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88751.

A metal M forms a compound M_2HPO_4 The formula of the metal sulphate is :

Answer»

`M_2SO_4`
`MSO_4`
`M(SO_4)_2`
`M_2(SO_4)_3`

ANSWER :A
Discussion
88752.

A metal M and its compounds can gives the following observable changes in a consequence of reaction

Answer»

Mg
Pb
Zn
Sn

Solution :`UNDERSET((M))"Zn"OVERSET(dil.HNO_(3))rarr underset(underset("solution")"Colouless")(Zn(ZO_3)_2)overset(NAOH)underset(aq.)rarr underset("White ppt.")(Zn(OH)_2)`
`overset("Excess")underset(NaOH)rarr underset("soluble")(Na_(2)[Zn(OH)_(4) overset(H_2S)rarrunderset("White ppt.")(ZNS)]`
Discussion
88753.

A metal M and its compound can give the following observable changes in a consequence of reactions

Answer»

Mg
Pb
Zn
Sn

Solution :`underset("M")(Zn)overet("dil" HNO_(3))(to)underset("colourless solution")(Zn(ZO_(3))_(2))overset(Ag)underset(NAOH)(to)underset("WHITE ppt.")(Zn(OH)_(2))overset("Excess")underset(NaOH)(to)underset("SOLUBLE")(Na_(2)[Zn(OH)_(4))overset(H_(2)S)(to)underset("White ppt.")(ZNS)`
Discussion
88754.

A metal is left exposed to the atmosphere for some time . It gets coated with green carbonate . The metal must be

Answer»

SILVER
COPPER
iron
zinc .

Solution :`CuCO_(3)` is green .
Discussion
88755.

A metal gets coated with a green basic carbonate when exposed to atmosphere. Metal is

Answer»

NICKEL
CHROMIUM
Copper
Zinc

Answer :C
Discussion
88756.

A metal is left exposed to air for some time. It becomes coated with basic green carbonate. The metal is:

Answer»

K
Cu
Zn
Al

Answer :B
Discussion
88757.

A metal is burnt in air and the ash on moistening smells of ammonia. The metal is:

Answer»

Na
Fe
Mg
Al

Answer :C
Discussion
88758.

A metal ion M^(n+) having d^(4) valenceelectronic configuration combines with three didentate ligands to form a complex compound. Assuming Delta_(@)gt P : (i) Draw the diagram showing d-orbital splitting during this comples formation. (ii) What type of hybridisation will M^(n+)have? (iii) Name the type of isomerism exhibited by this complex. (iv) Write the electronic configuration of metal M^(n+)

Answer»

<P>

SOLUTION :(i) If `Delta_(@)>p` then
Discussion
88759.

A metal ion M^(n+) having d^(4) valence electronic configuration combines with three didentate ligands to form a complex compound. Assuming Delta_(0) gt P (i) Draw the diagram showing d orbital splitting during this complex formation. (ii) Write the electronic configuration of the valence electrons of the metal M^(n+) ion in terms of t_(2g) and e_(g) (iii) What type of hybridisation will M^(n+) ion have ? (iv) Name the type of isomerism exhibited by this complex.

Answer»

SOLUTION :(i) If `Delta_(0) gt P`, pairing will occur in the `t_(2g)` orbitals and `e_(g)` orbitals will remain vacant

(ii) `t_(2g)^(4) e_(g)^(0)` (iii) `d^(2)SP^(3)` (as there are three didentate ligands to combine)
(iv) `[M(A A)_(3)]` type complexes show optical isomerism.
Discussion
88760.

A metal ion M^(n+) having d^(4) valence electrode configuration combines with three ligands to form a complex compound. Assuming Delta_(0)gtP. i) Draw the diagram assuming d-orbital splitting during the complex formation. ii) Write the electronic configuration of valence electrons of the metal M^(n+) ion in terms t_(2g) and e_(g). iii) What type of of hybrisation will M^(n+) ion have? iv) Name the type of isomerism exhibited by the complex.

Answer»

SOLUTION :i) N/A
ii) When `Delta_(0) gt P`, the electrons get paired up in the lower `t_(2g)` ORBITALS. The electronic configuration of `d^(4)` ion is: `t_(2g)^(2)e_(g)^(0)`.
iii) The COMPLEX is octahedral since the didenate ligands are attached to the `M^(n+)` ion which is `d^(2)sp^(3)` hybridised.
IV) The complex will exhibit optical isomerism.
Discussion
88761.

