94420 + Interview Questions in Chemistry in Class 12

1.	An element with atomic number 113 has been discovered.It will belong to which of the following block,group number,period and outershell electronic configuration?
Answer» s-block,group 2,period 7,` 7s^2` p-block,group 13,period 7, `7s^2 7p^1` p-block,group 13,period 6,` 6s^2 6p^1` d-block,group -12,period 6,` 5d^10 6s^2` ANSWER :B

Discussion

2.	An element whose IUPAC name is ununtrium (Uut) belongs to
Answer» s-block element p-block element d-block element transition element Solution :UNUNTRIUM for atomic number=113, for Z>86pd. No =7 `UNDERSET("p-block")(113[RN])7s^(2),5f^(14),6d^(10),7p^(1)`

Discussion

3.	An element which never has a positive oxidation number in any of its compounds is:
Answer» Boron Oxygen Chlorine Fluroine Answer :D

Discussion

4.	An element whose IUPAC name is ununtrium (Uut) belongs to :-
Answer» s-block p-block d-block f-block Answer :B

Discussion

5.	An element which is highly toxic for plants and animals is ?
Answer» Au Mn Hg Ca Answer :C

Discussion

6.	Mark the element which shows only one oxidation state in its compounds
Answer» Solution :An element shows only an UNIQUE oxidation STATE in `F_(2)` its ATOMICITY is 2. It is `F_(2) -.2.`.

Discussion

7.	An element undergoes a reaction as shown : X+e^(-) to X^(-) Energy released =30.876 eV The energy released, is used to dissociate 8 g of H_2 molecules equality into H^+ and H^, where H^ is in an excited state, in which the electron travels a path length equal to four times its debroglie wavelength. (a)Determine the least amount (moles) of 'X' that would be required. Given : I.E. of H =13.6eV/atom Bond energy of H_2 =4.526 eV/molecule. (b) Why is the amount of X calculated in the above question 'least' /
Answer» SOLUTION :(a)`X+e^(-)TOX^(-)` ENERGY released`=E.A_1=30.87` eV/atom Let no of MOLES of X be a `:. axxN_Axx30.87=4xxN_Axx4.526+4xxN_Axx13.6+4xxN_Axx12.75 implies a=4` moles

Discussion

8.	An element reacts with hydrogen to form a compound X which on treatment with water liberates hydrogen gas. The element can be
Answer» Fluorine Nitrogen Sodium Oxygen Answer :C

Discussion

9.	An element posses cubic lose packing structure. Calculate the radius (mu) of the atom in the unit cell [Edge length a = 252 nm].
Answer» 152 NM 89.36 nm 12.6 nm 109.1 nm Solution :In fcc, RADIUS `(r)= (sqrt2)/4 a implies (sqrt2)/4 XX 252`.

Discussion

10.	An element X (At,wt = 80 g//mol) having fcc structure, calculate the number of unit cells in 8g of X
Answer» `0.4xxN_(A)` `0.1xxN_(A)` `4xxN_(A)` NONE of these Answer :D

Discussion

11.	An element of group 2 forms covalent oxide, which is amphoteric ini nature and dissolves in water to give an amphoteric hydroxide. Identify the element.
Answer» Beryllium Magnesium Calcium Barium Answer :A

Discussion

12.	An element of atomic mass 90 occurs in fce structure with cell edge of 500 pm. Calculate the Avogadro's number if the density is 4.2g cm^(-3)
Answer» Solution :M=90, Z=4 for FCC structure, a = 500 pm, d=4.20g `cm^(-3)` `N_A=(Z xx M)/(d xx a^3)=(4 xx 90)/(4.2 xx (500)^3 10^(-30))=6.86 xx 10^(23) mol^(-1)`

Discussion

13.	An element of 3d-transition series two oxidation states x and y, differ by two units then:
Answer» compounds in OXIDATION state x are ionic if `XGTY` compounds in oxidation state x are ionic if `xlty` compounds in oxidation state y are covalents if `xlty` compounds in oxidation state y are covalents if `xgty` Answer :B::C

