The following XRD is recorded for the crystalline solid, having simple cubic crystal structure :PeakABCDEFGH(hkl)(100)(110)(111)(200)(210)(211)(220)(300)If this crystal would have been FCC, the following peaks would have missing :

The missing peaks would be A, B, E, F and H.In a face-centred cubic lattice, reflections appear only when h, k and l are all odd or all even (zero counts as even). Mixed indices give zero structure factor and are absent.Allowed: (111) C, (200) D, (220) G.Absent: (100) A, (110) B, (210) E, (211) F, (300) H.Every other option either keeps one of the mixed-index peaks or removes C, D or G, which FCC does show. For comparison, a body-centred lattice keeps only h + k + l even, which would remove A, C, E and H instead.

Solid State Chemistry Questions and Answers

Q: A dislocation with Burgers vector that equals one lattice spacing is called ............................. dislocation.

Such a dislocation is called a unit dislocation.A unit (perfect) dislocation has a Burgers vector equal to one full lattice translation, so the crystal on either side of the slipped region is perfectly matched and no stacking fault is left behind.A partial (imperfect) dislocation has a Burgers vector that is only a fraction of a lattice vector and bounds a stacking fault. A Frank partial is one such case, with its Burgers vector perpendicular to the slip plane.

Q: According to Debye theory, specific heat at high temperature is proportional to :

At high temperature the specific heat is independent of T.Debye theory treats a solid's vibrations as phonons up to a cut-off frequency. Well above the Debye temperature every mode is fully excited, each contributing kT, so the molar heat capacity levels off at 3R, about 25 J/(mol K). That is the Dulong-Petit law, which the theory reproduces.The T3 dependence is the Debye law at low temperature, the trap here. A linear T term is the electronic contribution in metals at low temperature, and T2 does not apply to three-dimensional solids.

Q: A compound contains two types of atoms X and Y. It crystallizes in a cubic lattice with atom X at the corners of the unit cell and Y at the body centre. The simplest possible formula of this compound is :

The formula is XY.An atom at a corner of a cubic unit cell is shared by eight cells, and an atom at the body centre belongs wholly to one cell:\[ \begin{aligned} \mathrm{X} &= 8 \times \tfrac{1}{8} = 1 \\ \mathrm{Y} &= 1 \times 1 = 1 \end{aligned} \]One X and one Y per cell gives XY, the arrangement of the cesium chloride structure.X2Y and X4Y come from counting corner atoms as more than one-eighth each, and XY4 reverses the roles of the two atoms. The same counting with face centres (each shared by two cells) is what gives XY3 in a face-centred arrangement.

Q: A compound formed by element X and Y crystallizes in cubic structure in which atoms of ‘X’ are at the corners while that of ‘Y’ are at the face centre. The formula of the compound is :

The formula is XY3.Count the atoms belonging to one unit cell. A corner atom is shared by eight cells and a face-centre atom by two:\[ \begin{aligned} \mathrm{X} &= 8 \times \tfrac{1}{8} = 1 \\ \mathrm{Y} &= 6 \times \tfrac{1}{2} = 3 \end{aligned} \]One X to three Y gives XY3, as in the Cu3Au type of ordered alloy.X8Y6 counts the positions without sharing them among neighbouring cells. X3Y and 3XY reverse or misplace the factor of three, which belongs to the face-centred Y atoms.

Q: In X-ray diffraction, the rays are scattered from :

X-rays are scattered by the electron cloud in the crystal.X-rays are electromagnetic waves; their oscillating electric field drives the electrons of each atom, which re-radiate coherently (Thomson scattering). That is why X-ray diffraction maps electron density, why heavy atoms with many electrons scatter strongly, and why hydrogen atoms are hard to locate.Nuclei are far too heavy to be driven appreciably by the field, so their scattering of X-rays is negligible; nuclei scatter neutrons, which is the basis of neutron diffraction. Phonons cause thermal diffuse scattering, a minor effect, and the electrostatic potential is what electrons, not X-rays, are diffracted by.

