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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.
| 201. |
In a compound,oxide ions are arranged in cubic close packing arrangement. Cations A occupy one-sixth of the tetrahdral voids and cations B occupy one-third of the octahedral voids. The formula of the compound isA. `A_(2)BO_(3)`B. `AB_(2)O_(3)`C. `A_(2)B_(2)O_(2)`D. `ABO_(3)` |
| Answer» Correct Answer - D | |
| 202. |
Which one of the following schemes of ordering closed packed sheets of equal sized spheres does not generate close packed latticeA. ABCABCB. ABACABACC. ABBAABBAD. ABCBCABCBC |
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Answer» Correct Answer - c In `ABB A ABB A` , there is no close packing as there are repeated planes adjecent to each other. |
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| 203. |
A solid between `A` and `B` has the following arrangement of atoms `:` `(i)` Atoms `A` are arranged in `c.c.p.` array. `(ii)` Atoms `B` occupy all the all the octahedral voids and half the tetrahedral voids. What is the formula of the compound ? |
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Answer» Correct Answer - `AB_(2)O_(4);` |
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| 204. |
Structure of a mixed oxide is cubic close packed the cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal.A and the octahedral voids are occupied by a monvalent metal `B`. The formula of the oxide is :A. `AB_(2)O_(2)`B. `ABO_(2)`C. `A_(2)BO_(2)`D. `A_(2)B_(3)O_(4)` |
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Answer» Correct Answer - 1 oxide ions`(O^(2-))` are arranged in fcc manner. Therefore number of oxide ions per unit cell is `8xx(1)/(8)+6xx(1)/(2)=4` Therfore the number of octahedral voids is `4` and number of tetrahedral voids is `8`. One fourth of the tetrahedral voids are occupied by divalent metal `A`. Therfore number of `A^(2+)` ions per unit cell is `(1)/(4)xx8=2` All the octahedral voids are occupied by a monovalent metal `B`.Therfore number of `B^(+)` ions per unit cell is `4`. Consequently the emipirical formula of mixed oxide is `A_(2)B_(4)O_(4)` or `AB_(2)O_(2)` |
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| 205. |
Structure of a mixed oxide is cubic close packed the cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal.A and the octahedral voids are occupied by a monvalent metal `B`. The formula of the oxide is :A. `ABO_(2)`B. `A_(2)BO_(2)`C. `A_(2)B_(3)O_(4)`D. `AB_(2)O_(2)` |
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Answer» Correct Answer - D No. of atoms `(O^(2-))` in ccp `=4` No. of tetrahedral voids `=2xxN=8` No. of `A^(2+)` ions `=8xx1/4=2` No. of octahedral voids =No of `B^(+)` ions =4 Ratio , `O^(2-),A^(2+):B^(+)=4:2:4=2:1:2` Formula of oxide `=AB_(2)O_(2)` |
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| 206. |
By X-ray studies, the packing atoms, in a crystal of gold is found to be in layers such that starting from any layer, every fourth layer is found to be exactly idenctical. The density of gold is found to be ` 19.4 g cm^(-3)` and its atomic mass is 197 a.m.u. The length of the edge of the unit cell will beA. 407 pmB. 189 pmC. 814 pmD. 204 pm |
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Answer» ` p = (Z xx M)/( a^(3) xx N_(0) xx 10^(-30)) or a^(3) = (Z xx M)/( p xx N_(0) xx 10^(-30))` For ccp , i.e, f.c.c , Z= 4 Hence , ` a^(3) = ( 4 xx 197)/( 19.4 xx 6.02 xx 10^(23) xx 10^(-30))` ` = 6.747 xx 10^(7) = 67.47 xx 10^(6)` ` a = (67.47)^(1//3) xx 10^(2)` ` 4.07 xx 10^(2) "pm" = 407 "pm"` |
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| 207. |
Structure of a mixed oxide is cubic closed - packed (ccp) .The cubic unit cell of mixed oxide is composed of oxide ions .One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovelent metal B .The formula of the oxide isA. `ABO_(2)`B. `A_(2)BO_(2)`C. `A_(2)B_(3)O_(4)`D. `AB_(2)O_(2)` |
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Answer» Correct Answer - D `A^(2+)=(1)/(4)xx8=2` `B^(+)=4xx1=4` `O^(2-)=8xx(1)/(8)+6xx(1)/(2)=4]A_(2)B_(4)O_(4) AB_(2)O_(2)` |
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| 208. |
