Column-$1$ Column-$2$ Column-$3$ $(\text{I) NaCl}$ (Rock salt structure) (i) Cations occupy octahedral voids (P) Effective number of anions in the unit cell $=4$ $(\text{II) ZnS}$ (Zinc Blende) (ii) Anions form FCC lattice (Q) $\dfrac{{{r}_{1}}}{r}=0.414-0.732$ $(\text{III) Ca}{{\text{F}}_{2}}$ (Fluorite structure) (iii) Cations occupy tetrahedral voids (R) $\dfrac{{{r}_{1}}}{r}=0.225-0.414$ $(\text{IV) CsCl}$ (iv) Anions occupy tetrahedral voids (S) Effective number of anions in the unit cell $=8$
Identify the correct set of combinations for the compound that involves $4:4$ coordination number compound.
A. (I) (i) (Q)
B. (II) (iii) (R)
C. (II) (iii) (Q)
D. (IV) (i) (Q)
Column-$1$ | Column-$2$ | Column-$3$ |
$(\text{I) NaCl}$ (Rock salt structure) | (i) Cations occupy octahedral voids | (P) Effective number of anions in the unit cell $=4$ |
$(\text{II) ZnS}$ (Zinc Blende) | (ii) Anions form FCC lattice | (Q) $\dfrac{{{r}_{1}}}{r}=0.414-0.732$ |
$(\text{III) Ca}{{\text{F}}_{2}}$ (Fluorite structure) | (iii) Cations occupy tetrahedral voids | (R) $\dfrac{{{r}_{1}}}{r}=0.225-0.414$ |
$(\text{IV) CsCl}$ | (iv) Anions occupy tetrahedral voids | (S) Effective number of anions in the unit cell $=8$ |
Answer
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Hint: In order to answer this question, you should have an idea about various types of crystal lattice. Keep in mind that; here coordination number is the total number of surrounding particles present around one particle.
Complete step by step solution:
According to the given question, we are given different compounds and we have to match the correct information given in different columns. For this, we have to know about the lattice type and the coordination number of these compounds. Let us look upon each option.
In $NaCl$, $N{{a}^{+}}$ is a cation and $C{{l}^{-}}$ is an anion. As we know cations are occupied in the voids and anions are present in the spheres. Therefore, here sodium ions are present at octahedral voids while chloride ions are face centered. The coordination number of the sodium atoms and the chlorine atoms are in the ratio $6:6$ and the cation-anion ratio ranges from $\dfrac{{{r}_{1}}}{r}=0.414-0.732$. Thus, the correct matching will be (I) (i) (Q) or (I) (ii) (Q) having $6:6$ coordination number.
In $ZnS$, the zinc ion (that is $Z{{n}^{2+}}$) is a cation where as the sulphide ion (that is ${{S}^{2-}}$) is an anion. In this compound, the sulphide ions (i.e. anion) are face centered which means that it has a coordination number $4$ while the zinc cation ions are present in the tetrahedral voids which are alternate and it has a coordination number $4$. Thus, the radius ratio rate of the cation to that of anion is $4:4$. As it has tetrahedral void the cation-anion radius ratio ranges from $0.225-0.414$. So, the matching will be (II) (iii) (R) or (II) (ii) (R)
The coordination number of $Ca{{F}_{2}}$ and $CsCl$ is in the ratio of $8:8$. We can see that only Zinc Blende ($ZnS$) has a coordination number of $4:4$.
Hence, the correct option is B.
Note: It is important to remember the cation-anion radius ratio for different types of voids. Tetrahedral voids have a coordination number of $4$ with radius ratio $0.225-0.414$, octahedral voids have coordination numbers of $6$ with radius ratio $0.414-0.732$.
Complete step by step solution:
According to the given question, we are given different compounds and we have to match the correct information given in different columns. For this, we have to know about the lattice type and the coordination number of these compounds. Let us look upon each option.
In $NaCl$, $N{{a}^{+}}$ is a cation and $C{{l}^{-}}$ is an anion. As we know cations are occupied in the voids and anions are present in the spheres. Therefore, here sodium ions are present at octahedral voids while chloride ions are face centered. The coordination number of the sodium atoms and the chlorine atoms are in the ratio $6:6$ and the cation-anion ratio ranges from $\dfrac{{{r}_{1}}}{r}=0.414-0.732$. Thus, the correct matching will be (I) (i) (Q) or (I) (ii) (Q) having $6:6$ coordination number.
In $ZnS$, the zinc ion (that is $Z{{n}^{2+}}$) is a cation where as the sulphide ion (that is ${{S}^{2-}}$) is an anion. In this compound, the sulphide ions (i.e. anion) are face centered which means that it has a coordination number $4$ while the zinc cation ions are present in the tetrahedral voids which are alternate and it has a coordination number $4$. Thus, the radius ratio rate of the cation to that of anion is $4:4$. As it has tetrahedral void the cation-anion radius ratio ranges from $0.225-0.414$. So, the matching will be (II) (iii) (R) or (II) (ii) (R)
The coordination number of $Ca{{F}_{2}}$ and $CsCl$ is in the ratio of $8:8$. We can see that only Zinc Blende ($ZnS$) has a coordination number of $4:4$.
Hence, the correct option is B.
Note: It is important to remember the cation-anion radius ratio for different types of voids. Tetrahedral voids have a coordination number of $4$ with radius ratio $0.225-0.414$, octahedral voids have coordination numbers of $6$ with radius ratio $0.414-0.732$.
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