52671 + Interview Questions in Chemistry in Class 11

1.	An element with mass number 81 contains 31.7% more neutrons as compared to protons.Assign the sysmbol to the element.
Answer» Solution :An element can be identified by its atomic NUMBER only.Let us find the number.<BR>Let the number of protons=x `:.`Number of neutrons `=x+(x xx32.7)/(100)=(x+0.317x)` Now,mass no of element=NO. of protons+ No. of neutrons `81=x+x+0.317x=2.317x` or x=(81)/(2.317)=35 `:.`NO.of protons=35,No.of neutrons=81-35=46 The element with atomic number(Z)35 is `"BROMINE"_(35)^(81)Br.`

Discussion

2.	An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol
Answer» Solution :Mass NUMBER = 81, i.e., `p + N = 81` If PROTONS = x, then neutrons `= x + (31.7)/(100) XX x = 1.317 x` `:. x + 1.317 x = 81 or 2.317 x = 81 or x = (81)/(2.317) = 35` Thus, Protons = 35, i.e., ATOMIC no. = 35 Hence, the symbol is `._(35)^(81)Br`

Discussion

3.	An element with electronic arrangement as 2, 8, 18, 1 will exhibit the following stable oxidation states
Answer» `+2 & +4` `+1 & +2` `+2` only `+1 only` ANSWER :B

Discussion

4.	An element with density 11.2 g cm^(-3) forms a f.c.c. lattice with the edge length of 4xx10^(-8) cm. Calculate the atomic mass of the element. (Given :N_A=6.022xx10^23 "mol"^(-1))
Answer» SOLUTION :`rho=(ZxxM)/(a^3xxN_A)` For element with f.c.c. lattice , Z=4 `THEREFORE M=(rhoxxa^3xxN_A)/Z=((11.2 g CM^(-3))(4XX10^(-8) cm)^3xx(6.022xx10^23 mol^(-1)))/4=107.9 "g mol"^(-1)`

Discussion

5.	An element with density 11.2g cm^(-3)forms a f.c.c lattic with edge lengh of4 xx 10^(-8)cm. calculate the ofatoms mass of the element( Given :N_(A) = 6.022 xx 10^(23) mol^(-1)
Answer» <P> Solution :`p (Z xx M)/( a^(3)xx N_(A)) ` For element with f.c.c. lattice, Z = 4 `M = ( p xx a^(3) xx N_(A))/ Z = (( 11.2 " G cm"^(-3)) ( 4xx 10^(-8) "cm")^(3) xx ( 6.022 xx 10^(23) "mol"^(-1)))/4 = 107.9 " g mol"^(-1)`

Discussion

6.	An element with atomic number 20 will be placed in which period of the periodic table?
Answer» 4 3 2 1 ANSWER :A

Discussion

7.	An element with atomic number 106 has beendiscovered recently. Which of the following electronic configuration will it posses
Answer» `[Rn]5f^(14)6d^(5)7S^(1)` `[Rn]5f^(14)6d^(5)7s^(2)` `[Rn]5f^(14)6d^(6)7s^(0)` `[Rn]5f^(14)6d^(1)7s^(2)7P^(3)` Answer :A

Discussion

8.	An elementwith atomic number106has beendiscoveredrecently. Which of the followingelectronicconfigurationwill itposses ?
Answer» `[Rn ] 5f^(14) 6D^(4)6d^(4) 7s^(2) ` `[Rn ] 5d^(14)6d^(5) 7s^(1)` `[Rn] 5f^(14) 6d ^(6)7s^(0)` `[Rn] 5f^(14) 6d^(1) 7s^(2) 7p^(3)` Solution :In heavierelements HUND's rule is notalwaysfollowed.Referto PeriodicTablegivenon thebackof thetitlepage.

