Q.
Match List I with List II
List I (Complexes)
List II (Hybridisation)
A
$ [Ni(CO)_4]$
I
$sp^{3}$
B
$ [Cu(NH_{3})_{4}]^{2+} $
II
$dsp ^{2} $
C
$ [Fe(NH_{3})_{6}]^{2+} $
III
$ sp^{3}d^2 $
D
$ [Fe(H_{2}O)_{6}]^{2+} $
IV
$d^{2}sp^{3} $
| List I (Complexes) | List II (Hybridisation) | ||
|---|---|---|---|
| A | $ [Ni(CO)_4]$ | I | $sp^{3}$ |
| B | $ [Cu(NH_{3})_{4}]^{2+} $ | II | $dsp ^{2} $ |
| C | $ [Fe(NH_{3})_{6}]^{2+} $ | III | $ sp^{3}d^2 $ |
| D | $ [Fe(H_{2}O)_{6}]^{2+} $ | IV | $d^{2}sp^{3} $ |
Solution:
For $\left[ Fe \left( NH _3\right)_6\right]^{+2}, \Delta_0< P$, hence the pairing of electrons does not occur in $t _{2 g }$. Therefore complex is outer orbital and its hybridisation is $sp ^3 d ^2$.
Match List I with List II
List I (Complexes)
List II (Hybridisation)
$ [Ni(CO)_4]$
$sp^{3}$
$ [Cu(NH_{3})_{4}]^{2+} $
$dsp ^{2} $
$ [Fe(NH_{3})_{6}]^{2+} $
$ sp^{3}d^2 $
$ [Fe(H_{2}O)_{6}]^{2+} $
$sp^{3} d^2$
| List I (Complexes) | List II (Hybridisation) |
|---|---|
| $ [Ni(CO)_4]$ | $sp^{3}$ |
| $ [Cu(NH_{3})_{4}]^{2+} $ | $dsp ^{2} $ |
| $ [Fe(NH_{3})_{6}]^{2+} $ | $ sp^{3}d^2 $ |
| $ [Fe(H_{2}O)_{6}]^{2+} $ | $sp^{3} d^2$ |