Discussion

14.	An element of 3d-transition series shows two oxidation states x and y, differe by two units then
Answer» Compounds in OXIDATION state x are ionic if `x gt y` Compounds in oxidation state x are ionic if `x LT y` Compounds in oxidation state y are COVALENT `x lt y` Compounds in oxidation state y are covalent ` y lt x` Solution :With increase in OXIDATIONS state, covalent nature increases

Discussion

15.	An element occurs in bcc structure with cell edge 288 pm. Its density is 7.2 g cm^(-3). Calculate the atomic mass of the element.
Answer» Solution :Z = 2 for bcc structure , a=288 PM, d = 7.2 g `CM^(-3)`, `N_A =6.023 xx 10^(23) MOL^(-1)` `M= (N_A xx d xx a^3)/(Z)=(6.023 xx 10^(23) xx 7.2 xx (288)^3 xx 10^(-30))/(2) = 51.79 g mol^(-1)`

Discussion

16.	An element occurs in bcc structure. It has a cell edge length of 250 pm. Calculate the molar mass if its density is 8.0 g cm^(-3). Also calculate radius of an atom of this element.
Answer» Solution :a = 250 pm `= 250 xx 10^(-10)` cm, d = 8 `cm^(-3)`, Z = 2 (for BCC), M = ? `d = (Z xx M)/(a^(3) xx N_(A))` `8 = (2 xx M)/((250 xx 10^(-10))^(3) (6.022 xx 10^(23)))` `N = ((250 xx 10^(-10))^(3) xx (6.022 xx 10^(23)))/(2) xx 8` `M = (9.409 xx 8)/(2) = 37.64 g MOL^(-1)` For bcc unit CELL, `4r = sqrt(3)a ` radius, `R = (sqrt(3)a)/(4) = (1.732 xx 250)/(4)` = 108.25 pm

Discussion

17.	An element in oxidationstate of +3 has the electronic configuratioin : [Ar] 3d^(3). Its atomic number is :
Answer» 24 26 20 23 Answer :A

Discussion

18.	An element having bcc structure has atomic mass 50 u and density 6.81 g cm^(-3). Calculate the edge length of the unit cell.
Answer» Solution :The following data is PROVIDED : `z=2, M=50u=50" g mol"^(-1), d="6.81 g cm"^(-3), a=?, N_(A)=6.02xx10^(23)mol^(-1)` Applying the following RELATION and substituting the values, we get `a^(3)=(zxxM)/(dxxN_(A))=(2xx50)/(6.81xx6.02xx10^(23))` `=(100)/(40.9962)xx10^(-23)` `"or"a^(3)=2.43xx10^(-23)=24.3xx106(-24)` `"or"a^(3)=x^(3)xx10^(-24)cm^(3)` `"or"a=x xx10^(-8)cm` `"or"a=2.896xx10^(-8)cm` `"or"a=289.6cm` `["To find out the cube root of 24.3, use logarithms as given below :"` Let `x^(3)=24.3` `"or3 log x "=log 24.3` `"or3 log x = 1.3856"` `" ORLOG x = 0.4618orx = Antilog 0.4618"` `"orx = 2.896."]`

Discussion

19.	An element having atomic mass 63.1 g/mol has face centred cubic unit cell with edge length 3.608xx10^(-8)cm. Calculate the density of unit cell. [Given : NA=6.022xx1023" atoms/mol"].
Answer» Solution :GIVEN M = 63.1 g/mol `Z=4` `a=3.609xx10^(-8)cm` `N_(A)=6.023xx10^(23)` `"DENSITY "=(zxxM)/(a^(3)xxN_(a))` `=(4xx63.1)/((3.608xx10^(-8))xx6.022xx10^(23))` `=8.924g//cm^(3)`

Discussion

20.	An element having atomic mass 60 amu. has fcc unit cell. The edge length of the unit cell is 4 xx 10^(2) pm. Find the density of the unit cell.
Answer» Solution :`d=(ZM)/(a^(3)N_(A))=(60xx4)/((4XX10^(2))xx6.023xx10^(23))`

