Question: In the NiAs structure, arsenic atoms form hexagonal close packing (HCP). What fraction of octahedral holes are occupied by Ni atoms?
Answer:
In HCP, number of octahedral holes = number of atoms in HCP
Since the formula is NiAs (1:1 ratio), all octahedral holes are occupied by Ni atoms.
Fraction occupied = 1 (or 100%)
Question: What is the coordination number of Ni and As in the niccolite structure?
Answer:
Coordination number of Ni = 6 (octahedral coordination by 6 As atoms)
Coordination number of As = 6 (trigonal prismatic coordination by 6 Ni atoms)
The coordination ratio is 6:6
Question: The niccolite structure shows metallic properties. What structural feature is responsible for this?
Answer:
The niccolite structure exhibits metal-metal bonding along the c-axis.
Ni atoms are close enough along this direction to form direct metal-metal bonds, giving the structure metallic conductivity.
This is a unique feature distinguishing NiAs from purely ionic structures.
Question: Nickel arsenide (NiAs) crystallizes in the hexagonal NiAs structure with lattice parameters a = 3.62 Å and c = 5.12 Å. The density is 7.78 g/cm³.
(a) Calculate the number of formula units per unit cell. For a hexagonal unit cell, volume = a²c sin(120°) × √3/2.
(b) In the NiAs structure, Ni atoms along the c-axis are at a distance equal to c/2. Calculate the Ni-Ni distance and compare it with twice the metallic radius of Ni (2.49 Å). Comment on the nature of Ni-Ni interaction.
(c) If the structure is transformed to a hypothetical rock salt type structure (cubic, FCC) maintaining the same Ni-As nearest neighbor distance, calculate the new lattice parameter and predict whether this transformation would be energetically favorable.
[Given: Atomic mass Ni = 58.7 u, As = 74.9 u, NA = 6.022 × 10²³ mol⁻¹, rNi(metallic) = 1.245 Å]
(a) Number of formula units per unit cell:
For hexagonal unit cell: V = a²c sin(120°) × √3/2
sin(120°) = √3/2, so V = a²c × (√3/2) × (√3/2) = (√3/2) × a²c
V = 0.866 × (3.62 × 10⁻⁸)² × (5.12 × 10⁻⁸)
V = 0.866 × 13.10 × 10⁻¹⁶ × 5.12 × 10⁻⁸
V = 58.08 × 10⁻²⁴ cm³ = 5.808 × 10⁻²³ cm³
M = 58.7 + 74.9 = 133.6 g/mol
Using ρ = (Z × M)/(V × NA)
Z = (ρ × V × NA)/M
Z = (7.78 × 5.808 × 10⁻²³ × 6.022 × 10²³)/133.6
Z = 272.03/133.6 = 2.04 ≈ 2 formula units
(b) Ni-Ni distance and metal-metal bonding:
Ni atoms are located at heights 0 and c/2 along the c-axis
Ni-Ni distance along c-axis = c/2 = 5.12/2 = 2.56 Å
Twice the metallic radius = 2 × 1.245 = 2.49 Å
Observed Ni-Ni distance (2.56 Å) > 2 × rmetallic (2.49 Å)
However, the difference is only 0.07 Å (2.8%)
Conclusion: The Ni-Ni distance is very close to the sum of metallic radii, indicating significant metal-metal bonding along the c-axis. This is characteristic of the NiAs structure and contributes to its metallic conductivity despite being formally an ionic compound. The slight expansion suggests weak but definite Ni-Ni interactions.
(c) Transformation to rock salt structure:
In NiAs structure, Ni-As nearest neighbor distance:
In hexagonal structure, nearest Ni-As ≈ a = 3.62 Å (approximately)
For rock salt (cubic) structure with same Ni-As distance:
In rock salt: nearest neighbor distance = acubic/2
Therefore: acubic = 2 × 3.62 = 7.24 Å
Energy considerations:
Rock salt structure volume: Vcubic = (7.24)³ = 379.7 ų
NiAs hexagonal volume per formula unit: 58.08/2 = 29.04 ų
Rock salt would have 4 formula units: 4 × 29.04 = 116.16 ų (for comparison)
The transformation would NOT be energetically favorable because:
1. Loss of metal-metal bonding (stabilization energy lost)
2. Lower packing efficiency in cubic structure for this radius ratio
3. NiAs structure better accommodates the size mismatch between Ni²⁺ and As³⁻
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