94420 + Interview Questions in Chemistry in Class 12

1.	In Hofmann bromamide reation, the carbonyl group is lost as______
Answer» SOLUTION :CARBONATE ION `(CO_(3)^(2-))`

Discussion

2.	In Goldschmidt aluminothermic process, thermite contains
Answer» 3 PARTS of `Al_(2)O_(3) and 4 `parts of AL 3 parts of `Fe_(2)O_(3) and 2` parts of Al 3 parts of `Fe_(2)O_(3)` and 1 part of Al 1 part of `Fe_(2)O_(3)` and 1 part of Al Solution :3 parts of `Fe_(2)O_(3)` and 1 part of Al.

Discussion

3.	In Hofmann's hypobromidereaction
Answer» Thealkylgroupin amide MIGRATE to oxygenatom The ALKYLGROUP in amidemigrateto oxygenatom There is a no migrationof alkylgroupofamide Hydrogenatom is migratedto carbonyloxgyen atom of amide ANSWER :A

Discussion

4.	In Hofmann bromamide degradation reaction R-overset(O)overset("\|\|")C-NH_(2)+KOH+Br_(2)rarr intermediates are:
Answer» `RCONHBr` `RNCO` `RNH_(2)` none of these Answer :A::B

Discussion

5.	In Hoffmann Bromamide reaction which substance is treated with Br_(2)" and "NaOH to obtain amine ?
Answer» Isonitrile Nitrile AMIDE Nitro Answer :C

Discussion

6.	In Hoffinann bromamide reaction, the reactive intermediate involved is a
Answer» CARBOCATION Carbanion CARBENE Nitrine. Solution :See A-Level INFORMATION.

Discussion

7.	In HO-overset(O)overset(\|\|)S-OH the oxidation states of S are
Answer» `+4,-2` `+4,0` `+2,-2` `+4,-4` ANSWER :A

Discussion

8.	In hexagonal systems of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism.Here, the top and bottom of the cell are regular hexagons and three atoms are sandwiched in between them.A space-fitting model of this structure, called hexagonal closed-packed (HCP), is constitution of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible Three spheres are then placed over the first layer so that they touch each other and represent the second layer.Each one of these three spheres touches three spheres of the bottom layer.Finally, the second layer is covered with third layer that is identical to the bottom layer in relative position.Assume radius of even sphere to be 'r'. In hcp lattice, if a plane is drawn parallel to layers A and B at a distance c/8 above the layer B, it passes through centres of x tetrahedral and y octahedral voids.A plane parallel to layer A at a distance c/8 just below it passes through centres of p tetrahedral and q octahedral voids.Find (x+y+p+q) c is the cell parameter of hcp lattice, the vertical edge length.
Answer» 10 9 6 14 Solution :x=7, y=0,p=3,q=0

Discussion

9.	In hexagonal systems of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism.Here, the top and bottom of the cell are regular hexagons and three atoms are sandwiched in between them.A space-fitting model of this structure, called hexagonal closed-packed (HCP), is constitution of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible Three spheres are then placed over the first layer so that they touch each other and represent the second layer.Each one of these three spheres touches three spheres of the bottom layer.Finally, the second layer is covered with third layer that is identical to the bottom layer in relative position.Assume radius of even sphere to be 'r'. The volume of this HCP unit cell is :
Answer» `12sqrt2 r^3` `24sqrt2 r^3` `16sqrt2 r^3` `64/(3sqrt3)r^3` Solution :`C=sqrt2/3` 4r=HEIGHT of the UNIT CELL BASE area`=6xxsqrt3/4(2r)^2` Volume of the hexagon=Area of base x Height `=6.sqrt2/3 a^2xxc=24sqrt2.r`

Discussion

10.	In hexagonal systems of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms are sandwiched in between them. A space-filling model of this structure, called hexagonal close-packed (hcp), is constitued of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible. Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Each one of these three spheres touched three spheres of the bottom layer. Finally, the second layer is convered with a third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be 'r'. The volume of this hcp unit cell is
Answer» `24 SQRT(2)r^(3)` `16 sqrt(2)r^(3)` `12 sqrt(2)r^(3)` `(64r^(3))/(3sqrt(3))` Answer :A