Discussion

21.	An element has successive ionization enthalpies as 940 (first), 2080, 3090, 4140, 7030, 7870, 16000 and 19500 kJ "mol"^(-1). To which group of the periodic table does this element belong ?
Answer» 14 15 16 17 Solution :`ubrace(underset(940)(IE_(1)) underset(2080)(IE_(II))underset(3090)(IE_(III))underset(4140)(IE_(IV))underset(7030)(IE_(V))underset(7870)(IE_(VI)))" "ubrace(underset(16000)(IE_(VII))underset(9500)(IE_(VIII)))` so valence electron is 6 THUS it belongs to gr. 16

Discussion

22.	An element has half-life 1600 years. The mass left after 6400 years will be
Answer» `1//16` `1//12` `1//4` `1//32` Solution :`T_(1//2) = 1600 yrs, N_(0) = 1, N = ?, T = 6400 yrs` `T = t_(1//2) XXN, or n = (6400)/(1600) = 4` `N = N_(0) xx ((1)/(2))^(n), N = 1 xx N = (1)/(16)`

Discussion

23.	An element has electronci configuration (Xe) 6s^(2) 4f^(13), 5d^(0). In which group it is placed
Answer» second THIRD FOURTH fifth ANSWER :B

Discussion

24.	An element has bcc structure with a cell edge of 208 pm. the density of the element is 7.2 gcm^(-3). How many atoms are present in 208g of the element.
Answer» Solution :An element has bcc STRUCTURE with a cell edge of 288 pm. The density of the element is `"7.2 gcm"^(-3)` For the Bcc structure, n = 2 `rho=(nM)/(a^(3)N_(A))` `"7.2 g cm"^(-3)=(2M)/((288xx10^(-10)cm)^(3)xx(6.023xx10^(23)" mol"^(-1)))` `"7.2 g cm"^(-3)=(2M)/((2.38xx10^(-23)cm^(3))xx(6.023xx10^(23)" mol"^(-1)))` `7.2g="0.140 M mol"` `M=(7.2g)/("0.140 mol")="51.42 g mol"^(-1)` By mole concept, 51.42 g of the element contains `6.023xx10^(23)` atom 208 g of the element will CONTAIN `=(6.023xx10^(23)xx208)/(51.42)" atoms"` `=24.14xx10^(23)" atoms (or) "2.417xx10^(24)" atoms"`

Discussion

25.	An element has a face centered cubic unit cell with a length of 352.4 pm along an edge. The density of the element is "8.9 gcm-3". How many atoms are present in 100 g of an element ?
Answer» Solution :`"MASS "=100g` `"Density"="8.9 g CM"^(-3)` `"Edge length"="352.4 pm"` `(a)=352.4xx10^(-10)cm` `"Volume of the unit cell,"` `a^(3)=(352.4xx10^(-10)cm)=4.37xx10^(-23)cm^(3)` `"Volume of 100 g of an element,"` `=("Mass")/("Density")` `=(100)/(8.9)cm^(3)=11.23cm^(3)` Therefore number of unit CELLS, `=(11.23)/(4.37xx10^(-23))=2.56xx10^(23)` SINCE each Fcc cube contains 4 ATOMS, therefore total number of atoms in 100 g. `=4xx(2.56xx10^(23))=10.24xx10^(23)" atoms"`

Discussion

26.	An element has a body-centred cubic (bcc) structure with a cell edge of 288 pm. The density of the element is 7.2 "g/cm"^3. How many atoms are present in 280 g of the element ?
Answer» Solution :VOLUME of UNIT cell `= a^3 xx 10^(-30) "cm"^3` `= (288)^3 xx 10^(-30) "cm"^3` No. of atoms PER unit cell (Z) = 2 `therefore M = (d xx N_A xx a^3 xx10^(-30))/Z` `therefore M = (7.2 xx 6.022 xx 10^(23) xx (288)^3 xx 10^(-30))/(2)` `therefore`M = 51.786 g/mol `therefore`Number of atoms in 208 g element `= 208/51.786 xx 6.022 xx 10^(23) = 24.18 xx 10^23` atoms.

