NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXII01:THE SOLID STATE
318886
Consider the given unit cell for an ionic compound made of cation \({\mathrm{(A)}}\) and anion (B): The formula of the ionic compound is written as \({{\text{A}}_{\text{x}}}{{\text{B}}_{\text{y}}}\). The value of 'y' is ____.
1 1
2 3
3 4
4 5
Explanation:
Every corner ion (B) is shared by eight unit cells. So, the contribution of ion (B) is \({\mathrm{1 / 8}}\) to the unit cell. Thus, the number of ions (B) per unit cell is 1 . The ion (A) present at body-centre contributes 1 to each unit cell as it is not shared by any other unit cell. So, formula is AB . \({\mathrm{\therefore}}\) Value of \({\mathrm{y=1}}\)
CHXII01:THE SOLID STATE
318887
In fcc arrangement of \(A\) and \(B\) atoms, where \(A\) atoms are at the corners of the unit cell and \(\mathrm{B}\) atoms are at the face centres, two atoms are missing from two corners in each unit cell, then the simplest formula of the compound is
1 \(A_{7} B_{24}\)
2 \(A_{7} B_{6}\)
3 \(A_{6} B_{7}\)
4 \(A B_{4}\)
Explanation:
Number of " \(A\) " atoms \(=\dfrac{1}{8} \times 6=\dfrac{3}{4}\) Number of B-atoms per unit cell \(=\dfrac{1}{2} \times 6=3\) \[ A: B=\dfrac{3}{4}: 3=1: 4 \] Thus, formula is \(A B_{4}\)
CHXII01:THE SOLID STATE
318888
A solid with formula \(A B C_{3}\) would probably have
1 A at corners, B at face centres, \(\mathrm{C}\) at body centre
2 A at body centre, B at face centres and C at corners of the cube
3 A at corners of cube, B at body centre, C at face centres
4 A at corners of hexagon, B at centres of the hexagon and \(\mathrm{C}\) inside the hexagonal unit cell
Explanation:
Number of atoms of \(\mathrm{A}=8 \times \dfrac{1}{8}=1\) Number of atoms of \(\mathrm{B}=1 \times 1=1\) number of atoms of \(\mathrm{C}=6 \times \dfrac{1}{2}=3\) So, Molecular formula \(=\mathrm{ABC}_{3}\)
CHXII01:THE SOLID STATE
318889
Match the column I having type of lattice point and its contribution to one unit cell in column II and mark the appropriate choice. Column I Column II A Corner P 1 B Edge Q 1/8 C Face centre R 1/4 D Body centre S 1/2
318886
Consider the given unit cell for an ionic compound made of cation \({\mathrm{(A)}}\) and anion (B): The formula of the ionic compound is written as \({{\text{A}}_{\text{x}}}{{\text{B}}_{\text{y}}}\). The value of 'y' is ____.
1 1
2 3
3 4
4 5
Explanation:
Every corner ion (B) is shared by eight unit cells. So, the contribution of ion (B) is \({\mathrm{1 / 8}}\) to the unit cell. Thus, the number of ions (B) per unit cell is 1 . The ion (A) present at body-centre contributes 1 to each unit cell as it is not shared by any other unit cell. So, formula is AB . \({\mathrm{\therefore}}\) Value of \({\mathrm{y=1}}\)
CHXII01:THE SOLID STATE
318887
In fcc arrangement of \(A\) and \(B\) atoms, where \(A\) atoms are at the corners of the unit cell and \(\mathrm{B}\) atoms are at the face centres, two atoms are missing from two corners in each unit cell, then the simplest formula of the compound is
1 \(A_{7} B_{24}\)
2 \(A_{7} B_{6}\)
3 \(A_{6} B_{7}\)
4 \(A B_{4}\)
Explanation:
Number of " \(A\) " atoms \(=\dfrac{1}{8} \times 6=\dfrac{3}{4}\) Number of B-atoms per unit cell \(=\dfrac{1}{2} \times 6=3\) \[ A: B=\dfrac{3}{4}: 3=1: 4 \] Thus, formula is \(A B_{4}\)
CHXII01:THE SOLID STATE
318888
A solid with formula \(A B C_{3}\) would probably have
1 A at corners, B at face centres, \(\mathrm{C}\) at body centre
2 A at body centre, B at face centres and C at corners of the cube
3 A at corners of cube, B at body centre, C at face centres
4 A at corners of hexagon, B at centres of the hexagon and \(\mathrm{C}\) inside the hexagonal unit cell
Explanation:
Number of atoms of \(\mathrm{A}=8 \times \dfrac{1}{8}=1\) Number of atoms of \(\mathrm{B}=1 \times 1=1\) number of atoms of \(\mathrm{C}=6 \times \dfrac{1}{2}=3\) So, Molecular formula \(=\mathrm{ABC}_{3}\)
CHXII01:THE SOLID STATE
318889
Match the column I having type of lattice point and its contribution to one unit cell in column II and mark the appropriate choice. Column I Column II A Corner P 1 B Edge Q 1/8 C Face centre R 1/4 D Body centre S 1/2
318886
Consider the given unit cell for an ionic compound made of cation \({\mathrm{(A)}}\) and anion (B): The formula of the ionic compound is written as \({{\text{A}}_{\text{x}}}{{\text{B}}_{\text{y}}}\). The value of 'y' is ____.
