List - I List - II (I) CO ( g ) + H 2 ( g ) Co Catalyst (P) All products soluble in H 2 O (II) NaH + H 2 O → (Q) One of the product is gas at room temperature (III) BaO 2 .8 H 2 O ( s ) +

Question

List - I List - II (I) CO ( g ) + H 2 ​ ( g ) Co Catalyst ​ (P) All products soluble in H 2 ​ O (II) NaH + H 2 ​ O → (Q) One of the product is gas at room temperature (III) BaO 2 ​ .8 H 2 ​ O ( s ) + H 2 ​ SO 4 ​ (aq.) → (R) One of the product have less boiling point than H 2 ​ O (IV) NaHCO 3 ​ Δ ​ (S) In one of the product H-bonding is possible (T) One of the product is found in liquid form at room temperature (U) One of the product contains peroxy linkage Which of the following option is CORRECT combination :

TestHub · Accepted Answer

I → P, R, S, T

II → Q, R, S

III → S, T, U

IV → P, Q, R, S, T

$CO (g) + 2H_{2} (g) Co Catalyst CH_{3} OH (ℓ)$

$NaH + H_{2} O \to NaOH + H_{2} (g)$

$BaO_{2} \cdot 8 H_{2} O (s) + H_{2} SO_{4} (aq .) \to BaSO_{4} (s) + H_{2} O_{2} (aq .) + 8 H_{2} O$

$2 NaHCO_{3} \to Na_{2} CO_{3} (s) + CO_{2} (g) + H_{2} O (ℓ)$

List - I	List - II
(I) $CO (g) + H_{2} (g) Co Catalyst$	(P) All products soluble in $H_{2} O$
(II) $NaH + H_{2} O \to$	(Q) One of the product is gas at room temperature
(III) $BaO_{2} .8 H_{2} O (s) + H_{2} SO_{4}$ (aq.) →	(R) One of the product have less boiling point than $H_{2} O$
(IV) $NaHCO_{3} Δ$	(S) In one of the product H-bonding is possible
	(T) One of the product is found in liquid form at room temperature
	(U) One of the product contains peroxy linkage

Chemistry - Complete s-block Question with Solution | TestHub

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