Consider the isoelectronic ions f− and na+. part a which ion is smaller? f− na+ submitmy answersgive up correct part b using the equation zeff=z−s and assuming that core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant, s, calculate zeff for the 2p electrons in both ions. express your answers as integers. enter your answers numerically separated by a comma. zeff f−, zeff na+ = 7,9 e− submitmy answersgive up correct part c repeat this calculation using slater's rules to estimate the screening constant, s. express your answers using two decimal places. enter your answers numerically separated by a comma.

Zeff is a good nuclear charge and is calculated by the formula:

Zeff = Z - S

Z refers to the quantity of protons and S is that the Slater constant.

Cations have 11 protons, the quantity of core electrons in sodium-ion is 2.

If the core electron accounts for 1.00 and also the negatron accounts for 0.00 then S = 2 for both ions.

The F⁻ ion has 9 protons, the quantity of core electrons within the F⁻ ion is 2. Therefore, for F⁻: Zeff = 9 - 2 = 7

On the premise of Slater's rule, for s and p orbitals, electrons within the same shell contribute 0.35 and electrons within the n-1 shell contribute 0.85.

S = (0.85 × 2) + (0.35 × 7)

= 4.15

For Na⁺, Zeff = 11 - 4.15 = 6.85

For F⁻, Zeff = 9 - 4.15 = 4.85

Thus, Zeff for Na⁺ and F⁻ calculated per Slater's rules is 6.85 and 4.85, respectively.

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Grade: College

Subject: Chemistry

Keyword: Zeff, cations, core

they include patterns; cause and effect; scale, proportion, and quantity; systems and system models; energy and matter; structure and function; and stability and change.

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hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.