
\[AX,{\text{ }}AY,{\text{ }}BX{\text{ }}and{\text{ }}BY\]have rock salt type structure with following internuclear distances:
Salt Anion-anion distance in Cation-anion distance in \[AX\] 2.4 1.7 \[AY\] 1.63 1.15 \[BX\] 2.66 1.88 \[BY\] 2.09 1.48
Ionic radii of \[{X^ - }\] and \[{Y^ - }\] in are, respectively:
\[
{A.{\text{ }}0.35\;and\;0.68} \\
{B.{\text{ }}0.68\;and\;0.35} \\
{C.{\text{ }}1.20\;and\;0.80} \\
{D.{\text{ }}0.80\;and\;1.20}
\]
| Salt | Anion-anion distance in | Cation-anion distance in |
| \[AX\] | 2.4 | 1.7 |
| \[AY\] | 1.63 | 1.15 |
| \[BX\] | 2.66 | 1.88 |
| \[BY\] | 2.09 | 1.48 |
Answer
583.5k+ views
Hint: We can understand that the given data is calculated from the inter-nuclear distance between a cation and its neighboring anion in a structure. So we can calculate ion radii by following the below-mentioned steps:
i. Find the radii of an anion atom.
ii. Find the internuclear distance between anion and cation.
iii. Use the Internuclear distance formula to find the radius of the cation atom.
Complete step by step solution:
From the data given in the question, we can say that Y- is smaller than X-.
Also, from the data, it is clear that the difference in inter-nuclear distance between AX and BX shows that the cation must not be touching all the anions in AX.
So AX contains smaller cations and larger anions. There must be an anion-anion contact in this compound.
Let’s consider, anion-anion distance will be equal to \[2{r^ - }\]
Now, radius of X- anion = $\dfrac{{2.40}}{2}$ = \[1.20{\text{ }}{A^o}\]
Now, we will calculate radius of B+ cation \[ = {\text{ }}1.80{\text{ }} - 1.20{\text{ }} = {\text{ }}0.68{\text{ }}{A^o}\]
Similarly, radius for Y- anion \[ = {\text{ }}1.48{\text{ }}-{\text{ }}0.68{\text{ }} = {\text{ }}0.80{\text{ }}{A^o}\]
Now, we will calculate radius of A+ cation \[ = {\text{ }}1.15{\text{ }} - 0.80{\text{ }} = {\text{ }}0.35{\text{ }}{A^o}\]
∴ Radius of X+ cation\[ = {\text{ }}1.48{\text{ }}-{\text{ }}0.68{\text{ }} = {\text{ }}0.80{\text{ }}{A^o}\]
Question has asked about the ionic radii of X- anion and Y- anion that is equal to \[1.20{\text{ }}{A^o}and{\text{ }}0.80{\text{ }}{A^o}\]
Hence, option C is the correct option.
Note: We must know that the cations are smaller in size compared to their neutral atoms while anions are larger. Cations are smaller because of their positive nuclear charge that holds the electrons closer whereas anions are larger because the electrons are not held as tightly that increases repulsions of electrons.
i. Find the radii of an anion atom.
ii. Find the internuclear distance between anion and cation.
iii. Use the Internuclear distance formula to find the radius of the cation atom.
Complete step by step solution:
From the data given in the question, we can say that Y- is smaller than X-.
Also, from the data, it is clear that the difference in inter-nuclear distance between AX and BX shows that the cation must not be touching all the anions in AX.
So AX contains smaller cations and larger anions. There must be an anion-anion contact in this compound.
Let’s consider, anion-anion distance will be equal to \[2{r^ - }\]
Now, radius of X- anion = $\dfrac{{2.40}}{2}$ = \[1.20{\text{ }}{A^o}\]
Now, we will calculate radius of B+ cation \[ = {\text{ }}1.80{\text{ }} - 1.20{\text{ }} = {\text{ }}0.68{\text{ }}{A^o}\]
Similarly, radius for Y- anion \[ = {\text{ }}1.48{\text{ }}-{\text{ }}0.68{\text{ }} = {\text{ }}0.80{\text{ }}{A^o}\]
Now, we will calculate radius of A+ cation \[ = {\text{ }}1.15{\text{ }} - 0.80{\text{ }} = {\text{ }}0.35{\text{ }}{A^o}\]
∴ Radius of X+ cation\[ = {\text{ }}1.48{\text{ }}-{\text{ }}0.68{\text{ }} = {\text{ }}0.80{\text{ }}{A^o}\]
Question has asked about the ionic radii of X- anion and Y- anion that is equal to \[1.20{\text{ }}{A^o}and{\text{ }}0.80{\text{ }}{A^o}\]
Hence, option C is the correct option.
Note: We must know that the cations are smaller in size compared to their neutral atoms while anions are larger. Cations are smaller because of their positive nuclear charge that holds the electrons closer whereas anions are larger because the electrons are not held as tightly that increases repulsions of electrons.
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