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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.
| 1251. |
Assertion : Conductivity of silicon increases by doping it with group-15 elements. Reason : Doping means introduction of small amount of impurities like P, As or Bi into the pure crystal.A. P is non-metal whereas Al is a metalB. P is a poor conductor while Al is a conductorC. P gives rise to extra electrons while Al gives rise to holesD. P gives rise to holes while Al gives rise to extra electrons |
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Answer» Correct Answer - C Doping of Si with P gives extra electrons while doping with Al gives rise to holes. |
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| 1252. |
Doping mens introduction of small amount of impurities like phosphorus , arsenic or boron into the pure crystal . In pure silicon , ther are four valenices used in bonding with other four adjacent silicon crystal is doped with a group -15 element ( with five valence electron ) such as P, As , or Bi , the structure of the crystal lattic remains unchanged . Out of the five valence electron of group -15 doped element four element are used in normal covalent bonding with silicon while fiffth electron is delcoasiled and thus conducts electricity Doping a silicon crystal with a group -13 element (with three valence electrons ) such as B, Al, Ga or In products a semiconductor with three electrons in in dopant . The place where fourth electron is missing is called an electron vacancy or hole . Such hole can move throught the crystal like a positive charge giving rise conduction of electricity. Silicon that has been dopend with group - 13 elements is called :A. `p`- type semicondouctorB. n-type semiconductorC. electron vancany or holeD. None of these |
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Answer» Correct Answer - A |
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| 1253. |
Doping silicon with boron produces a -A. n-type semiconductorB. p-type semiconductorC. MetaD. Insulator. |
| Answer» Correct Answer - B | |
| 1254. |
Doping mens introduction of small amount of impurities like phosphorus , arsenic or boron into the pure crystal . In pure silicon , ther are four valenices used in bonding with other four adjacent silicon crystal is doped with a group -15 element ( with five valence electron ) such as P, As , or Bi , the structure of the crystal lattic remains unchanged . Out of the five valence electron of group -15 doped element four element are used in normal covalent bonding with silicon while fiffth electron is delcoasiled and thus conducts electricity Doping a silicon crystal with a group -13 element (with three valence electrons ) such as B, Al, Ga or In products a semiconductor with three electrons in in dopant . The place where fourth electron is missing is called an electron vacancy or hole . Such hole can move throught the crystal like a positive charge giving rise conduction of electricity. Silicon that has been dopend with group - 15 elements is called :A. `p`- type semicondouctorB. n-type semiconductorC. electron vancany or holeD. None of these |
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Answer» Correct Answer - B |
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| 1255. |
Doping mens introduction of small amount of impurities like phosphorus , arsenic or boron into the pure crystal . In pure silicon , ther are four valenices used in bonding with other four adjacent silicon crystal is doped with a group -15 element ( with five valence electron ) such as P, As , or Bi , the structure of the crystal lattic remains unchanged . Out of the five valence electron of group -15 doped element four element are used in normal covalent bonding with silicon while fiffth electron is delcoasiled and thus conducts electricity Doping a silicon crystal with a group -13 element (with three valence electrons ) such as B, Al, Ga or In products a semiconductor with three electrons in in dopant . The place where fourth electron is missing is called an electron vacancy or hole . Such hole can move throught the crystal like a positive charge giving rise conduction of electricity. No. of valene electrons in silicon are :A. 3B. 4C. 5D. 6 |