By X-ray studies, the packing atoms, in a crystal of gold is found to be in layers such that starting from any layer, every fourth layer is found to be exactly idenctical. The density of gold is found to be ` 19.4 g cm^(-3)` and its atomic mass is 197 a.m.u. The fraction occupied by gold atoms in the crystal isA. 0.52B. 0.68C. 0.74D. `1.0` |
| Answer» In ccp , the fraction occupied is 0.74 . | |
| 209. |
Which of the following statement are correct for the ionic solids in which positive and negative ions are held by strong electrostatic attractive forces ?A. The radius `r^+//r^-` increases as coordination number increasesB. As the differnce in size of ions increases, coordination number increasesC. When coordination number is eight , the `r^+//r^-` ration lies between 0.225 to 0.414 .D. Coordinationumber of `Zn^(2+) " and " S^(2-)` respectively are 4 and 4 |
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Answer» Correct Answer - C when coordination number is eight , the radius ratio `r^+/r^-` lies between 0.732 to 1.000 |
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| 210. |
A solid is made up of two elements A and B . Atoms B are in ccp arrangement while atoms A occupy all the tetrahedral sites. Predict the formula of the compound .A. `A_2B`B. `AB_2`C. ABD. `AB_3` |
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Answer» Correct Answer - A The formula of the compound is `A_2B` |
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| 211. |
In a closed packed structure of mixed oxides, the lattice is composed of mixed oxides ions. One-eighth of tetrahedral voids are occupied by divalent cation `(A^(2+))` while one-half of octahedral voids are occupied by trivalent cations `(B^(3+))`. The fromula of mixed oxide is |
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Answer» Correct Answer - `AB;` |
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| 212. |
A closed packed structure of uniform spheres has the cell edge `=0.8mm`. Calculate the radius of molecule if it has `:` `(a)` simple cubic lattice, `(b)` `b.c.c.` lattice, `(c)` `f.c.c.` lattice. |
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Answer» Correct Answer - `(a)` 0.4mm, `(b)` 0.3464 mm, `(c)` 0.2828mm; |
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| 213. |
Structure of a mixed oxide is cubic closed - packed (ccp) .The cubic unit cell of mixed oxide is composed of oxide ions .One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovelent metal B .The formula of the oxide isA. `ABO_2`B. `A_2BO_2`C. `A_2B_3O_4`D. `AB_2O_2` |
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Answer» Correct Answer - D For ccp arrangement , number of `O^(2-)` ions = 4 Number of octahedral voids =4 Thus number of tetrahedral voids = 8 As `A^(2+)` (divalent matal) occupy 1/4 tetrahedral voids , Thus , number of `A^(2+)` ions `=1/4xx8=2` Again, `B^+` occupied all octahedral voids hence, number of B ions = 4 `therefore " Structure of oxide " = AB_2O_2` |
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| 214. |
By X-ray studies, the packing atoms, in a crystal of gold is found to be in layers such that starting from any layer, every fourth layer is found to be exactly idenctical. The density of gold is found to be ` 19.4 g cm^(-3)` and its atomic mass is 197 a.m.u. Assuming gold atom to be spherical , its radius will beA. 203.5 "pm"^(2)`B. 143.9 pmC. 176. 2 pmD. 287 . 8 pm |
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Answer» Correct Answer - c For fcc, ` r = a/(2sqrt2) = 0.3535 a` ` = 0.3535 xx 407 "pm" = 143.9" pm"` |
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| 215. |
Whenever two-demiensional square packing same layers are kept in the way so that the centres are aligned in all three dimensation, coordination numberof each sphere isA. 6B. 8C. 12D. 10 |
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Answer» Correct Answer - A a. The coordination number is `6. ("4 atoms"` in the plane `+ 1` atom above `+ 1` atoms below the particular atom). |
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| 216. |
