states of matter
identifying a molecule from its electrostatic potential map
this sketch of a neutral molecule is shaded red or blue wherever the electrostatic potential at the molecule’s surface isn’t zero. what could the chemical formula of the molecule be?
electrostatic potential map possible chemical formula (check all that apply?)
image of electrostatic potential map hi
hbr
icl
h₂
co

Question

states of matter
identifying a molecule from its electrostatic potential map
this sketch of a neutral molecule is shaded red or blue wherever the electrostatic potential at the molecule’s surface isn’t zero. what could the chemical formula of the molecule be?
electrostatic potential map	possible chemical formula (check all that apply?)
image of electrostatic potential map	hi
	hbr
	icl
	h₂
	co

Accepted Answer

# Brief Explanations:
1. Analyze the electrostatic potential map: The map shows a polar molecule with a region of negative (red - like) and positive (blue - like) charge, indicating a dipole (polar covalent bond or polar molecule with a dipole moment).
2. Analyze each option:
   - $HI$: $H - I$ is a polar covalent bond (iodine is more electronegative than hydrogen, but the difference is less than in some cases, but the molecule is polar). The electrostatic potential map would show a dipole (partial - + on H, partial - on I), which matches the map's polarity.
   - $HBr$: $H - Br$ is polar, but the electronegativity difference between H and Br is different from H and I. However, let's check other options too.
   - $ICl$: $I - Cl$ is polar (Cl is more electronegative than I), but the shape and the dipole direction might not match the given map as well as $HI$ or $CO$? Wait, no, $CO$ is a polar molecule with a dipole (C is less electronegative than O, so partial + on C, partial - on O). But the map shows a sort of "bulb" with one end more negative and the other more positive. Wait, $HI$: H is small, I is large. The electrostatic potential map for $HI$ would have a region of positive (near H) and negative (near I), which can be represented as a small positive - colored region and a larger negative - colored region, matching the given map (small blue, larger red - like? Wait, the color scale: blue is +, red is -, 0 is in between. So the molecule has a positive end (blue) and a negative end (red). So the positive end is the less electronegative atom, negative end is more electronegative.
   - $H_2$: Non - polar, so electrostatic potential map would be uniform (no red or blue regions except zero), so eliminate.
   - $CO$: Carbon and oxygen. Electronegativity of O is higher than C, so partial - on O, partial + on C. But the size of C and O: C is smaller than O? Wait, atomic radii: C (77 pm), O (66 pm). So the positive end (C) is a bit larger? No, maybe not. Wait, $HI$: H (37 pm), I (133 pm). So the positive end (H) is small, negative end (I) is large. The map shows a small blue (positive) region and a larger red (negative) region. So $HI$ fits. Also, $HBr$: H (37 pm), Br (114 pm). But the electronegativity of Br is higher than I? No, electronegativity: Br (2.96), I (2.66). So the polarity of $H - Br$ is more than $H - I$? Wait, no, the dipole moment depends on electronegativity difference and bond length. $H - I$ has a smaller electronegativity difference (2.66 - 2.20 = 0.46) than $H - Br$ (2.96 - 2.20 = 0.76), but $H - I$ has a longer bond length. The dipole moment ($\mu=\delta\times d$): for $HI$, $\delta = 0.46$, $d = 161$ pm; for $HBr$, $\delta=0.76$, $d = 141$ pm. $\mu(HI)=0.46\times161\approx74$ D - pm, $\mu(HBr)=0.76\times141\approx107$ D - pm. But the electrostatic potential map's appearance: the size of the atoms. $I$ is larger than $Br$, so the negative region (around I) would be larger, matching the map's larger red - like region. Also, $ICl$: I and Cl, both halogens. Cl is more electronegative than I, so partial - on Cl, partial + on I. But I is larger than Cl, so the positive region (I) would be larger, which does not match the map (small positive region). $CO$: C and O, O is more electronegative, but C and O are closer in size, so the regions would be more similar in size, not matching the small positive and large negative. So the best fit is $HI$. Also, let's re - check: the map has a small positive (blue) and large negative (red) region. $HI$: H (small, positive) and I (large, negative) – matches. $HBr$: H (small, positive) and Br (large, negative) – also possible? Wait, maybe I made a mistake. Wait, the electronegativity of Br is higher than I, so the negative charge on Br is more than on I. But the map's red (negative) region: if Br is more electronegative, the red region should be more intense? But the problem is about the shape (size of the regions). $I$ is larger than $Br$, so the negative region (around I) is larger, which matches the map's larger red region. So $HI$ is a good candidate. Also, $CO$: no, because C and O are both small, and the dipole is between them, but the map shows a sort of "bulb" with one end small (positive) and one end large (negative), which is more like a diatomic with a small and large atom, like $HI$ (H small, I large) or $HBr$ (H small, Br large). But let's check the options again. The question says "check all that apply". Wait, maybe $HI$ and $CO$? No, $CO$ has C and O, both in the second period, similar size. $HI$ has H (first period) and I (fifth period), big size difference. So the map's shape (small positive, large negative) matches $HI$ (and $HBr$?). Wait, maybe the answer is $HI$ (and maybe $HBr$, but let's see the electrostatic potential maps. For $HI$, the positive region (H) is small, negative (I) is large. For $HBr$, positive (H) small, negative (Br) large. But the electronegativity of Br is higher, so the negative region is more electronegative, but the size: I is larger than Br. So the map's negative region is larger, so $HI$ is better. Also, $ICl$: I is larger than Cl, so positive region (I) is larger, which does not match (map has small positive). $H_2$: non - polar, so no. $CO$: polar, but size of C and O are similar, so regions would be similar in size, not matching. So the correct option is $HI$.
# Answer:
$\boldsymbol{HI}$ (and also, upon re - evaluation, $HBr$ and $CO$ might be considered? Wait, no, let's re - check the electrostatic potential map. The map shows a molecule with a small positive (blue) and a large negative (red) region. $HI$: H (small, +) and I (large, -) – matches. $HBr$: H (small, +) and Br (large, -) – also matches. $CO$: C (slightly +) and O (slightly -), but C and O are similar in size, so the regions would be similar in size, not matching the small - large. $ICl$: I (slightly +) and Cl (slightly -), but I is larger than Cl, so the positive region is larger, not matching. $H_2$: non - polar. So the possible formulas are $HI$, $HBr$, and $CO$? Wait, no, $CO$ has a dipole moment, and the electrostatic potential map for $CO$ shows a partial positive on C and partial negative on O, with C and O having somewhat similar sizes (C: 77 pm, O: 66 pm), so the regions are not as different in size as $HI$ (H: 37 pm, I: 133 pm) or $HBr$ (H: 37 pm, Br: 114 pm). So the most probable are $HI$ and $HBr$, but the initial checked option is $HI$. So the answer is $HI$ (and maybe $HBr$ and $CO$, but based on the map's size difference, $HI$ is the best fit).

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