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Answer» Correct Answer - B |
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| 1256. |
The tetrahedral voids formed by ccp arrangement of `Cl^(-)` ions in rock salt structure are:A. Occupied by `Na^(+)` ionsB. Occupied by `Cl^(-)` ionsC. Occupied by either `Na^(+) " or " Cl^(-)` ionsD. Vacant |
| Answer» Correct Answer - D | |
| 1257. |
The tetrahedral voids formed by cup arrangement of `CI^(-)` ions in rock salt structure areA. Occupied by `Na^(+)` ionsB. Occupied by `CI^(-)` ionsC. Occupied by either `Na+ or CI^(-)` ionsD. Vacant |
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Answer» Correct Answer - d In rock salt stracture `CI^(-)` form (ccf) lattice and `Na^(+)` occupies octahdral voids so tertahedral volid are vacant. |
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| 1258. |
A compound AB has a rock salt type structure with A:B=1:1 . The formula weight of AB is 6.023 y amu and the closest A-B distance is `y^(1//3)` nm. Calculate the density of lattice (in `kg//m^(3))` |
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Answer» Correct Answer - 5 AB has rock slat (A:B::1:1) structure, i.e, f.c.c. structure (n=4) and formula weight of BB is 6.023 y amu having closet distance `A-B_(Y^(1//3))`nm. Therefore edge length of unit cell `=2(A^(+)+B^(-))=2xxy^(1//3)xx10^(-9)m` `therefore "Density of AB"(nxx"mol wt.")/(Av. no. xxV)` `=(4xx6.023xxyxx10^(-3))/(6.023xx10^(23)xx(2y^(1//3)xx10^(-9))^(3))` (mol. wt. in kg) =5.0kg `m^(-3)` |
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| 1259. |
Under which category iodine crystals are placed among the followingA. MetallicB. IonicC. MolecularD. Covalent |
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Answer» Correct Answer - C Iodine crystals are molcular. |
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| 1260. |
Which among the following will not show anisotropy ?A. `BaCl_(2)`B. `NaCl`C. GlassD. `KNO_(3)` |
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Answer» Correct Answer - C Crystalling soilds such as NaCl, `BaCl_(2)` ect. Will show anistropy. |
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| 1261. |
Amorphous solids areA. Solid substance in real senseB. Liquid in real senseC. Supercooled liquidD. Substance with difinite melting point |
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Answer» Correct Answer - C Amorphous solids neither have ordered arrangement (i.e. no difinite shape )nor have sharp melting point like crystals, but when heated, they become pliable unitl they assume the properties unually related to liquids. It is therefore they are regarded as super-cooled liquids. |
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| 1262. |
Define the term "amorphous". Give a few example of amorphous solids. |
| Answer» Amorphous or amorphous solids imply those solids in which the constituent particles have short range order. These have irregular shapes and are isotropic in nature. Apart from that they do not have sharp melting point. A few examples of amorphous solids are : glass, rubber, plastic, celluose etc. | |
| 1263. |
In NaCl is doped with `1 xx 10^(-3)` mol percent of `SrCl_(2)` , what is the concentration of cation vacancies ? |
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Answer» The addition of `SrCl_(2)` to NaCl , each `Sr^(+2)` ion replaces two `Na^(+)` ions and occupies only one lattice point in place of `Na^(+)` . Due to this one cation vacancy arised . The number of moles of cation vacancies in 100 moles of NaCl = `1 xx 10^(-3)` The number of moles of cation vacancies in 1 mole NaCl = `(1 xx 10^(3))/(100) = 10^(-5)` mole Total number of cation vacancies = `10^(-5) xx 6.023 xx 10^(23) = 6.023 xx 10^(18)`. |
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| 1264. |
Titanium (Ti) crystallizes in fcc lattice. It reacts with C or H intersititally and these elements occupy the voids of host lattice (i.e., Ti) radius that can be occupies in TVs without Casing strain in the host lattice. Calculate the formula of titanium hydride and titanium carbide.A. TiH,TiCB. `TiH_(2)` TiCC. Ti H , Ti` C_(2)`D. Ti `H_(2) Ti C_(2)` |
