Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Give suitable laboratory preparation for diborane.

Answer»

SOLUTION :`2NaBH_4 +I_2 to B_2H_6 + 2NaI + H_2`
Discussion
2.

Give suitable answers for the following questions : What are the main conditions for the formation of an ionic bond?

Answer»

SOLUTION :An ATOM with low ionization ENTHALPY and an atom with high ELECTRON affinityform - an ionic bond.
The lattice energy of formation should ALSO be high.
Discussion
3.

Give suitable answers for the following questions : C Cl_(4) is insoluble in water but NaCl is soluble Why ?

Answer»

SOLUTION :`C Cl_(4)` is a covalent compound and NaCl is an IONIC compound.
Discussion
4.

Give suitable answers for the following questions : Arrange NaCl, MgCl_(2)" and " AlCl_(3) in the increasing order of covalnet character

Answer»

SOLUTION :GREATER the sizeof cation , higher is the polarising power.
` :. AlCl_(3) gt MgCl_(2) gt NaCl`
Discussion
5.

Give suitable answers for the following questions : Among NaCl, KCl and RbCl which one has highest lattice enthalpy? Why?

Answer»

SOLUTION :NaCl has highest lattice enthalpy because for similary anions, The CATION with smaller radii EXHIBIT higher lattice enthalpy.
Discussion
6.

Give suitable answers for the following questions : Among Na^(+), Ca^(2+) , Mg^(2+) , Al^(3+) which has high polarisingpower ?

Answer»

Solution :`Al^(3+)` has higher POLARISING POWER because of large cationic SIZE.
Discussion
7.

Give structures of following : (i) Glycol (ii) Glyoxal (iii) Ethanal and Ethanol (iv) Propanal, Propanol and Propanone (v) Gem-dibromide (Two) (vi) Vicinal dibromide (Two)

Answer»

Solution :(i) Glycol : `CH_(2)OHCH_(2)OH` <BR> (ii) Glyoxal: `{:(CHO),("|"),(CHO):}`
(iii) Ethanal : `CH_(3)CHO`
Ethanol : `CH_(3)CH_(2)OH`
(iv) Propanal : `CH_(3)CH_(2)CHO`
PROPANOL : `CH_(3)CH_(2)CH_(2)OH`
Propanone : `CH_(3)COCH_(3)`
(v) Gem-dibromide: `CH_(3)CHBr_(2)` 1,1-dibromoethane
`""CH_(3)CH_(2)CHBr_(2)`
`""CH_(3)CBr_(2)CH_(3)` 2,2-dibromopropane
Vicinal dibromide :
`underset("1,2-Dibromoethane")(CH_(2)Br-CH_(2)Br)""underset("1,2-Dibromopropane")(CH_(3)CHBr-CH_(2)Br)`
Discussion
8.

Give structure of secondary and tertiary butyl.

Answer»

SOLUTION :(i) Secondary-butyl `(C_(4)H_(9)^(-))` :
`CH_(3) underset(|)(C)HCH_(2)CH_(3)` OR `CH_(3)CH_(2) underset(|)(C)HCH_(3)`
(ii) Tertiary-butyl `(C_(4)H_(9)^(-))` :
`CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-` OR `CH_(3)-underset(|)overset(CH_(3))overset(|)(C)-CH_(3)`
Discussion
9.

Give structure of nitro and sulphonic acd group.

Answer»

Solution :NITRO group `-NO_(2)` :
Sulphonic acid group `-SO_(3)H` :
Discussion
10.

Give structure of gaseous AlCl_3.

Answer»

Solution :
Use : Being a Lewis acid, it ACT as CATALYST in Friedal CRAFT alkylation and acylation and also in electrophilic aromatic SUBSTITUTION REACTION.
Discussion
11.

Give structure of following (i) Cyclohexane-1, 2-diol (ii) Vinylalcohol (iii) 2-bromo-4-methylaniline (iv) 3-hydroxy-1, 3-5-pentantrioicacid OR 3-hydroxy-1, 2, 3-propentricarboxylicacid

Answer»

Solution :
(II) `CH_(2)=CHOH`: Vinylalcohol

(iv) `HOO overset(1)(C )-overset(2)(C )H_(2) - underset(underset(COOH)(|))overset(overset(OH)(3|))(CH)-overset(4)(CH_(2))-overset(5)(C )OOH`
3-hydroxy-1, 3, 5-pentantrioicacid OR 3-hydroxy -1, 2-3-propentricarboxylicacid
Discussion
12.