Three element `A,B,C` crystallize into a cubic solid lattice.Atoms `A` occupy the corners `B` atoms the cube centres and atom `C` the edge .The formula of the compound isA. ABCB. `ABC_(2)`C. `A_(2)B_(3)`D. `A_(2)B_(5)` |
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Answer» Correct Answer - C Number of atoms of A per unit cell `8xx(1)/(8)` (at the corners)=1 Number of atoms of B per unit cell =1 (at the centre )=1 Number of atoms of C per unit cell `=12xx(1)/(4)` (at the edge centre)=3. So, formula of compound `ABC_(3)`. |
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| 217. |
In a closed packed structureA. tetrahedral voids and bigger than octahedralB. tetrahedral voids are smaller than octahedralC. tetradhedral voids are equal in size as octahedralD. None of the above |
| Answer» Correct Answer - 2 | |
| 218. |
which of the following statemets is not true about the hexagonal close packing ?A. The coordination number is 12B. It has 74% packing efficiencyC. Tetrahedral voids of the second layer are covered by the spheres of the third layerD. In this arrangement spheres of the fourth layer are exactly aligned with thoseof the first layer |
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Answer» Correct Answer - D In hexagonal close packing,it is the third layer,spheres of which are exactly aligned with those of the first layer. Other given statements are ture. |
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| 219. |
Hexagonal close packing is found in crystal lattice ofA. NaB. MgC. AlD. None of these |
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Answer» Correct Answer - B Mg has hexagonal close packing |
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| 220. |
Three elements `P,Q` and `R` crystallize in a cubic solid lattice. The `P` atoms occupy the corners. `Q` aroms the cube centres and `R` atoms the edges. The formula of the compound isA. `PQR`B. `PQR_(2)`C. `PQR_(3)`D. `PQ_(3)R` |
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Answer» Correct Answer - 3 `P=8xx(1)/(8)=1,Q=1=1,R=12xx(1)/(4)=3,`formula`=PQR_(3)` |
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| 221. |
Three element `A,B,C` crystallize into a cubic solid lattice.Atoms `A` occupy the corners `B` atoms the cube centres and atom `C` the edge .The formula of the compound isA. ABCB. `ABC_(2)`C. `ABC_(3)`D. `ABC_(4)` |
| Answer» Correct Answer - 3 | |
| 222. |
If three elements A, B and C crystallize in a cubic structure with atoms A at corners, B at the body centre and C at the face crntres, the formula of the compound will beA. ABCB. `ABC_(2)`C. `ABC_(3)`D. `AB_(2)C` |
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Answer» Correct Answer - B Atoms A per unit cell `=8xx(1)/(8)=1` Atomc B per unit cell = 1 Atomc C per unit cell `=6xx(1)/(8)=3` Ratio `A:BC=1:1:3` . Hence,formula is `ABC_(3)` . |
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| 223. |
Why an end-centred unit cell cannot he cubic? What is the highest possible symmetry for this type of unit cell? |
| Answer» A cubic unit cell must have all the six faces the same. The highest symmetry possible for a centred unit cell is tetragonal. | |
| 224. |
Assertion. Antiferromagnetic substance on heating temperature become paramagnetic. Raason. On heating, randomisation of spins occurs. |
| Answer» R is the correct explanation of A. | |
| 225. |
Why uncharged atoms or mulecular never crystallize in simple cubic structures? |
| Answer» Uncharged atoms and molecular pack more efficiently in close-packed structures. | |
| 226. |
Assertion. When 1.0 mol of NaCl is doped with `10^(-3) "mol " SrCl^(2)` . The number of cationic of cationic sites remaining vacant is `10^(-3)` . Each `SrCl_(2)` unit produces two cation vacancies. |
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Answer» Correct Answer - a Correct A. As each `SrCl_(2)` unit creates one cation vacancy, number of cation vacancies will be ` 10^(-3) " mol" = 10^(-3) xx ( 6.02 xx 10^(23)) = 6.02 xx 10^(20)` Correct R. Each `SrCl_(2)` produces one cation vacancy . |
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| 227. |
Calculate `lambda` of X-rays which give a diffraction angle `2 theta = 16.8^(@)` for crystal, if the interplanar distance in the crystal is `0.2 nm` and that only for the first-order diffraction is observed. Given `sin 8.40^(@) = 0.146`. |