Discussion

11.	In hexagonalprimitiveunit cell, the Corneris sharedby:
Answer» 4 UNIT cells 6 unit cells 3 unit cells 8 unit cells ANSWER :B

Discussion

12.	In hexacyanomanganate (II) ion the Mn atom assumes Id^2sp^3 hybrid state. The number of unpaired electrons in the complex is:
Answer» 1 2 3 zero Answer :A

Discussion

13.	In hexacyanomanganate (II) ion the Mn atom assumes d^2sp^3 hybrid state. The number of unpaired electrons in the complex is
Answer» 1 2 3 zero Answer :A

Discussion

14.	In hexacyanomanganate (II) ion the Mn atom assumes a^2sp^3-hybrid states. The number of unpaired electrons in the complex is
Answer» 1 2 3 0 Answer :A

Discussion

15.	In heterogeneous catalytic reactions involving solid catalyst and gaseous reactants, the catalysts, most generally used are
Answer» METALS METAL OXIDES only TRANSITION metals only transition metals and transition metal oxides ANSWER :D

Discussion

16.	In heterogeneous catalytic reactions, catalyst in the finely divided form possesses higher catalytic reactivity. This is because
Answer» SURFACE area of the finely DIVIDED CATALYST is large surface area of the catalyst in the LUMP form is large finely divided catalyst has more internal energy in it finely divided catalyst has stable surface Answer :A

Discussion

17.	In He^(o+) ion sample e^(-) is in ground state. If photon of energy 52.24 eV is given to the sample all the atom goes to higher energy state. It again falls back up to ground state. If it is not emitting any lines in Balmer series then what is the maximum possible number of spectral lines observed
Answer» ANSWER :6

Discussion

18.	In HCP or CCP constituent particles occupy 74% of the available space. The remaining space (26%) in between the spheres remains unoccupied and is called interstitial voids or holes.Considering the close packing arrangement, each sphere in the second layer rests on the hollow space of the first layer, touching each other. The void created is called tetrahedral void. If R is the radius of the spheres in the close packed arrangement then,R (radius of tetrahedral void) = 0.225 R In a close packing arrangement, the interstitial void formed by the combination of two triangular voids of the first and second layer is called octahedral coid. Thus, double triangular void is surrounded by six spheres. The centre of these spheres on joining, forms octahedron. If R is the radius of the sphere. in a close packed arrangement then,R (radius of octahedral void = 0.414 R).If the anions (A) form hexagonal close packing and cations (C ) occupy only 2/3rd octahedral voids in it, then the general formula of the compound is
Answer» AB `A_(3)B_(2)` `A_(2)B` `AB_(3)` ANSWER :B

Discussion

19.	In HCP or CCP constituent particles occupy 74% of the available space. The remaining space (26%) in between the spheres remains unoccupied and is called interstitial voids or holes.Considering the close packing arrangement, each sphere in the second layer rests on the hollow space of the first layer, touching each other. The void created is called tetrahedral void. If R is the radius of the spheres in the close packed arrangement then,R (radius of tetrahedral void) = 0.225 R In a close packing arrangement, the interstitial void formed by the combination of two triangular voids of the first and second layer is called octahedral coid. Thus, double triangular void is surrounded by six spheres. The centre of these spheres on joining, forms octahedron. If R is the radius of the sphere. in a close packed arrangement then,R (radius of octahedral void = 0.414 R).In Schottky defect
Answer» cations are missing from the lattice sites and OCCUPY the interstitial sites EQUAL number of cations and ANIONS are missing anions are missing and electrons are present in their place equal number of EXTRA cations and electrons are present in the interstitial sites Answer :B