Discussion

27.	An element forms two oxides, the weight-ratio composition in them is A : O = x : y in the first oxide and y : x in the second oxide. If the equivalent weight of A in the first oxide is 10.33, the equivalent weight of A in the second oxide is
Answer» 6.2 10.33 x/y y/x Answer :A

Discussion

28.	An element forms two oxides containing, 50% & 40% of oxygen respectively by weight of the element. Does these oxides illustrate the law of multiple proportions :-
Answer» Yes No NOTHING can be said Any one of the above Solution :40 g of element `to` 60g of oxygen , 1g `to` 60/40=1.5 g 50 g of element `to` 50 g of oxygen , 1 g `to` 50/50=1g THEREFORE the weight of oxygenthat combine with 1 g of the element in the two oxides are in the ratioof 1:1.5 or 2:3 which is a SIMPLER ratio . Hence it holds the law of multiple proportions

Discussion

29.	An element forms compounds of the type MCl_(3), M_(2)O_(5) and Ca_(3)M_(2) but does not form MF_(5). The element could be :
Answer» Al B N P Answer :C

Discussion

30.	An element forms a gaseous oxide which on dissolving in water gives an acid solution. The element is:
Answer» S Na P H Answer :A

Discussion

31.	An element exists in the body-centred cubic structure whose cell edge is 2.88 A. The density of the element is 7.20 g/cm^2. Calculate the number of atoms in 104 g of the element.
Answer» Solution :Volume of unit cell `=(2.88 XX 10^(-8))^(3) cc = 2.39 xx 10^(-23)` cc Volume of the element weighing 104 g = `("MASS")/("density")` `therefore` number of unit cells present in 104 g of the element `=(14.44)/(2.39 xx 10^(-23)) = 6.04 xx 10^(23)` Since each body-centred CUBIC cell contains 2 atoms, number of atoms `=2 xx 6.04 xx 10^(23) = 1.208 xx 10^(24)`

Discussion

32.	An element exist in nature in two isotopic forms: X^(30)(90%) and X^(32)(10%). What is the average atomic mass of element?
Answer» SOLUTION :AV. Atomic mass `=(sum(%"abundance"xx"atomic mass"))/(100)=(90xx30+10xx32)/(100)=30.2`

Discussion

33.	An element E loses one alpha and two beta-particles in three successive stages. The resulting element will be
Answer» an ISOBAR of E an ISOTONE of E an ISOTOPE of E E itself. Solution :`""_(Z)^(A)E overset(-alpha)(rarr) ""_(Z - 2)^(A-4)X overset(-beta)(rarr)""_(Z - 1)^(A- 4)Y overset(-beta)(rarr)""_(Z)^(A-4)W` both elements E and W have same atomic number but different MASS numbers. Hence, both are isotopes. So, the resulting element will be an isotope of E.

Discussion

34.	An element crystallizes into structure which may be described by a cube type of unit cell having one atom on each corner of the cube and two atoms on one of its diagonals. If the volume of this unit cell is 24 xx 10^(-24) cm^3 and density of the element is 7.2 g cm^(-3), calculate the number of atoms present in 200 g of the element.
Answer» Solution :We know that a simple CUBIC UNIT cell has atoms=1 Volume of element= `("Mass of element")/("density of element") = (200g)/(7.2 g CM^(-3))= 27.78 cm^3` Number of unit CELLS= `("Volume of element")/("Volume of unit cell")=(27.78 cm^3)/(24xx10^(-24)cm^3)= 1.157 xx 10^(24)` Number of atoms present in the given unit cell = 1+2 (given) = 3 `:.` Number of atoms present in `1.157 xx 10^(24)` unit cells = `3 xx 1.157 xx 10^(24) = 3.471 xx 10^(24)`

