Description : Sp3d2 hybridization of central atom of a molecule would lead to............ geometry. a) square planar b) Tetrahedral c) Trigonal bipyramidal d) Octahedral
Last Answer : d) Octahedral
Description : Among the following molecules, `(i)XeO_(3)(ii)XeOF_(4)(iii)XeF_(6)` those having same number of lone pairs on `Xe` are:
Last Answer : Among the following molecules, `(i)XeO_(3)(ii)XeOF_(4)(iii)XeF_(6)` those having same number of lone pairs on ... iii) only D. (D) (i),(ii) and (iii)
Description : The number of geometrical isomers for octahedral `[CoCl_(4)(NH_(3))_(2)]^(-)`, square planar `[AuBr_(2)Cl_(2)]^(-)` and `[PtCl_(2)(en)]` are
Last Answer : The number of geometrical isomers for octahedral `[CoCl_(4)(NH_(3))_(2)]^(-)`, square planar `[AuBr_(2)Cl_(2)]^(-) ... are A. 4,2 B. 2,2 C. 3,2 D. 2,3
Description : A coordination complex of type `MX_(2)Y_(2)`[M=metal ion, X,Y=monodentata ligands ], can have either a tetrahedral of a square planar geometry. The ma
Last Answer : A coordination complex of type `MX_(2)Y_(2)`[M=metal ion, X,Y=monodentata ligands ], can have either a tetrahedral ... 2 and 1 C. 1 and 3 D. 3 and 2
Description : The ammonium ion is (a) Square planar (b) Tetrahedral (c) Square pyramidal (d) Trigonal pyramidal
Last Answer : Ans:(b)
Description : Apart from tetrahedral geometry another possible geometry for CH4 is square planar with four hydrogen atoms at the corners of the square and the carbon atom at the centre.Explain why CH4 is not square planar?
Last Answer : Ans For square planar arrangement hybridization is dsp2 which is not possible with Carbon as it does not have d orbital with it
Description : `XeF_(4)` is a square planar molecule. The hybridisation of xenon atom in this molecule is
Last Answer : `XeF_(4)` is a square planar molecule. The hybridisation of xenon atom in this molecule is A. `dsp^(2)` B. `sp^(3)d` C. `sp^(3)d^(2)` D. `d^(2)sp^(3)`
Description : Find the number of compounds among the following whose hydrolysis is a non-redox reaction. `XeF_(2),XeF_(4),XeF_(6),XeO_(2)F_(2),XeOF_(4),Xe,XeO_(3)`
Last Answer : Find the number of compounds among the following whose hydrolysis is a non-redox reaction. `XeF_(2),XeF_(4),XeF_(6),XeO_(2)F_(2),XeOF_(4),Xe,XeO_(3)`
Description : Give the molecular structures of : `XeF_(2),XeF_(4),XeF_(6)` `XeOF_(4) " and " XeO_(3)`
Last Answer : Give the molecular structures of : `XeF_(2),XeF_(4),XeF_(6)` `XeOF_(4) " and " XeO_(3)`
Description : In `XeO_(3)` and `XeF_(6)` the oxidation state of `Xe` is
Last Answer : In `XeO_(3)` and `XeF_(6)` the oxidation state of `Xe` is A. `sp^(3)d^(2)` to `sp^(3)d` B. `sp^(3)d^(3)` ... ^(3)d^(2)` D. `sp^(3)d^(3)` to `sp^(3)d`
Description : Total number of lone pairs on Xe in `XeF_(2), XeO_(3)F_(2), XeF_(4), XeF_(6)` is t,u,v & w respectively. Then
Last Answer : Total number of lone pairs on Xe in `XeF_(2), XeO_(3)F_(2), XeF_(4), XeF_(6)` is t,u,v & w respectively. Then A. ... `v +w = 3` C. `u = 0` D. `w = 1`
Description : The hydrolysis of `XeF_(6)` takes place in the following steps : `XeF_(6) rarr A rarr B rarr XeO_(3)`. Then the correct statement regarding A and B is
Last Answer : The hydrolysis of `XeF_(6)` takes place in the following steps : `XeF_(6) rarr A rarr B rarr ... A is also obtained when `XeF_(6)` reacts with silica
Description : The crystl field splitting energy for octahedral complex `(Delta_(0))` and that for tetrahedral complex `(Delta_(1))` rae related as :
Last Answer : The crystl field splitting energy for octahedral complex `(Delta_(0))` and that for tetrahedral complex `(Delta_(1) ... ` D. `Deltat=(9)/(4)Delta_(0)`
Description : The size of the tetrahedral void in the closest packing of atoms is __________ that of the octahedral void. (A) Equal to (B) Greater than (C) Less than (D) Either (A), (B) or (C); depends on the size of atom
Last Answer : (C) Less than
Description : There are one octahedral void and __________ tetrahedral void in the closest packing of atoms. (A) One (B) Two (C) Three (D) None of these
Last Answer : Option B
Description : The shape of a carbon molecule is (a) Linear (b) Planar (c) Cubical (d) Tetrahedral
Last Answer : Ans:(d)
