COORDINATION COMPOUNDS NCERT SOLUTIONS PDF

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psychpadnarecea.ga - No.1 online tutoring company in India provides you Free PDF download of NCERT Solutions for Class 12 Chemistry Chapter 9 - Coordination. Question Explain the bonding in coordination compounds in terms of Werner's postulates. Answer. Werner's postulates explain the bonding in coordination. NCERT Solutions Class 12 Chemistry Chapter 9 Coordination Compounds Download in Pdf.


Coordination Compounds Ncert Solutions Pdf

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NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds. Write the formulas for the following coordination compounds. Ncert Solutions For Class 12 Chemistry Chapter 9 PDF Free Download . What are the various kinds of isomerism present in coordination compounds?. These NCERT Solutions for Class 12 of Chemistry subject includes detailed answers of all the questions in Chapter 9 – Coordination Compounds provided in .

As there are 6 ligands, hybridization has to be either sp3d2 or d2sp3 hybridization. Hence, the geometry of the complex is found to be octahedral.

It cannot cause the pairing of the 3d electrons.

Question 9. Answer: The splitting of the d orbitals in an octahedral field takes palce in such a way that , experience a rise in energy and form the eg level, while dxy, dyzand dzx experience a fall in energy and form the t2g level. Explain the difference between a weak field ligand and a strong field ligand.

Answer: A spectrochemical series is the arrangement of common ligands in the increasing order of their crystal-field splitting energy CFSE values. The ligands present on the R.

S of the series are strong field ligands while that on the L. S are weak field ligands. Also, strong field ligands cause higher splitting in the d orbitals than weak field ligands.

Therefore, it cannot cause the pairing of the 3d orbital electrons. As there are 6 ligands, hybridization has to be either sp3d2 or d2sp3 hybridization.

NCERT Solutions Class 12 Chemistry Chapter 9 Coordination Compounds

Hence, the geometry of the complex is found to be octahedral. It cannot cause the pairing of the 3d electrons. Page No Draw figure to show the splitting of d orbitals in an octahedral crystal field.

Answer: The splitting of the d orbitals in an octahedral field takes palce in such a way that , experience a rise in energy and form the eg level, while dxy, dyzand dzx experience a fall in energy and form the t2g level. Page No Question 9.

Q4.Providing two examples in each case,explain the followingterms :unidentate, ambidentate

Explain the difference between a weak field ligand and a strong field ligand. Answer: A spectrochemical series is the arrangement of common ligands in the increasing order of their crystal-field splitting energy CFSE values.

The ligands present on the R. S of the series are strong field ligands while that on the L. Ans: Q Differentiate between a strong field ligand and a weak field ligand. Ans: Series of common ligands in an ascending order of their crystal-field splitting energy CFSE is termed as Spectrochemical series.

Strong field ligands have larger values of CFSE. Whereas, weak field ligands have smaller values of CFSE.

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Define crystal field splitting energy. Ans: Crystal-field splitting energy is the difference in the energy between the two levels i.

Once the orbitals split up, electrons start filling the vacant spaces. An electron each goes into the three t2g orbitals, the fourth electron, however, can enter either of the two orbitals: 1 It can go to the eg orbital producing t2g3eg 1 like electronic configuration , or 2 it can go to the t2g orbitals producing t2g4eg 0 like electronic configuration.

Thus, it has d 8configuration.Explain why?

The bond are formed by overlap of atomic orbital of metal with that of C-atom of carbon monoxide in following sequence: The ions present in the solution of Ques 9. Define crystal field splitting energy.

UP Board Class 12 Chemistry Solutions रसायन विज्ञान

The crystal field is octahedral. As, NH3 is not a strong field ligand it does not cause the electrons in the 3d orbital to pair.

Thus, three ions are produced.