Give the electron configurations for the following ions: p, p. ar, and te plain what interaction is in the hydrogen fluoride (hf) and hydrogen chloride (hci. which one do you espect to have a higher bolling emperature? why?

Explanation:

Atomic number of P = 15

Common oxidation state of P = -3, +3 and +5

Electronic configuration of P^3+:

Electronic configuration of P^3-:

Electronic configuration of P^5+:

Atomic number of Ar = 18

Argon has stable octet, so it generally does not exist as ions.

Electronic configuration of Ar:

Atomic number of Te = 52

Common oxidation states of Te = -2, +2, +4, +6

Electronic configuration of Te^2-:

Electronic configuration of Te^2+:

Electronic configuration of Te^4+:

Electronic configuration of Te^6+:

Boiling point of HF is more as compared to HCl

Reason:

Molecular weight of HF is low, but is boiling point is high because of presence of hydrogen bonding. Whereas in case of HCl, its molecular weight is high but has only weak van der Waals intercations. As hydrogen bonding is stronger than van der Waals interactions, therefore, boiling point of HF is more.

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the answer for sodium is 0.186

the answer for rubidium is 0.251