Q: Due to symmetry consideration associated to crystals, some of the reflections in X-ray diffraction are absent, systematically when symmetry increases from triclinic to cubic, the number of systematic absence :

On the key's reading, the number of systematic absences increases.Systematic absences arise from lattice centring (I, F, C, A or B lattices) and from translational symmetry elements, screw axes and glide planes, which make the scattered waves from symmetry-related atoms cancel exactly for certain classes of reflections. A triclinic crystal has no such elements beyond a possible inversion centre (which causes no absences), whereas the higher-symmetry systems admit more centred lattices, screw axes and glide planes, so more reflection classes can be systematically extinguished.Saying there is no change, or that the trend is unpredictable, ignores that absences come directly from symmetry elements, and a decrease runs against the growth in available symmetry.

Q: A crystal is spatial periodic arrangement of large number of atom/ions. When size of crystal is reduced to nano-meter range the diffraction peaks :

The diffraction peaks broaden.Sharp diffraction peaks need many repeating lattice planes, so that waves scattered at angles slightly off the Bragg angle cancel. In a nanocrystal there are few planes, the cancellation is incomplete and the peak spreads out. The Scherrer equation relates the crystallite size τ to the peak width β (full width at half maximum, in radians):\[ \tau = \frac{K\lambda}{\beta\cos\theta} \]The smaller the crystallite, the larger β. So the peaks do not sharpen, their width does change, and reducing the size does not split them.

11.

A dislocation with Burgers vector that equals one lattice spacing is called ............................. dislocation.

A.Unit
B.Partial
C.Imperfect
D.Frank

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

Such a dislocation is called a unit dislocation.

A unit (perfect) dislocation has a Burgers vector equal to one full lattice translation, so the crystal on either side of the slipped region is perfectly matched and no stacking fault is left behind.

A partial (imperfect) dislocation has a Burgers vector that is only a fraction of a lattice vector and bounds a stacking fault. A Frank partial is one such case, with its Burgers vector perpendicular to the slip plane.

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12.

According to Debye theory, specific heat at high temperature is proportional to :

A.T
B.\(\mathrm{T^{2}}\)
C.\(\mathrm{T^{3}}\)
D.Independent of T

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

At high temperature the specific heat is independent of T.

Debye theory treats a solid's vibrations as phonons up to a cut-off frequency. Well above the Debye temperature every mode is fully excited, each contributing kT, so the molar heat capacity levels off at 3R, about 25 J/(mol K). That is the Dulong-Petit law, which the theory reproduces.

The T³ dependence is the Debye law at low temperature, the trap here. A linear T term is the electronic contribution in metals at low temperature, and T² does not apply to three-dimensional solids.

Discuss in Forum

13.

A compound contains two types of atoms X and Y. It crystallizes in a cubic lattice with atom X at the corners of the unit cell and Y at the body centre. The simplest possible formula of this compound is :

A.\(\mathrm{X_{2}Y}\)
B.\(\mathrm{X_{4}Y}\)
C.XY
D.\(\mathrm{XY_{4}}\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The formula is XY.

An atom at a corner of a cubic unit cell is shared by eight cells, and an atom at the body centre belongs wholly to one cell:

\[ \begin{aligned} \mathrm{X} &= 8 \times \tfrac{1}{8} = 1 \\ \mathrm{Y} &= 1 \times 1 = 1 \end{aligned} \]

One X and one Y per cell gives XY, the arrangement of the cesium chloride structure.

X₂Y and X₄Y come from counting corner atoms as more than one-eighth each, and XY₄ reverses the roles of the two atoms. The same counting with face centres (each shared by two cells) is what gives XY₃ in a face-centred arrangement.

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14.

A compound formed by element X and Y crystallizes in cubic structure in which atoms of ‘X’ are at the corners while that of ‘Y’ are at the face centre. The formula of the compound is :

A.3XY
B.\(\mathrm{X_{3}Y}\)
C.\(\mathrm{XY_{3}}\)
D.\(\mathrm{X_{8}Y_{6}}\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The formula is XY₃.

