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MHT-CET Chemistry · Formula sheet

Solid State formulas

14 formulas, 4 reference tables and 18 common traps for MHT-CET Chemistry Solid State, grouped by subtopic.

Full notes with worked examples

Types of Solids, Crystal Systems and Properties

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Seven Crystal Systems, Fourteen Bravais Lattices

Bravais count

3+2+4+2+1+1+1=14 lattices in 7 systems3 + 2 + 4 + 2 + 1 + 1 + 1 = 14 \text{ lattices in } 7 \text{ systems}

Crystalline Versus Amorphous Solids

PropertyCrystallineAmorphous
ArrangementLong-range order, regular and periodicShort-range order only
MeltingSharp, definite temperatureSoftens over a range of temperature
Directional propertiesAnisotropic — refractive index, conductivity vary with directionIsotropic — the same in every direction
'Crystalline solids are isotropic' is the planted false statement.
Heat of fusionDefiniteNot definite
ExamplesDiamond, NaCl, ice, graphite, ceramics, sodiumGlass, plastic, rubber, metallic glass
Ice and ceramics are crystalline; only glass-like solids are amorphous.
Isotropy belongs to amorphous solids; anisotropy to crystalline ones.

Ionic, Covalent Network, Molecular and Metallic Solids

ClassParticlesBinding forceExamples
IonicIonsElectrostaticNaCl, KCl, CaF2_2
Covalent networkAtomsCovalent bonds in a networkDiamond, silica (SiO2\text{SiO}_2), SiC, graphite
Silica is covalent, not ionic — Si–O bonds run through the whole crystal.
MolecularMoleculesDispersion, dipole–dipole or hydrogen bondsIce, dry ice, solid Ar, I2_2, naphthalene
Ice is a MOLECULAR solid: hydrogen bonds between H₂O molecules.
MetallicCations + mobile electronsMetallic bondCu, Fe, Na, Ag
Classify by the force between the particles, not by the element's name.

Common traps

Reading 'NOT true' as 'true'

Three of the four statements are correct properties of crystalline solids; the one that is wrong is always the isotropy line. Mark the odd one, not the first true one.

Ice as an ionic or covalent solid

Water has covalent O–H bonds inside each molecule, but the SOLID is held by hydrogen bonds between molecules — molecular. The bond inside the particle never decides the class.

Swapping 7 and 14

Both numbers are offered as options in the same question. Systems are the SHAPES (7); Bravais lattices are shapes × centring (14).

Unit Cells, Edge Length and Atomic Radius

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Particles Per Unit Cell and Coordination Number

Sharing fractions

n=18 ncorner+12 nface+14 nedge+nbodyn = \tfrac{1}{8}\,n_{\text{corner}} + \tfrac{1}{2}\,n_{\text{face}} + \tfrac{1}{4}\,n_{\text{edge}} + n_{\text{body}}

Edge Length From Radius: Where the Atoms Touch

Edge–radius relations

asc=2r,abcc=4r3,afcc=22 ra_{\text{sc}} = 2r,\qquad a_{\text{bcc}} = \frac{4r}{\sqrt{3}},\qquad a_{\text{fcc}} = 2\sqrt{2}\,r

Unit Cell Volume and the Volume of Its Atoms

Cell volume and atom volume

Vcell=a3,Vatoms=n⋅43πr3V_{\text{cell}} = a^3,\qquad V_{\text{atoms}} = n \cdot \tfrac{4}{3}\pi r^3

Common traps

Counting the eight corners as eight particles

Each corner particle belongs to eight cells, so the eight corners together contribute ONE. A simple cubic cell has 1 particle, not 8.

Inverting the bcc relation

a=3r/4a = \sqrt{3}r/4 and a=3/4⋅ra = \sqrt{3}/4 \cdot r are both offered beside the right a=4r/3a = 4r/\sqrt{3}. Check the size: aa must be LARGER than rr (about 2.3 times), so any option that makes aa smaller than rr is wrong.

