1. For HF and HBr, the partial positive charge on H atom is 0.29 and 0.09, respectively. Use electronegativities (EN) to explain why the partial charge on H in HF is more positive than the partial charge on H in HBr.

The partial charge on H in HF is more positive than the partial charge on H in HBr because the electronegativity of F is more than the electronegativity of Br.

The HF and HBr form ionic bonds. The bonds have been formed by the transfer of electrons between the two atoms.

The more the electronegative the atom, the more charge has been exerted by the molecule onto the adjoining element.

In the periodic table, the electronegativity decreases down the group. The electronegativity of F is much larger than the electronegativity of Br. Therefore the partial charge exerted to H in HF has been more than the charge exerted in the HBr molecule.

For more information about the partial charge, refer to the link:

link

Electro negativity decreases down the group

Explanation:

One of the known periodic trends is that electro negativity decreases down the group but increases across the period. The electro negativity of fluorine is 3.98 on the Pauling's scale while that of bromine is 2.96. Hence the magnitude of charge separation and degree of partial positive charge on hydrogen in HF must be much greater than that of HBr to a large extent due to the significant difference in electronegativity in HF compared to HBr.

it's filtration, but this is an assumption so my apologies if i got this

c. is the answer

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