Discussion

35.	An element crystallizes into a structure whichh may be described by a cubic layer of unit cell having one atom in each corner of the cube and two atoms on one of its face diagonals. If the volume of this unit cell is 24 xx 10^(-24) cm^(3) and density of the element is 7.20 gm//cm^(3), calcualte no. of atoms present in 200g of the element.
Answer» SOLUTION :Number of atoms contributed in onne unit CELL. = one atom from the eight corners = one atom from the two faces diagonals. =1+1=2 atoms. Mass of one unit cell = volume x its DENSITY `=24 xx 10^(-24) cm^(3) xx 7.2 gm cm^(3)` `=172.8 xx 10^(-24)` gm `therefore 172.8 xx 10^(-24) gm` is the mass of one - unit cell i.e., 2 atoms. `therefore 200` gm is the mass `=(2 xx 200)/(172.8 xx 10^(-24))` atoms `=2.3148 xx 10^(24)` atoms.

Discussion

36.	An element crystallizes in fec lattice with a cell edge of 300 pm. The density of the element is 10.8 g cm^(-3). Calculate the number of atoms in 108 g of the element.
Answer» SOLUTION :`d=(zm)/(a^(3)N)`, m = Mass of element, N = number of atoms Or `N=(108xx4)/(10.8xx27xx10^(-24))=1.48xx10^(24)` atoms `M=(a^(3)xxN_(a)xxd)/(Z)` `=(27xx10^(-24)xx6.022xx10^(23)xx10.8)/(4)=43.88"g mol"^(-1)` `43.88"g mol"^(-1)` CONTAINS `6.02xx10^(23)` atoms So, 108 g contains`=(6.02xx10^(23)xx108)/(43.88)=1.48xx10^(24)` atoms

Discussion

37.	An element crystallizes in f.c.c. lattice having edge length 400 pm. Calculate the maximum diameter of the atom which can be placed in the interstitial site without distorting the structure.
Answer» SOLUTION :R = 0.414 R and `R = (SQRT2 a)/(4)`117.1 PM

Discussion

38.	An element crystallizes in body - centred cubic structure. If the edge length of the unit cell is 400 pm . Calculate interatomic distance in the crystal.
Answer» ANSWER :346.4 PM

Discussion

39.	An element crystallizes in face centred cubic lattice. Calculate the length of the side of the unit cell if the radius of atom is 200 pm.
Answer» Solution :For face centred cubic lattice, radius, `R = a//2sqrt2` where a is the length of the SIDE of unit CELL. Substituting the values we get : `200 = a//(2 xx 1.4142)` or `a=200 xx 2 xx 1.4142` = 565.7 pm

Discussion

40.	An element crystallizes in a structure having fcc unit cell of an edge 200 pm. Calculate the density if 200 g of this element contains 24 xx 10^(23) atoms.
Answer» SOLUTION :Edge length = 200 pm Volumne of the unit cell `= (200 xx 10^(-10) cm)^(3)` `= 8 xx 10^(-24) cm^(3)` In a fcc unit cell there are four atoms PER unit cell. Mass of unit cell = `(200 xx 4)/(24 xx 10^(23))` `= 33.3 xx 10^(-23)` g Density `= ("Mass of unit cell")/("VOLUME of unit cell")` `= (33.3 xx 10^(-23) g)/(8 xx 10^(-24) cm^(3))` = 41.6 g `cm^(-3)`

Discussion

41.	An element crystallizes in a structure having fee unit cell of an edge 200 pm. Calculate the density if 200 g of this element contains 24 xx 10^(23) atoms.
Answer» SOLUTION :a = 200 pm = `200 xx 10^(-10)` CM, Z= `4.24 xx 10^(23)` atoms of the element are present in 200 g `6.022 xx 10^(23)` atoms of the element are present in mass `200 xx 6.022 xx 10^(23)` = 50.18 g `:.` Atomic mass of the element = 50.18 g `mol^(-1)` `d= (Z xx M)/(N_A xx a^3 "(in pm) "xx 10^(-30)) = (4 xx (50.18 g mol^(-)))/((6.022 xx 10^(23) mol^(-1)) xx (200)^3 X 10^(-30) cm^3) = 41.67 g cm^(-3)`