Cubing the picometres

4003 pm3400^3\ \text{pm}^3 is not a number any option shows. Convert to cm before cubing: (4×10−8)3=6.4×10−23(4 \times 10^{-8})^3 = 6.4 \times 10^{-23}; cubing 10−810^{-8} gives 10−2410^{-24}, and the mantissa's cube moves the exponent up.

Packing Efficiency and Voids

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Packing Efficiency: 52.4%, 68%, 74%

Packing efficiency

PE=n⋅43πr3a3×100:π6, 3π8, π32\text{PE} = \frac{n \cdot \tfrac{4}{3}\pi r^3}{a^3} \times 100:\quad \tfrac{\pi}{6},\ \tfrac{\sqrt{3}\pi}{8},\ \tfrac{\pi}{3\sqrt{2}}

Occupied Volume, Void Volume and the Volume Per Particle

Occupied and void volume

Vocc=PE⋅V,Vvoid=(1−PE) V,Vparticle=PE⋅VnV_{\text{occ}} = \text{PE}\cdot V,\qquad V_{\text{void}} = (1-\text{PE})\,V,\qquad V_{\text{particle}} = \frac{\text{PE}\cdot V}{n}

Tetrahedral and Octahedral Voids: 2N and N

Void counts

Ntet=2N,Noct=N,N=nmol×6.022×1023N_{\text{tet}} = 2N,\qquad N_{\text{oct}} = N,\qquad N = n_{\text{mol}} \times 6.022 \times 10^{23}

Formula of a Compound From the Voids the Cations Fill

Cations from void fraction

nA=ftet⋅2N+foct⋅N,formula=AnABNn_{\text{A}} = f_{\text{tet}} \cdot 2N + f_{\text{oct}} \cdot N,\qquad \text{formula} = \text{A}_{n_{\text{A}}}\text{B}_{N}

Common traps

Giving 74% for bcc

bcc is the middle value, 68%. 74% belongs to the two close-packed structures (fcc and hcp), which is why they are called close packed.

Reporting the occupied volume when ONE particle is asked

'Volume occupied by a particle in fcc' means one of the four: 0.74 V/4=0.185 V0.74\,V/4 = 0.185\,V. The option 0.74 V0.74\,V is always there for the student who skips the division.

Swapping the two counts

Tetrahedral is the SMALLER hole and the MORE numerous: 2 per atom. Octahedral is larger and fewer: 1 per atom. The option with the numbers reversed is always offered.

Forgetting the factor 2 on tetrahedral voids

13\tfrac{1}{3} of the tetrahedral voids is 13×2N=23N\tfrac{1}{3} \times 2N = \tfrac{2}{3}N, giving A2B3\text{A}_2\text{B}_3. Using 13N\tfrac{1}{3}N gives AB3\text{AB}_3, which is offered as an option every time.

Density and Crystal Structure Calculations

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The Density Formula and the Lumped Constants

Density of a cubic crystal

ρ=n Ma3 NA\rho = \frac{n\,M}{a^3\,N_A}

Molar Mass From Density

Molar mass

M=ρ a3NAnM = \frac{\rho\,a^3 N_A}{n}

Unit Cell Volume From Density and Molar Mass

Cell volume

a3=n Mρ NAa^3 = \frac{n\,M}{\rho\,N_A}

Which Structure? Solve for n

Particles per cell

n=ρ a3NAM=ρ a3matomn = \frac{\rho\,a^3 N_A}{M} = \frac{\rho\,a^3}{m_{\text{atom}}}

Counting Unit Cells and Atoms in a Mass or a Volume

Counting cells

Ncells=wρ a3=Va3=wM⋅NAn,Natoms=n NcellsN_{\text{cells}} = \frac{w}{\rho\,a^3} = \frac{V}{a^3} = \frac{w}{M}\cdot\frac{N_A}{n},\qquad N_{\text{atoms}} = n\,N_{\text{cells}}

Common traps

Using n = 4 for bcc

The wrong n doubles or halves the answer, and that wrong answer is always among the options. bcc is 2; fcc is 4; simple cubic is 1.

