JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Coordination Compounds

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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, Y which has a Z geometry.

X, Y

and Z , respectively, are

(A) $X = [Co (H_{2} O)_{6}]^{2 +}, Y = [CoCl_{4}]^{2 -}, Z =$ Tetrahedral✓ Correct
(B) $X = [Co (H_{2} O_{6})]^{2 +}, Y = [CoCl_{6}]^{3 -}, Z =$ Octahedral
(C) $X = [Co (H_{2} O)_{6}]^{3 +}, Y = [CoCl_{6}]^{3 -}, Z =$ Octahedral
(D) $X = [Co (H_{2} O)_{4} Cl_{2}]^{+}, Y = [CoCl_{4}]^{2 -}, Z =$ Tetrahedral

Answer: (A)

Solution

Dissolving cobalt(II) chloride in water gives the pink octahedral aqua complex

[Co (H X_{2} O) X_{6}]^{2 +}

, which is X. Adding concentrated hydrochloric acid provides a high chloride concentration that displaces the water ligands to form the deep blue chloro complex

[CoCl X_{4}]^{2 -}

, which is Y, and with four ligands about a cobalt(II) centre this complex is tetrahedral. Thus X is

[Co (H X_{2} O) X_{6}]^{2 +}

, Y is

[CoCl X_{4}]^{2 -}

and the geometry Z is tetrahedral.

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Single correct option

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JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Coordination Compounds

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