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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
Naphrthalene is a/anA. ionicsolidB. covalent solidC. metallic solidD. molecular solid |
| Answer» Correct Answer - 4 | |
| 102. |
When molten zinc is cooled to solid state, it assumes `hcp` structure. Then the number of nearest neighbours of zine atom will beA. `4`B. `6`C. `8`D. `12` |
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Answer» Correct Answer - D hcp has `12` nearest neighbours. |
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| 103. |
To get `n`-type doped semiconductor, impurity to be added to silicon should have the following number of valence electronsA. `2`B. `5`C. `3`D. `1` |
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Answer» Correct Answer - B For `n-`type impurity added to silicoln should have more than `4` valence electrons. |
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| 104. |
Radius ratio of an ionic compound is 0.93. the structure of the above ionic compound is ofA. NaCl typeB. CsCl typeC. ZnS typeD. None of these |
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Answer» Correct Answer - B Radius ratio range 0.732-0.99 signifies co-ordination number 8. In CsCl, co-ordination number ratio is 8:8 |
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| 105. |
The greater the value of `r_(+)//r_(-)`.A. The lower will be the C.N.B. The higher the value of C.N.C. The higher will be the number of cationsD. The lower will be the number of anions |
| Answer» Correct Answer - 2 | |
| 106. |
The range of radius ratio (cationic to anionic) for an octahedral arrangement of ions in an ionic solid isA. `0.155-0.225`B. `0.225-0.414`C. `0.414-0.732`D. `0.732-1.000` |
| Answer» Correct Answer - C | |
| 107. |
The ratio of cationic radius to anionic radius in an ionic crystal is greater than `0.732` its coordination number isA. 190pmB. 368pmC. 181 pmD. 276pm |
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Answer» Correct Answer - B When radius ratio between 0.732-1, then co-ordination number is 8 and structural arrangement is body-centred cubic. |
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| 108. |
The ratio of cationic radius to anionic radius in an ionic crystal is greater than `0.732` its coordination number isA. 6B. 8C. 1D. 4 |
| Answer» Correct Answer - 2 | |
| 109. |
A spinel is an important class of oxide consisting two types of metal ions withs the oxide ions arranged in ccp layers . The normal spinel has one -eight of the tetrahendral holes occupied by one type of metal ions and one- half of the octaherdral holes occupied by another type of metal ion. Such a spine is formed by `Mg^(2+), Al^(3+) and O^(2-)` . The netutrality of the crystal is benig maintained. Type of hole occupied by `Al^(3+)` ions is:A. tetrahedralB. octahedralC. both (a) amd(b)D. None of these |
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Answer» Correct Answer - B |
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| 110. |
Which of the following are the correct axial distance and axial angles for rhombohedral system ?A. `a = b = c, alpha = beta = gamma != 90^(@)`B. `a = b != c, alpha = beta = gamma = 90^(@)`C. `a!= b != c, alpha = beta = gamma = 90^(@)`D. `a != b != c, alpha != beta != gamma != 90^(@)` |
| Answer» Correct Answer - A | |
| 111. |
`{:(" Comlumn I (type of packing)", "Column II ( Metal possessing it/Space occupied)"),("(A) Hexagonal cubic packing (hcp)" , "(p) Iron"),("(B) Cubic close packing (ccp) " , "(q) 52%"),("(C) Body centred cubic (bcc)", "(r) 68%"),("(D) Simple cubic" , "(s) 74%"):}` |
| Answer» Correct Answer - A-p,s; B-p,q ; C-q, D-q,r | |
| 112. |
Calculate the value of `(Z)/(10). Where z = co-ordination number of 2D-square close packing + Co-ordination number of 2D-hcp + Co-ordination number of 3D-square close packing + Co-ordination number of 3D, ABCABC........packing + Co-ordiantional number of 3D, ABAB.......packing . |
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Answer» Correct Answer - 4 |
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| 113. |
