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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
Solid State

NTA Abhyas 2020 · Chemistry · Solid State

Which of the following expression is correct for packing fraction...

Which of the following expression is correct for packing fraction of NaCl if a−a^{-} ions from one face and one diagonal are removed?

Options

  1. A133πr−3+163πr+38(r++r−)3\dfrac{\dfrac{13}{3}\pi{r}_{-}^{3} + \dfrac{16}{3}\pi{r}_{+}^{3}}{8 ( {r}_{+} + {r}_{-}{)}^{3}}
  2. B133πr−3+43πr+38(r++r−)3\dfrac{\dfrac{13}{3}\pi{r}_{-}^{3} + \dfrac{4}{3}\pi{r}_{+}^{3}}{8 ( {r}_{+} + {r}_{-}{)}^{3}}
  3. C163πr−3+133πr+38(r++r−)3\dfrac{\dfrac{16}{3}\pi{r}_{-}^{3} + \dfrac{13}{3}\pi{r}_{+}^{3}}{8 ( {r}_{+} + {r}_{-}{)}^{3}}
  4. D43πr−3+133πr+38(r++r−)3\dfrac{\dfrac{4}{3}\pi{r}_{-}^{3} + \dfrac{13}{3}\pi{r}_{+}^{3}}{8 ( {r}_{+} + {r}_{-}{)}^{3}}

Correct answer

A. 133πr−3+163πr+38(r++r−)3\dfrac{\dfrac{13}{3}\pi{r}_{-}^{3} + \dfrac{16}{3}\pi{r}_{+}^{3}}{8 ( {r}_{+} + {r}_{-}{)}^{3}}

Step-by-step solution

Effective number of remaining Cl−=4−[2×18+12]=134Cl^{-} = 4 - \left[2 \times \dfrac{1}{8} + \dfrac{1}{2}\right] = \dfrac{13}{4}
Effective number of Na+=4Na^{+} = 4
Packing fraction =OccupiedvolumeTotal and volume= \dfrac{Occupied volume}{Total \text{ and } volume}
=134×43πr−3+4×43πr+38(r++r−)3= \dfrac{\dfrac{13}{4} \times \dfrac{4}{3}\pi{r}_{-}^{3} + 4 \times \dfrac{4}{3}\pi{r}_{+}^{3}}{8 ( {r}_{+} + {r}_{-}{)}^{3}}
133πr−3+163πr+38(r++r−)3\dfrac{\dfrac{13}{3}\pi{r}_{-}^{3} + \dfrac{16}{3}\pi{r}_{+}^{3}}{8 ( {r}_{+} + {r}_{-}{)}^{3}}
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