As you go down a group on the periodic table, atomic radii tend to increase because elements with larger atomic numbers have more occupied electron levels which take up more space surrounding the nucleus.
I hope this helps.
In order to solve this, we need to know the standard cell potentials of the half reaction from the given overall reaction.
The half reactions with their standard cell potentials are:
<span>2ClO−3(aq) + 12H+(aq) + 10e- = Cl2(g) + 6H2O(l)
</span><span>E = +1.47
</span>
<span>Br(l) + 2e- = 2Br-
</span><span>E = +1.065
</span>
We solve for the standard emf by subtracting the standard emf of the oxidation from the reducation, so:
1.47 - 1.065 = 0.405 V
Only one molecule have <u>sp2 </u>hybridization on central atom and that is <u>SO₂</u>.
<u>XeCl₂</u> have <u>sp3d </u>hybridization.
<u>OCl₂</u> have <u>sp3 </u>hybridization.
<u>HCN </u>have <u>sp </u>hybridization.
Hybridization is defined as the concept of mixing two atomic orbitals to give rise to a new type of hybridized orbitals.
Hybridization intermixing usually results in the formation of hybrid orbitals having entirely different energies, shapes, etc.
Learn more about hybridization here:- brainly.com/question/22765530
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