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Answer» Correct Answer - A For fcc lattice, `Z_(ff)=4Ti` atoms/unit cell Number of TVs=Number of H-atoms=8/unit cell (i) Formula of titanium hydride `=Ti_(4)H_(8)rArTiH_(2)` (ii) Number of Ovs=Number of C-atom=4/unit cell Formula of titanium carbide`=Ti_(4)C_(4)rArrTiC` |
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| 1265. |
The number of atoms contained in a fcc unit cell of a monoatomic substance isA. 1B. 2C. 4D. 6 |
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Answer» Correct Answer - C In fcc, contribution of each atom present at the corner = `1/8` Contribution of each atom present at the face centre `=1/2` `threrefore` Total number of atoms in fcc `8xx1/8+6xx1/2` =1+3=4 |
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| 1266. |
Number of unit cells in 4g of X(atomic mass=40). Which crystallises in bcc pattern in `(N_(0)="Avogadro number")`A. `0.1N_A`B. `2xx0.1N_A`C. `(0.1N_A)/(2)`D. `2xxN_A` |
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Answer» Correct Answer - C `40g=N_A` atoms `4g=0.1N_A` atoms 2 atoms form 1 unit cell in bcc crystal `therefore 0.1xxN_A "atoms"=(0.1xxN_A)/(2)` unit cells |
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| 1267. |
Number of unit cells in 4g of X(atomic mass=40). Which crystallises in bcc pattern in `(N_(0)="Avogadro number")`A. `0.1 N_(0)`B. `2xx0.1N_(0)`C. `(0.1N_(0))/(2)`D. `2xxN_(0)` |
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Answer» Correct Answer - 3 40 grams=N atoms 4 grams=0.1N atoms 2 atoms form 1 unit cell in bcc 0.1 N atoms `=(0.1xxN)/(2)` |
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| 1268. |
Number of unit cells in 4g of X(atomic mass=40). Which crystallises in bcc pattern in `(N_(0)="Avogadro number")`A. 0.1 `N_(A)`B. `2xx0.1N_(A)`C. `(0.1N_(A))/(2)`D. `2xxN_(A)` |
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Answer» Correct Answer - C 40 g = `N_(A)` atoms 2 atoms form 1 unit cell in bcc crystal `:.0.1xxN_(A)` atom `=(0.1xxN_(A))/(2)` unit cells |
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| 1269. |
Which point defect decreases the density of crystals ? |
| Answer» Correct Answer - Schottky defect | |
| 1270. |
Define the following terms in relation to crystalline solids : (i) Unit cell (ii) Coordination number. |
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Answer» Correct Answer - (a) p-type (b) Ferromagnetism (c ) Impurity defect producing cation vacancies |
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| 1271. |
First three nearestneighbour distances for body centered cubic lattice are respectively :A. `sqrt(2a),a,sqrt(3a)`B. `(a)/sqrt(2),a,sqrt(3a)`C. `sqrt(3a)/(2),a,sqrt(2)`D. `sqrt(3a)/(2),a,sqrt(3a)` |
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Answer» Correct Answer - C |
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| 1272. |
Given : The unit cell structure o fcompound is show below . The formula of compound is :A. `A_(8)B_(12)C_(15)`B. `AB_(2)C_(3)`C. `A_(2)B_(2)C_(5)`D. `ABC_(5)` |
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Answer» Correct Answer - B |
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| 1273. |
Which of the following is true about the value of refrective of quartz glass ?A. positiveB. neutralC. negativeD. depends on concentration of p impurity |
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Answer» Correct Answer - D All p-type semiconductors are nautral. P only indicates that move like positive charge, i.e, towards cathode . |
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| 1274. |
Sodium crystallizes in a bcc unit cell. Calcuate the approximate number of unit cells in 9.2 g of sodium (Atomic mass of Na=23) |
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Answer» Correct Answer - `7.64 xx 10^(20)` BCC has 2 atoms per unit cell. Proceed as in Solved problem 3 . |
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| 1275. |
Explain how vacancies are introduced in an ionic solid when a cation of higher valence is aded as an impurity in it ? |
| Answer» When a cation of higher vacancies is added as an impurity to an ionic solid , some vacancies are created . This can be explained with the help of an example . When strontium , chloride `(SrCl_(2))` is added as an impurity to ionic solid sodium chloride (NaCl) , two vacant sites are created by removal of one `Na^(+)` ion . One vacant site is replaced by `Sr^(2+)` ion but the other remains vacant . The reason is that the crystal as a whole is to remain electrically neutral. | |