1 1
2 3
3 4
4 5
Explanation:
Every corner ion (B) is shared by eight unit cells. So, the contribution of ion (B) is \({\mathrm{1 / 8}}\) to the unit cell. Thus, the number of ions (B) per unit cell is 1 . The ion (A) present at body-centre contributes 1 to each unit cell as it is not shared by any other unit cell. So, formula is AB . \({\mathrm{\therefore}}\) Value of \({\mathrm{y=1}}\)
CHXII01:THE SOLID STATE
318887
In fcc arrangement of \(A\) and \(B\) atoms, where \(A\) atoms are at the corners of the unit cell and \(\mathrm{B}\) atoms are at the face centres, two atoms are missing from two corners in each unit cell, then the simplest formula of the compound is
1 \(A_{7} B_{24}\)
2 \(A_{7} B_{6}\)
3 \(A_{6} B_{7}\)
4 \(A B_{4}\)
Explanation:
Number of " \(A\) " atoms \(=\dfrac{1}{8} \times 6=\dfrac{3}{4}\) Number of B-atoms per unit cell \(=\dfrac{1}{2} \times 6=3\) \[ A: B=\dfrac{3}{4}: 3=1: 4 \] Thus, formula is \(A B_{4}\)
CHXII01:THE SOLID STATE
318888
A solid with formula \(A B C_{3}\) would probably have
1 A at corners, B at face centres, \(\mathrm{C}\) at body centre
2 A at body centre, B at face centres and C at corners of the cube
3 A at corners of cube, B at body centre, C at face centres
4 A at corners of hexagon, B at centres of the hexagon and \(\mathrm{C}\) inside the hexagonal unit cell
Explanation:
Number of atoms of \(\mathrm{A}=8 \times \dfrac{1}{8}=1\) Number of atoms of \(\mathrm{B}=1 \times 1=1\) number of atoms of \(\mathrm{C}=6 \times \dfrac{1}{2}=3\) So, Molecular formula \(=\mathrm{ABC}_{3}\)
CHXII01:THE SOLID STATE
318889
Match the column I having type of lattice point and its contribution to one unit cell in column II and mark the appropriate choice. Column I Column II A Corner P 1 B Edge Q 1/8 C Face centre R 1/4 D Body centre S 1/2
318886
Consider the given unit cell for an ionic compound made of cation \({\mathrm{(A)}}\) and anion (B): The formula of the ionic compound is written as \({{\text{A}}_{\text{x}}}{{\text{B}}_{\text{y}}}\). The value of 'y' is ____.
1 1
2 3
3 4
4 5
Explanation:
Every corner ion (B) is shared by eight unit cells. So, the contribution of ion (B) is \({\mathrm{1 / 8}}\) to the unit cell. Thus, the number of ions (B) per unit cell is 1 . The ion (A) present at body-centre contributes 1 to each unit cell as it is not shared by any other unit cell. So, formula is AB . \({\mathrm{\therefore}}\) Value of \({\mathrm{y=1}}\)
CHXII01:THE SOLID STATE
318887
In fcc arrangement of \(A\) and \(B\) atoms, where \(A\) atoms are at the corners of the unit cell and \(\mathrm{B}\) atoms are at the face centres, two atoms are missing from two corners in each unit cell, then the simplest formula of the compound is
1 \(A_{7} B_{24}\)
2 \(A_{7} B_{6}\)
3 \(A_{6} B_{7}\)
4 \(A B_{4}\)
Explanation:
Number of " \(A\) " atoms \(=\dfrac{1}{8} \times 6=\dfrac{3}{4}\) Number of B-atoms per unit cell \(=\dfrac{1}{2} \times 6=3\) \[ A: B=\dfrac{3}{4}: 3=1: 4 \] Thus, formula is \(A B_{4}\)
CHXII01:THE SOLID STATE
318888
A solid with formula \(A B C_{3}\) would probably have
1 A at corners, B at face centres, \(\mathrm{C}\) at body centre
2 A at body centre, B at face centres and C at corners of the cube
3 A at corners of cube, B at body centre, C at face centres
4 A at corners of hexagon, B at centres of the hexagon and \(\mathrm{C}\) inside the hexagonal unit cell
Explanation:
Number of atoms of \(\mathrm{A}=8 \times \dfrac{1}{8}=1\) Number of atoms of \(\mathrm{B}=1 \times 1=1\) number of atoms of \(\mathrm{C}=6 \times \dfrac{1}{2}=3\) So, Molecular formula \(=\mathrm{ABC}_{3}\)
CHXII01:THE SOLID STATE
318889
Match the column I having type of lattice point and its contribution to one unit cell in column II and mark the appropriate choice. Column I Column II A Corner P 1 B Edge Q 1/8 C Face centre R 1/4 D Body centre S 1/2