Give structure of following (a) 3, 4-dimethyl-hexane (b) 4, 5-diethyl -5-methylnonane (c ) Methylcyclopentane (d) 1, 3, 5 triethyloclohexane (e ) Buta -1, 3-diene (f) Buta -1, 3-dyene (g) Hexa-1, 3-diene -5-yne

Answer»

Solution :(a) `overset(1)(C )H_(3) - overset(2)(C )H_(2) - underset(underset(CH_(3))(|))overset(3)(C H)- underset(underset(CH_(3))(|))overset(4)(CH)- overset(5)(C H_(2)) - overset(6)(CH_(3))`
(b) `overset(9)(C H_(3))- overset(8)(CH_(2))- overset(7)(CH_(2)) - overset(6)(CH_(2))- underset(underset(CH_(2)CH_(3))(|))overset(overset(CH_(3))(5|))(C )- overset(overset(CH_(2)CH_(3))(4|))(C H)- overset(3)(C )H_(2) - overset(2)(C )H_(2) - overset(1) (CH_(3))`

(E ) `overset(1)(C )H_(2) = overset(2)( C)H- overset(3)(C )H= overset(4)( C)H_(2)` (f) `H overset(1)(C )-= overset(2)(C )- overset(3)(C )-= overset(4)(C )H` (G) `overset(1)(C )H_(2)= overset(2)(C )H- overset(3)(C )H= overset(4)(C )H- overset(5)(C )-= overset(6)(C )H`
Discussion
13.

Give structure is neo-pentile group.

Answer»

Solution :Neo-pentile MEANS `C_(5)H_(11)` and structure is : `H_(3)C-underset(CH_(3))underset(|)OVERSET(CH_(3))overset(|)(C)-CH_(2)-`
Discussion
14.

Give structure and name of one double bond containing cyclic compound

Answer»

SOLUTION :
Discussion
15.

Give structure and IUPAc name of any one isomer of compound possessing only hydrogen with molecular formula C_(5)H_(12).

Answer»

Solution :`underset("2,2-dimethyl PROPANE")(H_(3)-underset(H_(3))underset(|)overset(CH_(3))overset(|)(C)-CH_(3))`
Discussion
16.

Give structure and formula of hexachloro benzene.

Answer»

Solution :CHEMICL FORMULA : `C_(6)Cl_(6)`
STRUCTURE :
Discussion
17.

Give stability order of carbocations : (i) (CH_(3))_(3)overset(+)(C), (CH_(3))_(2)overset(+)(C)H, CH_(3)overset(+)(C)H_(2), overset(+)(C)H_(3)

Answer»

Solution :`UNDERSET(3^(@))((CH_(3))_(3))OVERSET(+)(C) gt underset(2^(@))((CH_(3))_(2))overset(+)(C)H gt underset(1^(@))(CH_(3)overset(+)(C)H_(2) gt overset(+)(C)H_(3)`
Discussion
18.

Give steps and enthalpy of formation of NaCl (Sodium chloride).

Answer»

SOLUTION :Ionization :
`Na_((G)) RARR Na_((g))^(+) E , Delta_(i) H = ` 495.8 KJ `mol^(-1)`
`Cl_((g)) + e^(-) rarr Cl_((g))^(-) , Delta_(e) H = - 348.7 " kJ" mol^(-1)`
Crystallization :
` Na_((g))^(+) + Cl_((g))^(-) rarr NaCl_((s)) , Delta_(L) H = - 788 " kJ " mol^(-1)`
Total `Delta_(f) H ("NaCl") = Delta_(i) H + Delta_(e) H + Delta_(L) H `
= (495.8 - 348.7 - 788 ) =- 640.9 kJ `mol^(-1)`
Discussion
19.

Give some functional groups, its examples, prefix, suffix and class of organic compounds

Answer»

SOLUTION :
Discussion
20.

Give one example for spontaneous combustion.

Answer»

Solution :BURNING wood.Fire works and ALKALI METALS areadded to water . When a radioactiveatom splits up , it released energy , this a spontaneous , exothermic nuclear reaction .
Discussion
21.

Give some examples for metals other than beryllium and aluminium that becomes passive on. treatment with concentrated nitric acid.

Answer»

Solution :CHROMIUM, IRON, COBALT and nickel are passive to conc. `HNO_(3)`.
Discussion
22.

Give similarity and difference of dissociation of Hydrochloric acid and acetic acid in water.