Discussion

20.	In HCP or CCP constituent particles occupy 74% of the available space. The remaining space (26%) in between the spheres remains unoccupied and is called interstitial voids or holes.Considering the close packing arrangement, each sphere in the second layer rests on the hollow space of the first layer, touching each other. The void created is called tetrahedral void. If R is the radius of the spheres in the close packed arrangement then,R (radius of tetrahedral void) = 0.225 R In a close packing arrangement, the interstitial void formed by the combination of two triangular voids of the first and second layer is called octahedral coid. Thus, double triangular void is surrounded by six spheres. The centre of these spheres on joining, forms octahedron. If R is the radius of the sphere. in a close packed arrangement then,R (radius of octahedral void = 0.414 R).Mark the false statement :
Answer» CsCl CRYSTAL shows Schottky DEFECT Crystals having F - centres are COLOURED and paramagnetic Photosensitivity of AGBR is due to the presence of Frenkel defect in this crystal. None of the above ANSWER :D

Discussion

21.	In HCP or CCP constituent particles occupy 74% of the available space. The remaining space (26%) in between the spheres remains unoccupied and is called interstitial voids or holes.Considering the close packing arrangement, each sphere in the second layer rests on the hollow space of the first layer, touching each other. The void created is called tetrahedral void. If R is the radius of the spheres in the close packed arrangement then,R (radius of tetrahedral void) = 0.225 R In a close packing arrangement, the interstitial void formed by the combination of two triangular voids of the first and second layer is called octahedral coid. Thus, double triangular void is surrounded by six spheres. The centre of these spheres on joining, forms octahedron. If R is the radius of the sphere. in a close packed arrangement then,R (radius of octahedral void = 0.414 R).In the spinel structure, oxide ions are cubic close packed whereas 1/8th of tetrahedral voids are occupied by A^(2+) cations and 1/2 of octahedral voids are occupied by B^(3+) cations. The general formula of the compound having spinel structure is
Answer» `A_(2)B_(2)O_(4)` `AB_(2)O_(4)` `A_(2)B_(4)O_(2)` `A_(4)B_(2)O_(2)` ANSWER :B

Discussion

22.	In HCP or CCP constituent particles occupy 74% of the available space. The remaining space (26%) in between the spheres remains unoccupied and is called interstitial voids or holes.Considering the close packing arrangement, each sphere in the second layer rests on the hollow space of the first layer, touching each other. The void created is called tetrahedral void. If R is the radius of the spheres in the close packed arrangement then,R (radius of tetrahedral void) = 0.225 R In a close packing arrangement, the interstitial void formed by the combination of two triangular voids of the first and second layer is called octahedral coid. Thus, double triangular void is surrounded by six spheres. The centre of these spheres on joining, forms octahedron. If R is the radius of the sphere. in a close packed arrangement then,R (radius of octahedral void = 0.414 R).In the figure given below, the site marked as S is a
Answer» tetrahedral VOID CUBIC void OCTAHEDRAL void None of the above ANSWER :C

Discussion

23.	In HCI molecule, expected value of dipole momentis 6.12 D but experimental value is 1.03 D. Then, the percentage ionic character will be
Answer» `16.13` `15.14` `6.02` `18.9` ANSWER :A

Discussion

24.	In hcp arrangement the coordination number is
Answer» 6 12 8 10 Answer :B

Discussion

25.	In the HCOO^- the two carbon-oxygen bonds are found to be of equal length. What is the reason for it?
Answer» The anion is obtained by the removal of a PROTON from the acid MOLECULE Electronic orbitals of carbon atoms are hybridised The C=O bond is weaker than C-O bond The anion `"HCOO"^(-)` has two resonating STRUCTURES SOLUTION :In `"HCOO"^(-)`, the two carbon oxygen bond are of equal LENGTH because the anion `"HCOO"^(-)` has two resonating structure.

Discussion

26.	In HCHO carbon atom has hybridisation:
Answer» SP `sp^2` `sp^2` None Answer :B

Discussion

27.	In halogentionof nitroalkane
Answer» all halogonare replacedby HALOGENES all `ALPHA` -H are REPLACED byhalogens all `beta` - H are replaced byhalogens ONLYONE `alpha` -Hreplacedby halogaen Answer :B

Discussion

28.	In halogens, which of the following decreases from fluorine to iodine?
Answer» BOND length Electronegativity The IONIZATION ENERGY of the element Oxidizing power Answer :B::C::D

Discussion

29.	In halogens, which of the following increases from iodine to fluorine?
Answer» BOND LENGTH electronegativity the ionization energy of the element oxidising power Solution :In group 17, electronegativity, IE and oxidising power DECREASES from top to bottom. HENCE, all these properties increases from iodine to FLUORINE.

