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
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
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