`{:(,"Column-I(Reactions)",,,"Column-II(Product)"),((A),"Borax",overset(Delta)rarr,(p),"BN"),((B),B_(2)H_(6)+H_(2)O,overset(Delta)rarr,(q),B_(2)H_(6)),((C),B_(2)H_(6)NH_(3)("Excess"),overset(Delta)rarr,(r),H_(3)BO_(3)),((D),BCl_(3)+LiAlH_(4),overset(Delta)rarr,(s),NaBO_(2)+B_(2)O_(3)):}`A. A-p,B-r,C-q,D-sB. A-t,B-p,C-q,D-sC. A-r,B-p,C-s,D-qD. A-r,B-q,C-p,D-s |
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Answer» Correct Answer - 2 (A) `r_(Mg^(2+))//r_(O^(2-))=65//140=0.464` (lies in the range o.414-0.731)`rarr` Octahedral. Coord no. =6 (B) `r_(B^(3+))//r_(O^(2-))=27//140=0.193` (lies in the range 0.155-0.225)`rarr` Planar triangular coord no.=3. (C) `r_(Mg^(2+))//r_(S^(2-))=65//184=0.353` (lies in the range 0.225-0.414) `rarr` Tetrahedral. Coord no.=4 (D) `r_(Cs^(+))//r_(Cl^(-))=169//181=0.934` (lies in the range 0.732-1) `rarr` BC C. Coord .no.=8 |
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| 114. |
A solid `AB` has `CsCl`-type structure. The edge length of the unit cell is `404` pm. Calculate the distance of closest approach between `A^(o+)` and `B^(Θ)` ions.A. `4.0A^(0)`B. `6.4A^(0)`C. `3.5A^(0)`D. `7.0A^(0)` |
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Answer» Correct Answer - 3 (A) Simple cubic and FC C have a=b=c, `alpha=beta=gamma=90^(@)` and belong to the same crystal system (cubic). (B) Cubic `(a=b=c,alpha=beta=gamma=90^(@))` Thombohedral `(a=b=c,alpha=beta=gammacancel=90^(@))`. These are two different crystal systems. (C) Cubic and tetragonal are two different crystal systems. (D) Hexagonal: `alpha=beta=90^(@)(gamma=120^(@))` Monoclinic : `alpha=gamma=90^(@)cancel=beta`, i.e., each has two crystallographic angles of `90^(@)`. Further, these are two different crystal systems. |
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| 115. |
`{:("Column -I","Column -II"),("Tupe of packing","void volume"),(A.hcp,p . 0.38),(B. c cp,q . 0.48),(C. b c c ,r 0.22),(D. sc ,s. 0.66),("E. dx (Diamond cubic)",):}` |
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Answer» Correct Answer - `A rarr r; B rarr r; C rarr p; D rarr q; E rarr s` (a-r,b-r) hcp and ccp, closest packed structure, Volume occuped =0.78 Void volume=100.78=0.22 (c-p) For bcc, volume occupied=0.62 void volume=1-0.62=0.38 (d-q) For sc, volume occupied=0.52 void volume =1-0.52=0.48 (e-s) For dc, volume occupied=0.34 void volume=1-0.34=0.66 |
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| 116. |
Assertion . Hexagonal close packing is equally closely packed than cubic close packing . Reason. Hexagonal close packing has a corrdination number of 12 whereas cubic close packing has a coordination number of 8. |
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Answer» Correct Answer - d Correct A. Hexagonal close packing and cubic close packing are equally close packed (Space occpied = 74%) Correct R. Both have a coordination number of 12. |
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| 117. |
Tungsten has a density of 19.35 g `cm^(-3)` and the length of the side of the unit cell is 316 pm. The unit cell is a body centred unit cell. How many atoms does 50 grams of the element contain? |
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Answer» Correct Answer - `1.64xx10^(23)` atoms Step I. Calcualtion of mass of an atom of tungsten. Let the mass of the atom of tungsten metal = m gram. Mass of the unit cell `=2xxm=2` m gram. Edge length (a) = 316 pm = `316xx10^(-10)cm` Volume of the unit cell `(a^(3))=(316xx10^(-10))^(3)cm^(3)=31.55xx10^(-24)cm^(3)` `"density of unit cell"=("Mass of unit cell")/("Volume of unit cell")or (19.35"g cm"^(-3))=(("2 m gram"))/((31.55xx10^(-24)cm^(3)))` `m=1//2xx(19.35"g cm"^(-3))xx(31.55xx10^(-24)cm^(3))=30.529xx10^(-23)g.` Step II. Calculation of the number of atoms in 50 grams of tungsten. `"No. of atoms"=("Mass of tungsten")/("Mass of ne atom of tungsten")=((50 g))/((35.529xx10^(-23)g))=1.64xx10^(23)"atoms."` |
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| 118. |
Name the non-stoichiometric point defect responsible for the colour of alkali metals. |
| Answer» It is known as metal excess defect. The electrons responsible for the colours of the alkali metals are known as F-centres. | |
