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Question

Select the correct order of hydrated size of ion.

A

L i_{(a q .)}^{+} > N a_{(a q .)}^{+} > K_{(a q .)}^{+}

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B

B e_{(a q .)}^{2 +} < M g_{(a q .)}^{2 +} < C a_{(a q .)}^{2 +}

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C

A l_{(a q .)}^{3 +} < M g_{(a q .)}^{2 +} < S i_{(a q .)}^{4 +}

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D

C s_{(a q .)}^{+} > R b_{(a q .)}^{+} > K_{(a q .)}^{+}

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Solution

The correct option is A $L i_{(a q .)}^{+} > N a_{(a q .)}^{+} > K_{(a q .)}^{+}$
Hydration depends on charge density so it means if some species has same charge so the smaller among them will have highest charge density and hence highest hydration. So, the correct order will be $L i_{(a q .)}^{+} > N a_{(a q .)}^{+} > K_{(a q .)}^{+}$ .

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Similar questions

Q. Arrange the following in the iIncreasing hydrated ion size:

{C s}_{a q}^{+} < {R b}_{a q}^{+} < K_{a q}^{+} < {N a}_{a q}^{+}

If true enter 1, else enter 0.

ionic species having largest size:

A. Li+(g)

B.Na+ (aq)

C. Rb+(aq)

D. Li+(aq)

Q. Statement-1: The ionic mobility of

N a_{(a q)}^{+}

is lower than that of

K_{(a q)}^{+}

r_{N a^{+} (a q)} > r_{k^{+} (a q)}

Q. Consider a galvanic cell using solid

C u

F e

metals with their corresponding solutions.
What is the

E_{c e l l}^{\circ}

Standard Potential (V)	Reduction Half-Reaction
$2.87$	$F_{2} (g) + 2 e^{-} \to 2 F^{-} (a q)$
$1.51$	$M n O_{4}^{-} (a q) + 8 H^{+} (a q) + 5 e^{-} \to M n^{2 +} (a q) + 4 H_{2} O (l)$
$1.36$	$C l_{2} (a q) + 3 e^{-} \to 2 C l^{-} (a q)$
$1.33$	$C r_{2} O_{7}^{2 -} (a q) + 14 H^{+} (a q) + 6 e^{-} \to 2 C r^{3 +} (a q) + 7 H_{2} O (l)$
$1.23$	$O_{2} (g) + 4 H^{+} (a q) + 4 e^{-} \to 2 H_{2} O (l)$
$1.06$	$B r_{2} (l) + 2 e^{-} \to 2 B r^{-} (a q)$
$0.96$	$N O_{3}^{-} (a q) + 4 H^{+} (a q) + 3 e^{-} \to N O (g) + H_{2} O (l)$
$0.80$	$A g^{+} (a q) + e^{-} \to A g (s)$ $
$0.77$	$F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q)$
$0.68$	$O_{2} (g) + 2 H^{+} (a q) + 2 e^{-} \to H_{2} O_{2} (a q)$
$0.59$	$M n O_{4}^{-} (a q) + 2 H_{2} O (l) + 3 e^{-} \to M n O_{2} (s) + 4 O H^{-} (a q)$
$0.54$	$I_{2} (s) + 2 e^{-} \to 2 I^{-} (a q)$
$0.40$	$O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} \to 4 O H^{-} (a q)$
$0.34$	$C u^{2 +} (a q) + 2 e^{-} \to C u (s)$
$0$	$2 H^{+} (a q) + 2 e^{-} \to H_{2} (g)$
$- 0.28$	$N i^{2 +} (a q) + 2 e^{-} \to N i (s)$
$- 0.44$	$F e^{2 +} (a q) + 2 e^{-} \to F e (s)$
$- 0.76$	$Z n^{2 +} (a q) + 2 e^{-} \to Z n (s)$
$- 0.83$	$2 H_{2} O (l) + 2 e^{-} \to H_{2} (g) + 2 O H^{-} (a q)$
$1.66$	$A l^{3 +} (a q) + 3 e^{-} \to A l (s)$
$- 2.71$	$N a^{+} (a q) + e^{-} \to N a (s)$
$- 3.05$	$L i^{+} (a q) + e^{-} \to L i (s)$

Q. Consider a spontaneous electrochemical cell between

C u

A l

. Predict what would happen if excess concentrated

N a O H

were added to the cell with copper ions and a precipitate forms.

Standard Potential (V)	Reduction Half-Reaction
$2.87$	$F_{2} (g) + 2 e^{-} \to 2 F^{-} (a q)$
$1.51$	$M n O_{4}^{-} (a q) + 8 H^{+} (a q) + 5 e^{-} \to M n^{2 +} (a q) + 4 H_{2} O (l)$
$1.36$	$C l_{2} (a q) + 3 e^{-} \to 2 C l^{-} (a q)$
$1.33$	$C r_{2} O_{7}^{2 -} (a q) + 14 H^{+} (a q) + 6 e^{-} \to 2 C r^{3 +} (a q) + 7 H_{2} O (l)$
$1.23$	$O_{2} (g) + 4 H^{+} (a q) + 4 e^{-} \to 2 H_{2} O (l)$
$1.06$	$B r_{2} (l) + 2 e^{-} \to 2 B r^{-} (a q)$
$0.96$	$N O_{3}^{-} (a q) + 4 H^{+} (a q) + 3 e^{-} \to N O (g) + H_{2} O (l)$
$0.80$	$A g^{+} (a q) + e^{-} \to A g (s)$ $
$0.77$	$F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q)$
$0.68$	$O_{2} (g) + 2 H^{+} (a q) + 2 e^{-} \to H_{2} O_{2} (a q)$
$0.59$	$M n O_{4}^{-} (a q) + 2 H_{2} O (l) + 3 e^{-} \to M n O_{2} (s) + 4 O H^{-} (a q)$
$0.54$	$I_{2} (s) + 2 e^{-} \to 2 I^{-} (a q)$
$0.40$	$O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} \to 4 O H^{-} (a q)$
$0.34$	$C u^{2 +} (a q) + 2 e^{-} \to C u (s)$
$0$	$2 H^{+} (a q) + 2 e^{-} \to H_{2} (g)$
$- 0.28$	$N i^{2 +} (a q) + 2 e^{-} \to N i (s)$
$- 0.44$	$F e^{2 +} (a q) + 2 e^{-} \to F e (s)$
$- 0.76$	$Z n^{2 +} (a q) + 2 e^{-} \to Z n (s)$
$- 0.83$	$2 H_{2} O (l) + 2 e^{-} \to H_{2} (g) + 2 O H^{-} (a q)$
$1.66$	$A l^{3 +} (a q) + 3 e^{-} \to A l (s)$
$- 2.71$	$N a^{+} (a q) + e^{-} \to N a (s)$
$- 3.05$	$L i^{+} (a q) + e^{-} \to L i (s)$

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