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Valence Bond Theory - (Concept)
According to this theory, the metal atom or ion under the influence of ligands can use its (n-1)d or nd orbitals along with its ns and np for hybridisation to yield a set of equivalent orbitals of definite geometry such as octahedral, tetrahedral, square planar and so on. These hybridised orbitals are allowed to overlap with ligand orbitals that can donate electron pairs for bonding. The different types of hybridisation and their respective shapes are given below
| Coordination Number | Type of Hybridisation | Shape |
| 4 | sp3 | Tetrahedral |
| 4 | dsp2 | Square Planar |
| 5 | sp3d | Trigonal Bipyramidal |
| 6 | sp3d2 | Octahedral |
| 6 | d2sp3 | Octahedral |
| Exam | Chapter |
| JEE MAIN | Co-ordination Compounds |
Complete removal of both the axial ligands (along the z-axis) from an octahedral complex leads to which of the following splitting patterns? ( relative orbital energies not on scale).
| A. |
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| B. |
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| C. |
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| D. |
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The molecule in which hybrid molecular orbitals involve only one d-orbital of the central atom is:
| A. |
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| B. |
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| C. |
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| D. |
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According to the valence bond theory the hybridization of central metal atom is for which one of the following compounds?
| A. |
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| B. |
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| C. |
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| D. |
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3 moles of metal complex with formula gives 3 moles of silver chloride on treatment with excess of silver nitrate. The secondary valency of CO in the complex is_______.
(Round off to the nearest integer)
| A. |
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| B. |
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| C. |
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| D. |
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The sum of oxidation states of two silver ions in complex is ___________.
| A. |
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| B. |
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| C. |
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| D. |
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In Wilkinson’s catalyst, the hybridization of central metal ion and its shape are respectively :
| A. |
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| B. |
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| C. |
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| D. |
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Identify the pair in which the geometry of the species is T-shape and squarepyramidal, respectively :
| A. |
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| B. |
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| C. |
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| D. |
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The geometry of and
are respectively :
| A. |
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| B. |
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| C. |
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| D. |
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In the complex hybridisation is present. Geometry of the complex is.
| A. |
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| B. |
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| C. |
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| D. |
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An octahedral complex is formed, when hybrid orbitals of the following type are involved
| A. |
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| B. |
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| C. |
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| D. |
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Which complex has square planar structure?
| A. |
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| B. |
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| C. |
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| D. |
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is
| A. |
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| B. |
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| C. |
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| D. |
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hybridisation leads to
| A. |
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| B. |
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| C. |
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| D. |
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The hybridisation of the complex is
| A. |
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| B. |
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| C. |
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| D. |
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In the species , the hybiridisation of Ni are respectively
| A. |
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| B. |
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| C. |
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| D. |
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One mole of the complex compound gives 3 moles of ions on dissolution in water. One mole of the same complex reacts with two moles of
solution to yield two moles of
. The structure of the complex is
| A. |
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| B. |
|
| C. |
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| D. |
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The coordination geometry around the manganese in decacarbonyldimanganese ( 0 ) is
| A. |
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| B. |
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| C. |
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| D. |
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$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ and $\left[\mathrm{CoF}_6\right]^{3-}$ are respectively known as :
| A. |
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| B. |
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| C. |
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| D. |
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According to this theory, the metal atom or ion under the influence of ligands can use its (n-1)d, ns, np or ns, np, nd orbitals for hybridisation to yield a set of equivalent orbitals of definite geometry such as octahedral, tetrahedral, square planar and so on (Table 9.2).
