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Select the larger member of each pair. (a) $\mathrm{Ca}$ and $\mathrm{Ca}^{2+}$ (b) $\mathrm{Sr}$ and $\mathrm{Sr}^{2+}$ (c) $\mathrm{F}$ and $\mathrm{F}^{-}$ (d) $\mathrm{Cu}^{+}$ and $\mathrm{Cu}^{2+}$

In this problem, in this problem, SIH three all three and and and C C L t. D and c c l T. Are the molecules are the molecules in this problem as I edged three whole three N and CCL three are the molecules involves B, bye, deep I bonding. So according to the option, often a n C are correct.

Hi friends answer of this problem is seek because the resultant of addition of Quebec to never be less than he might not speak and never be greater than a plus B. So answer. She must be correct that self. Thanks.

So using the fact that cat islands are going to be smaller um than their um parent atom because you lose an energy level of worth of electrons. Then the larger member of the pair, when we have Canadians involved is going to be the Adams, calcium rather than calcium two plus. Strong team rather than the strong team two plus. And then when we're dealing with copper, um the more electrons you take away, the greater the effect of like protons pulling the remaining electrons a higher proton to electron ratio. And the smaller the atom. That means someone with less charge is going to be bigger. Mhm. And then when we consider an ions, you're adding electrons that generates more repulsion. And then the item, the ion can be more spread out in space. So the iron will be larger than the atom. So F minus is the larger out of F and F minus.