A metal ion from the first transition has a magnetic moment of 3.87 B.M. How many unpaired electrons are expected to be present in the ion

Answer»


SOLUTION :`M^(+2) rArr MU = sqrt(n(n+2)) = 3.87 BM rArrn = 3`
Discussion
88762.

A metal ion give chocolate coloured ppt with K_(4)[Fe(CN)_(6)] What is the oxidation state of that metal in its ion ?

Answer»


ANSWER :2
Discussion
88763.

A metal- insoluble salt electrode consists of

Answer»

A PIECE of metal placed in a solution containing a SPARINGLY SOLUBLE salt.
Crystals of an insoluble salt coated with a metal.
A piece of metal coated with one of its insoluble salts.
A metal fixed with an insoluble salt at high temperature.

Solution :Factual statement
Discussion
88764.

A metal having negative reduction potential when dipped in the solution of its own ions, has a tendency

Answer»

to pass into the solution
to be DEPOSITED from the solution
to BECOME electrically positive
to REMAIN neutral

Solution :Metal with `- ve E_(RP)^(@) ` or `+ve E_(OP)^(@)` possesses the TENDENCY to get itself oxidised .
Discussion
88765.

A metal has fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm^(-3). The molar mass of the metal is

Answer»

30 G `mol^(-1)`
27 g `mol^(-1)`
20 g `mol^(-1)`
40 g `mol^(-1)`

ANSWER :B
Discussion
88766.

A metal has face centred cubic arrangement . If length of the edge of the cell is x pm and M is its atomic mass, then density will be equally to ( N_(0) is Avogadro number )

Answer»

`( M xx 10^(30))/( x^(3) xx N_(0)) g cm^(-3)`
`(M xx N_(0))/( x^(3)) g cm^(-3)`
`( 4M )/( x^(3) xx N_(0)) xx 10^(30) g cm^(-3)`
`( M ) /( 4x^(3) xx N_(0)) g cm^(-3)`

ANSWER :C
Discussion
88767.

A metal having negative reduciton potential when dipped in the solution of its ow ions, has a tendency:

Answer»

To PASS into the solution
To be deposited from the solution
To BECOME electrically positive
To REMAIN neutral

Answer :A
Discussion
88768.

A metal has bcc structure and the edge length of its unit cell is 3.04 A. The volume of the unit cell in cm^(3) will be

Answer»

`1.6xx10^(-21) cm^(3)`
`2.81 xx10^(-23) cm^(3)`
`6.02 xx10^(-23) cm^(3)`
`6.6 xx10^(-24) cm^(3)`

Answer :B
Discussion
88769.

A metal has bcc structure and the edge length of its unit cell is3.04 Å . The volume of the vnit cell incm^(3)willbe

Answer»

`1.6xx10^(-21)CM^(3)`
`2.81xx10^(-23)cm^(3)`
`6.02xx10^(-23)cm^(3)`
`6.6xx10^(-24)cm^(3)`

Solution :VOLUME of the unit CELL `=a^(3)`
`=(3.04xx10^(-8)cm)^(3)=2.81xx10^(-23)cm^(3)`
Discussion
88770.

A metalhas bcc structure and the edge length of its unit cell is3.04 Å. The volume of the unit cell cm^(3)will be

Answer»

`1.6xx10^(23)CM^(3)`
`1.6xx10^(-23)cm^(3)`
`6.02xx10^(-23)cm^(3)`
`6.6xx10^(-24)cm^(3)`

Solution :Volume of UNIT CELL (V) `=a^(3)`
`=(3.04xx10^(-8)cm)^(3)`
`=2.81xx10^(-23)cm^(3)`
Discussion
88771.

A metal has a high concentration into the earth crust and whose oxides cannot be reduced by carbon. The most suitable method for the extraction of such metal is :

Answer»

Alumino THERMITE PROCESS
ELECTROLYSIS process
Van-Arkel’s process
Cupellation

Answer :B
Discussion
88772.