Discussion

42.	An element crystallizes in a fcc lattice and the edge length of the unit cell is 0.559 nm . The density of crystal is 3.19 g / cm^3 . Find atomic weight of the element .
Answer» `100.6` `75.9` `95.8` `83.9` ANSWER :D

Discussion

43.	An element crystallizes in a fee lattice with cell edge of 400 pm. The density of the element is 7 g//cm3. How many atoms one present in 280 g of the element ?
Answer» SOLUTION :`"Volume of unit cell"=a^(3)"(a = edge length)"` `="400 pm"` `=(400xx10^(-12)m)^(3)` `=(400xx10^(-10)cm)^(3)=64xx10^(-24)cm^(3)` `"Volume of 208 g of the element"=(208)/(7g//cm^(3))=29.71cm^(3)` `"Number of unit cells in this volume"=("VOL. of given amount")/("Vol of ONE unit cell")` `=(29.71)/(64xx10^(-24))=0.46xx10^(24)` Since each f.c.c. unit cell containss 4 atoms therefore, `"Total number "=4xx0.46xx1024` `=1.84xx"1024 atoms."`

Discussion

44.	An element crystallized in fcc lattice and edge length of unit cell is 400pm. If density of unit cell is 11.2 g cm^(-3), then atomic mass of the element is
Answer» `215.6 g mol^(-1)` `431.2 g mol^(-1)` `107.8 g mol^(-1)` `98.6 g mol^(-1)` Solution :Edge LENGTH of unit cell, a = 400 PM `= 400 xx 10^(-10) CM` Density = 11.2 `g cm^(-3)` Z = 4 ( for fcc ) `d = ( Z xx M )/( a^(3) xx N_(O))` `11.2= ( 4 xx M )/( ( 400 xx 10^(-10))^(3) xx ( 6.022 xx 10^(23)))` or `M = ( 11.2 xx ( 400 xx 10^(-10))^(3) xx ( 6.022xx 10^(23)))/( 4)` `= 107.8 g mol^(-1)`

Discussion

45.	An elementcrystallises inthe fcccrystallatticeand hasa densityof 10 g cm^(-3)withunitcell edgelengthof 100 pm . Calculatenumberof atomspresentin 1 gof crystal.
Answer» Solution :Given : Densityof CRYSTAL = d= 10 `g cm^(-3)` Edgelength =a= 100 pm `=1xx 10^(-8)cm` Avogadronumber `= N_(A) = 6.022 xx 10^(23) mol^(-1)` Numberof atom = s PRESENTIN 1 gcrystal = ? LETN be theatomicmass of theelement . Mass of one atom `= (M )/(6.022 xx 10^(23))` Fcc typeunitcellcontains 4 atoms. `:. ` Massof unitcell = Massof 4 atoms `= 4 xx (M)/(6.022 xx 10^(23))` Volumeof unitcell `=a^(3) = (1 xx 10^(-8))^(3) =1 xx 10^(-24) cm^(3)` Densityof the crystal =d= `("Mass ofunit CELL" )/("volumeof unit cell" )` `10= 4 xx M //6.022 xx (10^(23))/(1xx 10^(-24))` `= (4 xx M)/(6.022 xx 10^(23) xx 1 xx 10^(-24)) = (4 M)/(0.6022)` `:. M = (10 xx0.6022)/(4) =1.5 g mol^(-4)` Now1 gramatom= 1mole ofelement=1.5g `:.` 1.5 gelementcontains6.022 `xx 10^(23)` atoms `:.` 1 g element willcontains.` (6.022 xx 10^(23))/(1.5) =4 xx 10^(23) ` atoms

Discussion

46.	An element crystallises infccstructure. If theatomicradius is 2.2 Å , whatwill be theedgelengthof the unitcell ?
Answer» Solution :GIVEN : `r= 2.2Å = 2.2 xx 10^(-8) cm , a = ?` `a= 2 SQRT(2) xx r` `2XX 1.414 xx 2.2 xx10^(-8)` `=6.22 xx 10^(-8) cm`