| 119. |
Assertion : A crystal having `fcc` structure is more closely packed then a crystal having `bcc` structure Reason: packing fraction for `fcc` structure id double than that of `bcc`structureA. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true bur reason is falseD. If assertion is false bur reason is true |
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Answer» Correct Answer - c Packing fraction in `fcc = 74%` Packing fraction in `fcc = 47%` |
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| 120. |
Assertion :The close packing of atoms in cubic structure is in the order `fcc gt bcc gt sc` Reason: Packing density`= ("Volume of unit cell")/(a^(3))` |
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Answer» Correct Answer - c Explanation is correct reason for statement. |
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| 121. |
Tungsten crystallises in a body centred cubic unit cell. If the edge of the unit cell is 316.5 pm, what is the radius of tungsten atom ? |
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Answer» Correct Answer - 137 pm For a body centred cubic unit cell (bcc) Radius `(r)=(axxsqrt3)/(4)=((316.5"pm")xxsqrt3)/(4)="137 pm".` |
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| 122. |
Sodium metal crystallises in body centred cubic lattic with cell edge 427 pm .Thus radius of sodium atom isA. 429 pmB. `214.5` pmC. 186 pmD. 185 pm |
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Answer» Correct Answer - d in bcc stracture radius `(r )` is related to edge length by `r = sqrt(3)/(4) a = sqrt(3)/(4) xx 427` pm `= 185 = 185.67 = 185` pm |
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| 123. |
How many atoms are present in a cubic unit cell having one atom at each corner and two atoms on each body diagonal ? |
| Answer» The total contribution by the atoms at four body diagonals present in the cubic unit cell is `8(4xx2=8)` and the contribution by the atoms on the corners s 1. Therefore, total no. of atoms present are 9. | |
| 124. |
The ratio of packing fraction in fcc, bcc, and cubic structure is, respectively,A. 0.92:0.70:1B. 0.70:0.92:1C. 1:0.92:0.70D. 1:0.70:0.92 |
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Answer» Correct Answer - C Packing fraction for fcc, bcc and Cs are 0.74 , 0.68 and 0.52 , respectively `therefore " Ratio" = (0.74)/(0.74):(0.68)/(0.74):(0.52)/(0.74)` =1:0.92:0.70 |
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| 125. |
Iron crystallises in a body centred cubic structure. Calculate the radius of ioun atom if edge length of unit cell is 286 pm.A. r = 124 pmB. r = 128 pmC. `r = 124 Å`D. `r = 128 Å` |
| Answer» Correct Answer - A | |
| 126. |
The number of atoms in a cubic based unit cell having one atom on each corner and two atoms on each body diagonal isA. 8B. 6C. 4D. 9 |
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Answer» Correct Answer - D There are four body diagonals. Atoms on the body diagonals are not shared by any other unit cell . Contribution by atoms on corners `=8xx(1)/(8)=1` and contribution by atoms on body `"diagonal"=2xx4=8,8+1=9`. |
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| 127. |
Calculate the number of atoms in a cubic-shared unit cell having one atom on each corner and two atoms one each diagonal. |
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Answer» Contibution by two atoms present on one body diagonal =2 Contribution by four atoms present on four body diagonals =8 Contribution by the atoms present on the corners =1 Total no. of atoms per unit cell`=8+1=9` |
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| 128. |
Which of the following statements is /are correc ? I. For a simple cubic lattice , d= a and for fcc it is `d=a//sqrt2` II. The density of the unit cell for cubic crystals for element is `rho=(ZM)/(a^3N_Axx10^(-3)) gcm^(-3)` III. In case of ionic compound `A^+B^-` having fcc structures as NaCl edge (a) = 2 x distance between `A^+ " and " B^-` ions .A. I and IIB. II and IIIC. I and IIID. All of these |