Discussion

30.	In halogen family which element forms strongest hydrogen bonds ?
Answer» SOLUTION :FLUORINE

Discussion

31.	In haloarenes C-Cl bond has partial double bond character, which is due to
Answer» INDUCTIVE effect electromertic RESONANCE STERIC effect Answer :C

Discussion

32.	In Hall's process, the ore is mixed with :
Answer» Coke Sodium carbonate Sodium hydroxide Calcium carbonate Answer :B

Discussion

33.	In Hall's process, the main reagent mixed with…
Answer» `NAF` `Na_3AlF_6` `AlF_3` `AlF_6` ANSWER :B

Discussion

34.	In Hall-Herold process, act as an anode.
Answer» CARBON blocks hydrogen copper rods Zinc rods Answer :A

Discussion

35.	In Hall and Heroult's process, products liberated at anode and cathode are _________.
Answer» OXYGEN and aliminium respectively carbon MONOXIDE and aluminium respectively carbon DIOXIDE and aliminium respectively All of these Answer :D

Discussion

36.	In hall and Heroult's process , the molten electrolyte is covered with the layer of powdered coke, which help to _________
Answer» PREVENT oxidation prevent redcution prevent loss of heat due to ratiation both (A) and (C) ANSWER :D

Discussion

37.	In halides, halogen atom is attached to
Answer» `SP^(3)` hybridised carbon atom `sp^(2)` hybridised carbon atom sp hybridised carbon atom sp-d hybridised carbon atom Answer :B

Discussion

38.	In haemoglobin the metal ion present is .
Answer» `FE^(2+)` `ZN^(2+)` `CO^(2+)` `CU^(2+)` Answer :A

Discussion

39.	In haemoglobin the iron shows oxidation state:
Answer» `+2` `+3` `+1` `+4` ANSWER :A

Discussion

40.	In Haber's process of manufacture of ammonia, the Fe catalyst is poisoned by the pressure of .............
Answer» Mo Co `H_2S` `As_2O_3` SOLUTION :`H_2S`

Discussion

41.	In haemoglobin the iron shows oxidation state :
Answer» `+2` `+3` `+1` `+4` Answer :B

Discussion

42.	In Haber's process, hydrogen is obtained by reacting methane with steam in presence of Nio as catalyst. The process is known as steam reforming. Why is it necessary to remove CO when ammonia is obtained by Haber's process ?
Answer» Solution :CO acts as a poison for the catalyst in the manufacture of AMMONIA by Haber.s PROCESS. Hence, it is NECESSARY to remove it.

Discussion

43.	In Haber's process, hydrogen is obtained by reacting methane with steam in presence of Nio as catalyst. The process is known as steam reforming. Why is it necessary to remove CO when ammonia is obtained by Haber's process?
Answer» Solution :It is necessary to REMOVE CO while obtaining ammonia from Heber PROCESS because, produced CO gas EFFECTS on the reactivity of iron CATALYST.

Discussion

44.	In Haber's process for manufacture of ammonia which metals acts as a promotor for iron ?
Answer» Cu Zn Mo As Answer :C

Discussion

45.	In Haber's process for NH_3 the temperature and pressure are respectively
Answer» `STP` `200-250^0C`, 450 ATM `450^0C` and 200-250 atm `100^0C` and 25-50 atm Answer :C

Discussion

46.	In haber process 30 litres of dihydrogen and 30 littres of dinitorgen were taken for reaction which yieided only 50% of the expected product. What will be the composition of gaseous mixture under these condition in the end
Answer» 20 litres ammonia, 25 litres NITROGEN, 15 litres hydrogen 20 litres ammonia, 20 litres nitrogen, 20 litres hydrogen 10 litres ammonia, 25 litres notrogen, 15 litres hydrogen 20n litres ammonia, 10 litres notrogen, 30 litres hydrogen Solution :`N_(2)+2H_(2)hArr2NH_(3)` `{:(30,30,0),(30-x,30-3x,2x):}` `{:(,N_(2)=30-5=25 litre),(,H_(2)=30-3xx5=15litre),(,NH_(3)=2xx5=10 litre):}`