Discussion

47.	An element crystallizes in fcc lattice. If the edge length of the unit cell is 408.6 pm and the density is 10.5 g cm^(-3). Calculate the atomic mass of the element.
Answer» Solution :`a=408.6 xx 10^(-12)m = 408.6 xx 10^(-10)cm`. `d=10.5 g cm^(-3)` `N_(A)= 6.022 xx 10^(23)"mol"^(-1)` Z = 4 `:.` The CRYSTAL is fcc lattice `M=(DA^(3)N_(A))/(Z)=(10.5 g cm^(-3)xx (408.6xx10^(-10) cm)^(3) xx 6.022 xx 10^(23) "mol"^(-1))/(4)` `=107.8"g mol"^(-1)` * Atomic MASS of the element = 107.8 u.

Discussion

48.	An element crystallises in fcc crystal lattice and has a density of 10 g cm 3 with unit cell edge length of 100 pm. Calculate number of atoms present in 1 g of crystal.
Answer» ANSWER :NUMBER of atoms in 1G of crystal `=4xx10^(23)`

Discussion

49.	An element belongsto group 15 and 3^("rd")period of the periodic table, its electronicconfiguration .
Answer» `1s^(2)2s^(2)2p^(4)` `1s^(2)2s^(2)2p^(3)` `1s^(2)2s^(2)2p^(6)3S^(2)3p^(2)` `1s^(2)2s^(2)2p^(6)3s^(2)3p^(3)` Solution :`1s^(2)2s^(2)2p^(6)3s^(2)3p^(3)`

Discussion

50.	An element belongs to group 15 and 3 rd period of the periodic table, its electrons configuration would be …………………………
Answer» `1s^(2)2S^(2)3s^(4)` `1s^(2)2s^(2)sp^(3)` `1s^(2)2s^(2)2P^(6)3s^(2)3p^(2)` `1s^(2)2s^(2)2p^(6)3s^(2)3p^(3)` SOLUTION :`1s^(2)2s^(2)2p^(6)3s^(2)3p^(3)`

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

An element with atomic number 113 has been discovered.It will belong to which of the following block,group number,period and outershell electronic configuration?

An element whose IUPAC name is ununtrium (Uut) belongs to

An element which never has a positive oxidation number in any of its compounds is:

An element whose IUPAC name is ununtrium (Uut) belongs to :-

An element which is highly toxic for plants and animals is ?

Mark the element which shows only one oxidation state in its compounds

An element reacts with hydrogen to form a compound X which on treatment with water liberates hydrogen gas. The element can be

An element posses cubic lose packing structure. Calculate the radius (mu) of the atom in the unit cell [Edge length a = 252 nm].

An element X (At,wt = 80 g//mol) having fcc structure, calculate the number of unit cells in 8g of X

An element of group 2 forms covalent oxide, which is amphoteric ini nature and dissolves in water to give an amphoteric hydroxide. Identify the element.

An element of atomic mass 90 occurs in fce structure with cell edge of 500 pm. Calculate the Avogadro's number if the density is 4.2g cm^(-3)

An element of 3d-transition series two oxidation states x and y, differ by two units then:

An element of 3d-transition series shows two oxidation states x and y, differe by two units then

An element occurs in bcc structure with cell edge 288 pm. Its density is 7.2 g cm^(-3). Calculate the atomic mass of the element.

An element occurs in bcc structure. It has a cell edge length of 250 pm. Calculate the molar mass if its density is 8.0 g cm^(-3). Also calculate radius of an atom of this element.

An element in oxidationstate of +3 has the electronic configuratioin : [Ar] 3d^(3). Its atomic number is :

An element having bcc structure has atomic mass 50 u and density 6.81 g cm^(-3). Calculate the edge length of the unit cell.

An element having atomic mass 63.1 g/mol has face centred cubic unit cell with edge length 3.608xx10^(-8)cm. Calculate the density of unit cell. [Given : NA=6.022xx1023" atoms/mol"].

An element having atomic mass 60 amu. has fcc unit cell. The edge length of the unit cell is 4 xx 10^(2) pm. Find the density of the unit cell.