Discussion

9.	An element with atomic no.20 will be placed in which period of the periodic table ?
Answer» 4 3 2 1 Answer :a

Discussion

10.	An element with a mass number of 81 contains 31.7% more neutrons as compared to protons. Identify the element
Answer» SC BA Br I Answer :C

Discussion

11.	An element with 1s^2 2s^2 2p^6 3s^2 electronic arrangement will be forming
Answer» ACIDIC oxide Basic oxide Neutral oxide Amphoteric oxide Answer :B

Discussion

12.	An element which lies in the same group of the periodic table as mercury is
Answer» CADMIUM GOLD TIN THALLIUM ANSWER :A

Discussion

13.	An element which is recently discovered is placed in 7th period and 10th group. IUPAC name of the element will be
Answer» Unnilseptium Ununnilium UNUNBIUM NONE ANSWER :B

Discussion

14.	An element of I^(st) transition series X^(+3) have highest magnetic moment in series and X^(+2) have non mangnetic nature as low spin complex. If atomic number of X is 'a' and number of unpaired e^(-) in free state X^(+2) and X^(+3) is 'b' and 'c' then calculate a+b+2c.
Answer» ANSWER :B::D

Discussion

15.	An element of group 2 forms covalent oxide which is amphoteric in nature and dissolves in water to give an amphoteric hydroxide . Identify the element and write chemical reactions of the hydroxide of the element with an alkali and an acid .
Answer» Solution :Because of small size and comparatively high electronegativity of Be , BeO is covalent . It is INSOLUBLE in water . It is amphoteric in nature and DISSOLVES in acids as well as alkalies . `BeO + 2 HCL to BeCl_(2) + H_(2)O` `BeO + 2NaOH to underset("Sod. beryllate")(Na_(2)BeO_(2) +) H_(2)O`

Discussion

16.	An element of group-2 forms covalent oxide which is amphoteric in nature and dissolves in water to give an amphoteric hydroxide. Identify the element and write chemical reactions of the hydroxide of the element with an alkali and an acid.
Answer» Solution :In group 2, Be is the only ELEMENT which gives COVALENT oxide Bel, which is AMPHOTERICIN nature. The rest elements of this group give ionic oxides which are BASIC in nature. Beo dissolves in water and gives sparingly soluble hydroxide which reacts with acid and base to give salt `BeO+H_(2)O to underset("Beryllium hydroxide")(Be(OH)_(2))` `Be(OH)_(2)`is AMPHOTERIC hydroxide. So it is soluble in both acid and base. In produces tetra hydroxide beryllate `(Z^(-))` ionon salublizing in NaOH like alkali solution. `2NaOH_((aq)) +Be(OH)_(2(s)) to underset("hydroxidoberyllate")underset("Sodium tetra")(Na_(2)Be(OH)_(4(aq)))` It produces beryllium salt on reaction with acid. `Be(OH)_(2) +H_(2)SO_(4) underset("Beryllium sulphate")(to BeSO_(4)+2H_(2)O)`

Discussion

17.	An element of group 14 forms two oxides one of which is highlty poisionous and neutral. Other oxide can be easily liquefied and compressed to give a solid which is used as a refrigerant under the name of drikold. The element and the oxides are
Answer» `Si,SiO,SiO_(2)` `PB,PbP,PbO_(2)` `C,CO,CO_(2)` `Sn,SnO,SnO_(2)` Solution :C forms two oxides CO and `CO_(2)`. CO is NEUTRAL and poisnonous `CO_(2)` can be compressed to a refrigerant dry is ice or solid `CO_(2)`. SiO is UNSTABLE. SnO and PbO are amphoteric in nature.

Discussion

18.	Anelementof group 13 whichformsa basicoxideis ………
Answer» SOLUTION :THALLIUM , TI

Discussion

19.	An element of 5f-series but has no electrons filled in 5f-sub shell
Answer» AC Ce Th U ANSWER :C

Discussion

20.	An element occupies group number 1 and period number 3. This element heated in air gives compound A. With water it gives compound B - a strong base. With ammonia gives compound C, which is used as a reducing agent in organic chemistry. Identify the element A,B and C.
Answer» Solution :(i) As per the position in the PERIODIC table this element occupying group NUMBER 1 and period number 3 is sodium. (ii) When heated in air, it forms sodium peroxide (A) `4 Na +O_2 rArr 2 Na_2O_2` (iii) With water, sodium gives sodium hydroxide (B) `2Na +2H_2O rArr 2NaOH +H_2` (iv) With ammonia, sodium gives sodamide (I) `2Na + 2NH_(3) overset(570-670 K)rarr 2 NaNH_2 +H_2` The element is sodium. The compound A is sodium peroxide. The compound B is sodium hydroxide. The compound C is sodamide.