Answer»

SOLUTION :
Discussion
23.

Give SI unit of Luminous intensity.

Answer»



SOLUTION :CANDELA (CD)
Discussion
24.

Give similarities between beryllium and aluminum metal.

Answer»

Solution : The ionic RADIUS of`Be^(2+)` is estimated to be 31pm, the charge/radius ratio is nearly the same as that of the`Al^(3+)`ION.Hence beryllium resembles aluminium in some ways. Some of the similarities are:
(i) Like aluminium, beryllium is not readily attacked by acids because of the presence of an oxide film on the surface of the metal.
(II) Beryllium hydroxide dissolves in excess of alkali to give a beryllate ion, `[Be(OH)_(4)]^(2-)`just as aluminium hydroxide gives aluminate ion, `[Al(OH)_(4)]^(-)`.
(iii) The chlorides of both beryllium and aluminium have `Cl^(-)`bridged chloride structure in vapour phase. Both the chlorides are soluble in organic solvents and are strong Lewis acids. They are used as Friedel Craft catalysts.
(iv) Beryllium and aluminium ions have strong tendency to FORM complexes`[BeF_(4)]^(2-), [AlF_(6)]^(3-)`.
Discussion
25.

Give short explanations on salt of oxo - acids of alkali metal elements.

Answer»

Solution : Oxo-acids are those in which the acidic proton is on a hydroxyl group with an oxo group ATTACHED to the same ATOM.
e.g., carbonic acid, `H_(2)CO_(3)[OC(OH)_(2)]`, sulphuric acid `H_(2)SO_(4)(O_(2)S(OH)_(2))`.
The alkali metals form salts with all the oxoacids. They are generally soluble in water and thermally stable. Their carbonates `(M_(2)CO_(3))`and in most cases the hydrogen carbonates `(MHCO_(3))`also are highly stable to heat.
As the electropositive character increases down the group, the stability of the carbonates and hydrogen carbonates increases. LITHIUM carbonate is not so stable to heat, lithium being very small in size POLARISES a large `CO_(3)^(2-)`ion leading to the formation of more stable `Li_(2)O` and `CO_(2)`. Its hydrogen carbonate does not exist as a solid.
Discussion
26.

Give shape and bond angle of the compounds of questions.

Answer»

Solution :ETHANE : Tetrahedral shape in three dimension and bond angle of `109.5^(@)`.
Ethene : All atoms POSSESS planner shape and bond angle of `120^(@)`.
Ethyne : It is LINEAR in shape and have bond angle of `180^(@)`.
Benzene : All atoms are arrange in ONE planner and all bond angles are of `120^(@)`.
Discussion
27.

GiveRydbergequationfor all linesin hydrogenspectrum

Answer»

Solution :The Swedishspectroscopist JohannesRydbergnotedthat all series of linein thehydrogenspectrumcouldbe described by thefollowingexpression.
where `n_(1)=1,2,3`
`n_(2)= (n_(1) +1)(n_(1) +2) (n_(1) +3)`
Rydbergconstantfor hydrogen `=109677 CM^(-1) = R_(H)`
Theseriesof hydrogen spectrumare knowby`n_(1)`value.

Thefigureare asundershows thelymanBalmerandpaschenseriesof transitions foratom hasteh simplest LIEN spectrum
Hydrogenspectrum is a line EMISSIONSPECTRUM
Discussion
28.

Give rules of nomenclature of organic compound containing one or more functional group

Answer»

Solution :(i) First of all, the functional group present in the molecule is identified which DETERMINES the choice of appropriate suffix.
(ii) Longest parent chain: The longest chain of carbon ATOMS CONTAINING the functional group is numbered in such a way that the functional group is attached at the carbon atom possessing lowest possible number in the chain.
(iii) Polyfunctional compounds: In the case of polyfunctional compounds, one of the functional groups is chosen as the principal functional group and the compound is then named on that basis. The remaining functional groups, which are subordinate functional groups, are named as substituents using the appropriate prefixes.
(iv) The choice of principal functional group is made on the basis of order of perference. The order of decreasing priority for some functional groups is:
`-COOH, -SO_(3)H, - COOR` (R= alkyl group)
`-COCl, -CONH_(2), -CN, -HC= O, gt C= O, -OH`
`-NH_(2), gt C= C lt, - C -= C -`
(v) `-R, C_(6)H_(5)-`, halogens `(-F, -Cl, -Br, -I)- NO_(2)` -alkoxy (-OR) etc. are always prefix substituents.
(iv) A compound containing both an alcohol and a keto group is named as hydroxyalkanone since the keto group is preferred to the hydroxyl group. e.g `HOCH_(2) (CH_(2))_(3)CH_(2)COCH_(3)` will be named as 7-hydroxyheptan-2-one
(vii) If more than one functional group of the same type are present, their number is indicated by adding di, tri, etc. before the class suffix. In such cases the full name of the parent ALKANE is written before the class suffix. e.g. `CH_(2)(OH)CH_(2)(OH)` is named as ethane -1, 2-diol.
(viii) However, the ending -ne of the parent alkane is dropped in the case of compounds having more than one double or triple bond. For e.g. `CH_(2)= CH-CH= CH_(2)` is named as buta-1, 3-diene
Discussion
29.

