The charges x and y on the following ions are (i) [Co(NH3)2Cl4]x (ii) [Fe(CN)6]y (Oxidation state of Co is +3 and Fe is +2 in their respective complexes.)

Question

The charges x and y on the following ions are (i) [Co(NH3)2Cl4]x (ii) [Fe(CN)6]y (Oxidation state of Co is +3 and Fe is +2 in their respective complexes.) (A) x=+1, y=-1 (B) x=-1, y=+3 (C) x=-1, y=-4 (D) x=-2, y=-3

Tardigrade Admin · Accepted Answer

In [Co (NH3)2Cl4]x, oxidation state of Co = +3 Hence, x = + 3 + 2 × 0 + 4 × (-1) = -1 In [Fe(CN)6]y, oxidation state of Fe = +2 Hence, y = + 2 + 6 × (-1) = - 4

Q. The charges $x$ and $y$ on the following ions are
(i) $[C o (N H_{3})_{2} C l_{4}]^{x}$
(ii) $[F e (CN)_{6}]^{y}$
(Oxidation state of Co is $+ 3$ and $F e$ is $+ 2$ in their respective complexes.)

$x = + 1, y = - 1$

$x = - 1, y = + 3$

$x = - 1, y = - 4$

$x = - 2, y = - 3$

In $[C o (N H_{3})_{2} C l_{4}]^{x}$ , oxidation state of Co $= + 3$
Hence, $x = + 3 + 2 \times 0 + 4 \times (- 1) = - 1$
In $[F e (CN)_{6}]^{y}$ , oxidation state of $F e = + 2$
Hence, $y = + 2 + 6 \times (- 1) = - 4$

Q. The charges x and y on the following ions are (i) [Co(NH3​)2​Cl4​]x (ii) [Fe(CN)6​]y (Oxidation state of Co is +3 and Fe is +2 in their respective complexes.)

x=+1,y=−1

x=−1,y=+3

x=−1,y=−4

x=−2,y=−3