A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm^(-3). The molar mass of the metal is (N_(A) Avogadro's constant=6.02xx10^(23)mol^(-1))

Answer»

30 g `mol^(-1)`
27 g `mol^(-1)`
20 g `mol^(-1)`
40 g `mol^(-1)`

Solution :Density is given by
`d=(zxxM)/(N_(A)a^(3))`, where z=number of formula units
present in unit cell, which is 4 for fcc
a=edge length of unit cell. M=molecular MASS.
`2.72=(4xxM)/(6.02xx10^(23)XX(404xx10^(-10))^(3))`
`(because 1"pm"=10^(-10)CM)`
`M=(2.72**6.02**(404)^(3))/(4**10^(7))=26.99`
=27g `mol^(-1)`
Discussion
88773.

A metal halide X on treating with copper sulphate solution yields a brown precipitate Y. Y turns colourless on adding with hypo. What are X and Y?

Answer»

SOLUTION :`X=I^(-), Y=I_(2)`
Discussion
88774.

A metal gives two different chlorides 'A' and 'B'. The chloride 'A' gives a white precipitate with NH_(4)OH solutionwhile 'B' gives a black precipitate. With KI solution, 'A' gives a red precipitate which dissolves in excess of KI solution. The two chlorides are :

Answer»

`HgCl_(2)" and "HgCl`
`HgCl_(2)" and "Hg_(2)Cl_(2)`
`Hg_(2)Cl_(2)" and "HgCl_(2)`
NONE of these.

Solution :(b) 'A' is `HgCl_(2)` and 'B' is `Hg_(2)Cl_(2)`
Discussion
88775.

A metal gives two chlorides Aand B.A gives black precipitate with NH_4OH and B gives white. With KI,B gives a red precipitate soluble in excess of KI.Aand B are respectively:

Answer»

`HgCl_2` and `Hg_2Cl_2`
`Hg_2Cl_2` and `HgCl_2`
`HgCl_2` and `ZnCl_2`
`ZnCl_2` and `HgCl_2`

ANSWER :B
Discussion
88776.

A metal froms a volatile carbonylis:

Answer»

Iron
Nickel
Cobalt
TITANIUM

ANSWER :B
Discussion
88777.

A metal forms two oxides. The higher oxide contains 80% metal. 0.72 g of the lower oxide gave 0.8 g of higher oxide when oxidiesd. Then the ratio of the weight of oxygen that combines with the fixed weight of the metal in the two oxides will be

Answer»

`2 : 3`
`1 : 2`
`4 : 5`
`3 : 2`

ANSWER :B
Discussion
88778.

A metal forms two oxides. One contains 46.67% of the metal and another, 63.94% of the metal. Show that these results are in accordance with the law of multiple proportions.

Answer»

Solution :MASSES of METAL that combine with `1.0` g of OXYGEN are 0.875 g and 1.773 g, the RATIO is `1:2`.
Discussion
88779.

A metal forms a compound MHPO_4. What should be the formula of the metal chloride :

Answer»

`MCl_2`
`MCl_3`
MCL
`MCl_4`

ANSWER :A
Discussion
88780.

A metal forms hexagonal close-packed structure? How many voids are present in 0.5 mol of it?

Answer»

Solution :Number of ATOMS in 0.5 mol of the given metal = `3 xx 10^(23)`
If there are .N. atoms, number of tetrahedral VOIDS is .2n. and number of octabhedral voids is .n.
Number of tetrahedral voids present in 0.5 mol of metal `= 6 xx 10^(23)`
Number of octahedral voids present in 0.5 mol of metal `= 3 xx 10^(23)`
Total number of voids present in the given metal `= 9 xx 10^(23 )`
Discussion
88781.

A metal crystallizes with a face-centred cubic lattice. The edge of the unit cell is 408 pm.The diameter of the metal atom is

Answer»

144 pm
204 pm
288 pm
408 pm

Answer :C
Discussion
88782.

A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 408 pm. The diameter of the metal atom is

Answer»

288 PM
408 pm
144 pm
204 pm

SOLUTION :For CCP`sqrt(2)a=4r`
`sqrt(2)xx408=4r`
But Diameter = 2 R
`:.`Diameter`=(sqrt(2)xx408)/(2)=288.5`
Discussion
88783.

A metal crystallizes in a simple cubic unit cell. The length of the edge of the unit cell, a, is 6.22 A. Find the radius of each atom of the metal.