Count the atoms belonging to one unit cell. A corner atom is shared by eight cells and a face-centre atom by two:

\[ \begin{aligned} \mathrm{X} &= 8 \times \tfrac{1}{8} = 1 \\ \mathrm{Y} &= 6 \times \tfrac{1}{2} = 3 \end{aligned} \]

One X to three Y gives XY₃, as in the Cu₃Au type of ordered alloy.

X₈Y₆ counts the positions without sharing them among neighbouring cells. X₃Y and 3XY reverse or misplace the factor of three, which belongs to the face-centred Y atoms.

Discuss in Forum

15.

In X-ray diffraction, the rays are scattered from :

A.atomic/ionic nuclei periodically arranged in crystal
B.Lattice vibration (phonon) in the crystal
C.Electrons cloud present in the crystal
D.Electrostatic potential among ions in the crystal

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

X-rays are scattered by the electron cloud in the crystal.

X-rays are electromagnetic waves; their oscillating electric field drives the electrons of each atom, which re-radiate coherently (Thomson scattering). That is why X-ray diffraction maps electron density, why heavy atoms with many electrons scatter strongly, and why hydrogen atoms are hard to locate.

Nuclei are far too heavy to be driven appreciably by the field, so their scattering of X-rays is negligible; nuclei scatter neutrons, which is the basis of neutron diffraction. Phonons cause thermal diffuse scattering, a minor effect, and the electrostatic potential is what electrons, not X-rays, are diffracted by.

Discuss in Forum

16.

Due to symmetry consideration associated to crystals, some of the reflections in X-ray diffraction are absent, systematically when symmetry increases from triclinic to cubic, the number of systematic absence :

A.increases
B.decreases
C.no change
D.difficult to predict

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

On the key's reading, the number of systematic absences increases.

Systematic absences arise from lattice centring (I, F, C, A or B lattices) and from translational symmetry elements, screw axes and glide planes, which make the scattered waves from symmetry-related atoms cancel exactly for certain classes of reflections. A triclinic crystal has no such elements beyond a possible inversion centre (which causes no absences), whereas the higher-symmetry systems admit more centred lattices, screw axes and glide planes, so more reflection classes can be systematically extinguished.

Saying there is no change, or that the trend is unpredictable, ignores that absences come directly from symmetry elements, and a decrease runs against the growth in available symmetry.

Discuss in Forum

17.

A crystal is spatial periodic arrangement of large number of atom/ions. When size of crystal is reduced to nano-meter range the diffraction peaks :

A.Sharpen
B.Broaden
C.No change in HWFM
D.Splits

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The diffraction peaks broaden.

Sharp diffraction peaks need many repeating lattice planes, so that waves scattered at angles slightly off the Bragg angle cancel. In a nanocrystal there are few planes, the cancellation is incomplete and the peak spreads out. The Scherrer equation relates the crystallite size τ to the peak width β (full width at half maximum, in radians):

\[ \tau = \frac{K\lambda}{\beta\cos\theta} \]

The smaller the crystallite, the larger β. So the peaks do not sharpen, their width does change, and reducing the size does not split them.

Discuss in Forum

18.

The following XRD is recorded for the crystalline solid, having simple cubic crystal structure :

Peak	A	B	C	D	E	F	G	H
(hkl)	(100)	(110)	(111)	(200)	(210)	(211)	(220)	(300)

If this crystal would have been FCC, the following peaks would have missing :

A.A, B, C, F, G H
B.B D, F H A
C.A B E F H
D.F G H C D

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The missing peaks would be A, B, E, F and H.

In a face-centred cubic lattice, reflections appear only when h, k and l are all odd or all even (zero counts as even). Mixed indices give zero structure factor and are absent.

Allowed: (111) C, (200) D, (220) G.
Absent: (100) A, (110) B, (210) E, (211) F, (300) H.

Every other option either keeps one of the mixed-index peaks or removes C, D or G, which FCC does show. For comparison, a body-centred lattice keeps only h + k + l even, which would remove A, C, E and H instead.

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