| 1276. |
A compound `M_(p)X_(q)` has cubic close packing `(ccp)` arrangement of `X`. Its unit cell structure is shown below. The empirical formula of the compound is A. MXB. `MX_(2)`C. `M_(2)X`D. `M_(5)X_(14)` |
| Answer» Correct Answer - B | |
| 1277. |
A compound `M_(p)X_(q)` has cubic close packing `(ccp)` arrangement of `X`. Its unit cell structure is shown below. The empirical formula of the compound is A. `MX`B. `MX_(2)`C. `M_(2)X`D. `M_(5)X_(14)` |
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Answer» Correct Answer - B Number of `M` atoms `= 1/4 xx 4 + 1 = 1 + 1 = 2` Number of `X` atoms `= 1/2 xx 6 + 1/8 xx 8 = 3 + 1 = 4` So, formula of compound is `M_(2)X_(4) = MX_(2)` |
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| 1278. |
The second order Bragg diffraction of X-rays with `lambda=1.0 Å` from a set of parallel planes in a metal occurs at an angle `60^(@)`. The distance between the scattering planes in the crystals isA. `0.575 overset(@)A`B. `1.00 overset(@)A`C. `2.00 overset(@)A`D. `1.15 overset(@)A` |
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Answer» Correct Answer - C Quartz is a covalent solid in which constituent particles are atoms which are held together by covalent bond forces. |
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| 1279. |
The second order Bragg diffraction of `X` rays with `lambda=Å` form a set of parallel planes in a metal occurs at an angle `60^(@)`. the distance between the scattering planes in the crystal isA. `0.575 overset(@)A`B. `1.00 overset(@)A`C. `2.00overset(@)A`D. `1.15overset(@)A` |
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Answer» Correct Answer - 4 `nlambda=2d sin theta` |
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| 1280. |
An example of covalent solid is:A. MgOB. MgC. SiCD. `CaF_(2)` |
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Answer» Correct Answer - C SiC is a covalent solid |
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| 1281. |
An element crystallising in body centred cublic lattice has edge length of 500 pm. If the density is 4 g `cm^(-3)`, the atomic mass of the element `("in g mol"^(-1))` is (consider `N_(A)=6xx10^(23))`A. 100B. 250C. 125D. 150 |
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Answer» Correct Answer - D `M=(rhoxxN_(A)xxa^(3)xx10^(-30))/Z` `M=((4 g cm^(-3))xx(6xx10^(23)mol^(-1))xx(500)^(3)xx(10^(-30)cm^(3)))/2` `=150 g " mol"^(-1)` |
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| 1282. |
A compound `M_(p)X_(q)` has cubic close packing `(ccp)` arrangement of `X`. Its unit cell structure is shown below. The empirical formula of the compound is A. `MX`B. `MX_(2)`C. `M_(2)X`D. `M_(5)X_(14)` |
| Answer» Correct Answer - B | |
| 1283. |
The first `-` order diffraction os `X-` rays from a certain set of crystal planes occurs an angle of `11.8^(@)` from the planes. If the planes are `0.281 nm` apart, what is the wavelength of `X-` rays? |
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Answer» According to Bragg equation `nlambda=2dsintheta` `sintheta=sin11.8^(@)=0.2045,d=0.281nm,n=1` `lambda=(2dsintheta)/(n)=(2xx(0.281nm)xx0.2045)/(1)=0.115nm` |
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| 1284. |
Analysis shows that an oxide ore of nickel has formula `Ni_(0.98) O_(1.00)`.The percentage of nickel as `Ni^(3+)` ions is nearlyA. 96B. 4C. 7D. 93 |
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Answer» Correct Answer - 2 `3x+1.96-2x=2` `x+1.96=2, x=0.04, %=4` |
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| 1285. |
A metal oxide has empirical formula `M_(0.96)O_(1.00)` What will be the percentage of `M^(2+)` ion in the crystal ?A. `90.67`B. `91.67`C. `8.33`D. `9.33` |
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Answer» Correct Answer - B Let the amount of `M^(2+)` ions in the `M_(0.96)O_(1.00)=x` The amount of `M^(3+)` ions will be `=(0.96-x)` As the metal oxide is neural, `(+2)x+(+3)(0.96-x)+(-2)(1.00)=0` or `2x+2.88-3x-2.00=0` or `x=0.88` `%` of `M^(2+)` ions in the crystal `=0.88/0.96xx100` `=91.67%` |
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| 1286. |