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Answer» Correct Answer - `5.84xx10^(-11)m;` |
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| 228. |
What will be the wavelength of the X-rays which give a diffraction angle `2theta` equal to `16.80^(@)` for a crystal ? The interplanar distance in the crystal is `0.200` nm and only the diffraction of the first order is absorbed. |
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Answer» According to Bragg equation, `nlambda=2dsinthetaor lambda=(2dsintheta)/(n)` `2theta=16.80^(@)and theta=8.40^(@),n=1` (First order diffraction) `d=0.2nm=0.2xx10^(-9)m` `lambda=(2xx(0.2xx10^(-9))xx(sin8.4^(@)))/(1)` `=(2xx(2xx10^(-10)m)xx(0.146^(@)))/(1)=0.0584xx10^(-9)m=0.0584nm` |
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| 229. |
The figure below shown a unit cell of the mineral perovskite (the titanium atom is at the centre of the cube), What of the following is a correct chemical formula for this mineral? A. `Ca_(8)TiO_(6)`B. `CaTiO`C. `Ca_(2)TiO_(3)`D. `CaTiO_(3)` |
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Answer» Correct Answer - d Effective number of `Ca= 8 xx (1)/(8) = 1` Effective number of `Ti= 1 xx 1 = 1` Effective number of `O= 6 xx (1)/(2) = 3` Formula `= CaTIO_(3)` |
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| 230. |
Consider the following statements I. InSb, AIP , GaAs are the compound of groups 13-15 II. ZnS, CdS , HgTe are the compounds of groups 12-16 III. `ReO_3` is like metallic copper in its conductivity IV. Metallic or insulating properties of certain oxides like VO, `VO_2, VO_3 " and " TiO_3` are independent on temperature . Which of the above statements is /are incorrect ?A. I and IIB. II and IIIC. Only IIID. Only IV |
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Answer» Correct Answer - D All the above three statements are correct but IV is wrong because metallic oxides like `VO,VO_2, VO_3 " and " TiO_3` are dependent of temperature. |
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| 231. |
Perovskite, a mineral of titanium is found to contain calcium atoms at the corners, oxygen atoms at the face centres and titanium atoms at the centre of the cubic. Oxidation number of titanium in the mineral isA. `+2`B. `+3`C. `+4`D. `+1` |
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Answer» Correct Answer - C Ca atoms per unit cell =` 8xx 1/8 =1 ` O atoms per unit cell =` 6 xx 1/2 =3` Ti atom per unit cell =1 Hence, the formula of the mineral is ` CaTiO_(3)` Suppose ox. No. of Ti =x . Then + 2 + x + 3 (-2) = - or x = +4 |
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| 232. |
In terms of dielectric properties, barium titanate is…… |
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Answer» Correct Answer - curie temperature. |
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| 233. |
Perovaskite, a mineral containing calcium, oxygen `&` titanium crystallizes in the given unit cell The oxidation number of titanium in the perovskite isA. `+5`B. `+4`C. `+3`D. `+2` |
| Answer» `CaTiO_(3),2+x-6=0,x=4` | |
| 234. |
Analysis shows that nickel oxide has the formula `Ni^(0.98) 0 ,1.00` , what fractions of nickel exist as `Ni^(2+) and Ni^(3+)` ions ? |
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Answer» In pure nickel oxide (NiO) the ratio of Ni and O atoms = `1 : 1` Let x be the no. of Ni (II) atoms replaced by Ni (III) atoms in the oxide `therefore` Number of Ni (II) atoms present = `0.98 - x` Total charge on Ni atoms = charge on oxygen atom (`because` the oxide is neutral) `2(0.98 - x) + 3x = 2` `1.96 - 2x + 3x = 2` `x = 2 - 1.96 = 0.04` % of Ni (III) atoms in Nickel oxide = `("No. of Ni (iii) atoms")/("Total No. of Ni atoms") xx 100` `= (0.04)/(0.98) = 4.01`% % of Ni (II) atoms in nickel oxide = `100 - 4.01 = 95.99 %` |
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| 235. |
Analysis show that nickel oxide consists of nickel ion with `96%` ions having `d^(8)` configuration and `4%` having `d^(7)` configuration. Which amongst the following best represents the formula of the oxide?A. `Ni_(0.98)O_(0.98)`B. `Ni_(1.02)O_(1.02)`C. `Ni_(1.02)O_(1.00)`D. `Ni_(0.98)O_(1.00)` |
