Which of the following options has the correct combination considering List-I and List-II?

I I \to P; I V \to U

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I I \to S; I V \to R

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I I \to U; I V \to T

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I I \to P; I V \to R

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Solution

The correct option is A $I I \to P; I V \to U$

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Similar questions

\begin{matrix} List-I & List-II (I) & Teflon & (P) & Electrically insulated buttons (II) & Vinyl Rubber & (Q) & Pipe Insulation (III) & Lexan & (R) & Rubber Erasers (IV) & PVC & (S) & Bullet proof glass (T) & Foam thermocole (U) & Non-stick kitchenware \end{matrix}

\begin{matrix} List - I & List - II (I) Dissociation of N_{2} O_{4} (g) : & (P) Increases with N_{2} O_{4} (g) ⇌ 2 N O_{2} (g); & temperature Δ H = + 57.0 k J (II) Oxidation of N H_{3} (g) : & (Q) Decreases with 4 N H_{3} (g) + 5 O_{2} ⇌ 4 N O (g) + 6 H_{2} O (g); & pressure Δ H = - 900.0 k J (III) Oxidation of nitrogen: & (R) Increases with N_{2} (g) + O_{2} (g) ⇌ 2 N O (g); & addition of inert Δ H = + 180.0 k J & gas at constant pressure (IV) Formation of N O_{2} (g) : & (S) Decreases with N O (g) + O_{3} (g) ⇌ N O_{2} (g) + O_{2} (g); & temperature Δ H = - 200.0 k J (T) K_{p} = K_{c} (U) K_{p} = K_{c} (R T)^{1} \end{matrix}

\begin{matrix} List - I & List - II (Compound) & (Hybridisation) (I) [F e (E D T A)]^{-} & (P) s p^{3} d^{2} (II) [N i (C N)_{4}]^{2 -} & (Q) s p^{3} (III) [C u C l_{4}]^{2 -} & (R) d^{3} s (IV) [F e (H_{2} O)_{6}]^{2 -} & (S) d^{2} s p^{3} (T) d s p^{2} (U) s p^{3} d \end{matrix}

(H N O_{3}),

H N O_{3}

\begin{matrix} L i s t - I & L i s t - I I (I) & Ca with dilute and cold H N O_{3} & (P) & Metal nitrate (I I) & As with concentrated H N O_{3} & (Q) & N O (I I I) & Zn with dilute H N O_{3} & (R) & N O_{2} (I V) & Cu with dilute H N O_{3} & (S) & N_{2} O (T) & Tribasic acid (U) & Hydrogen gas \end{matrix}

\begin{matrix} List-I & List-II (I) & Molarity (mol/L) & (P) & 20 (I I) & Molality (mol/kg) & (Q) & 0.67 (I I I) & % mass by mass & (R) & 1.2 (I V) & % mass by volume (g/ml) & (S) & 1 (T) & 25 (U) & 1.25 \end{matrix}

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