The composition of a sample of Wustite is `Fe_(0.93)O_(1.00)`. What fraction of iron is present in the from of `Fe^(2+)` ion ? |
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Answer» The ratio of the Fe and O atoms in pure iron oxide (FeO) = `1:1` Let x atoms of Fe(II) be replaced by Fe(III) in Wustite. `therefore` No. of Fe(II) atoms present = `(0.93-x)` Since the oxide is neutral, Total charge on iron atoms = Charge on oxygen atom `2(0.93-x)+3x=2` `1.86-2x+3x=2` `x=2-1.86=0.14` No. of `Fe^(3+)` ions in the sample = 0.14 No. of `Fe^(2+)` ions in the sample = `0.93-0.14=0.79` Fraction of `Fe^(2+)` ions in the sample `=(0.79)/(0.93)=0.85` |
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| 1287. |
A group-14 element is to be converted into n-type semiconductor by doping it with a suitalbe impurity. To which group this impurity belong? |
| Answer» n-type semiconductors are conducting due to presence of excess of negatively charged electrons. In order to convert group 14 elements (e.g. Si, Ge) into n-type semi-conductors, doping is done with some elements of group 15 (e.g. P, As) | |
| 1288. |
A group 14 elements is to be converted into n- type semiconductor by doping in with a suitable impurity. To which group shouyld theis impurity belong ? |
| Answer» n-type semiconductor means increase in conductivity due to precence of excess of electrons. Therefore , a 14 group element should be doped with a 15-group element Eg. Arsenic or phosphorsus. | |
| 1289. |
Analysis shows that a metal oxide has the empirical formula `M_(0.96)O_(1.00)`. Calculate the percentage of `M^(2+)` and `M^(3+)` ions in the sample. |
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Answer» The ratio of `M^(2+)` and `O^(2-)` ions in the pure sample of metal oxide = `1:1`. Let x ions `M^(2+)` be replaced by `M^(3+)` ions in the given sample. No. of `M^(2+)` ions present = `(0.96-x)` Since the oxide is neutral in nature, Total charge on M atoms = Charge on oxygen atoms `2(0.96-x)+3x=2` `1.92-2x+3x=2` `x=2-1.92=0.08` `%"of M"^(3+)"ions in the metal oxide"=("No. if M"^(3+)"ions")/("Total no. of M atoms")xx100=(0.08)/(0.96)xx100=8.33%` `%"of M"^(2+)"ions in the metal oxide"=(100-8.33)=91.67%` |
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| 1290. |
Consider the following statements I. NaCl and KCl show metal excess defect due to anionic vacancies II. F- centres are the anionic site occupied by unpaired electrons III. F-centres is derived from the germen word farbenzenter for colour centre . F-centres impart yellow colour to the NaCl crystals Which of the above statements is / are correct ?A. II, III and IVB. I, III and IVC. I, III and IIID. All of these |
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Answer» Correct Answer - D Alkali metals like KCl and NaCl show metal deficiency defect. Due to loss of unpaired electron, Na changes to `Na^+ " and combine with " Cl^-` to form NaCl These unpaired electrons occupy anionic sites which is called F-centres that imparts yellow colour in the crystal. |
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| 1291. |
Analysis shows that nickel oxide has the formula `Ni_(0.98)O_(1.00)`. What fractions of nickel "exist" as `Ni^(2+)` and `Ni^(3+)` ions? |
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Answer» `Ni_(0.98)O_(1.00)` Let `Ni^(2+)` be `a` and `Ni^(3+)` be `(1-a)` , then average oxidation no. of `Ni`. `axx2+(1-a)xx3=(2xx100)/(0.98)` `:. a=0.959` `:. Ni^(2+)=0.959,Ni^(3+)=0.041` The percentage of `Ni^(2+)` is `95.9%~~96%` and that of `Ni^(3+)=4%` |
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| 1292. |
`Al^(3+)` ions replace `Na^(+)` ions at the edge centres of NaCl lattice. The number of vacancies in one mole NaCl is found to be x ` 10^(23)`. The value of x approximately is |
| Answer» Correct Answer - 2 | |
| 1293. |
KCl crystallises in the same type of lattice as does NaCl. Given `(r_(Na^(+)))/(r_(Cl^(-)))=0.5` and `(r_(Na^(+)))/(r_(K^(+)))=0.7` The ratio of the side of the unit cell for NaCl to that for KCl isA. `1:1.172`B. `1:1.1143`C. `1:1.1413`D. `1:1.732` |
