<p>Elements H and J lie in the same period. If the atoms of H are smaller than the atoms of J, then compared to atoms of J, atoms of H are most likely to</p>
<p>A)have a greater positive charge in their nuclei.
B)have a lesser positive charge in their nuclei.</p>
<p>I haven’t taken chem in 2 years. IDK</p>
<p>The atoms of H are smaller than the atoms of J which means that the atomic # of H is smaller than the atomic number of J. Say atomic H was lithium and atomic J was boron since lithium’s atomic # is less than boron. Lithium has a +1 charge and boron has a +3 charge. If boron’s charge is larger, then atom J should have a larger positive charge. Hope that’s correct :P</p>
<p>NOPE ~10char. This was a pretty tricky question due to size not being thought of as mass.</p>
<p>Since someone already got it wrong, then of course I would get the right answer.
So here’s the reason: the atoms on the left side of the table are actually larger because of fewer number of protons holding the electrons, thus the distance between the electrons and the protons are farther and thus making the atom itself bigger. So therefore, atom H are more strongly charged in the nuclei.</p>
<p>Fairly ambiguous if you asked me. We considering size in terms of atomic radius or what?
I’m going to assume H is smaller than J in terms of Atomic Radius, and going across a period, we notice an increase in Nuclear Charge while no extra shells are being added to an atom, therefore H must have a greater positive charge in its nuclei.</p>
<p>The answer is A because the extra positive charge brings the electrons in closer. If there were more electrons, there would be more repulsion, resulting in a larger atomic radius.</p>
<p>The problem states that we are dealing with atoms which is good (ionic radii can make this a bit weird). I assume that H and J represent elements on the periodic table.</p>
<p>Atomic radii decrease going from left to right of a period. So H has to be to the right of J on the periodic table because it was stated that H was smaller in the question. Elements on the right side of a period have a bigger atomic number than elements on the left side. Atomic number denotes the number of protons, so H, which is on the right side, has more protons in its nucleus compared to J. The only other particles in the nucleus besides protons are neutrons, which do not affect the charge of the nucleus. So H, having more protons, has a greater positive charge in its nucleus, which is answer choice A.</p>
<p>A I think. Because less electrons = positive charge?</p>
<p>Have a nice day</p>
<p>It’s A because as you go right across a period, atom radius decreases.</p>
<p><em>does a dance</em> I picked A. </p>
<p>Why? Smaller atomic size attributes to a greater positive charge in the nucleus due to the increase of protons. Those protons are able to draw in those electron, hold them in tighter. If J has a large atomic radius then H then that means J has less protons to attract those electron floating all up in that electron cloud. </p>
<p>@That You better learn your trends, boy. (Atomic radius, Electron affinity, Ionization energy…)</p>
<p>It’s in the same period so you’re not gaining any shells (principle quantum number is not increasing), you’re just gaining a proton and an electron as Z goes up.
Because it’s possible to have a greater positive charge yet get bigger (when you’re adding in shells)! the protons are in the middle and electrons are surrounding them, the effect of the positive charge > effect of the negative charge. So the electron cloud is drawn in closer = atom decreases in size.</p>