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Answer» Correct Answer - D For a simple cubic lattice , d=a and for fcc , `d=a//sqrt2`. For a cubic crystal of elements , the density of the unit cell is `rho=(ZM)/(a^3N_Axx10^(-30))gcm^(-3)` In case of ionic compounds `A^+ B^-` having fcc structures as NaCl , edge (a) = 2 x distance between `A^+ " and " B^-` ions . Hence , all the statements are correct . |
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| 129. |
Polonium crystallises in a simple cubic structure. If the edge length of the unit cel is 336 pm, calculate the atomic radius of polonium. |
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Answer» Correct Answer - 168 pm Edge length (a) = 336 pm. For a simple cubic unit cell, Atomic radius (r) `=(a)/(2)=(336)/(2)=168 "pm."` |
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| 130. |
Calculate the number of atoms in a cubic based unit cell having one atome on each corner and two atoms on each body diagonal. |
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Answer» It may be noted that there are four body diagonals passing through the centre and connecting opposity corners. The atoms present on these diagonals are not shared by the surrounding unit cells. Contribution by two atoms present on one body diagonal = 2 Contribution by four atoms present on one body diagonal = `2xx4=8` Contribution by the atoms present on corners = 1 Total number of atoms present per unit cell `=8+1=9` |
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| 131. |
A cubic solid is made up of two atoms A and B. Atoms A are present at the corners and B at the centre of the body. What is the formula of the unit cell ? |
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Answer» Contribution by the atoms A present at eight corners = 1 Contribution by the atom B present in the centre of the body = 1 Thus, the ratio of atoms of A : B = 1 : 1 Formula of unit cell = AB |
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| 132. |
Out of Xe, `CH_(3)Cl` and HF which has : (i) the smallest dipole-pole forces (ii) the largest hydrogen bond forces (c) the larges dispersion forces? |
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Answer» (i) Smallest dipole-pole forces exist in the Xe atoms since the noble gas is inert. (ii) The largest hydrogen bond forces are present in HF molecules since the element F is highly electronegative. (iii) The largest dispersion forces are present in Xe atoms. |
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| 133. |
The correct statement (s) for cubic close packed (ccp) three dimensional structure is (are)A. The number of the nearest neighbours of an atom present in the topmost layer is 12B. The packing efficiency of atom is `74%`C. The number of octahedral and tetrahedral voids per atom are 1 and 2, respectivelyD. The unit cell edge length is `2sqrt(2)` times the radius of the atom |
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Answer» Correct Answer - B::C::D |
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| 134. |
What types of semi-conductors will be formed when silicon is doped with Indium ? |
| Answer» p-type semiconductors. | |
| 135. |
Which of the following statements are ture about semi-conductors ?A. Silicon doped with electron rich impurity is a p-type semi-conductorB. Silicon doped with an electron rich impurity is an n-type semi-conductorC. Delocalised electrons increase the conductivity of doped siliconD. An electron vacancy increases the conductivity of a-type semi-conductor. |
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Answer» Correct Answer - C (b,c) are corrcet statements. (a) Silicon doped with electron rich impurity is an n-type semiconductor. Therefore, staement (a) is false. ltbRgt (b) The conductivity of n-type semi-conductors increases due to extraelectrons and not because of electron valency. Therefore statement (b) is false. |
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| 136. |
If the unit cell of a mineral has cubic close packed (ccp) array of oxygen atoms with m fraction of octahedral holes occupied by aluminium ions and n fraction of tetrahedral holes occupied by magnesiums ions, m and n respectively, areA. `(1)/(2),(1)/(8)`B. `1,(1)/(4)`C. `(1)/(2),(1)/(2)`D. `(1)/(4),(1)/(8)` |
