NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXII01:THE SOLID STATE
318956
Assertion : In \(\mathrm{NaCl}\) crystal, all the octahedral voids are occupied by \(\mathrm{Na}^{+}\)ions. Reason : The number of octahedral voids is equal to the number of \(\mathrm{Cl}^{-}\)ions in the packing.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Correct explanation. \(\mathrm{Na}^{+}\)ions can be placed only in the octahedral voids but not double in size.
CHXII01:THE SOLID STATE
318957
The radii of \(\mathrm{Na}^{+}\)and \(\mathrm{Cl}^{-}\)ions are \(95 \mathrm{pm}\) and \(181 \mathrm{pm}\) respectively. The edge length of \(\mathrm{NaCl}\) unit cell is
1 \(276 \mathrm{pm}\)
2 \(138 \mathrm{pm}\)
3 \(552 \mathrm{pm}\)
4 \(415 \mathrm{pm}\)
Explanation:
In a \(f c c\) lattice, the distance between the cation and anion is equal to the sum of their radii, which is equal to half of the edge length of unit cell, i.e. \(r^{+}+r^{-}=\dfrac{a}{2} \quad\) (where \(\mathrm{a}=\) edge length) \[ r^{+}=95 p m, r^{-}=181 \mathrm{pm} \] Edge length \(=\) \[ \begin{aligned} 2 r^{+}+2 r^{-} & =(2 \times 95+2 \times 181) \mathrm{pm} \\ & =(190+362) \mathrm{pm}=552 \mathrm{pm} \end{aligned} \]
KCET - 2006
CHXII01:THE SOLID STATE
318958
\(\mathrm{CsCl}\) crystallises in body centred cubic lattice. If ' \(a\) ' is its edge length then which of the following expressions is correct?
318959
Which among the following crystal lattices occupies all of the cubic holes by cations?
1 \(\mathrm{UO}_{2}\)
2 \(\mathrm{CsCl}\)
3 \(\mathrm{CaF}_{2}\)
4 \(\mathrm{SrCl}_{2}\)
Explanation:
CsCl crystal lattices occupied all of the cubic holes by cations. An atom at centre of cube belongs only to this unit cell and there is only one body centre in the unit cell. Thus, the formula of compound is \(\mathrm{CsCl}\) for body centered cubic cell. In this type of cell, the particles are present at the corner of the cube as well as one particle is present at the centre with in the body
318956
Assertion : In \(\mathrm{NaCl}\) crystal, all the octahedral voids are occupied by \(\mathrm{Na}^{+}\)ions. Reason : The number of octahedral voids is equal to the number of \(\mathrm{Cl}^{-}\)ions in the packing.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Correct explanation. \(\mathrm{Na}^{+}\)ions can be placed only in the octahedral voids but not double in size.
CHXII01:THE SOLID STATE
318957
The radii of \(\mathrm{Na}^{+}\)and \(\mathrm{Cl}^{-}\)ions are \(95 \mathrm{pm}\) and \(181 \mathrm{pm}\) respectively. The edge length of \(\mathrm{NaCl}\) unit cell is
1 \(276 \mathrm{pm}\)
2 \(138 \mathrm{pm}\)
3 \(552 \mathrm{pm}\)
4 \(415 \mathrm{pm}\)
Explanation:
In a \(f c c\) lattice, the distance between the cation and anion is equal to the sum of their radii, which is equal to half of the edge length of unit cell, i.e. \(r^{+}+r^{-}=\dfrac{a}{2} \quad\) (where \(\mathrm{a}=\) edge length) \[ r^{+}=95 p m, r^{-}=181 \mathrm{pm} \] Edge length \(=\) \[ \begin{aligned} 2 r^{+}+2 r^{-} & =(2 \times 95+2 \times 181) \mathrm{pm} \\ & =(190+362) \mathrm{pm}=552 \mathrm{pm} \end{aligned} \]
KCET - 2006
CHXII01:THE SOLID STATE
318958
\(\mathrm{CsCl}\) crystallises in body centred cubic lattice. If ' \(a\) ' is its edge length then which of the following expressions is correct?
318959
Which among the following crystal lattices occupies all of the cubic holes by cations?
1 \(\mathrm{UO}_{2}\)
2 \(\mathrm{CsCl}\)
3 \(\mathrm{CaF}_{2}\)
4 \(\mathrm{SrCl}_{2}\)
Explanation:
CsCl crystal lattices occupied all of the cubic holes by cations. An atom at centre of cube belongs only to this unit cell and there is only one body centre in the unit cell. Thus, the formula of compound is \(\mathrm{CsCl}\) for body centered cubic cell. In this type of cell, the particles are present at the corner of the cube as well as one particle is present at the centre with in the body
318956
Assertion : In \(\mathrm{NaCl}\) crystal, all the octahedral voids are occupied by \(\mathrm{Na}^{+}\)ions. Reason : The number of octahedral voids is equal to the number of \(\mathrm{Cl}^{-}\)ions in the packing.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Correct explanation. \(\mathrm{Na}^{+}\)ions can be placed only in the octahedral voids but not double in size.