Description : What is the hybridization and geometry of the carbonyl carbon in carboxylic acids and their derivatives? (a) sp3, tetrahedral (b) sp2, trigonal planar (c) sp2, tetrahedral (d) sp3, trigonal planar
Last Answer : sp2, trigonal planar
Description : What is the molecular geometry of XeO2 (Xenon Dioxide)? w) linear x) angular y) tetrahedral z) trigonal planar
Last Answer : ANSWER: X -- ANGULAR
Description : The molecular shape of formaldehyde, H2CO, is: w) linear x) trigonal planar y) bent z) tetrahedral
Last Answer : ANSWER: X -- TRIGONAL PLANAR
Description : The structure of an ammonia molecule can best be described as: w) linear x) tetrahedral y) pyramidal z) triagonal planar
Last Answer : ANSWER: Y -- PYRAMIDAL
Description : The triple bond between the carbon atoms causes acetylene, C2H2, to have which of the following shapes? w) trigonal planar (pron: try-gon-al) x) linear y) tetrahedral z) trigonal bipyramidal
Last Answer : ANSWER: X -- LINEAR
Description : Which of the following describes the orientation of bonds in an sp3 hybridized atom? w)triagonal x)linear y)tetrahedral z)planar
Last Answer : ANSWER: Y -- TETRAHEDRAL
Description : The number of geometric isomers that can exist for square planar `[Pt(Cl)(py)(NH_(3))(NH_(2)OH)]^(+)` is (py=pyridine)
Last Answer : The number of geometric isomers that can exist for square planar `[Pt(Cl)(py)(NH_(3))(NH_(2)OH)]^(+)` is (py=pyridine) A. 2 B. 3 C. 4 D. 6
Description : Which one of the following has a square planar geometry? `(Co=27, Ni=28, Fe=26, Pt=78)`
Last Answer : Which one of the following has a square planar geometry? `(Co=27, Ni=28, Fe=26, Pt=78)` A. `[NiCl_(4)]^(2-) ... `[CoCl_(4)]^(2-)` D. `[FeCl_(4)]^(2-)`
Description : Cis-trans isomerism is found in square planar complexes of molecular formula: (a and b are monodentate ligands)
Last Answer : Cis-trans isomerism is found in square planar complexes of molecular formula: (a and b are monodentate ligands) A. ` ... . `Ma_(2)b_(2)` D. `Mab_(3)`
Description : Apart from tertrahedral geometry ,another possible geometry for CH4 is square planar with the four H atoms at the corners of square and C atom at its centre. Explain why CH4 is not square planar.
Last Answer : Ans. In square planar geometry , the bond angle will be 90 ° which is less than bond angle in tetrahedral geometry ( 109.5 °).Therefore repulsive forces in square planar will be more and it will be less stable as compared to tetrahedral geometry.
Description : Match the two lists given below: List-I a. London b. Vatican City c. Moscow d. New York List-II (i) St. Peter's Square (ii) Times Square (iii) Trafalgar Square (iv) Red Square (1) a-(ii), b-(iv), c-(iii), d- ... ) a-(iv), b-(ii), c-(i), d-(iii) (4) a-(i), b-(iii), c-(ii), d-(iv)
Last Answer : a-(iii), b-(i), c-(iv), d-(ii)
Description : Oxidation state of Xe in `Ba_(2)[XeO_(6)]` is
Last Answer : Oxidation state of Xe in `Ba_(2)[XeO_(6)]` is A. 4 B. 6 C. 7 D. 8
Description : `XeO_(6)^(4-)` contains
Last Answer : `XeO_(6)^(4-)` contains A. Eight bond pairs and no lone pairs at Xe B. Three bond pairs and three ... Xe D. Four bond pairs and four lone pairs at Xe
Description : Explain the following : (i) Low spin octahedral complexes of nickel are not known. (ii) The π-complexes are known for transition elements only. (iii) CO is a stronger ligand than NTL, for many metals. -Chemistry
Last Answer : (i) The electronic configuration of Ni is [Ar] 3d8 4s2 which shows that it can only form two types of complexes i.e. square planar (dsp2) in presence of strong ligand and tetrahedral (sp3) in ... more splitting of d-orbitals and moreover it is also able to form π bond due to back bonding.
Description : A list of species having the formula of `XZ_(4)` is given below `XeF_(4), SF_(4), SiF_(4), BF_(4)^(-), BrF_(4)^(-), [Cu(NH_(3))4]^(2+),[FeCl_(4)]^(2-)
Last Answer : A list of species having the formula of `XZ_(4)` is given below `XeF_(4), SF_(4), SiF_ ... the total number of species having a square planar shape is
Description : Planar Platinum compounds are being investigated in the treatment of cancer. Name the isomers of dichlorodiamineplatinum(II) (pron: die-klor-o-di-a-MIN-e-plat-i-num two)?