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Answer» Correct Answer - A |
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| 137. |
If the unit cell of a mineral has cubic close packed (ccp) array of oxygen atoms with m fraction of octahedral holes occupied by aluminium ions and n fraction of tetrahedral holes occupied by magnesiums ions, m and n respectively, areA. `1/2, 1/8`B. `1, 1/4`C. `1/2, 1/2`D. `1/4, 1/8` |
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Answer» Correct Answer - A For ccp arrangement , `Z=4` `:.` No. of oxygen (O) atoms present `=4` No. of octahedral voids `=4` No. of tetrahedral voids `=2xx4=8` No. `Al^(3+)` ions `=mxx4` No. of `Mg^(2+)` ions `=nxx8` The formula of mineral `=Al_(4m)Mg_(n8)O_(4)` `4m(+3)+8n(+2)+4(-2)=0` or `3m+4n-2=0` or `3m+4n=2` The possible values of m and n are `1//2` and `1//8` respectively Substituting these values `(3xx1/2)+(4xx1/8)=2` |
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| 138. |
What are semi-conductors ? Describe the two main types of semi- conductors. |
| Answer» Semi-conductors are the solids whose conductivity lies in between those of conductors and insulators. It normally ranges between `10^(-6)` to `10^(4)ohm^(-1)cm^(-1)`. Two main types of semi-conductors are : n-type semi-conductors , p-type semi-conductors. | |
| 139. |
KF has ccp structure. Calculate the radius of the unit cell if the edge length of the unit cell is 400 pm. How many `F^(-)` ions and octahedral voids are there in the unit cell ? |
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Answer» For ccp unit cell, radius (r) `=(sqrt2a)/(4)` `=(sqrt2xx(400"pm"))/(4)=(1.414xx(400"pm"))/(4)` There are four `F^(-)` ions and four octahedral voids per unit cell. |
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| 140. |
KF has ccp structure. Calculate the radius of unit cell if the side of the cube or edge length is 400 pm. How many ` Fe^(-)` ions and octahedral voids are there in the unit cell ? |
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Answer» ccp is same as fcc, hence, ` r= a/ (2sqrt2) = 0.3535 a = 0.3535 xx 400 "pm" = 141.4 ` pm. In fcc, struture, ther are ` 4 F^(-)` ions in the packing and hence 4 octahedral voids. |
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| 141. |
A metal crystallizes into two cubic phases, face- centred cubic (fcc) and body -centred cubic (bcc) whose unit cell lengths are 3.5 and 3.0 Å respectively. Calculate the ratio of the densities of fcc and bcc. |
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Answer» Denstiy ` (p) = ( Z xx M)/(a^(3) xx N_(0)) ` For Z = 4, a = 3.5 Å = 3.5 Å = 3.5 ` xx 10^(-8)` cm . ` p_("fcc")= (4xxM) /((3.5 xx 10^(-8))xx N_(0)) ` for bcc, Z =2 , a = 3.0 Å = 3.0 ` xx 10^(-8)` cm `p_("bcc")= ( 2xx M)/((3.0 xx10^(-8))^(3) xx N_(0)) therefore p_("fcc")/p_("bcc") = ( 4xx (3.0 xx 10^(-8))^(3))/(2xx (3.5 xx10^(-8))^(3) ) =1.259` |
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| 142. |
A meteal crystallises into two cubic phases, face-centred (FCC) and body - centred cubic (BCC) whose unit cell lengths are 3.5 and 3.0 Å respectively. Calculate the ratio of the densities of FCC and BCC. |
| Answer» `(p_("FCC"))/(p_("BCC")) = (Z_("FCC"))/((a^(3)) _("FCC")) xx ((a^(3))_("BCC"))/(Z_("BCC")) = (4 xx (3.0)^(3))/(2 xx (3.5)^(3)) = 1.26 ` | |
| 143. |
X- ray diffraction studies show that copper crystallizes in an fcc unit cell with cell edge of ` 3.608 xx 10^(-8) ` cm. In a separate experiment, copper is deteremined to have a density of 8.92 g/`cm^(3)` calculate the atomic mass of copper. |
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Answer» ` P = (Z xx M)/(a^(3) xx N_(0)) or M = ( p xx a^(3) xx N_(0))/Z` for fcc lattice , Z= 4. Hence , ` M = (( 8.92 " g cm" ^(-3)) ( 3.608 xx 10^(-8) cm) ^(3) ( 6.022 xx 10^(23) " atoms mol"^(-1))) = 63.1 " g mol ^(-1)` Atomic mass of copper = `63.1 "g mol"^(-1)` |
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| 144. |
`X`-rays diffraction studies show that copper crystallizes in an fcc unit cell with cell edge of `3.608 xx 10^(-8) cm`. In a separte experiment, copper is determined to have a density of `8.92 g cm^(3)`. Calculate the atomic mass of copper. |