Dividing by N_A when the lump already contains it

a3NAa^3 N_A in cm³ mol⁻¹ has Avogadro's number built in. Multiply ρ\rho by it and divide by n — nothing else. A second NAN_A sends the exponent off by 23.

Trusting a printed exponent over the order of magnitude

One 2024 paper printed dNA=120×1021d N_A = 120 \times 10^{21}; the working gives 6×10−216 \times 10^{-21}, the key says 6.00×10−236.00 \times 10^{-23}. Match the mantissa to the options and let the sanity range (10−2310^{-23}) settle the exponent.

Stopping at n and not naming the cell

Half these questions ask for the STRUCTURE, not the number. n = 2 is 'body-centred cubic'; n = 4 is 'face-centred cubic'. hcp is never the answer of a cubic-cell calculation.

Counting cells when atoms are asked

'Number of atoms in 0.3 g of a bcc metal' is cells × 2. The cell count 102110^{21} is offered as option (A) for the student who stops one step early.

Crystal Defects, Magnetic Properties and Semiconductors

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Band Gap, n-Type and p-Type Doping

Dopant rule

Group 15 (P, As, Sb)→n-type;Group 13 (B, Al, Ga, In)→p-type\text{Group 15 (P, As, Sb)} \to n\text{-type};\qquad \text{Group 13 (B, Al, Ga, In)} \to p\text{-type}

Point Defects: Schottky, Frenkel, Impurity and Non-Stoichiometric

DefectWhat happensEffect on densityExamples
SchottkyEqual numbers of cations and anions missingDecreasesNaCl, KCl, CsCl, AgBr
FrenkelIon leaves its site for an interstitial positionUnchangedZnS, AgCl, AgBr, AgI
AgBr shows BOTH Schottky and Frenkel.
Substitutional impurityForeign atom replaces a host atomDependsBrass (Zn in Cu), Cd²⁺ in AgCl
Interstitial impuritySmall foreign atom in a voidIncreasesStainless steel (C in Fe)
Brass is SUBSTITUTIONAL, steel is INTERSTITIAL — size decides.
Non-stoichiometricCation:anion ratio ≠ formulaVaries( ext{Fe}_{0.95} ext{O}) (deficiency), NaCl with F-centres (excess)
Missing → Schottky, displaced → Frenkel, foreign → impurity, wrong ratio → non-stoichiometric.

Diamagnetic, Paramagnetic, Ferromagnetic

BehaviourElectronsIn a fieldExamples
DiamagneticAll pairedWeakly repelledNaCl, H₂O, C₆H₆, N₂
ParamagneticUnpaired, randomWeakly attracted, temporaryO₂, Cu²⁺, Fe³⁺, NO
O₂ is paramagnetic, never ferromagnetic.
FerromagneticUnpaired, aligned in domainsStrongly attracted, permanentFe, Co, Ni, Gd, CrO₂
CrO₂ is the one ferromagnetic COMPOUND the paper lists.
AntiferromagneticAligned opposite, equalNet zeroMnO
FerrimagneticAligned opposite, unequalWeakly attractedFe₃O₄, ferrites
Count unpaired electrons, then ask whether they line up.

Common traps

Calling every missing-ion picture Frenkel

Frenkel needs the ion to REAPPEAR in an interstitial site. If the ions are simply gone in equal numbers, it is Schottky. Read for the words 'interstitial position'.

Confusing 'n' with 'negative charge on the dopant'

n-type means the CARRIER is a negative electron, supplied by a Group 15 donor. The dopant itself is neutral. p-type means positive holes from a Group 13 acceptor. In, Ga, B and Al are always p-type.

Picking O₂ because it is magnetic

O₂ has two unpaired electrons and is attracted to a magnet — but weakly and only while the field is on. That is paramagnetism. Ferromagnetic needs domains: Fe, Co, Ni, CrO₂.

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