Discussion

21.	An element occupies group no.13 and period number 2 is an representative element of that group reacts with carbondioxide and forms an oxide (A). (A) reacts with CuSO_(4) give blue beads (B). Identify the element compound (A) and (B). Write the reaction.
Answer» Solution :As per the position in the periodic table this element which occupies group number 13 and period number 2 is BORON and it reacts with `CO_(2)` to form boric oxide 1) `3CO_(2)+4Brarr2B_(2)O_(3)+3C` `therefore` the COMPOUND A is `B_(2)O_(3)` 2) `B_(2)O_(3)` reacts with `CuSO_(4)` to form `underset(("Red"))((CuBO_(2))_(2))` `CuSO_(4)+B_(2)O_(3)rarrCu(BO_(2))_(2)+SO_(3)` `therefore` the compound B is copper BEAD. `Cu(BO_(2))_(2)`

Discussion

22.	An element M reacts with chlorine to from a compound X. The bond angle in X si 120^(@). What is M?
Answer» Be B Mg N Answer :B

Discussion

23.	An element M has an atomic number 9 and atomic mass 19. Its ion will be represented by (1) M (2) M^(2+) (3) M^(-) (4) M^(2-)
Answer» Solution :Since its atomic number is 9, hence it is some halogen (e.g. F). Thus, its ION should EXIST as `M^(-)` ion as given in CHOICE (3).

Discussion

24.	An element in its lowest oxidation state can act only as an oxidising agent.
Answer» ANSWER :F

Discussion

25.	An element having electronic configuration [Rn]6s^(1) will:
Answer» Form BASIC oxide Can be used in photolectric cell Form acidic oxide Has HIGH ionisation enthalpy Solution :The ELEMENT is Cs. Due to larger size and low ionisation enthalpy of `[Rn]6s^(1)` (i.e. Cs).

Discussion

26.	An element having electronic configuration- [Ar] 3d^(3) 4s^(2)belongs to ........
Answer» s-block p-block d-block f-block Solution :`[Ar] 3D^(3) 4s^(2)` electronic CONFIGURATION means Z = 23 (V). In the element LAST electron entres in 3d-orbital. "Elements in which the last electron is FILLED in d-orbital are known as d-block element.

Discussion

27.	An element has nine positive charges in its nucleus. Its common oxidation state is
Answer» `+7` `+5` `-1` `+1` ANSWER :C

Discussion

28.	An element has electronic configuration 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(4). Predict their period , group and block :
Answer» <P>Period = `3^(rd)`, block = p, group = 16 Period = `5^(th)`, block = s, group = 1 Period = `3^(rd)` , block = p, group = 10 Period = `4^(th)` , block = d, group = 12 Solution : Period = `3^(rd)`, block = p. group = 16 By observing PRINCIPAL quantam NUMBER (n), Orbital (s,p, d,F)and equating no. of `e^(-)`, s we are able to find the period, block and group of element in periodic table.

Discussion

29.	An element has atomic mass 93 "g mol"^(-1) and density 11.5 "g cm"^(-3). If the edge length of its unit cell is 300 pm, identify the type of unit cell
Answer» Solution : `rho=(ZxxM)/(a^3xxN_0), Z=(rhoxxa^3xxN_0)/M =((11.5 g CM^(-3))(300xx10^(-10)cm)^3(6.023xx10^23 mol^(-1)))/(93 g mol^(-1))=2.01 approx2` HENCE, the type of unit cell is BCC.

Discussion

30.	An element has atoic mass93 " g mol" ^(-1)and density11.5 " g cm" ^(-3). If the edge length of its unit cell is 300 pm,identify the type of unit cell.
Answer» SOLUTION :`p = (Z xx M)/(a^(3)xxN_(0)) , Z = ( p xx a^(3) xx N_(0))/M = (( 11.5 " g cm"^(-3)) (300xx 10^(-10)"cm")^(3)(6.023 xx10^(23) "mol"^(-1)))/(93" mol"^(-1)) = 2.01 =2 ` Hence, the TYPE of unit cell in BCC.