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Answer» Correct Answer - B NaCl crystallises in the face centred cubic unit cell such that `r_(Na^(+))+r_(Cl^(-))=(a)/(2)` Given, `(r_(Na^(+)))/(r_(Cl^(-)))=0.5,(r_(Na^(+)))/(r_(K^(+)))=0.7` Thus we have , `(r_(Na^(+))+r_(Cl^(-)))/(r_(Cl^(-)))=1.5` `(r_(K^(+)))/(r_(Cl^(-)))=(r_(K^(+)))/(r_(Na^(+))//0.5)=(0.5)/(r_(Na^(+))//r_(K^(+)))=(0.5)/(0.7)` Therefore, `(r_(K^(+))+r_(Cl^(-)))/(r_(Na^(+))+r_(Cl^(-)))=(1.2)/(0.7)xx(1)/(1.5)` `:.(a_(KCl//2))/(a_(NaCl//2))=(1.2)/(0.7xx1.5)=(a_(KCl))/(a_(NaCl))=1.143` or `(a_(NaCl))/(a_(KCl))=1:1.143` |
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| 1294. |
Analysis shows that nickel oxide has the formula `Ni_(0-98) O_(1.00)` . What fractions of the nickel exist as ` Ni^(2+) and Ni^(3+) ` ions ? |
| Answer» `Ni^(2+) = 96 % , Ni^(3+) = 4%` | |
| 1295. |
If ` Al^(3+) ` replaces ` Na^(+)` at the edge centre of NaCl lattice then calculate that vacanicles in 1 mole NaCl. |
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Answer» 1 mole of NaCl contains 1 mole of ` Na^(+)` ions. ` 6.023 xx 10^(23) Na^(+)` ions . NaCl has fcc arragement of `Cl^(-)` ions and ` Na^(+)` ions are present at eh edge centres and body - centre. As there are 12 edge centre and each edge centre is shared by 4 shared by 4 unit cells, their contribution per unit cell ` = 1/4 xx 12 = 3` Conribution of ` Na^(+) ` ion at the body -centre =1 Thus, for every ` 4 Na^(+)` ions, the ions present at the edge centres = 3 . This means that `N^(+)` ions which have been replaced ` = 3/4 xx 6.023 xx 10^(23) = 4.517 xx 10^(23)` ` 1Al^(3+)` ion will replace ` 3Na^(+)` ions to maintain electrical neutrality . one vacancy will be occupied by ` Al^(3+)` ion and the remaining 2 will be vacent. The means that ` 1/3` rd of these positions will be occupied by `Al^(3+)` ions and ` 2/3` rd will remain vacant. Hence, no. of vanancies in 1 mole of ` NaCl = 2/3 xx 4.517 xx 10^(23) = 3.01 xx 10^(23)` |
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| 1296. |
Non-stoichiometric cuprous oxidem, ` Cu_(2)O` can be prepared in the laboratory. In this oxide, copper to oxygen ratio is slightly less than 2 :1 . Can you account for the fact that this substances is a a p -type semiconductor ? |
| Answer» The ratio less than 2:1 in ` Cu_(2)O` shows that some cuprous `(Cu^(+))` ions have been replaced by cupric ` (Cu^(2+))` ions. To maintion electrical neutrality , every two ` Cu^(+)` ions will be due to presence of these positive holes, hence it is a p-type semiconductor. | |
| 1297. |
How many lattice points are there in one unit cell of each of the following lattice? a. Face-centred cubic b. Face-centred tetragonal c. Body-centred |
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Answer» In face centred cubic arrangement, Lattice point located at the corners of the cube = 8 Lattice point located at the corners of each face = 6 Total no. of lattice points = 8 + 6 = 14 (b) In face centred tetragonal, the number of lattice points is also the same i.e., 8 + 6 = 14. (In both the cases, particles per unit cell `=8xx1//8+6xx1//2=4)` (c) In body centred cubic arrangement, Lattice point located at the corners of the cube = 8 Lattice points located in the centre of the body = 1 Total no. of lattice points = `8+1=9` (the total number of particles per unit cell `=8xx1//8+1=2`) |
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| 1298. |
A particular solid is very hard and has a very high melting point.In solid state it is nonconductor and its melt is a conductor of electricity. Classify the solid.A. MetallicB. MolecularC. NetworkD. Ionic. |
| Answer» Correct Answer - D | |
| 1299. |
A particular solid is very hard and has a very high melting point.In solid state it is nonconductor and its melt is a conductor of electricity. Classify the solid.A. MetallicB. MolecularC. NetworkD. Ionic |
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Answer» Correct Answer - 4 In ionic solids the constituent particles are positive and negative ions.Because of strong electrostatic forces of attraction the ions are closely packed and hence ionic solids are hard. They are electrical insulators in the solid state because their ions are not free to move about ,However, in the aqueous solutions or in the molten state,they are good conductors of electricity because ions becomes free. |
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| 1300. |
What type of solids are electrical conductors , malleable and ductile ? |
| Answer» Correct Answer - Metallic solids . | |