Match the complexed listed column `-I` with type of hybridisation listed in column `-II` and select the correct answer using the code given below the lists `:`
`{:(,"Column"-I,,"Column"-II,),((a),[AuF_(4)]^(-),,(p)" "dsp^(2)"hybridisation",),((b),[Cu(CN)_(4)]^(3-),,(q)" "sp^(3)"hybridisation",),((c ),[Co(NH_(3))_(6)]^(3+),,(r)" "sp^(3)d^(2)"hybridisation",),((d),[Fe(H_(2)O)_(5)NO]^(2+),,(s)" "d^(2)sp^(3)"hybridisation",):}`

Options

A. `{:(,"(a)","(b)","(c)","(d)"),(,p,q,s,r):}`
B. `{:(,"(a)","(b)","(c)","(d)"),(,p,s,q,r):}`
C. `{:(,"(a)","(b)","(c)","(d)"),(,q,p,r,s):}`
D. `{:(,"(a)","(b)","(c)","(d)"),(,s,q,p,r):}`

Answer

1

Explanation

`(a)[AuF_(4)]^(-),Au` is in `+3` oxidation state with `5d^(8)` configuration , so complex is diamagnetic with `dsp^(2)` hybridisation.
`(b)[Cu(CN)_(4)]^(3-),Cu` is in `+1` oxidation state with `3d^(10)` configuration , so complex is diamagnetic with `sp^(3)` hybridisation.
`(c ) [Co(NH_(3))_(6)]^(3+),Co^(3+)` has `d^(5)` configuration which has higher `CFSE`, so hybridisation of `d^(2)sp^(3)` and complex is diamagnetic.
`(d)[Fe(H_(2)O)_(5)NO]^(2+),Fe` is in `+1` oxidation state having `3d^(7)` configuration, so hybridisation of complex is `sp^(3)d^(2)` and paramagnetic with three unpaired electrons.
`[Fe(H_(2)O)_(5)NO]^(2+):`