The mechanism of a hypothetical reaction X2 + Y2 → 2XY is given below:
(i) X2 → X + X (Fast)
(ii) X + Y2 ⇄ XY + Y (slow)
(iii) X + Y → XY (Fast)
The overall order of the reaction will be:
1. 2
2. 0
3. 1.5
4. 1
The element Z = 114 has been discovered recently. Family/group and electronic configuration from which it will belong are:
1. Carbon family and [Rn] 5f14 6d10 7s2 7p2
2. Oxygen family and [Rn] 5f14 6d10 7s2 7p4
3. Nitrogen family and [Rn] 5f14 6d10 7s2 7p6
4. Halogen family and [Rn] 5f14 6d10 7s2 7p5
The heating of phenylmethyl ether with HI produces:
1. Iodobenzene
2. Phenol
3. Benzene
4. Ethyl chloride
What is the correct order of acidity among the following compounds?
| 1. | \(\small\mathrm{CH \equiv CH > CH_3 - C \equiv CH }~ \mathrm{ > CH_2 = CH_2 > CH_3 - CH_3}\) |
| 2. | \(\small\mathrm{CH \equiv CH > CH_2 = CH_2 }~ \mathrm{ > CH_3 - C \equiv CH > CH_3 - CH_3}\) |
| 3. | \(\small\mathrm{CH_3 - CH_3 > CH_2 = CH_2 }~ \mathrm{ > CH_3 - C \equiv CH > CH \equiv CH }\) |
| 4. | \(\small\mathrm{ CH_2 = CH_2 > CH_3 - CH_3 } ~\mathrm{ > CH_3 - C \equiv CH > CH \equiv CH}\) |
1. \(A :\) \(H_{3} C - \underset{OH}{\underset{\left|\right.}{C}} = CH_{2}\) \(\) \(;\) \(B :\) \(\) \(H_{3} C - \underset{HSO_{4}}{\underset{\left|\right.}{C}} = CH_{2}\)
2. \(A :\) \(H_{3} C - \underset{O}{\underset{\left|\right. \left|\right.}{C}} - CH_{3}\) \(\) \(;\) \(B :\) \(\) \(H_{3} C - C \equiv CH\)
3. \(A :\) \(H_{3} C - \overset{OH}{\overset{\left|\right.}{C}} = CH_{2}\) \(\) \(;\) \(B :\) \(\) \(H_{3} C - \overset{O}{\overset{\left|\right. \left|\right.}{C}} - CH_{3}\)
4. \(A :\) \(H_{3} C - \underset{HS O_{4}}{\underset{\left|\right.}{C}} = CH_{2}\) \(\) \(;\) \(B :\) \(\) \(H_{3} C - \underset{O}{\underset{\left|\right. \left|\right.}{C}} - CH_{3}\)
Consider the given equilibrium constants:
\(\mathrm{{N}_2+3 {H}_2 ~\rightleftharpoons 2 {NH}_3 ~~~~~~~{K}_1 \\ {N}_2+{O}_2 ~~~\rightleftharpoons 2 {NO} ~~~~~~~~{K}_2 \\ {H}_2+\frac{1}{2} {O}_2 \rightleftharpoons {H}_2 {O} ~~~~~~~~{K}_3}\)
The equilibrium constant (K) of the following reaction will be:
\(2 \mathrm{NH}_3+\frac{5}{2} \mathrm{O}_2 \stackrel{\mathrm{~K}}{\rightleftharpoons} 2 \mathrm{NO}+3 \mathrm{H}_2 \mathrm{O}\)
| 1. | \( \dfrac{K_{2}K_{3}^{3}}{K_{1}}\) | 2. | \( \dfrac{K_{2}K_{3}}{K_{1}}\) |
| 3. | \( \dfrac{K_{2}^{3}K_{3}}{K_{1}}\) | 4. | \( \dfrac{K_{3}^{3}K_{1}}{K_{2}}\) |
The correct increasing order of basic strength of the given molecules is:
 |
 |
 |
| (I) | (II) | (III) |
1. III < I < II
2. III < II < I
3. II < I < III
4. II < III < I
Ionic mobility of which of the following alkali metal ions is lowest when aqueous solution of their salts are put under an electric field?
1. K
2. Rb
3. Li
4. Na
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