Practicing Success
Match List-I with List-II
Choose the correct answer from the options given below: |
(A)-(I), (B)-(II), (C)-(III), (D)-(IV) (A)-(II), (B)-(III), (C)-(IV), (D)-(I) (A)-(III), (B)-(IV), (C)-(I), (D)-(II) (A)-(IV), (B)-(III), (C)-(I), (D)-(II) |
(A)-(IV), (B)-(III), (C)-(I), (D)-(II) |
The correct answer is (4) (A)-(IV), (B)-(III), (C)-(I), (D)-(II). (A) \([CoF_6]^{3-}\): The atomic number of Co is 27 and its electronic configuration is \(_{18}Ar]4s^2 3d^7\) Here, \(Co\) is in \(+3\) oxidation state, so the electronic configuration will be since \(F^-\) is a weak field ligand it will form an outer orbital complex so, the hybridization will be Hence the hybridization is \(sp^3d^2\) hybridization. (B) \([Co(NH_3)_6]^{3+}\) The atomic number of Co is 27 and its electronic configuration is \(_{18}Ar]4s^2 3d^7\) Here, \(Co\) is in \(+3\) oxidation state, so the electronic configuration will be Ammonia is placed in the middle of the spectrochemical series. It is a weak ligand but in presence of octahedral complexes containing a central metal atom belonging to 3d series has a coordination number of 3, ammonia acts as a strong ligand. For example, in the complex \([Co(NH_3)6]^{3+}\), Cobalt belongs to 3d series and has the oxidation number 3, in this case ammonia acts as a strong ligand and leads to pairing. Hence the hybridization is \(d^2sp^3\) hybridization. (C) \([NiCl_4]^{2-}\) The atomic number of Ni is 28 and its electronic configuration is \(_{18}Ar]4s^2 3d^8\) Here, \(Ni\) is in \(+2\) oxidation state, so the electronic configuration will be As chlorine is a weak field ligand, it is not able to pair the electrons in Ni and thus d orbitals do not participate in hybridization. So, only 4s and 4p orbitals participate. So, the hybridization is \(sp^3\) (D) \([Ni(CN)_4]^{2-}\) The atomic number of Ni is 28 and its electronic configuration is \(_{18}Ar]4s^2 3d^8\) Here, \(Ni\) is in \(+2\) oxidation state, so the electronic configuration will be In presence of strong field \(CN^-\) ions, all the electrons are paired up. The empty 3d, 3s and two 4p orbitals undergo \(dsp^2\) hybridization. So, the hybridization is \(dsp^2\). Hence the correct answer is (4) (A)-(IV), (B)-(III), (C)-(I), (D)-(II). |