CBSE 2024 · Region 4 · Set 1 · Q29 · 4 marks
The nature of bonding, structure of the coordination compound can be explained to some extent by valence bond theory. The central metal atom/ion makes available a number of vacant orbitals equal to its coordination number. The appropriate atomic orbitals (s, p and d) of the metal hybridise to give a set of equivalent orbitals of definite geometry such as square planar, tetrahedral, octahedral and so on. A strong covalent bond is formed only when the orbitals overlap to the maximum extent. The d-orbitals involved in the hybridisation may be either inner d-orbitals i.e. $\displaystyle (\mathrm{n}-1) \mathrm{d}$ or outer d-orbitals i.e. nd. The complexes formed are called inner orbital complex (low spin complex) and outer orbital complex (high spin complex) respectively. Further, the complexes can be paramagnetic or diamagnetic in nature. The drawbacks of this theory are that this involves number of assumptions and also does not explain the colour of the complex. Answer the following questions :(a)Predict whether $\displaystyle \left[\mathrm{CoF}_{6}\right]^{3-}$ is diamagnetic or paramagnetic and why ? [Atomic number : $\displaystyle \mathrm{Co}=27$ ](b)What is the coordination number of Co in $\displaystyle \left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}$?(c)Write the IUPAC name of the given complex : $\displaystyle \left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]^{2+}$(ii)Explain $\displaystyle \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ is an inner orbital or outer orbital complex. $\displaystyle 1+1$Using valence bond theory, deduce the shape and hybridisation of $\displaystyle \left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ [Atomic number of Ni = $\displaystyle 28$]
The nature of bonding, structure of the coordination compound can be explained to some extent by valence bond theory. The central metal atom/ion makes available a number of vacant orbitals equal to its coordination number. The appropriate atomic orbitals (s, p and d) of the metal hybridise to give a set of equivalent orbitals of definite geometry such as square planar, tetrahedral, octahedral and so on. A strong covalent bond is formed only when the orbitals overlap to the maximum extent. The d-orbitals involved in the hybridisation may be either inner d-orbitals i.e. $\displaystyle (\mathrm{n}-1) \mathrm{d}$ or outer d-orbitals i.e. nd. The complexes formed are called inner orbital complex (low spin complex) and outer orbital complex (high spin complex) respectively. Further, the complexes can be paramagnetic or diamagnetic in nature. The drawbacks of this theory are that this involves number of assumptions and also does not explain the colour of the complex. Answer the following questions :
(a)
Predict whether $\displaystyle \left[\mathrm{CoF}_{6}\right]^{3-}$ is diamagnetic or paramagnetic and why ? [Atomic number : $\displaystyle \mathrm{Co}=27$ ]
(b)
What is the coordination number of Co in $\displaystyle \left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}$?
(c)
Write the IUPAC name of the given complex : $\displaystyle \left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]^{2+}$
(ii)
Explain $\displaystyle \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}$ is an inner orbital or outer orbital complex. $\displaystyle 1+1$
Using valence bond theory, deduce the shape and hybridisation of $\displaystyle \left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$ [Atomic number of Ni = $\displaystyle 28$]
Marking-scheme solution
Paramagnetic, \(\displaystyle \mathrm{F}^{-}\)does not cause pairing of electrons and hence unpaired electrons are left.
(b) $\displaystyle 6$
(c) (i) diamminedichloridoplatinum(IV) ion
(ii) It uses inner d orbitals because \(\displaystyle \mathrm{NH}_{3}\) causes pairing of electrons
Ni(II)
\(\displaystyle \mathrm{Ni}(\mathrm{II})\) in \(\displaystyle \left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\)
Shape: Octahedral ; Hybridization : \(\displaystyle \mathrm{sp}^{3} \mathrm{~d}^{2}\)
Coordination CompoundsBonding in Coordination CompoundsApplycase_studymedium
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