Discussion

47.	In Haber process, 30 litres of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only 50% of the expected product. What will be the composition of the gaseous mixture under the aforesaid condition in the end?
Answer» 20 litres `NH_3`, 25 litres `N_2`, 20 litres `H_2` 10 litres `NH_3`, 25 litres `N_2`, 15 litres `H_2` 20 litres `NH_3`, 10 litres `N_2`, 30 litres `H_2` 20 litres `NH_3`, 25 litres `N_2`, 15 litres `H_2` Solution :`N_(2)(g) + 3H_(2)(g) 1 L of `N_(2)` REACTS with 3 L of `H_(2)` to form 2 L of `NH_(3)`. Thus, `N_(2)` is the limiting reacant. 10 L `N_(2)` will react with 30 L `H_(2)` to form 20 L `NH_(3)`. As actual yield is 50% of the EXPECTED value, `NH_(3)` formed = 10 L, `N_(2)` reacted = 5L, `H_(2)` reacted = 15 L `THEREFORE` Mixture will contain 10 L `NH_(3), 25 L N_(2), 15 L H_(2)`

Discussion

48.	In Haber process, 30 litres of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only 50% of the expected product. What will be the compositon of the gaseous mixture under the aforesaid condition in the end?
Answer» 20 litres `NH_(3)`, 25 litres `N_(2)`, 20 litres `H_(2)` 10 litres `NH_(3)`, 25 litres `N_(2)`, 15 litres `H_(2)` 20 litres `NH_(3)`, 10 litres `N_(2),` 30 litres `H_(2)` 20 litres `NH_(3)`, 25 litres `N_(2)`, 15 lites `H_(2)` SOLUTION :`N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g)` 1 L of `N_(2)` reacts with 3 L of `H_(2)` to form 2 L of `NH_(3)`. Thus, `N_(2)` is the limiting reactant. 10 L `N_(2)` will REACT with 30 L `H_(2)` to form 20 L `NH_(3).` As actual yield is `50%` of hte expected value, `NH_(3)`formed = 10 L, `N_(2)` REACTED = 5 L, `H_(2)` reacted = 15 L `therefore" Mixture will contain 10 L NH"_(3),25 L N_2),` `15L H_(2)`.

Discussion

49.	In H_(3)PO_(4), which of the following is true
Answer» `K_(a)=K_(a_(1))xxK_(a_(2))xxK_(a_(3))` `K_(a_(1))ltK_(a_(2))ltK_(a_(3))` `K_(a_(1))gtK_(a_(2))ltK_(a_(3))` `K_(a_(1))=K_(a_(2))=K_(a_(3))` SOLUTION :`H_(3)PO_(4)hArrH^(+)+H_(2)PO_(4)^(-)Ka_(1)=([H^(+)][H_(2)PO_(4)^(-)])/([H_(3)PO_(4)])""...(i)` `h_(2)PO_(4)^(-)hArrH^(+)+HPO_(4)^(2-)K_(a_(2))=([H^(+)][HPO_(4)^(-2)])/([H_(2)PO_(4)^(-)])""...(ii)` `HPO_(4)^(2-)hArrH^(+)+PO_(4)^(3-)K_(a_(3))=([H^(+)][HPO_(4)^(2-)])/([HPO_(4)^(2-)])""...(iii)` `H_(3)PO_(4)hArr3H^(+)+P_(4)^(3-)K_(a)=([H^(+)]^(3)[PO_(4)^(3-)])/([H_(3)PO_(4)])""...(iv)` Multiplying (i),(ii), (ii) and equate with (iv) `K_(a)=K_(AQ)xxK_(a_(2))xxK_(a_(3)),therefore(a)` is correct `1^(st)` acid dissociation constant is highest because dissociation is maximum in FIRST step, less in `2^(nd)` step and MINIMUM in `3^(rd)` step because it is difficult for a negatively changed ion tolose `H_(+)` ion. `K_(a_(1))gtK_(a_(2))gtK_(a_(3)), therefore(c)` is correct, (b) and (d) are ruled out.