An element has successive ionization enthalpies as 940 (first), 2080, 3090, 4140, 7030, 7870, 16000 and 19500 kJ "mol"^(-1). To which group of the periodic table does this element belong ?

An element has half-life 1600 years. The mass left after 6400 years will be

An element has electronci configuration (Xe) 6s^(2) 4f^(13), 5d^(0). In which group it is placed

An element has bcc structure with a cell edge of 208 pm. the density of the element is 7.2 gcm^(-3). How many atoms are present in 208g of the element.

An element has a face centered cubic unit cell with a length of 352.4 pm along an edge. The density of the element is "8.9 gcm-3". How many atoms are present in 100 g of an element ?

An element has a body-centred cubic (bcc) structure with a cell edge of 288 pm. The density of the element is 7.2 "g/cm"^3. How many atoms are present in 280 g of the element ?

An element forms two oxides, the weight-ratio composition in them is A : O = x : y in the first oxide and y : x in the second oxide. If the equivalent weight of A in the first oxide is 10.33, the equivalent weight of A in the second oxide is

An element forms two oxides containing, 50% & 40% of oxygen respectively by weight of the element. Does these oxides illustrate the law of multiple proportions :-

An element forms compounds of the type MCl_(3), M_(2)O_(5) and Ca_(3)M_(2) but does not form MF_(5). The element could be :

An element forms a gaseous oxide which on dissolving in water gives an acid solution. The element is:

An element exists in the body-centred cubic structure whose cell edge is 2.88 A. The density of the element is 7.20 g/cm^2. Calculate the number of atoms in 104 g of the element.

An element exist in nature in two isotopic forms: X^(30)(90%) and X^(32)(10%). What is the average atomic mass of element?

An element E loses one alpha and two beta-particles in three successive stages. The resulting element will be

An element crystallizes in fec lattice with a cell edge of 300 pm. The density of the element is 10.8 g cm^(-3). Calculate the number of atoms in 108 g of the element.

An element crystallizes in f.c.c. lattice having edge length 400 pm. Calculate the maximum diameter of the atom which can be placed in the interstitial site without distorting the structure.

An element crystallizes in body - centred cubic structure. If the edge length of the unit cell is 400 pm . Calculate interatomic distance in the crystal.

An element crystallizes in face centred cubic lattice. Calculate the length of the side of the unit cell if the radius of atom is 200 pm.

An element crystallizes in a structure having fcc unit cell of an edge 200 pm. Calculate the density if 200 g of this element contains 24 xx 10^(23) atoms.

An element crystallizes in a structure having fee unit cell of an edge 200 pm. Calculate the density if 200 g of this element contains 24 xx 10^(23) atoms.

An element crystallizes in a fcc lattice and the edge length of the unit cell is 0.559 nm . The density of crystal is 3.19 g / cm^3 . Find atomic weight of the element .

An element crystallizes in a fee lattice with cell edge of 400 pm. The density of the element is 7 g//cm3. How many atoms one present in 280 g of the element ?

An element crystallized in fcc lattice and edge length of unit cell is 400pm. If density of unit cell is 11.2 g cm^(-3), then atomic mass of the element is

An elementcrystallises inthe fcccrystallatticeand hasa densityof 10 g cm^(-3)withunitcell edgelengthof 100 pm . Calculatenumberof atomspresentin 1 gof crystal.

An element crystallises infccstructure. If theatomicradius is 2.2 Å , whatwill be theedgelengthof the unitcell ?

An element crystallizes in fcc lattice. If the edge length of the unit cell is 408.6 pm and the density is 10.5 g cm^(-3). Calculate the atomic mass of the element.

An element crystallises in fcc crystal lattice and has a density of 10 g cm 3 with unit cell edge length of 100 pm. Calculate number of atoms present in 1 g of crystal.

An element belongsto group 15 and 3^("rd")period of the periodic table, its electronicconfiguration .

An element belongs to group 15 and 3 rd period of the periodic table, its electrons configuration would be …………………………