Chemistry Test - 5

\(\lambda=\frac{h}{p}\)

or \(\Delta x=\frac{h}{p}\)

By uncertainty principle \(\Delta x \cdot \Delta p=\frac{h}{4 \pi}\)

or \(\frac{h}{p} \times \Delta p=\frac{h}{4 \pi}\)

\(\frac{\Delta p}{p}=\frac{1}{4 \pi}\)

\(\frac{m \times \Delta v}{m \times v}=\frac{1}{4 \pi}\)

\(\frac{\Delta v}{v}=\frac{1}{4 \pi}\)

\(\%\) uncertainty in velocity \(=\) \(\frac{\Delta v}{v} \times 100=\frac{1}{4 \pi} \times 100=8 \%\)

Hence option B is the correct answer.

During this process, chromyl chloride is formed which is deep red coloured fuming liquid.

\(4 N_{\alpha} C l+K_{2} C r_{2} O_{7}+H_{2} S O_{4} \longrightarrow 2 C r O_{2} C l_{2}+4 N_{\alpha} H S O_{4}+2 K H S O_{4}+3 H_{2} O\)

The vapour of \(\mathrm{CrO}_{2} \mathrm{Cl}_{2}\) when passed into \(\mathrm{NaOH}\) solution gives a yellow solution of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) \(\mathrm{CrO}_{2} \mathrm{Cl}_{2}+4 \mathrm{NaOH} \rightarrow \mathrm{Na}_{2} \mathrm{CrO}_{4} \mathrm{O}+2 \mathrm{NaCl}+2 \mathrm{H}_{2} \mathrm{O}\)

Hence option D is the correct answer.

Step 1:

HCN attacks the carbonyl and forms hydrocyanin compound.

Step 2

LiAlH \(_{4}\) reduces \(-C N\) to \(-C H_{2}-N H_{2}\)

Step 3:

\(N a N O_{2} /\)

\(H^{8}\)

creates carbocation which then expands and aet stabilized.

Hence, option (A) is correct.

\(\operatorname{Agcl}(s) \rightleftharpoons A g^{\ominus}+C l^{\ominus}\)

\(\Delta H^{\circ}=\Delta_{f} H^{\circ}\left(A g^{+}\right)+\Delta_{f} H^{\circ}\left(C l^{-}\right)-\Delta_{f} H^{\circ}(A g C l)\)

\(=25.3-40+30.36=15.7 k\) cal \(/\) mol

\(\Delta S^{\circ}=S_{A g^{+}}^{\circ}+S_{C^{\prime}}^{0}-S_{A g}^{\circ}\)

\(\Delta S=17.7+13.2-23.0=7.9\) cal \(/\) mol

\(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}=13350\) cal

\(\Delta G^{\circ}=-2.303 R T \log k\)

\(-\log k=9.79\)

\(\log k=-9.79\)

\(\log k=-10+0.21\)

\(k=\) Antilog \((-10+0.21)\)

\(k=(\) Antilog 0.21\() \times 10^{-10}\)

\(k=1.6 \times 10^{-10}\)

Hence option A is the correct answer.

The following two factors are responsible for increase in the rate of surface-catalysed reactions:

1. Adsorption is an exothermic process and the heat of adsorption of the reactants on the catalyst helps the reactant molecules to acquire the activation energy for the reaction.

2. Adsorption increases the local concentration of reactant molecules on the surface of the catalyst.

Hence option C is the correct answer.

Cinnabar (HgS) is Ore of Hg.

Siderite is FeCO₃

Malachite is Cu₂CO₃(OH)₂

Hornsilver is AgCl

Hence, option (B) is correct.

Iron does not form amalgams. Amalgam is an alloy of mercury and other metal like zinc, copper and magnesium.

Alloys are substances that possess metallic character and are obtained by mixing metal with another metal or another element. Either the mixture/alloy of mercury \(({Hg})\) with any other metal is called an amalgam whereas Tungsten, Platinum, Iron, and Tantalum are exceptions that do not form an amalgam.

Frenkel defect is shown by ionic solids. The smaller ion (usually cation) is dislocated from its normal site to an interstitial site. Thus, it is a dislocation effect and is usually favoured by large size difference between cation and anion.

F-centre is formed when anionic vacancies are occupied by electrons.

Schottky defect is a vacancy defect, where equal number of cations and anions are missing. The presence of Schottky defect decreases the density of the crystals.

Hence option B is the correct answer.

In BF₃, due to F back bonding to B, least Lewis acid character is observed. So out of above four, BF₃ will have least tendency to accept electron from NH₃.

Hence, option (D) is correct.

RbCl has non-directional bonds.

Ionic bonds are non-directional in nature because they are formed by the positive charge of one species and the negative charge of another species. As charges are non-directional in nature, so the bonds formed by them are also non-directional. RbCl forms an ionic bond that is formed between Rb⁺ and Cl^- ions.

So, RbCl has non-directional bonds.

Hence the correct option is (B).

Statement (2), (3) and (4) are correct as:

(2) F and E, G are tautomers. These are isomers of organic compounds that readily interconvert in their aqueous solutions by the movement of the α-hydrogen. The phenomenon is called tautomerisation.

E is the keto form and F and G are the en-ol forms.

(3) F and G are geometrical isomers.

F is the cis isomer and G is the trans isomer.

(4) Diastereomers are the stereo-isomers which are not mirror images of each other.

F and G are geometrical isomers and are diastereomers.

Hence option D is the correct answer.

\(O S O_{4}\) is used for syn addition of \(-O H\) groups across the double bond.

Let

\(-O H\) is at position (1) is above the plane, \(s o-O H\) addition will take above the plane. In this compound, attack on double bond will be in syn manner and from below the plane. This will result in dash configuration which would further avoid steric hindrance.

Hence option A is the correct answer.

\(K O_{2}\) is paramagnetic due to \(O_{2}^{\ominus}\)

(superoxide ion) \(2 K O_{2}+S \rightarrow\)

\(K_{2} S O_{4} \frac{B a C l_{2}}{W h i t e p p t .}+B a S O_{4} \downarrow+2 K C l\)

Hence, (C) is correct options.

3-buten-2-one

(i)

(ii) Intramolecular old of happens

Hence option C is the correct answer.

\(K_{4}\left[F e(C N)_{6}\right]+6 H_{2} S O_{4}+6 H_{2} O \rightarrow 2 K_{2} S O_{4}+F e S O_{4}+3\left(N H_{4}\right)_{2} S O_{4}+6 C O\)

\(K_{4}\left[F e(C N)_{6}\right]+3 H_{2} S O_{4} \longrightarrow 2 K_{2} S O_{4}+F e S O_{4}+6 H C N \uparrow\)

Hence option B is the correct answer.