Valence Bond Theory
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(I) [Fe(CN)6]4−
(II) [Fe(CN)6]3−
(III) [Cr(NH3)6]3+
(IV) [Ni(H2O)6]2+
- I<II<III<IV
- IV<III<II<I
- II<III<I<IV
- I<II<IV<III
- Acidic in nature
- Basic in nature
- Neither acidic nor basic
- Some are acidic some are basic
- Both are square planar
- Tetrahedral and square planar
- Both are tetrahedral
- Square planar and tetrahedral
- Both complex have square planar geometries
- Both complexes have tetrahedral geometries
- [NiCl4]2− has a square planar geometry while [Ni(CN)4]2− has a tetrahedral geometry
- [NiCl4]2− has a tetrahedral geometry while [Ni(CN)4]2− has a square planar geometry
According to valence bond theory, a bond between two atoms is formed when
half-filled atomic orbitals overlap
fully-filled atomic orbitals overlap
non-bonding atomic orbitals overlap
electrons of the two atoms overlap
- 4
- 1
- 2
- 3
- 0
- 2
- 3
- 4
Column-IColumn-IIa. [Ni(CN)4]2−p. d2sp3b.[Ni(CO)4]q. sp3d2c. [Co(en)3]3+r. dsp2d. [Co(Br)3Cl3]3−s. sp3t. Inner orbital complex u. Outer orbital complex
- (a−r, t), (b−s, u)
(c−p, t), (d−p, t) - (a−s, t), (b−r, u)
(c−p, t), (d−p, t) - (a−r, t), (b−s, u)
(c−q, t), (d−p, t) - (a−s, t), (b−s, u)
(c−q, u), (d−q, u)
- 2
- 4
- 6
Complexes :
A . [CoF6]3−
B . [Co(H2O)6]2+
C . [Co(NH3)6]3+
D . [Co(en)3]3+
Choose the correct option :
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- C < D < B < A
- A < B < C < D
- B < C < D < A
- B < A < C < D
- 2
- 4
- 5
- 1
- dsp2
- sp3d
- dsp3
- sp3
- d2 sp3
- sp3 d2
- dsp2
- sp3
- The equivalence of the bonds in most of the compounds
- The stereochemistry of the molecules
- The better overlapping of the orbitals
- None of these
- O2+2 is expected to have a longer bond length than O2
- N+2 and N−2 have the same bond order
- He+2 has the same energy as two isolated He atoms
- C2−2 is expeced to be diamagnetic
- 2
- 3
- 4
- 5
- four 2C−2e bonds and two 3C−2e bonds
- four 2C−2e bonds and four 3C−2e bonds
- two 2C−2e bonds and two 3C−2e bonds
- two 2C−2e bonds and four 3C−2e bonds
Which among the following are the repulsive forces acting between two atoms A and B?
If eA, eB represents electrons of atom A and B respectively and NA, NB represents nuclei of atom A and B respectively.
- eA − eB
- NA − NB
- NA − eB
- NB − eA
- (n + 1)s
- (n+2)p
- (n+1)d
- (n+2)s
- [Cu(CN)4]3−
- [Cr(NH3)6]3+
- [Fe(CN)6]4−
- [Ni(CO)4]
- 2
- 6
- 4
- 8
An AB2 type structure is found in
NaCl
Al2O3
CaF2
N2O
Addition of excess aqueous ammonia to a pink coloured aqueous solution of MCl2.H2O(x) and NH4Cl gives an octahedral complex Y in the presence of air. In aqueous solution, complex Y behaves as 1:3 electrolyte. The reaction of X with excess HCl at room temperature results in the formation of a blue coloured complex Z. The calculated spin only magnetic moment of X and Z is 3.87 B.M., whereas it is zero for complex Y.
Among the following options, which statemet(s) is (are) correct?
The hybridization of the central metal ion in Y is d2sp3
Addition of silver nitrate to Y gives only two equivalents of silver chloride
When X and Z are in equilibrium at 0∘, the colour of the solution is pink
Z is a tetrahedral complex
- pz and pz
- s and pz
- s and px
- dx2−y2 and dx2−y2
- [CoCl4]2−
- [FeCl4]2−
- [NiCl4]2−
- [PtCl4]2−
- The hybridization of the central metal ion in Y is d2sp3
- Z is a tetrahedral complex
- Addition of silver nitrate to Y gives only two equivalents of silver chloride
- When X and Z are in equilibrium at 0∘C, the colour of the solution is pink
- He+2 has the same energy as two isolated He atoms
- O2+2 is expected to have a longer bond length than O2
- C2−2 is expeced to be diamagnetic
- N+2 and N−2 have the same bond order
- +25Δt
- +35Δt
- −25Δt
- −45Δt
- CN−
- [Pt(CN)4]2−
- Na+
- Pt2+