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Answer» In case fo fcc lattice, number of atoms per unit cell, z = 4 atoms. Therefore, `M=(pN_(0)a^(3))/(Z)` `=((8.92"g cm"^(-3))xx(6.022xx10^(23)"atoms mol"^(-1))xx(3.608xx10^(-8)"cm")^(3))/(("4 atmos"))=63.1"g/mol"`. Atomic mass of copper = 63.1 u. |
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| 145. |
X-ray diffraction studies show that copper crystallises in an fcc unit cell with cell edge of `3.608 xx 10^(-8) ` cm . In a separate experiment , copper is determined to have a density of 8.92 g/`cm^(3)` , calculate the atomic mass of copper . |
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Answer» In case of fcc lattice , number of atomic per unit cell z = 4 atoms Therefore ,M = `(dN_(A) a^(3))/(z)` `= (8.92 g cm^(-3) xx 6.022 xx 10^(23) "atoms" mol^(-1) xx (3.608 xx 10^(-8) cm)^(3))/(4 "atoms") = 63.1 ` g/mol . Atomic mass of copper = 63.1 u |
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| 146. |
Silver froms ccp lattice and `X`-ray studies of its crystals show that the edge length of its unit cell is `408.6` pm. Calculate the density of silver (atomic mass `= 107.9 u`). |
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Answer» Since the lattice is ccp , the number of silver atoms per unit cell = z = 4 Molar mass of silver = `107.9 g mol^(-1) = 107.9 xx 10^(-3) kg mol^(-1)` Edge length of unit cell = a = 408. 6 pm = 408.6 `xx 10^(-12)`m Density , d = `(zM)/( a^(3) . N_(A))` `= (4 xx (107.9 xx 10^(-3) kg mol^(-1)))/((408.6 xx 10^(-12) m)^(3) (6.022 xx 10^(23) mol^(-1)))` `=10.5 xx 10^(3) kg m^(-3)` `= 10.5 g cm^(3)`. |
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| 147. |
Atoms of element B form hcp lattic and those of the element A occupy 2/3 of tetrahedral voids. What is the formuala fo the compound formed by these elements A and B ? |
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Answer» Suppose number of atoms B in hcp lattic = n As number of tetrahedral voids is double the number of atoms in the close packing . Therefore, number of tetrachedral voids = 2n. As atoms A occupy 2/3rd of the tetrahedral voids, therefore, number of atoms A in the lattic `= 2/3 xx 2n = ( 4n) /3` Ratio of A : B = `(4n)/2 : n = 4/3 : 1= 4:3 ` Hence, the formula of the compound ` A_(4)B_(3) ` |
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| 148. |
Classify the following solids in different based on the nature of intermolecular forces operating in them : Potassium sulphate, tin, benzene, urea, ammonia, water,zine sulphide, graphite, rubidium, argon, silicon carbide. |
| Answer» Potassium sulphate = Ionic, Tin = Metallic , Benzene = Molecular ( non-polar) , Urea = Molecular ( Polar) , Ammonia = Molecular ( Hydrongen bonded ), water = Molecular ( Hydrongen bonded ) , Zine sulphide = Ionic , Graphite = Covalent or Network , Rubidium = Metallic, Argon= Molecular ( Non- Polar), Silicon carvbide = Covalent or Network. | |
| 149. |
Distinguish between : (i) Hexagonal and monoclinic unit cells. (ii) Face-centred and end-centred unit cells. |
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Answer» (i) For hexongonal unit cell , ` a = b ne c, alpha = beta = 90^(@) , gamma= 120^(@)` for monoclinic unit , ` a ne b ne c, alpha = gamma = 90^(@) , beta ne 90^(@)` , (ii) A face -centred unit cell has one constituent paticel present at the centre of each face in addition to the particles present at the corners. An end-face centred has one constituent particle each at the centre of any two opposite faces in addition to the particles present at the corners. |
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| 150. |
Classify the following solids in different categories based on the nature of intermolecular forces operating in them: ltbr. Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide,graphite, rubidium , argon, silicon carbide. |
| Answer» Ionic, metallic, molecular, molecular, molecular (hydrogen bonded), molecular (hydrogen bonded), ionic, covalent, metallic, molecular, covalent (net work). | |