Discussion

31.	An element has a body-centred cubic (bcc) structure with cell edge of 288 pm. The density of the element is 7.2 g//cm^3 . How many atoms are present in 208 g of the element ?
Answer» Solution :For the BCC structure, Z=2 EDGE of the unit cell , a=288 pm , DENSITY of the element , `rho=7.2 g//cm^3` Subtituting the values in the expression `rho=(ZxxM)/(a^3xxN_0)` `7.2 g cm^(-3) =(2xxM)/((288xx10^(-10) cm)^3 XX(6.02xx10^23 mol^(-1))) "or" M=51.8 g mol^(-1)` By mole concept, 51.8 g of the element contains =`6.02xx10^23` atoms `therefore` 208 g of the element contains =`(6.02xx10^23)/51.8xx208` atoms =`24.17xx10^23` atoms

Discussion

32.	An element has 2 electrons in its K-shell. 8 electrons in L-shell, 13 electrons in M-shell and one electron in N-shell. The element is
Answer» Cr Fe V Ti Answer :A

Discussion

33.	An element has 18 electrons in the outer most shell. The element is
Answer» Transition METAL RARE earth metal ALKALINE earth metal Alkali metal Answer :A

Discussion

34.	An element forms two oxides containing 50% and 40% of the element by mass. Prove that the results are in agreement with the law of multiple proportions.
Answer» Solution :In the first oxide, the PERCENTAGE of element is 50. This means that 50 g of element combine with 100 - 50 = 50 g of oxygen. Let us calculate the amount of oxygen that combines with a fixed amount (say 100 g) of the element. The amount of oxygen that combines with 100 g of element = `50/50 XX 100 = 100 g` In the second oxide, the percentage of the element is 40. This means that 40 g of element combine with 60 g of oxygen. `therefore`The amount of oxygen that combines with 100 g of element `=60/40 xx 100 = 150 g` Now, the amounts of oxygen that combine with 100 g of element in the TWO cases are respectively 100 g and 150 g. The ratio in these amounts is 2 : 3 which is a simple whole number ratio. Thus, the results are in agreement with the LAW of multiple proportions.

Discussion

35.	An element (density 6.8 g cm^(-3)) occurs in the BCC structure with cell edge of 290 pm. Calculate the number of atoms present in 200 g of the element.
Answer» SOLUTION :Calculate the ATOMIC mass, M. Then M GRAMS contains `6.023xx10^23` atoms. Calculate atoms PRESENT in 200 g.

Discussion

36.	An element (density 6.8 gcm^(-3)and the length of the side of the unit cell is 316 pm. The unit cell in the most important crystalline form of tungsten is the body centred unit cell. How many atoms of theelementdoes 50 gof the element contain ?
Answer» SOLUTION :CALCULATE the atomic mass. M's Then grams CONTAIN ` 6.023 xx 10^(23) ` ATOMS. Calculate atoms present in 200g.

Discussion

37.	An element crytallizes in the cubic lattice and the edge of the unit cell is 430 pm. Calculation the number of atoms in a unit cell. [ Atomic mass of Na = 23.0amu. Density of sodium = 0.9623 gcm^(-3) ,N_(A)= 6.023 xx 10^(23)mol^(-1)
Answer» ANSWER :FCC

Discussion

38.	An element crystallizes into a structure which may be described by a cubic type of unit cell having one atom on each corner of the cube and two atoms on one of its body diagonals. If the volume of this unitis24 xx 10^(-24)"cm"^(3)and density of element is7.2 g cm ^(-3),calculate the number of atoms present in 200 g of the element.
Answer» Solution :No. of atoms per unit cell ` = 8 XX 1/8 + 2 = 3 ,i.e. , Z =3 ` Volume of the unit cell = ` a^(3) = 24 xx 10^(-24) " cm" ^(3)` ` p = ( Z xx M)/ ( a^(3) xx N_(0))` ` 7.2 = (3xx M) / (( 24 xx 10^(-24) ) xx ( 6.023 xx 10^(23)) )or M = 34.69` Thus, 34.69g of the element have to atoms =`6.023 xx 10^(23)` 200 G of the element will have atoms ` = ( 6.023 xx 10^(23))/(34.69) xx 200 = 3.4722 xx 10^(24)` atoms

Discussion

39.	An element crytallizes in fcc lattice having edge length 400 pm. Calculate the maximum diameter which can be placed in interestital sites without disturing the structure.
Answer» Solution :Interstitial sites can be either tetrahedral or orctahdral or octahedral voids. As octahedral voids are bigger than tetrahedral voids, maximum diameter can fit into octahedral voids. If R is the R is the size of atoms in the fcc packing and r is the size of the octahedral void , r= 0.414 R. if fcc lattice, face diagonal =` SQRT2 a ` As in fcc, atoms along the face diagonal are tourching each other. ` 4R =sqrt2 a or R = (sqrt2a)/4` Required diameter of the interestitial site = 2R = ` 2xx 0.414 R = 2 xx 0.414 xx sqrt2/4 xx 400`pm. 117.1 pm.