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Answer» Correct Answer - D E.C of `Ni^(2+)=3d^8 " and " Ni^(3+) = 3d^7` Hence, 96% ions of `Ni^(2+)` and 4% ions of `Ni^(3+) are present . Let numbers of `O^(2-) ion present in the crystal = x Applying electroneutrality rule, Total positive charge = Total negative charge `96xx2+4xx3=xxx2` `therefore 96xx2+4xx3-2x=0 rArr x=1.02` So, formula of crystal `Ni_(100)O_(1.02) " or " Ni_(0.98)O_(1.00)` |
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| 236. |
Barium titanate crystallizes in perovskite structure `(ABO_(3))` which is a cubic lattice with barium ions occupying the corners of the unit cell ,oxide ions occupying the face centres and titanium ion occupying centes of unit cell.if `Ti^(4+)` ions are described as occupying the holes in `Ba-O` lattice. then type of hole and fraction of these holes occupied by these ions areA. `100%` of tetrahedral holesB. `25%` of octahedral holesC. `100%` of octahedral holesD. `25%` of tetrahedral holes |
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Answer» Correct Answer - 2 Since the centre of an fcc unit cell is an octahedral void,`Ti^(4+)` ion is present in the octahedral void. Total number of octahedral voids per fcc unit cell is `4`. Since only one of the voids is captured we have `(1)/(4)xx100%=25%` occupany. |
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| 237. |
A solid compound contains X,Y and Z atoms in a cubic lattice with X atoms occupying the corners,Y atoms in the body centred positions and Z atoms at the centres of faces of the unit cell. What is the empirical formula of the compoundA. `XY_2Z_3`B. `XYZ_3`C. `X_2Y_2Z_3`D. `X_8YZ_6` |
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Answer» Correct Answer - B Number of X atoms per unit cell `=1/8xx8=1` and number of Z atoms per unit cell `=1/2xx6=3` Hence , the formula of compound `=XYZ_3` |
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| 238. |
A solid compound contains X,Y and Z atoms in a cubic lattice with X atoms occupying the corners. Y atoms in the body centered positions and Z atoms at the centres of faces of the unit cell. What is the empirical formula of the compound.A. `XY_(2)Z_(1)`B. `XYZ_(3)`C. `X_(2)Y_(2)Z_(3)`D. `X_(8)YZ_(6)` |
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Answer» Correct Answer - B Atoms of X per unit cell `=8xx(1)/(8)=1` Atoms of Y per unit cell `=1` Atoms of Z per unit cell `=6xx(1)/(2)=3` Hence the formula is `XYZ_(3)`. |
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| 239. |
The number of atoms per unit cell in a simple cube, face `-` centred cube and body`-` centred cube are respectively `:`A. `1,4,2`B. `1,2,4`C. `8,14,9`D. `8,4,2` |
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Answer» Correct Answer - A The no. of atoms per unit in `s.c., f.c.c.` and `b.c.c.` are `1,4,2,` respectively. |
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| 240. |
In a unit cell, atoms `(A)` are present at all corner lattices, `(B)` are present at alternate faces and all edge centres. Atoms `(C)` are present at face centres left from `(B)` and one at each body diagonal at disntance of `1//4th` of body diagonal from corner. Formula of given solid isA. `A_(3)B_(8)C_(7)`B. `AB_(4)C_(6)`C. `A_(6)B_(4)C_(8)`D. `A_(2)B_(9)C_(11)` |
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Answer» Correct Answer - B Number of A atom `= 8("corner") xx 1/8 ("per corner share")` `= 1` Number of B atom `= 2("Alternate face") xx 1/2 ("face center share ")+12("edge")` `xx 1/4 ("edge centre share")` `= 1 + 3 = 4` Number of `C` atom `= 4 ("face centres of left from B")` `xx 1/2 ("face centre share")` `+1 ("at body diagonal")` `xx 4 ("body diagonal")` `= 2 + 4 = 6` Formula : `AB_(4) C_(6)`. |
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| 241. |
A unit cell consists of a cube in which anions (B) are present at each corner and cations (A) at centre of the alternate faces of the unit cell. What is the simplest formula of the compound ? |
