You should take note that the question is about stability. A compound is stable if it does not easily react with other elements. Hence, its reactivity must be low. As you move down the group, reactivity decreases. So, the halide at the very bottom is the least reactive. It would then be logical that the most stable conjugate base is I⁻ and the least stable conjugate base is the most reactive which is F⁻.
A contains 38.5 g of tin for each 12.3 g of fluorine:
<span>mole ratio: </span>
<span>(38.5 g)/(118.71 g/mol):(12.3 g)/(18.998 g/mol) = 0.324:0.647 = 1:2 ⇒ SnF₂ </span>
<span>B contains 56.5 g of tin for each 36.2 g of fluorine: </span>
<span>mole ratio: </span>
<span>(56.5 g)/(118.71 g/mol):(36.2 g)/(18.998 g/mol) = 0.476:1.905 = 1:4 ⇒ SnF₄
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
</span>
Because the calcium ion donates two electrons to achieve the stable state. So the layer of electrons decrease form 4 layers(2,8,8,2) to 3 layers(2,8,8). Thus, the radius of a calcium ion is smaller than a calcium atom.
PH = −log [H+] = − log [5.4 × 10−3] ≈ 2.27 or 2.3.
or basically 2
I’m pretty sure it’s D. Hope that helps! Correct me if I am wrong.
