Question

Which of the following is a low - spin (spin - paired) complex?

• A. $\left[Ni\right(N{H}_3{)}_6{]}^{2+}$
• B. $\left[Ti\right(H_{2}O{)}_6{]}^{3+}$
• C. $\left[Cr\right(N{H}_3{)}_6{]}^{3+}$
• D. $\left[Fe\right(N{H}_2{)}_6{]}^{3+}$

Answer 1

Answer: (B)

$\left[Ti\right(H_{2}O{)}_6{]}^{3+}$

(A) $N{i}^{2+}=4s^{0}3d^8$

(Refer to Image 1)

$s{p}^3{d}^3$ hybridization

Outer d-orbital used hence high spin complex.

(B) $T{i}^{3+}⟶4s^{0}3d^1$

(Refer to Image 2)

$d^{2}s{p}^3$ hybridization

Inner d-orbital used in hybridization hence it is a low spin complex.

(C) $C{r}^{3+}⟶4s^{0}3d^3$

(Refer to Image 3)

$s{p}^3{d}^2$ hybridization

Outer d-orbital used hence high spin complex.

(D) $F{e}^{3+}⟶4s^{0}3d^5$

(Refer to Image)

$s{p}^3{d}^2$ hybridization

Outer orbital is used hence high spin complex.

$\therefore$ Answer is option B.