Last Answer : ANSWER: CIS AND TRANS (DICHLORODIAMINEPLATINUM (II)
Description : The number of unpaired electrons in `d^(6)`, low spin, octahedral complex is :
Last Answer : The number of unpaired electrons in `d^(6)`, low spin, octahedral complex is : A. 4 B. 2 C. 1 D. 0
Description : `XeO_(3)` forms xenate ion in alkaline medium. `XeO_(3) +NaOH rarr Na[HXeO_(4)]` But the xenate ions slowly disproportionate in alkaline solution as `
Last Answer : `XeO_(3)` forms xenate ion in alkaline medium. `XeO_(3) +NaOH rarr Na[HXeO_(4)]` But the xenate ions slowly ... `Na_(4)XeO_(6)` D. `Na_(4)XeO_(4)`
Description : The structure of `XeO_(2)F_(2)` is
Last Answer : The structure of `XeO_(2)F_(2)` is A. Square pyramidal B. Trigonal pyramidal (see-sea) C. Octahedral D. Tetrahedral
Description : `XeO_(4)` contains :
Last Answer : `XeO_(4)` contains : A. four `pi` - bonds, and the remaining three electron pairs form ... , and the remaining four electron pairs form a tetrahedron
Description : The nature of `pi` - bonds in `XeO_(3)` :
Last Answer : The nature of `pi` - bonds in `XeO_(3)` : A. two `(p pi- p pi)` and one `(p pi- d pi)` B. one `(p ... C. three `(p pi- d pi)` D. three `(p pi- p pi)`
Description : The shape of `XeO_(3)` molecule is
Last Answer : The shape of `XeO_(3)` molecule is A. planar triangle B. pyramid C. linear D. square planar
Description : `XeO_(3)` has
Last Answer : `XeO_(3)` has A. Three double bonded O-atoms B. Trigonal pyramidal geometry C. One lone pair and `sp^(3)` hybridisation D. All of these
Description : The number of s and p bonds in `XeO_(3)` molecule are
Last Answer : The number of s and p bonds in `XeO_(3)` molecule are A. 1s , 2p B. 3s , 3p C. 3s , 0p D. 2s , 1p
Description : The hybridisation of Xe in `XeO_(3)` is
Last Answer : The hybridisation of Xe in `XeO_(3)` is A. `sp^(2)` B. `sp^(3)d` C. `sp^(3)` D. `sp^(3)d^(2)`
Description : `(A)+SbF_(5)to[XeF_(3)]^(+)[SbF_(6)]^(-)` Compound (A) is
Last Answer : `(A)+SbF_(5)to[XeF_(3)]^(+)[SbF_(6)]^(-)` Compound (A) is A. II and III only B. I,II and IV only C. III and IV only D. I,II,III and IV
Description : `XeF_(6)` on reaction with CsF gives:
Last Answer : `XeF_(6)` on reaction with CsF gives: A. `Cs[XeF_(7)]` B. `[XeF_(4)][CsF_(3)]` C. `XeF_(8)` D. `[XeF_(5)] [CsF_(2)]`
Description : `XeF_(6)` dissolves in anhydrous `HF` to give a good conducting solution which contains:
Last Answer : `XeF_(6)` dissolves in anhydrous `HF` to give a good conducting solution which contains: A. `H^(+)` and ` ... +)` and `F^(-)` ions D. none of these
Description : `XeF_(6)` reacts with silica to form xenon compound X. The oxidation state of Xe in X is :
Last Answer : `XeF_(6)` reacts with silica to form xenon compound X. The oxidation state of Xe in X is :
Description : Assertion : `XeF_(6)` cannot be stored in the dry glass bottles Reason : `XeF_(6)` attacks the glass
Last Answer : Assertion : `XeF_(6)` cannot be stored in the dry glass bottles Reason : `XeF_(6)` attacks the glass
Description : Which of the following does not exist ? a) `XeOF_(4)` b) `NeF_(2)` c) `XeF_(2)` d) `XeF_(6)`
Last Answer : Which of the following does not exist ? a) `XeOF_(4)` b) `NeF_(2)` c) `XeF_(2)` d) `XeF_(6)`
Description : `XeF_(6)` on hydrolysis gives
Last Answer : `XeF_(6)` on hydrolysis gives A. `XeOF_(4)` B. `XeO_(2)F_(2)` C. `XeO_(3)` D. `XeO_(4)`
Description : `XeF_(6)` can acts as
Last Answer : `XeF_(6)` can acts as A. Fluoride donor only B. Fluoride acceptor only C. Either fluoride donor or acceptor D. Catalyst in nuclear reactions
Description : Which of the following compound will not form during the hydrolysis of `XeF_(6)` ?
Last Answer : Which of the following compound will not form during the hydrolysis of `XeF_(6)` ? A. `XeO_(3)` B. `XeO_(4)` C. `XeOF_(4)` D. `XeO_(2)F_(2)`