Discussion

40.	An element crystallizes into a structure which may be described by a cubic type of unit cell having one atom on each corner of the cube and two atoms on one of its body diagonals. If the volume of this unit cell is 24xx10^(-24) cm^3 and density of element is 7.2 g cm^(-3) , calculate the number of atoms present in 200 g of the element.
Answer» Solution :No. of ATOMS PER unit cell =`8xx1/8+2=3` i.e., Z=3 VOLUME of the unit cell =`a^3 = 24xx10^(-24) cm^3` `RHO=(ZxxM)/(a^3xxN_0)` `therefore 7.2=(3xxM)/((24xx10^(-24))xx(6.023xx10^23))` or M=34.69 Thus , 34.69 g of the element have atoms =`6.023xx10^23` `therefore` 200 g of element will have atoms =`(6.023xx10^23)/34.69xx200=3.4722xx10^24` atoms

Discussion

41.	An element crystallizes in fcc lattice having edge length 400 pm. Calculate the maximum diameter which can be placed in interstitial sites without disturbing the structure .
Answer» Solution :Interstitital sites can be either tetrahedral or octahedral voids. As octahedral voids are bigger than tetrahedral voids, MAXIMUM diameter can fit into octahedral voids. If R is the size of atoms in the fcc packing and r is the size of the octahedral void, r=0.414 R In fcc lattice, face diagonal =`sqrt2a` As in fcc, atoms along the face diagonal are touching each other . `4R=sqrt2a "or" R=(sqrt2a)/4` REQUIRED diameter of the interstitial site =2 r =2 x 0.414 R =2 x 0.414 x `sqrt2/4` x 400 PM =117.1 pm (`because` a=400 pm )

Discussion

42.	An element crystallizes in a f.c.c. lattice with cell edge of 250 pm. Calculate the density if 300 g of this element contains 2 xx 10^24 atoms
Answer» Solution :MOLAR mass of the element (M)=`300/(2xx10^24)xx6.02xx10^23=90.3 g mol^(-1)` Density`(RHO)=(ZxxM)/(a^3xxN_0)=(4xx90.3)/((250xx10^(-10))^3xx(6.02xx10^23))=38.4 g cm^(-3)`

Discussion

43.	An element crystallize in f.c. lattice with cell edge of 250 pm. Calculate the density if 300 g of this element contains2 xx 10^(24)atoms.
Answer» Solution :MOLAR mass of the emlment (M) = `( 300)/(2xx 10^(24)) xx 6.02xx 10^(23)= 90.3 " G MOL"^(-1)` Density `(p)= (Z xx M)/(a^(3)xxN_(0))= (4xx90.3)/((250xx10^(-10))^(3) xx (6.02 xx 10^(23))) = 38.4 " g mol" ^(-3)`

Discussion

44.	An element crystallises in a faces centered cubic lattice. Hence, its unit cell contains
Answer» 14 ATOMS of the ELEMENT and 8 of them belong to the unit cell 14 atoms of the element and 4 of them belongs to the unit cell 8 atoms of the unit cell and only 1 of them belongs to the cell 8 atoms of the unit cell and only 2 of them belong to the cell Answer :B

Discussion

45.	An element belongs to 3^(rd) period and group 13 of the periodic table. Which of the following properties will be shown by the element ?
Answer» GOOD conductor of electricity Liquid, metallic Solid, metallic Solid, non-metallic Solution :All the ELEMENTS of GROUP 13, except boron are metallic in Nature. Being metallic in nature, ALUMINIUM behaves as a good conductor of electricity

Discussion

46.	An element belongto 3rdperiodand group- 13 of theperiodictable. Which of thefollowingproperties will beshown by theelement gt
Answer» Goodconductor of electricity Liquid , metallic solid metallic Solid NON metalic Solution :The elementwhich BELONGS to 3rdperiodand group 13 isaluminium (A1). It is solidmetallicand goodconductorof electricity. Thusoptions(a) and ( c)are correct.

