हिंदी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

Atomic number of Mn, Fe and Co are 25, 26 and 27 respectively. Which of the following inner orbital octahedral complex ions are diamagnetic? (i) [Co(NHX3)X6]X3+ (ii) [Mn(CN)X6]X3− (iii) [Fe(CN)X6]X4− - Chemistry

Advertisements
Advertisements

प्रश्न

Atomic number of \[\ce{Mn}\], \[\ce{Fe}\] and \[\ce{Co}\] are 25, 26 and 27 respectively. Which of the following inner orbital octahedral complex ions are diamagnetic?

(i) \[\ce{[Co(NH3)6]^{3+}}\]

(ii) \[\ce{[Mn(CN)6]^{3-}}\] 

(iii) \[\ce{[Fe(CN)6]^{4-}}\]

(iv) \[\ce{[Fe(CN)6]^{3-}}\]

टिप्पणी लिखिए
Advertisements

उत्तर

(i) \[\ce{[Co(NH3)6]^{3+}}\]

(iii) \[\ce{[Fe(CN)6]^{4-}}\] 

Explanation:

(i) Molecular orbital electronic configuration of \[\ce{Co^{3+}}\] in \[\ce{[Co(NH3)6]^{3+}}\] is 

Number of unpaired electron = 0

Magnetic property = Diamagnetic

(ii) Molecular orbital electronic configuration of \[\ce{Mn^{3+}}\] in \[\ce{[Mn(CN)6]^{3-}}\]

Number of unpaired electron = 2

Magnetic property = Paramagnetic

(iii) Molecular orbital electronic configuration of \[\ce{Fe^{3+}}\] in \[\ce{[Fe(CN)6]^{4-}}\] is

Number of unpaired electron = 0

Magnetic property = Diamagnetic

(iv) Molecular orbital electronic configuration of \[\ce{Fe^{3+}}\] in \[\ce{[Fe(CN)6]^{3-}}\]

Number of unpaired electron = 1

Magnetic property = Paramagnetic

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Coordination Compounds - Exercises [पृष्ठ १२२]

APPEARS IN

एनसीईआरटी एक्झांप्लर Chemistry [English] Class 12
अध्याय 9 Coordination Compounds
Exercises | Q II. 15. | पृष्ठ १२२

संबंधित प्रश्न

On the basis of crystal field theory, write the electronic configuration for d4 ion if Δ0 > P.


Draw figure to show the splitting of d orbitals in an octahedral crystal field.


 Write the electronic configuration of Fe(III) on the basis of crystal field theory when it forms an octahedral complex in the presence of (i) strong field, and (ii) weak field ligand. (Atomic no.of Fe=26) 


Draw the structures of the following :
(1) XeF6
(2) IF7


Complete and balance the following reactions:

\[\ce{P4 + H2SO4 ->}\] ______ + ______ + ______


The colour of the coordination compounds depends on the crystal field splitting. What will be the correct order of absorption of wavelength of light in the visible region, for the complexes, \[\ce{[Co(NH3)6]^{3+}}\], \[\ce{[Co(CN)6]^{3-}}\], \[\ce{[Co(H2O)6]^{3+}}\]


Atomic number of \[\ce{Mn, Fe, Co}\] and Ni are 25, 26, 27 and 28 respectively. Which of the following outer orbital octahedral complexes have same number of unpaired electrons?

(i) \[\ce{[MnCl6]^{3-}}\]

(ii) \[\ce{[FeF6]^{3-}}\]

(iii) \[\ce{[CoF6]^{3-}}\]

(iv) \[\ce{[Ni(NH3)6]^{2+}}\]


An aqueous pink solution of cobalt (II) chloride changes to deep blue on addition of excess of HCl. This is because:

(i) \[\ce{[Co(H2O)6]^{2+}}\] is transformed into \[\ce{[CoCl6]}^{4-}\]

(ii) \[\ce{[Co(H2O)6]^{2+}}\] is transformed into \[\ce{[CoCl4]}^{2-}\]

(iii) tetrahedral complexes have smaller crystal field splitting than octahedral complexes.

(iv) tetrahedral complexes have larger crystal field splitting than octahedral complex.


\[\ce{CuSO4 . 5H2O}\] is blue in colour while \[\ce{CuSO4}\] is colourless. Why?


Match the complex ions given in Column I with the hybridisation and number of unpaired electrons given in Column II and assign the correct code:

Column I (Complex ion) Column II (Hybridisation, number of unpaired electrons)
A. \[\ce{[Cr(H2O)6]^{3+}}\] 1. dsp2, 1
B. \[\ce{[Co(CN)4]^{2-}}\] 2. sp3d2, 5
C. \[\ce{[Ni(NH3)6]^{2+}}\] 3. d2sp3, 3
D. \[\ce{[MnF6]^{4-}}\] 4. sp3, 4
  5. sp3d2, 2

Why are different colours observed in octahedral and tetrahedral complexes for the same metal and same ligands?


[Ni(H2O)6]2+ (aq) is green in colour whereas [Ni(H2O)4 (en)]2+ (aq)is blue in colour, give reason in support of your answer.


Considering crystal field theory, strong-field ligands such as CN:


What is crystal field splitting energy?


The correct order of intensity of colors of the compounds is ______.


For octahedral Mn(II) and tetrahedral Ni(II) complexes, consider the following statements:

  1. Both the complexes can be high spin.
  2. Ni(II) complex can very rarely below spin.
  3. With strong field Ligands, Mn(II) complexes can be low spin.
  4. Aqueous solution of Mn (II) ions is yellow in colour.

The correct statements are:


On the basis of Crystal Field theory, write the electronic configuration for the d5 ion with a strong field ligand for which Δ0 > P.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×