Barium ions, CN -and Co 2+ form an ionic complex. If that is supposed to be 75 % ionised in water with van't Hoff factor (i) equal to four, then the coordination numbers of Co 2+ in the complex can be:

Question

Barium ions, CN -and Co 2+ form an ionic complex. If that is supposed to be 75 % ionised in water with van't Hoff factor (i) equal to four, then the coordination numbers of Co 2+ in the complex can be: (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

α=(i-1/x-1)=(3/4)=(4-1/x-1) ∴ x=5 ∴ Complex is Ba 3[ Co ( CN )5]2

Q. Barium ions, $C N^{-}$ and $C o^{2 +}$ form an ionic complex. If that is supposed to be $75%$ ionised in water with van't Hoff factor $(i)$ equal to four, then the coordination numbers of $C o^{2 +}$ in the complex can be:

$α = \frac{i - 1}{x - 1} = \frac{3}{4} = \frac{4 - 1}{x - 1}$
$∴ x = 5$
$∴$ Complex is $B a_{3} [C o (CN)_{5}]_{2}$

Q. Barium ions, CN−and Co2+ form an ionic complex. If that is supposed to be 75% ionised in water with van't Hoff factor (i) equal to four, then the coordination numbers of Co2+ in the complex can be:

α=x−1i−1​=43​=x−14−1​ ∴x=5 ∴ Complex is Ba3​[Co(CN)5​]2​

Q. Barium ions, $C N^{-}$ and $C o^{2 +}$ form an ionic complex. If that is supposed to be $75%$ ionised in water with van't Hoff factor $(i)$ equal to four, then the coordination numbers of $C o^{2 +}$ in the complex can be:

$α = \frac{i - 1}{x - 1} = \frac{3}{4} = \frac{4 - 1}{x - 1}$
$∴ x = 5$
$∴$ Complex is $B a_{3} [C o (CN)_{5}]_{2}$