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If the position of Na+ and Cl- are interchanged in NaCl having FCC arrangement of Cl- ions then what change we can observe in the unit cell of NaCl. I am confused as how the contribution will change is it due to the difference in the size of the ions or something else. Please give answer in terms of the reduction or addition of number of ions of sodium and chloride ions Thanks

The correct answer is Na+ ions will increase by 1 and Cl- ions will decrease by 1.

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    $\begingroup$ Basically anions are present in Crystal arrangement and cations in voids. But we can change their position for visualization purposes. It still remain FCC. $\endgroup$ – Manu Apr 26 at 3:00
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    $\begingroup$ No. of ions per unit cell remains same. As now $Na^+$ forms FCC arrangement so their effective number is 4. And $Cl^-$ is present in octahedral voids, so their effective number is 1(at centre of unit cell)+ 3(at edges)=4. So their effective number remains same, as arrangement remain same, only point of view changes. $\endgroup$ – Manu Apr 26 at 4:01
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    $\begingroup$ One can visualize this way, consider two unit cells having $Na^+$ at voids and $Cl^-$ in FCC. Now join these two unit cells and see only 1st half and second half which are joined together (not that half which are free). Now this arrangement has $Na^+$ in FCC arrangement and $Cl^-$ in voids. This good only for visualization point of view but generally the cations are present in voids due to there small size. $\endgroup$ – Manu Apr 26 at 4:06
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    $\begingroup$ Your correct answer isn't. $\endgroup$ – Ivan Neretin May 22 at 21:51
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    $\begingroup$ Why, the answer given by James Gaidis is quite correct. Your answer isn't. $\endgroup$ – Ivan Neretin May 22 at 21:55
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Reverse engineering

I googled the answer given by the OP to find out what the question might be

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Source: 42 Years (1978-2019) JEE Advanced (IIT-JEE) + 18 yrs JEE Main (2002-2019) Topic-wise Solved Paper Chemistry 15th Edition, Disha Experts

It makes note of two chloride ions and one sodium ion on each edge of the unit cell.

The question that would have the correct answer

How do the number of chloride and sodium ions on each edge of the unit cell change when you move the origin from the chloride ion (conventional choice) to the sodium ion (unconventional choice)?

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  • $\begingroup$ So in this question you means to say that at a particular edge we are shifting the edge from the chlorine to the sodium atom which is causing the increase of one sodium atom and decrease of one chlorine atom. $\endgroup$ – Bitthal Maheshwari May 23 at 3:07
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    $\begingroup$ Yes but just along the edge. The total number per cell is unchanged. @BitthalMaheshwari $\endgroup$ – Karsten Theis May 23 at 9:51
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NaCl has a cubic unit cell. It is best thought of as a face-centered cubic array of anions with an interpenetrating fcc cation lattice (or vice-versa). The cell looks the same whether you start with anions or cations on the corners. Each ion is 6-coordinate and has a local octahedral geometry. http://www.ilpi.com/inorganic/structures/nacl/index.html

So you have two interpenetrating FCC lattices. If you exchange the cations with the anions, you get: two interpenetrating FCC lattices! The unit cell will have to have the same number of ions, just exchanged, one for the other.

Maybe this was a trick question to get you thinking about FCC lattices.

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  • $\begingroup$ The answer is not the same as you are telling. Actually I know the answer but I don't know the reason. If you want I can tell u the answer just tell me in the comments. $\endgroup$ – Bitthal Maheshwari Apr 27 at 4:04
  • $\begingroup$ @Bitthal Maheshwari: Please do. I just copied mine from the reference. Maybe there is a better answer. $\endgroup$ – James Gaidis Apr 27 at 13:09
  • $\begingroup$ Sorry for delay, the answer is Na+ ions will increase by 1 while Cl- ions will decrease by 1. $\endgroup$ – Bitthal Maheshwari Apr 30 at 5:28

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