Q. 384.0( 1 Vote )

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


Code:

A. A (3) B (1) C (5) D (2)

B. A (4) B (3) C (2) D (1)

C. A (3) B (2) C (4) D (1)

D. A (4) B (1) C (2) D (3)

Answer :

• In the case of A. [Cr(H2O)6]3 +: The central metal atom is Cr and oxidation state x=+3 as water molecules are neutral.


Therefore , having a d3 configuration (1s2 2s2 2p6 3s2 3p6 3d3).


Cr has +3 oxidation state with d3 configuration and left with 2 empty d orbitals and 1 empty 4s orbital which in interaction with 3,4p orbitals and attain the hybridisation d2sp3 and the 6 H2O electrons will occupy these 6 hybridised orbitals. And the hybridisation scheme will be :1.PNG


The orbital energy diagram of [Cr(H2O)6]3+is :



Hence, there will be 3 unpaired electrons.


In case of B. [Co(CN)4]2– : The central metal atom is Co and oxidation state x= -4 (-1) – 2=+2,


Therefore, having a d7configuration (1s2 2s2 2p6 3s2 3p6 3d7).Hence, the Co2+ is left with 1 3d orbital,1 4s orbital which in interaction with 2, 4p orbitals and attains hybridisation of dsp2and the 4CN- electrons will occupy these 4 hybridised orbitals and the scheme will be:



The orbital energy diagram will be:


hence, there will be 1 unpaired electron.


In the case ofC. [Ni(NH3)6]2+: The central metal atom is Ni and oxidation state x=+2,


Therefore, having a d8 configuration (1s2 2s2 2p6 3s2 3p6 3d8).Hence, the Ni2+ is left with 2, 3d orbital,1 4s orbital which in interaction with 2, 4p orbitals and attain hybridisation of sp3d2and the 6, NH3 electrons will occupy these 6 hybridised orbitals and the scheme will be:



The orbital energy diagram will be:



hence, there will be 2 unpaired electrons.


In the case of D. [MnF6]4–: The central metal atom is Mnand oxidation state x=-6(-1) – 4 =+2


Therefore, having a d5configuration (1s2 2s2 2p6 3s2 3p6 3d5).Hence, the Mn2+ is left with 1 4s, 2 of 4d (outer orbital complex) which in interaction with 3, 4p orbitals and attain hybridisation of sp3d2and the 6, F-electrons will occupy these 6 hybridised orbitalsand the scheme will be:



hence, there will be 5 unpaired electrons as it a high spin complex.


The correct answer will be the option (i)A (3) B (1) C (5) D (2)


Hence, matching the two columns the right combination will be:


A. [Cr(H2O)6]3+ - 3. d2sp3, 3


B. [Co(CN)4]2–-1. dsp2, 1.


C. [Ni(NH3)6]2+-5. sp3d2, 2


D. [MnF6]4– -2. sp3d2, 5

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