Barium ions, CN - and Co 2+ form an ionic complex. If this complex is 75 % ionised in aqueous solution with van't Hoff factor (i) equal to four and paramagnetic moment is found equal to 1.73 BM (due to spin only) then, the hybridisation state of Co (II) in the complex will be

Question

Barium ions, CN - and Co 2+ form an ionic complex. If this complex is 75 % ionised in aqueous solution with van't Hoff factor (i) equal to four and paramagnetic moment is found equal to 1.73 BM (due to spin only) then, the hybridisation state of Co (II) in the complex will be (A) s p3 d (B) d2 s p3 (C) s p3 d2 (D) d s p3

Tardigrade Admin · Accepted Answer

α=(i-1/(n-1)) ; (4-1/(n-1))=(3/4) So, n=5, hence given complex must dissociate into five ions = Ba 3[ Co ( CN )5]2 Paramagnetic moment indicates one unpaired electron <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/aba08352c0c341be7c7eafaa861429cd-.png" />

Q. Barium ions, $C N^{-}$ and $C o^{2 +}$ form an ionic complex. If this complex is $75%$ ionised in aqueous solution with van't Hoff factor $(i)$ equal to four and paramagnetic moment is found equal to $1.73 BM$ (due to spin only) then, the hybridisation state of $C o (II)$ in the complex will be

$s p^{3} d$

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

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

$d s p^{3}$

$α = \frac{i - 1}{( n - 1 )}; \frac{4 - 1}{( n - 1 )} = \frac{3}{4}$

So, $n = 5,$ hence given complex must dissociate into five ions $= B a_{3} [C o (CN)_{5}]_{2}$

Paramagnetic moment indicates one unpaired electron

Q. Barium ions, CN− and Co2+ form an ionic complex. If this complex is 75% ionised in aqueous solution with van't Hoff factor (i) equal to four and paramagnetic moment is found equal to 1.73BM (due to spin only) then, the hybridisation state of Co(II) in the complex will be

sp3d

d2sp3

sp3d2

dsp3