Answer»

SOLUTION :We have, for SIMPLE CUBIC STRUCTURE,
RADIUS `=a/2 = 6.22/2 = 3.11 A`
Discussion
88784.

A metal crystallizes into a lattice containing a sequence of layers of atoms of ABABAB. . . . .. .. . What percentage by volume of this lattice has empty space.

Answer»

74
26
20
16

Solution :ABAB. . . . . Is HEXAGONAL EMPTY space=26%
Discussion
88785.

A metal crystallizes in a closed packed structure, theatoms stack together occupying maximum space and leaving minimum vacant space. Each sphere in the first layer is in contact with six neighbours. Now while arranging the second layer on the first layer, spheres are placed in depression of first layer. Spheres of third layer lie in the depression of second layer. This repeating arrangement of atoms produces a giant lattice. If sphere of third layer do not lie directly over the atoms of first layer, then density of unit cell is, (Given: atomic weight of metal is 197u and a=4.07Å)

Answer»

`86.5 G//"cc"`
`19.7 ` g/cc
`56.2` g/cc
`40.5` g/cc

Solution :Above information REPRESENT cubic close packing
`RHO=(197xx4)/((4.07)^(3)xx10^(-24)xx6.02xx10^(23))`
`rho=19.41`g/cc
Discussion
88786.

A metal crystallizes in a closed packed structure, theatoms stack together occupying maximum space and leaving minimum vacant space. Each sphere in the first layer is in contact with six neighbours. Now while arranging the second layer on the first layer, spheres are placed in depression of first layer. Spheres of third layer lie in the depression of second layer. This repeating arrangement of atoms produces a giant lattice. If one of the edge length of unit cell is 'a' what is the shortest distance between two atoms?

Answer»

`(SQRT(3)a)/4`
`a/4` or `a/3`
`a` or `a/(sqrt(2))`
`a/2`

Solution :Above PACKING represent two possible ARRANGEMENTS, hexagonal close packing or cubic close packing, hence ,MINIMUM distance `a` or `a/(sqrt(2))`
Discussion
88787.

A metal crystallizes in 2 cubic phases fcc and bee whose unit cell lengths are 3.5 Å and 3.0Å respectively. The ratio of their densities is

Answer»

0.72
2.04
1.26
3.12

Solution :`(d_(1))/(d_(2))=((a_(2))^(3))/((a_(1))^(3))xx(z_(1))/(z_(2))=((3)/(3.5))^(3)xx(4)/(2)=1.26`
Discussion
88788.

A metal crystallized in fcc lattcie and edge of the unit cell is 620 pm. The radius of metal atom is

Answer»

265.5pm
219.2pm
310pm
438.6pm

Solution :`R = ( a)/( 2 SQRT( 2)) = ( 620)/( 2 XX 1.414) = 219 . 20 `pm
Discussion
88789.

A metal crystallizeds in cubic close packed ( ccp). The atomic radius of metal is 144 pm and its atomic mass is 197 a.m.u. Its density is

Answer»

19.4 `g cm^(-3)`
`1.94 g cm^(-3)`
`29.4 g cm^(-3)`
`2.94 g cm^(-3)`

Solution :For ccp arrangement
`a = 2 SQRT( 2) r= 2 xx 1.414 xx 144 `
`= 407.2`PM = `407.2 xx 10^(-10)`cm
`d = ( Z xx M )/( a^(3) xx N_(O))`
`= ( 4 xx 197)/((407.2 xx 10^(-10))^(3) xx ( 6.02 2 xx 10^(23)))`
`= 19.4g cm^(-3)`
Discussion
88790.

A metal crystallises with a face-centred cubic lattice. The adge of the unit cell is 408 pm. The diameter of the metal atom is :

Answer»

288 PM
408 pm
144 pm
204 pm

Answer :A
Discussion
88791.

A metal crystallises into a lattice containing a sequence of layers of atoms of AB AB AB "………". What is the percentage by volume of this lattice having empty space ?

Answer»

74
26
20
16

Solution :AB AB AB `"……………"` is hcp arrangement in which space occupied is 74%
`:.` EMPTY space = 26%.
Discussion
88792.

A metal crystallises in a simple cubic unit cell of edge length 6.22 Å. The radius of the metal atom

Answer»

`1.55A^(@)`
`3.11A^(@)`
`6.22A^(@)`
`2.45A^(@)`

ANSWER :B
Discussion
88793.