Give rules for writing resonance structure

Answer»

Solution :(i) The RESONANCE structures have the same positions of nuclei i.e.., REMAIN CONSTANT (II) The resonance structures have the same number of UNPAIRED electrons (iii) Among the resonance structures, the one which has more number of covalent bonds, all the atoms with octet of electrons, less separation of opposite charge and more dispersal of charge, is more stable than others.
Discussion
30.

Give resonance structures of following : (i) Phenol (ii) Nitrobenzene

Answer»

SOLUTION :
Discussion
31.

Give requirements for chemical synthesis and give the changes in equilibrium and laws.

Answer»

Solution :One of the principal goals of chemical synthesis is (i) to maximize the products (ii) minimizing the expenditure of energy.
This implies maximum yield of products at mild temperature and pressure CONDITIONS. If it does not happen, then the experimental conditions need to be adjusted. For example, in the Haber process for the synthesis of ammonia from `N_2` and `H_2`, the CHOICE of experimental conditions is of real economic importance.
Equilibrium constant, `K_c` is independent of initial concentrations. But if a system at equilibrium is subjected to a change in the concentration of one or more of the reacting substances, then the system is no longer at equilibrium, and net reaction takes place in some DIRECTION until the system returns to equilibrium once again.
If a change in temperature or pressure of the system may ALSO alter the equilibrium.
In order to decide what course the reaction adopts and make a qualitative prediction about the effect of a change in conditions on equilibrium Le-Chatelier.s principle is used.
Le-Chatelier.s principle: "Change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to counteract the effect of the change." This is applicable to all physical and chemical equilibria.
Discussion
32.

Give resonance structure of nitromethane and explain its real structure

Answer»

Solution :The nitromethane `(CH_(3)NO_(2))` which can be represented by two LEWIS structures, (I) and (II). These are its resonance STRUCTURE

There are two types of N-O bonds in these structures. However, it is known that the two N-O bonds of nitromethane are of the same length (intermediate between a N-O single bond ADN a N= O double bond). The actual structure of nitromethane is therefore a resonance HYBRID of the two canonical forms (I) and (II)
Discussion
33.

Give relation of polarity of polyatomic molecules and its bond polarity.

Answer»

SOLUTION :`({:("Polarity of ")/("MOLECULE"):}) = ({:("Vactor addition of DIPOLE "),("moment and BOND base of "),("on shape of molecules "):})`
Discussion
34.

Give relation between u_(rms), u_(av) and u_(mp).

Answer»

SOLUTION :`{:(u_(RMS),gt,u_(AV),gt,u_(mp),,),("RATIO",:,u_(mp),:,u_(av),:,u_(rms)),(,,1,,1.128,,1.224),(,or,0.82,,0.92,,1.00):}`
Discussion
35.

Give relation between partial pressure and mole fraction.

Answer»