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Answer» Correct Answer - `A_(3)B_(2)` Contribution made by anion (B) at one corner = 1/8 Condtribution made by anions (B) at eight corners `=1//8xx8=1` Since cations (A) are present at the centres of all alternate faces only, this means that only three face centres of the cubic unit cell have cation (A). Contribution made by cation (A) at one face = 1/2 Contribution made by cations (B) at three faces `=3xx1//2=3//2` `therefore` formula of the compound : `A_((3//2))B or A_(3)B_(2)`. |
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| 242. |
CsBr has bcc like structures with edge length `4.3Å`. The shortest inter ionic distance in between `Cs^(+)` and `Br^(-)` is:A. 3.72pmB. 1.86pmC. 7.44pmD. 4.3pm |
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Answer» Correct Answer - A |
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| 243. |
What is the total number of atoms per unit cell in a face centred cubic (fcc) structure ?A. 6 at edge centres and8 along body diagonalsB. 12 at edge centres and one at body centreC. 8 along body diagonal and6 at edge centtresD. all the edge centres only . |
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Answer» Correct Answer - B In fcc or ccp, 12 octahedral voids are present at edge centres and one at the body centre. |
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| 244. |
What is the total number of atoms per unit cell in a face centred cubic (fcc) structure ?A. 1B. 2C. 0.5D. 2.5 |
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Answer» Correct Answer - C Contribution of each atom per unit cell located at face centre in a fcc unit cell is 0.5 or `1/2` |
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| 245. |
Which kind of defets are intorduced by doping ?A. Disolocation defectB. Schottky defectC. Frenkel defectD. Electronic defects |
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Answer» Correct Answer - B Doping is done with electron rich or electron deficit impuities. Hence, it introduces electronic defects. |
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| 246. |
What is the effect of Schottky defect on the density of crystal ? |
| Answer» Correct Answer - It decreases. | |
| 247. |
Consider the figure of a defective crystal : (a) Name the defect shown by this diagram. (b) What is the effect of this defect on the solid ? (c) Name as ionic compound which can show this type of defect. |
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Answer» (a) Frenkel defect since cation size is smallar than that of the anion. (b) No change in density since cation size is smaller than that of the anion. (c) ZnS. |
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| 248. |
An element has a bcc structure with a celledge of `288` pm. The density of the element is `7.2 g cm^(-3)`. How many atins are present in `208 g` of the element? |
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Answer» We know that `rho=(ZxxM)/(a^(3)xxN_(0)xx10^(-30))orN_(0)=(ZxxM)/(a^(3)xxrhoxx10^(-30))` Edge length of the unit cell (a) = 288 pm = 288 No. of atoms per unit cell (Z) = 2 Density of the unit cell `(rho)=7.2g//cm^(3)` Mass of the element (M) = 208g Number of atoms (N) present may be calculated as : `N=(2xx(208g))/((288)^(3)xx(7.2"g mol"^(-3))xx(10^(-3)cm^(3)))=24.18xx10^(23)` |
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| 249. |
The coordination number of Al in the crystalline of ` AlCl_(3)` is |
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Answer» Correct Answer - 8 At low temperature . `AlCl_(3)` exists as a close paked lattice of ` Cl^(-)` ions with ` Al^(3+)` ions occupying octahedral voids. Thus, ` Al^(3+)` ions has 6 `Cl^(-)` ions around it. Hence, its corrdination number is 6. |
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| 250. |
Consider a Body Centred Cube (BCC) arrangement, let `d_(e), d_(fd), d_(bd)` be the distances between successive atoms located along the edge, the face-diagonal, the body diagonal respectively in a unit cell. Their order is given as:A. `d_(e) lt d_(fd) lt d_(bd)`B. `d_(fd) gt d_(bd) gt d_(e)`C. `d_(fd) gt d_(e) gt d_(bd)`D. `d_(bd) gt d_(e) gt d_(fd)` |
| Answer» Correct Answer - C | |