How do you find the effective nuclear charge of lanthanides?
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You do not have to understand this throughout your chemistry curriculum (or at least, I never had to), because it is the result of multiple complicated competing trends. The effective nuclear charge in general is not predictable for the Lanthanides.
Due to scalar relativistic effects, the 6s orbital contracts and the 4f and 5d orbitals (slightly) expand, which we call the "lanthanide contraction". What that says for the effective nuclear charge is that it is not a good approximation to try to use Slater's rules to predict Zeff for the Lanthanides.
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Due to scalar relativistic effects, the 6s orbital contracts and the 4f and 5d orbitals (slightly) expand, which we call the "lanthanide contraction". What that says for the effective nuclear charge is that it is not a good approximation to try to use Slater's rules to predict Zeff for the Lanthanides.
hope it help
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