Discussion

47.	An element (atomic mass=250 u) crystallizes in a simple cubic. If the density of the unit cell is 7.2 g cm^(-3),what is the radius of the element ?
Answer» `1.93times10^(-3)CM` `1.93times10^(-8) cm` `1.93times10^(-8)Å` `1.93times10^(-8)m` SOLUTION :`d=(ZtimesM)/(a^(3)timesN_(0))therefore7.2=(1times250)/(a^(3)times6.02times10^(23))` or`a^(3)=57.6times10^(-24)` or`a=3.86times10^(-8)` For simple cubic, `r=a//2=1.93times10^(-8)` cm

Discussion

48.	An element ("atomic mass"=100 g//mol)having BCC structure has uniti cell edge 400 pm. The density of the element is
Answer» 2.144g`CM^(-3)` 5.188g`cm^(-3)` 7.289g`cm^(-3)` 10.376g`cm^(-3)` Solution :Density, `d(G cm^(-3))`=`(Z timesM)/(a^(3)timesN_(A) times10^(-30))` `=(2times100)/((400)^(3)times6.02times10^(23)times10^(-30))` `= 200/(64times6.02times10^(-1))= 5.188 cm^(-3)`

Discussion

49.	An element A in conmpound ABD has an oxidation no. A^(n-). It is oxidisides by Cr_(2)O_(7)^(2-) in acid medium. In an experiment 1.68xx10^(-3) mole of K_(2) Cr_(2)O_(7) was required for 3.26xx10^(-3) moel of the compound ABD. Calculate new oxidation state of A.
Answer» ANSWER :`3-n`

Discussion

50.	An element A in a compound AB has oxidationnumber -n. IT is oxidized by Cr_(2)O_(7)^(2-) in acid medium. In the experiment 1.68xx10^(-3) moles of K_(2)Cr_(2)O_(7) was usedfor 3.36xx10^(-3) moles of AB. The new oxidation number of A after oxidation is
Answer» 3 3 - n n - 3 `+n` Solution :`n= (1.68xx10^(-3))/(6xx 3.2 xx10^(-3))` `A^(n-)OVERSET(+n)rarrA^(X)` `1/(x+n)` moles `A^(n-),` `overset(+12)(Cr_2)overset(2-)(O_7)overset((6))rarr2cv^(3+)` `-=1/6` moles `Cr_(2)O_(7)^(2-)=1.68 xx10^(-3)` `(x+n)Cr_(2)O_(7)^(2-)-=6A^(n-)` `1.6 8810^(-3)""?` `=(1.68 xx10^(-3)xx6)/(x+n)=3.26xx10^(-3)` x = 3 - n

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

An element with mass number 81 contains 31.7% more neutrons as compared to protons.Assign the sysmbol to the element.

An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol

An element with electronic arrangement as 2, 8, 18, 1 will exhibit the following stable oxidation states

An element with density 11.2 g cm^(-3) forms a f.c.c. lattice with the edge length of 4xx10^(-8) cm. Calculate the atomic mass of the element. (Given :N_A=6.022xx10^23 "mol"^(-1))

An element with density 11.2g cm^(-3)forms a f.c.c lattic with edge lengh of4 xx 10^(-8)cm. calculate the ofatoms mass of the element( Given :N_(A) = 6.022 xx 10^(23) mol^(-1)

An element with atomic number 20 will be placed in which period of the periodic table?

An element with atomic number 106 has beendiscovered recently. Which of the following electronic configuration will it posses

An elementwith atomic number106has beendiscoveredrecently. Which of the followingelectronicconfigurationwill itposses ?

An element with atomic no.20 will be placed in which period of the periodic table ?

An element with a mass number of 81 contains 31.7% more neutrons as compared to protons. Identify the element

An element with 1s^2 2s^2 2p^6 3s^2 electronic arrangement will be forming

An element which lies in the same group of the periodic table as mercury is

An element which is recently discovered is placed in 7th period and 10th group. IUPAC name of the element will be

An element of I^(st) transition series X^(+3) have highest magnetic moment in series and X^(+2) have non mangnetic nature as low spin complex. If atomic number of X is 'a' and number of unpaired e^(-) in free state X^(+2) and X^(+3) is 'b' and 'c' then calculate a+b+2c.

An element of group 2 forms covalent oxide which is amphoteric in nature and dissolves in water to give an amphoteric hydroxide . Identify the element and write chemical reactions of the hydroxide of the element with an alkali and an acid .

An element of group-2 forms covalent oxide which is amphoteric in nature and dissolves in water to give an amphoteric hydroxide. Identify the element and write chemical reactions of the hydroxide of the element with an alkali and an acid.