Solution :Suppose at the temperature (T), THREE gases, enclosed in the volume (V), EXERT partial PRESSURE `p_(1), p_(2)` and `p_(3)` respectively then, Ideal gas equation pV = nRT therefore,
(i) `p_(1)=(n_(1)RT)/(V) ""` …..(Eq. -i)
(ii) `p_(2)=(n_(2)RT)/(V) ""`.......(Eq.-ii)
(iii) `p_(3)=(n_(3)RT)/(V) ""` ......(Eq. -iii)
where, `n_(1), n_(2)` and `n_(3)` are number of moles of these gases.
According to Dalton.s Law, equation of total pressure is written as,
`p_("total")=(p_(1)+p_(2)+p_(3))`
`= ((n_(1)RT)/(V))+((n_(2)RT)/(V))+((n_(3)RT)/(V))`
`p_("total")=(n_(1)+n_(2)+n_(3))(RT)/(V) ""` .....(Eq. -iv)
Mole fraction by ratio of partial pressure and total pressure `(CHI)` : Partial pressure is divided by total pressure, (Eq. -i / Eq. -iv)
`(p_(1))/(p_("total"))=((n_(1)RT)/(V))(V)/((n_(1)+n_(2)+n_(3))RT)`
`=((n_(1))/(n_(1)+n_(2)+n_(3)))((RTV)/(VRT))`
`therefore (p_(1))/(p_("total"))=(n_(1))/(n_(1)+n_(2)+n_(3)) ""` ......(Eq.-v)
but, `(n_(1)+n_(2)+n_(3))=` Total mole (n)
`(p_(1))/(p_("total"))=(n_(1))/(n) ""`.....(Eq. -vi)
Mole fraction of `1^(st)` gas `(chi)=(n_(1))/(n)`
`therefore (p_(1))/(p_("total"))=chi_(1)` and `p_(1)=chi_(1)p_("total")` .....(Eq. -vii)
Similarly for gas 2 and 3,
`p_(2)=chi_(2)p_("total")` and `p_(3)=chi_(3)p_("total")`
So general equation is `p_(i)=chi_(1)p_("total")` ......(Eq. - viii)
Partial pressure = mole fraction `xx` total pressure
Where, `p_(i)=` Partial pressure of gas
`chi_(i)=` Mole fraction of gas
If total pressure of a mixture of gases is known, the (Eq. - viii) can be used to find out pressure cxerted by individual gases.
Discussion
36.

Give relation between mass volume of gas.

Answer»

SOLUTION :In 1811 Italian scientist Amedeo Avogadro tried to combine conclusions of Dalton.s atomic theory and Gay Lussac.s law of combining volumes which is now known as Avogadro law.
Avogadro.s Law : It states that equal volumes of all gases under the same conditions of temperature and pressure CONTAIN equal number of molecules.
Mathematically Formula : According to Avogadro.s Law as the temperature and pressure remain constant, the volume depends UPON number of molecules of the gas or in other words amount of the gas.
`V prop n` (constant T and p).....(EQ. -i)
where, n = The number of MOLES of the gas
`therefore V=k_(4)n ""`.....(Eq. -ii)
Molar volume at STP = 22.7 L
Atoms in one mole `= 6.022xx10^(23)`
The relation of volume and density of gas can be obtained by `M = k_(4)d`.
Discussion
37.

Give relation between molecular mass density and volume.

Answer»

SOLUTION :Relation between molar MASS, density and volume is given.
`M prop (m)/(V) prop d prop N`
Molecular mass of GAS direactly proportional to density.
Density of gas is invertly proportional to its volume.
`M prop (1)/(V) prop d`
Discussion
38.

Give relation between K_(sp) of Ca_3(PO_4)_2 and S .

Answer»

SOLUTION :`K_(sp)=[Ca^(2+)]^3 [PO_4^(3-)]^2=(3S)^3(2S)^2=108 S^5`
Discussion
39.

Give relation between different units of pressure.

Answer»

Solution :(i) 1 Pa = 101.325 k Pa
(ii) 1 atm = 101.325 k Pa = 101325 Pa = 101325 `Nm^(-2)`
1 atm `= 10^(5)` Pa = `10^(2)` k Pa
(iii) 1 bar = 0.987 atm = 14.509 psi
1 atm = 1.01325 bar
1 bar `= 1xx10^(5)Nm^(-2)=10^(2)kPa = 10^(5)` Pa
(Note : Some times it is taking 1 bar = 1 atm, but it is not complete RIGHT)
(iv) 1 atm = 76 CM Hg
= 760 mm Hg
= 760 torr
1 torr = 1 mm Hg
Discussion
40.

Giverelationbetweenfollowingpairof species

Answer»

Solution :(i)ISOBAR (ii)ISOTOPES
(III)Isotopes (IV)Isotopes
Discussion
41.

Give rection which shows hydro-halogenation and hydration of alkyne.

Answer»

Solution :(a) Hydro-halogenation of alkyne or addition of hydrogen halide in alkyne : The electrophilic addition reaction of alkyne compounds occurs while it reacts with hydrogen halide (HCl, HBr, HI). In this reaction two molecule of hydrogen halide added in alkyne compound and form gem-dihalide. In gem-dihalide two halogen atoms arc attached on same carbon atoms.


