The f-block elements are those in which the differentiating electrons enters the (n-2)f orbitals. There are two series of f-Block elements corresponding to filling of 4f and 5f-orbitals. The series of 4f-orbitals is called lanthanides. Lanthanides show different oxidation states depending upon stability of f⁰, f⁷ and f¹⁴ configurations, though the most common oxidation states is +3. There is a regular decrease in the size of lanthanides ions with increase in atomic number which is known as lanthanides contraction.

Which of the following is not the configuration of lanthanide?

[Xe]4d¹⁴5d¹⁰6s²

The correct answer is option 3. \([Xe] 4d^{10} 5d^{10} 6s^2\).

Let us analyze each given electron configuration to determine which one is not representative of a lanthanide.

1. \([Xe] 4f^{10} 6s^2\)

This is a valid configuration for a lanthanide. It represents the element dysprosium (Dy, Z=66).

2. \([Xe] 4f^1 5d^1 6s^2\)

This is a valid configuration for a lanthanide. It represents the element cerium (Ce, Z=58) in its ground state configuration, where one electron occupies the 4f orbital and one occupies the 5d orbital.

3. \([Xe] 4d^{14} 5d^{10} 6s^2\)

This configuration is incorrect for a lanthanide. Lanthanides involve the filling of the 4f orbitals, not the 4d orbitals. This configuration does not correspond to any lanthanide element but could be representative of an element in the post-lanthanide series, such as an element in the transition metals or another series.

4. \([Xe]4f^7 5d^1 6s^2\)

This is a valid configuration for a lanthanide. It represents the element gadolinium (Gd, Z=64), which has a half-filled 4f orbital configuration plus one electron in the 5d orbital.

Summary: The configuration \([Xe] 4d^{14} 5d^{10} 6s^2\) is not a configuration of a lanthanide. This configuration does not involve the 4f orbitals and thus cannot represent any of the lanthanide elements.