CHXII01:THE SOLID STATE
318957
The radii of \(\mathrm{Na}^{+}\)and \(\mathrm{Cl}^{-}\)ions are \(95 \mathrm{pm}\) and \(181 \mathrm{pm}\) respectively. The edge length of \(\mathrm{NaCl}\) unit cell is
1 \(276 \mathrm{pm}\)
2 \(138 \mathrm{pm}\)
3 \(552 \mathrm{pm}\)
4 \(415 \mathrm{pm}\)
Explanation:
In a \(f c c\) lattice, the distance between the cation and anion is equal to the sum of their radii, which is equal to half of the edge length of unit cell, i.e. \(r^{+}+r^{-}=\dfrac{a}{2} \quad\) (where \(\mathrm{a}=\) edge length) \[ r^{+}=95 p m, r^{-}=181 \mathrm{pm} \] Edge length \(=\) \[ \begin{aligned} 2 r^{+}+2 r^{-} & =(2 \times 95+2 \times 181) \mathrm{pm} \\ & =(190+362) \mathrm{pm}=552 \mathrm{pm} \end{aligned} \]
KCET - 2006
CHXII01:THE SOLID STATE
318958
\(\mathrm{CsCl}\) crystallises in body centred cubic lattice. If ' \(a\) ' is its edge length then which of the following expressions is correct?
318959
Which among the following crystal lattices occupies all of the cubic holes by cations?
1 \(\mathrm{UO}_{2}\)
2 \(\mathrm{CsCl}\)
3 \(\mathrm{CaF}_{2}\)
4 \(\mathrm{SrCl}_{2}\)
Explanation:
CsCl crystal lattices occupied all of the cubic holes by cations. An atom at centre of cube belongs only to this unit cell and there is only one body centre in the unit cell. Thus, the formula of compound is \(\mathrm{CsCl}\) for body centered cubic cell. In this type of cell, the particles are present at the corner of the cube as well as one particle is present at the centre with in the body
318956
Assertion : In \(\mathrm{NaCl}\) crystal, all the octahedral voids are occupied by \(\mathrm{Na}^{+}\)ions. Reason : The number of octahedral voids is equal to the number of \(\mathrm{Cl}^{-}\)ions in the packing.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Correct explanation. \(\mathrm{Na}^{+}\)ions can be placed only in the octahedral voids but not double in size.
CHXII01:THE SOLID STATE
318957
The radii of \(\mathrm{Na}^{+}\)and \(\mathrm{Cl}^{-}\)ions are \(95 \mathrm{pm}\) and \(181 \mathrm{pm}\) respectively. The edge length of \(\mathrm{NaCl}\) unit cell is
1 \(276 \mathrm{pm}\)
2 \(138 \mathrm{pm}\)
3 \(552 \mathrm{pm}\)
4 \(415 \mathrm{pm}\)
Explanation:
In a \(f c c\) lattice, the distance between the cation and anion is equal to the sum of their radii, which is equal to half of the edge length of unit cell, i.e. \(r^{+}+r^{-}=\dfrac{a}{2} \quad\) (where \(\mathrm{a}=\) edge length) \[ r^{+}=95 p m, r^{-}=181 \mathrm{pm} \] Edge length \(=\) \[ \begin{aligned} 2 r^{+}+2 r^{-} & =(2 \times 95+2 \times 181) \mathrm{pm} \\ & =(190+362) \mathrm{pm}=552 \mathrm{pm} \end{aligned} \]
KCET - 2006
CHXII01:THE SOLID STATE
318958
\(\mathrm{CsCl}\) crystallises in body centred cubic lattice. If ' \(a\) ' is its edge length then which of the following expressions is correct?
318959
Which among the following crystal lattices occupies all of the cubic holes by cations?
1 \(\mathrm{UO}_{2}\)
2 \(\mathrm{CsCl}\)
3 \(\mathrm{CaF}_{2}\)
4 \(\mathrm{SrCl}_{2}\)
Explanation:
CsCl crystal lattices occupied all of the cubic holes by cations. An atom at centre of cube belongs only to this unit cell and there is only one body centre in the unit cell. Thus, the formula of compound is \(\mathrm{CsCl}\) for body centered cubic cell. In this type of cell, the particles are present at the corner of the cube as well as one particle is present at the centre with in the body