NOTE : Addition of HX molecule in alkene or alkyne follows Markovnikov.s rule, and `H^(+)` (electrophilic) attach with those carbon having more hydrogen and more double bonds.
(b) Hydration of alkynes or addition of water into alkyne : Hydration of alkyne means addition of water `(H_(2)O)`, which follows electrophilic addition reaction. In which, FIRSTLY it follows addition of electrophilic `H^(+)` of `H_(2)O (H^(+)OH^(-))` according to Markovnikov rule in first step andform more stable carbocation. In second step, they form `alpha`-akenal by addition. In second step, they form `alpha`-akenal by addition of `OH^(-)` ions

In above reaction only one ATOM of water is added and gives aldehyde or ketone as a product, which possesses group.
In this reaction, addition of `H^(+)` ion of dilute sulphuric acid and water molecule is talking place at 333K temperature and in presences of `Hg^(2+)` ions of mercuric sulphate CATALYST and gives a product.
(i) Ethanol is formed by the reaction of ethyne.
Discussion
42.

Give reasons why hydrogen resembles alkali metals ?

Answer»

Solution :HYDROGEN resembles alkali metals, i.e. Li, Na, K, Rb, Cs and Fr of group 1 of the periodic table in the following respects.
(i) Like alkali metals, hydrogen also contain one electron in its OUTERMOST orbit. It like alkali metal it gave + 1 oxidation state.
(ii) Like alkali metals, hydrogen also loses its only electron to form hydrogen ion, i.e., `H^+`.
(iii) Like alkali metals, hydrogen combines with ELECTRONEGATIVE elements such as oxygen, halogens and sulphur forming their oxides, halides and SULPHIDE respectively.
(iv) Like alkali metals, hydrogen also acts as a strong reducing agent.
Discussion
43.

Give reasons (vi) Aluminium utensils cannot be kept in water overnight

Answer»


Answer :AL rescts with water and the dissolved oxygen in it form an OXIDE layer on the surface of utensil. A small amount of oxide will dissolve in water to form `Al^(3+)` ions which are HARMFUL to human HEALTH.
Discussion
44.

Give reasons (vii) Aluminium wire is used to make transmission cables.

Answer»


Answer :Al is a GOOD conductor of ELECTRICITY and is LIGHTER than copper.
Discussion
45.

Give reasons: The stability of +3 oxidation state of 13 group elements decreases down the group.

Answer»


Solution :The stability of +3 oxidation state of group 13 elements decreases down the group DUE to the inert pair effect on ns2 electrons. This is because of the POOR SHIELDING effect of electron in d and f orbitals.
Discussion
46.

Give reasons (v) Al alloys are used to make aircraft body

Answer»


Answer :Aluminium alloys are light, TOUGH and not CORRODED EASILY.
Discussion
47.

Give reasons: (i)In stoichiometric defects, NaCl exhibits Schottky defect and not Frenkel defect (ii)Silicon on doping with phosphorus forms n-type semiconductor. (iii)Ferrimagnetic substances show better magnetism than antiferromagnetic substances

Answer»

Solution :(i)Schottky defect is shown by those ionic compounds which have small difference in the size of cation and anion whereas Frenkel defect is shown by those compounds which have large DIFFERENCES in the size of cation and anion. As the difference in the size of `Na^+` and `Cl^-` ions is small, HENCE it shows Schottky defect. (ii)Silicon belongs to Group 14 whereas phosphorus belongs to Group 15. As phosphorus has one extra electron than silicon, this extra electron makes silicon a semiconductor. Hence , we GET n-type semiconductor.
(iii)Ferrimagnetic substances have an unequal NUMBER of electrons with SPIN in the opposite direction. with spin in the opposite direction. Hence, their net magnetic moment is zero. For this reason, ferrimagnetic substances show better magnetism than antiferromagnetic substances .
Discussion
48.

Give reasons (ii) A mixture of Al pieces and dilute NaOH is used to open drains

Answer»


Answer :`H_2` GAS is formed with high PRESSURE by the action of NAOH and Al. The pressure so formed OPENS up the drain.
Discussion
49.

Give reasons (iv) Diamond is used as an abrasive

Answer»


ANSWER :DIAMOND is very HARD
Discussion
50.

Give reasons (iii) Graphite is used as a lubricant

Answer»


Answer :The DIFFERENT layers of carbon are HELD by WEAK van DER WALL's forces and hence one layer slides over the other.
Discussion