Discussion

50.	In H_(3)PO_(3) molecule
Answer» P - atom is SURROUNDED by THREE - OH groups P - atom is tetrahedrally surrounded by TWO - OH groups , one oxygen atom and one hydrogen atom P - atom is surrounded by four - OH groups P - atoms is surrounded by two - H atoms Answer :B

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

In Hofmann bromamide reation, the carbonyl group is lost as______

In Goldschmidt aluminothermic process, thermite contains

In Hofmann's hypobromidereaction

In Hofmann bromamide degradation reaction R-overset(O)overset("||")C-NH_(2)+KOH+Br_(2)rarr intermediates are:

In Hoffmann Bromamide reaction which substance is treated with Br_(2)" and "NaOH to obtain amine ?

In Hoffinann bromamide reaction, the reactive intermediate involved is a

In HO-overset(O)overset(||)S-OH the oxidation states of S are

In hexagonalprimitiveunit cell, the Corneris sharedby:

In hexacyanomanganate (II) ion the Mn atom assumes Id^2sp^3 hybrid state. The number of unpaired electrons in the complex is:

In hexacyanomanganate (II) ion the Mn atom assumes d^2sp^3 hybrid state. The number of unpaired electrons in the complex is

In hexacyanomanganate (II) ion the Mn atom assumes a^2sp^3-hybrid states. The number of unpaired electrons in the complex is

In heterogeneous catalytic reactions involving solid catalyst and gaseous reactants, the catalysts, most generally used are

In heterogeneous catalytic reactions, catalyst in the finely divided form possesses higher catalytic reactivity. This is because

In He^(o+) ion sample e^(-) is in ground state. If photon of energy 52.24 eV is given to the sample all the atom goes to higher energy state. It again falls back up to ground state. If it is not emitting any lines in Balmer series then what is the maximum possible number of spectral lines observed

In HCI molecule, expected value of dipole momentis 6.12 D but experimental value is 1.03 D. Then, the percentage ionic character will be

In hcp arrangement the coordination number is

In the HCOO^- the two carbon-oxygen bonds are found to be of equal length. What is the reason for it?

In HCHO carbon atom has hybridisation:

In halogentionof nitroalkane

In halogens, which of the following decreases from fluorine to iodine?

In halogens, which of the following increases from iodine to fluorine?

In halogen family which element forms strongest hydrogen bonds ?

In haloarenes C-Cl bond has partial double bond character, which is due to

In Hall's process, the ore is mixed with :

In Hall's process, the main reagent mixed with…

In Hall-Herold process, act as an anode.

In Hall and Heroult's process, products liberated at anode and cathode are _________.

In hall and Heroult's process , the molten electrolyte is covered with the layer of powdered coke, which help to _________

In halides, halogen atom is attached to

In haemoglobin the metal ion present is .

In haemoglobin the iron shows oxidation state:

In Haber's process of manufacture of ammonia, the Fe catalyst is poisoned by the pressure of .............

In haemoglobin the iron shows oxidation state :

In Haber's process, hydrogen is obtained by reacting methane with steam in presence of Nio as catalyst. The process is known as steam reforming. Why is it necessary to remove CO when ammonia is obtained by Haber's process ?

In Haber's process, hydrogen is obtained by reacting methane with steam in presence of Nio as catalyst. The process is known as steam reforming. Why is it necessary to remove CO when ammonia is obtained by Haber's process?

In Haber's process for manufacture of ammonia which metals acts as a promotor for iron ?

In Haber's process for NH_3 the temperature and pressure are respectively

In haber process 30 litres of dihydrogen and 30 littres of dinitorgen were taken for reaction which yieided only 50% of the expected product. What will be the composition of gaseous mixture under these condition in the end

In Haber process, 30 litres of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only 50% of the expected product. What will be the composition of the gaseous mixture under the aforesaid condition in the end?

In Haber process, 30 litres of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only 50% of the expected product. What will be the compositon of the gaseous mixture under the aforesaid condition in the end?

In H_(3)PO_(4), which of the following is true

In H_(3)PO_(3) molecule