A Metalcrystallisesinto twocubicfacesnamelyface centered(fcc)and bodycentered(bcc). Whoseunitcelledge lengthare 3.5 Å respectively.Findthe ratioof thedensitiesof fcc andbcc.

Answer»


Solution :Edge lengthof unitcell of fcc METAL`=3.5 Å = 3.5 Å = 3.5 xx 10^(-8)` cm
Edgelengthof unitcell of thebcc metal`=3 Å = 3 xx 10^(-8) cm`
Density `d= (z xx M)/(a^(3) xx N_(A))`
Where z= NUMBER of the Fe atomsin theunitcell
M= ATOMICMASS ofmetal
a= Edgelengthof unitcell
`N_(A) =` Avogadro number
`:.` For fcc unitcell = z =4
Forbcc unitcell= z= 2
`:. ("Densityof fccunit cell ")/("Densityof bccunit cell") = (d_("fcc"))/(d_("bcc"))`
`(z_("fcc")xx M)/(a_("fcc")^(3)) xx (N_(A))/(z_("bcc")xxM)(a_("bcc")^(3))xx N_(A)`
`=(z_("fcc"))/(z_("bcc")).(a_("bcc"))/(a_("fcc")^(3))`
`=(4)/(2) (3xx10^(-8))/(3.5xx 10^(-8))`
`= 1.26`
Discussion
88794.

A metal crystallises ina face -centred cubic structure. If the edge length of its unit cell is 'a' the closest approach between two atoms in metallic crystal will be ………

Answer»

`sqrt(2)a`
`a/(SQRT2)`
`2a`
`2sqrt(2)a`

ANSWER :B
Discussion
88795.

A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is 'a' the closest approach between two atoms in metallic crystal will be

Answer»

`2sqrt(2)a`
`sqrt(2)a`
`(a)/sqrt(2)`
`2a`

SOLUTION :For FCC, `sqrt(2)a=4R` So, `2R=(a)/sqrt(2)`
Discussion
88796.

A metal crystallises in a simple cubic unit cell. If the edge length of the unit cell is 565.6 pm then, radius of metal atom is

Answer»

282.8 PM
400 pm
200 pm
245 pm

Solution :We know that for SIMPLE cubic STRUCTURE RADIUS `(r)=a/2=565.6/2=282.8`pm
Discussion
88797.

A metal crystallises in a face centred cubic structure . If the edge length of its unit cell is 'a' , the closest approach between two atoms in metallic crystal will be -

Answer»

`2sqrt(2)a`
`sqrt2a`
`(a)/(SQRT(2))`
2a

Answer :C
Discussion
88798.

A metal crystallises in a face centred cubic structure. If the edge length of its cell is 'a', the closest approach between two atoms in metallic crystal will be:

Answer»

2a
`2sqrt(2)a`
`SQRT(2)a`
`a/sqrt(2)`

SOLUTION :for f.c.c, `r=sqrt(2)/4a`, closest approach = 2R
Discussion
88799.

A metal complex having the composition or (NH_(3))_(4)Cl_(2)Br has been isolated in two form A and B. The form A reacts with AgNO_(3) to give a white precipitate readily soluble in dilute aqueous ammonia whereas B gives a pale yellow precipitate soluble in concentrated ammonia . The hybridization of Cr in the complexes A and B respectively is

Answer»

`d^(2)SP^(3) and sp^(3)d^(2)`
`sp^(3)d^(2) and d^(2)sp^(3)`
`sp^(3)d and DSP^(3)`
`d^(2)sp^(3) and d^(2)sp^(3)`

ANSWER :D
Discussion
88800.

A metal complex having the composition or (NH_(3))_(4)Cl_(2)Br has been isolated in two form A and B. The form A reacts with AgNO_(3) to give a white precipitate readily soluble in dilute aqueous ammonia whereas B gives a pale yellow precipitate soluble in concentrated ammonia . The formula of the complex A is

Answer»

`[CR(NH_(3))_(4)BR]Cl_(2)`
`[Cr(NH_(3))_(4)BrCl]Cl`
`[Cr(NH_(3))_(4)Cl_(2)]Br`
Both A and B are possible.

Answer :B
Discussion