Question

Select correct statement(s) regarding given complexes :

• A. $\left[Fe\right(CO{)}_5]$, the orbitals used for hybridization in $Fe$ atom are $s,p_x,p_y,p_z,d_{z^2}$ and it is high spin complex
• B. $\left[Pt\right(N{H}_3{)}_{2}C{l}_2]$, the orbitals used for hybridization on $Pt$ atom are $s,p_x,p_y,d_{x^2-y^2}$ and it is low spin complex
• C. $\left[Cr\right(H_{2}O{)}_6{]}^{3+}$, the orbitals used for hybridization in $Cr$ atom are $s,p_x,p_y,p_z,p_{x^2},d_{x^2-y^2}$ and it is high spin complex
• D. $Ni(CO{)}_4$, the orbitals used for hybridization in $Ni$ atom are, $s,p_x,p_y,p_z$, and it is low spin complex

Answer 1

The correct option is D $Ni(CO{)}_4$, the orbitals used for hybridization in $Ni$ atom are, $s,p_x,p_y,p_z$, and it is low spin complex

(A) $\left[Fe\right(CO{)}_5]$ involves $ds{p}^3$ hybridization. The orbitals involved are $d_{z^2},s,p_x,p_{y}and{p}_z$
It is low spin complex. Thus option A is incorrect.
(B) $\left[Pt\right(N{H}_3{)}_{2}C{l}_2]$ undergoes $ds{p}^2$ hybridization to form high spin complex. The orbitals involved in the hybridization are $s,p_x,p_y,d_{x^2-y^2}$.
(C) In $\left[Cr\right(H_{2}O{)}_6{]}^{3+}$, the $C{r}^{3+}\left(d^3\right)$ ion undergoes $s{p}^3{d}^2$ hybridization to form outer orbital complex with 3 unpaired electrons.
The orbitals involved in the hybridization are $s,p_x,p_y,p_z,d_{x^2},d_{x^2-y^2}$
Thus option C is incorrect.
(D) $Ni(CO{)}_4$ involves $s{p}^3$ hybridization to form low spin complex. The orbitals involved in the hybridization are $s,p_x,p_y,p_z$.
Thus option D is correct.