Lewis structure xeh4? the hydrogen atom is not actually electronegative enough to form bonds to xenon. were the xenon-hydrogen bond to exist, what would be the structure of xeh4?

 Xenon has 8 valence electrons, each hydrogen atom has 1, therefore you have 8+4=12 valence electrons. 8 electrons are used to make 4 bonds between the Xe atom and the 4 hydrogen atoms. That leaves 12-8=4 left over electrons, which will occur in two lone pairs of electrons on the center xenon atom. There are 4 bonding pairs and two lone pairs of electrons and they will be arranged in an octahedral arrangement around the center atom, with the lone pairs across the molecule (180 degrees from each other). The resulting molecular structure will be square planar.

8+4=12 valence electrons. 8 electrons are used to make 4 bonds between the Xe atom and the 4 hydrogen atoms. That leaves 12-8=4 left over electrons, which will occur in two lone pairs of electrons on the center xenon atom. There are 4 bonding pairs and two lone pairs of electrons and they will be arranged in an octahedral arrangement around the center atom, with the lone pairs across the molecule . The resulting  structure is square planar.

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Hbr dissociates to form h+ and br-ca(oh)2 dissociates to form ca2+ and 2oh-the h+ and oh- react to form waterassuming the reactants are dissolved in water, the net reaction is h+ + oh- —> h2oif you’re asked to show all components of the reaction, the complete reaction is 2hbr + ca(oh)2 —> cabr2 + 2h2o