An element of group 14 forms two oxides one of which is highlty poisionous and neutral. Other oxide can be easily liquefied and compressed to give a solid which is used as a refrigerant under the name of drikold. The element and the oxides are

Anelementof group 13 whichformsa basicoxideis ………

An element of 5f-series but has no electrons filled in 5f-sub shell

An element occupies group number 1 and period number 3. This element heated in air gives compound A. With water it gives compound B - a strong base. With ammonia gives compound C, which is used as a reducing agent in organic chemistry. Identify the element A,B and C.

An element occupies group no.13 and period number 2 is an representative element of that group reacts with carbondioxide and forms an oxide (A). (A) reacts with CuSO_(4) give blue beads (B). Identify the element compound (A) and (B). Write the reaction.

An element M reacts with chlorine to from a compound X. The bond angle in X si 120^(@). What is M?

An element M has an atomic number 9 and atomic mass 19. Its ion will be represented by (1) M (2) M^(2+) (3) M^(-) (4) M^(2-)

An element in its lowest oxidation state can act only as an oxidising agent.

An element having electronic configuration [Rn]6s^(1) will:

An element having electronic configuration- [Ar] 3d^(3) 4s^(2)belongs to ........

An element has nine positive charges in its nucleus. Its common oxidation state is

An element has electronic configuration 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(4). Predict their period , group and block :

An element has atomic mass 93 "g mol"^(-1) and density 11.5 "g cm"^(-3). If the edge length of its unit cell is 300 pm, identify the type of unit cell

An element has atoic mass93 " g mol" ^(-1)and density11.5 " g cm" ^(-3). If the edge length of its unit cell is 300 pm,identify the type of unit cell.

An element has a body-centred cubic (bcc) structure with cell edge of 288 pm. The density of the element is 7.2 g//cm^3 . How many atoms are present in 208 g of the element ?

An element has 2 electrons in its K-shell. 8 electrons in L-shell, 13 electrons in M-shell and one electron in N-shell. The element is

An element has 18 electrons in the outer most shell. The element is

An element forms two oxides containing 50% and 40% of the element by mass. Prove that the results are in agreement with the law of multiple proportions.

An element (density 6.8 g cm^(-3)) occurs in the BCC structure with cell edge of 290 pm. Calculate the number of atoms present in 200 g of the element.

An element (density 6.8 gcm^(-3)and the length of the side of the unit cell is 316 pm. The unit cell in the most important crystalline form of tungsten is the body centred unit cell. How many atoms of theelementdoes 50 gof the element contain ?

An element crytallizes in the cubic lattice and the edge of the unit cell is 430 pm. Calculation the number of atoms in a unit cell. [ Atomic mass of Na = 23.0amu. Density of sodium = 0.9623 gcm^(-3) ,N_(A)= 6.023 xx 10^(23)mol^(-1)

An element crytallizes in fcc lattice having edge length 400 pm. Calculate the maximum diameter which can be placed in interestital sites without disturing the structure.

An element crystallizes in fcc lattice having edge length 400 pm. Calculate the maximum diameter which can be placed in interstitial sites without disturbing the structure .

An element crystallizes in a f.c.c. lattice with cell edge of 250 pm. Calculate the density if 300 g of this element contains 2 xx 10^24 atoms

An element crystallize in f.c. lattice with cell edge of 250 pm. Calculate the density if 300 g of this element contains2 xx 10^(24)atoms.

An element crystallises in a faces centered cubic lattice. Hence, its unit cell contains

An element belongs to 3^(rd) period and group 13 of the periodic table. Which of the following properties will be shown by the element ?

An element belongto 3rdperiodand group- 13 of theperiodictable. Which of thefollowingproperties will beshown by theelement gt

An element (atomic mass=250 u) crystallizes in a simple cubic. If the density of the unit cell is 7.2 g cm^(-3),what is the radius of the element ?

An element ("atomic mass"=100 g//mol)having BCC structure has uniti cell edge 400 pm. The density of the element is

An element A in conmpound ABD has an oxidation no. A^(n-). It is oxidisides by Cr_(2)O_(7)^(2-) in acid medium. In an experiment 1.68xx10^(-3) mole of K_(2) Cr_(2)O_(7) was required for 3.26xx10^(-3) moel of the compound ABD. Calculate new oxidation state of A.

An element A in a compound AB has oxidationnumber -n. IT is oxidized by Cr_(2)O_(7)^(2-) in acid medium. In the experiment 1.68xx10^(-3) moles of K_(2)Cr_(2)O_(7) was usedfor 3.36xx10^(-3) moles of AB. The new oxidation number of A after oxidation is