NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXII01:THE SOLID STATE
318735
In a crystalline solid anions \(B\) are arranged in cubic close packing. Cation A are equally distributed between octahedral and tetrahedral voids. If all the octahedral voids are occupied, the formula for the solid is
1 \(A_{2} B_{3}\)
2 \(A_{2} B\)
3 \(A B_{2}\)
4 \(\mathrm{AB}\)
Explanation:
Number of anions " \(B\) " \(=4\) Number of cations "A" \(=8\) Formula of the solid \(=\mathrm{A}_{8} \mathrm{~B}_{4}\) or \(\mathrm{A}_{2} \mathrm{~B}\)
CHXII01:THE SOLID STATE
318736
The ratio of closed packed atoms to the tetrahedral holes in cubic close packing is
1 \(1: 1\)
2 \(1: 2\)
3 \(1: 3\)
4 \(2: 1\)
Explanation:
Number of tetrahedral voids \(=2 \times\) Number atoms per unit cell in cubic close packing
CHXII01:THE SOLID STATE
318737
The number of octahedral void (s) per atom present in a cubic close-packed structure is
1 2
2 4
3 1
4 3
Explanation:
Number of octahedral voids \(=\) Number of atoms in the close packed structure. As number of atoms \(=1\) \(\therefore\) Number of octahedral voids \(=1\)
CHXII01:THE SOLID STATE
318738
A crystal is made of particles A and B. A forms FCC packing and \(\mathrm{B}\) occupies all the octahedral voids. If all the particles along the plane touches opposite corners through body center are removed, then the formula of the crystal would be
318735
In a crystalline solid anions \(B\) are arranged in cubic close packing. Cation A are equally distributed between octahedral and tetrahedral voids. If all the octahedral voids are occupied, the formula for the solid is
1 \(A_{2} B_{3}\)
2 \(A_{2} B\)
3 \(A B_{2}\)
4 \(\mathrm{AB}\)
Explanation:
Number of anions " \(B\) " \(=4\) Number of cations "A" \(=8\) Formula of the solid \(=\mathrm{A}_{8} \mathrm{~B}_{4}\) or \(\mathrm{A}_{2} \mathrm{~B}\)
CHXII01:THE SOLID STATE
318736
The ratio of closed packed atoms to the tetrahedral holes in cubic close packing is
1 \(1: 1\)
2 \(1: 2\)
3 \(1: 3\)
4 \(2: 1\)
Explanation:
Number of tetrahedral voids \(=2 \times\) Number atoms per unit cell in cubic close packing
CHXII01:THE SOLID STATE
318737
The number of octahedral void (s) per atom present in a cubic close-packed structure is
1 2
2 4
3 1
4 3
Explanation:
Number of octahedral voids \(=\) Number of atoms in the close packed structure. As number of atoms \(=1\) \(\therefore\) Number of octahedral voids \(=1\)
CHXII01:THE SOLID STATE
318738
A crystal is made of particles A and B. A forms FCC packing and \(\mathrm{B}\) occupies all the octahedral voids. If all the particles along the plane touches opposite corners through body center are removed, then the formula of the crystal would be
318735
In a crystalline solid anions \(B\) are arranged in cubic close packing. Cation A are equally distributed between octahedral and tetrahedral voids. If all the octahedral voids are occupied, the formula for the solid is
1 \(A_{2} B_{3}\)
2 \(A_{2} B\)
3 \(A B_{2}\)
4 \(\mathrm{AB}\)
Explanation:
Number of anions " \(B\) " \(=4\) Number of cations "A" \(=8\) Formula of the solid \(=\mathrm{A}_{8} \mathrm{~B}_{4}\) or \(\mathrm{A}_{2} \mathrm{~B}\)
CHXII01:THE SOLID STATE
318736
The ratio of closed packed atoms to the tetrahedral holes in cubic close packing is
1 \(1: 1\)
2 \(1: 2\)
3 \(1: 3\)
4 \(2: 1\)
Explanation:
Number of tetrahedral voids \(=2 \times\) Number atoms per unit cell in cubic close packing
CHXII01:THE SOLID STATE
318737
The number of octahedral void (s) per atom present in a cubic close-packed structure is
1 2
2 4
3 1
4 3
Explanation:
Number of octahedral voids \(=\) Number of atoms in the close packed structure. As number of atoms \(=1\) \(\therefore\) Number of octahedral voids \(=1\)
CHXII01:THE SOLID STATE
318738
A crystal is made of particles A and B. A forms FCC packing and \(\mathrm{B}\) occupies all the octahedral voids. If all the particles along the plane touches opposite corners through body center are removed, then the formula of the crystal would be
318735
In a crystalline solid anions \(B\) are arranged in cubic close packing. Cation A are equally distributed between octahedral and tetrahedral voids. If all the octahedral voids are occupied, the formula for the solid is
1 \(A_{2} B_{3}\)
2 \(A_{2} B\)
3 \(A B_{2}\)
4 \(\mathrm{AB}\)
Explanation:
Number of anions " \(B\) " \(=4\) Number of cations "A" \(=8\) Formula of the solid \(=\mathrm{A}_{8} \mathrm{~B}_{4}\) or \(\mathrm{A}_{2} \mathrm{~B}\)
CHXII01:THE SOLID STATE
318736
The ratio of closed packed atoms to the tetrahedral holes in cubic close packing is
1 \(1: 1\)
2 \(1: 2\)
3 \(1: 3\)
4 \(2: 1\)
Explanation:
Number of tetrahedral voids \(=2 \times\) Number atoms per unit cell in cubic close packing
CHXII01:THE SOLID STATE
318737
The number of octahedral void (s) per atom present in a cubic close-packed structure is
1 2
2 4
3 1
4 3
Explanation:
Number of octahedral voids \(=\) Number of atoms in the close packed structure. As number of atoms \(=1\) \(\therefore\) Number of octahedral voids \(=1\)
CHXII01:THE SOLID STATE
318738
A crystal is made of particles A and B. A forms FCC packing and \(\mathrm{B}\) occupies all the octahedral voids. If all the particles along the plane touches opposite corners through body center are removed, then the formula of the crystal would be
