xe] 4f¹⁴ 5d⁹ 6s¹. For heavier atoms the (n − 1)d, (n − 2)f and ns levels lie very close together, so one electron sits in d. These are the configurations found by experiment and printed in the NCERT tables.
  4. Write the subshells in order of n: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 4f¹⁴ 5s² 5p⁶ 5d⁹ 6s¹.
  5. Replace the xenon core with [Xe]: [Xe] 4f¹⁴ 5d⁹ 6s¹.
At a glance

Platinum (Pt): the numbers

Atomic number78
Atomic mass195.084 u
Period6
Group10
Blockd block
Valence electrons10
Unpaired electrons2
Magneticparamagnetic
Last electron5d: n = 5, l = 2, m = +1, s = −1/2
Questions

Platinum (Pt): quick answers

What is the electronic configuration of platinum?

Platinum has 78 electrons: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 4f¹⁴ 5s² 5p⁶ 5d⁹ 6s¹, or [Xe] 4f¹⁴ 5d⁹ 6s¹ in short.

Why is platinum [Xe] 4f¹⁴ 5d⁹ 6s¹ and not [Xe] 4f¹⁴ 5d⁸ 6s²?

For heavier atoms the (n − 1)d, (n − 2)f and ns levels lie very close together, so one electron sits in d. These are the configurations found by experiment and printed in the NCERT tables.

What are the quantum numbers of the last electron of platinum?

The last electron goes into 5d: n = 5, l = 2, m = +1, s = −1/2. Orbitals fill from m = −l up, spin +½ first.

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