Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- $X =\left[ Co \left( H _2 O \right)_4 Cl _2\right]^{+}, Y =\left[ CoCl _4\right]^{2-}, Z =$ Tetrahedral
- $X =\left[ Co \left( H _2 O \right)_6\right]^{2+}, Y =\left[ CoCl _6\right]^{3-}, Z =$ Octahedral
- $X =\left[ Co \left( H _2 O \right)_6\right]^{2+}, Y =\left[ CoCl _4\right]^{2-}, Z =$ Tetrahedral
- $X =\left[ Co \left( H _2 O \right)_6\right]^{3+}, Y =\left[ CoCl _6\right]^{3-}, Z =$ Octahedral
The Correct Option is C
Solution and Explanation
So, the correct answer is (C): X=[Co(H2O)6]2+,Y=[CoCl4]2−,Z= Tetrahedral
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Concepts Used:
D and F Block Elements
The d-block elements are placed in groups 3-12 and F-block elements with 4f and 5f orbital filled progressively. The general electronic configuration of d block elements and f- block elements are (n-1) d 1-10 ns 1-2 and (n-2) f 1-14 (n-1) d1 ns2 respectively. They are commonly known as transition elements because they exhibit multiple oxidation states because of the d-d transition which is possible by the availability of vacant d orbitals in these elements.
They have variable Oxidation States as well as are good catalysts because they provide a large surface area for the absorption of reaction. They show variable oxidation states to form intermediate with reactants easily. They are mostly lanthanoids and show lanthanoid contraction. Since differentiating electrons enter in an anti-penultimate f subshell. Therefore, these elements are also called inner transition elements.
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