Chemistry Test - 1

No. of different transition from \(n=n_{1}\) to \(n=n_{2}\) is given by \({ }^{n} 1 C_{n_{2}}\)

So for 6 different wavelength, no. of different transition \(=6\)

\({ }^{n}{ }_{1} C_{n_{2}}=6\) then \(n_{1}=n_{2}+3\)

If \(\mathrm{n}\) is the principal quantum number of energy level corresponding to energy \(\mathrm{X}\), the principal quantum number of energy level \(\mathrm{Y}\) is \((\mathrm{n}+\)

3) as it emit 6 wavelengths.

Now Energy of an orbit is directly proportional to square of the orbit no.(principlal quantum no.)

So, \(\frac{x}{Y}=\frac{(n+3)^{2}}{n^{2}} \Rightarrow \sqrt{\frac{X}{Y}}=1+\frac{3}{n}\)

Hence, The correct answer is A.

\(\mathrm{A}(\mathrm{g})+3 \mathrm{B}(\mathrm{s}) \Leftrightarrow 2 \mathrm{C}(\mathrm{g})+\mathrm{D}(\mathrm{s}),+\) Heat

\(K_{p}=\frac{(P c)^{2}}{P_{A}}\)

\(\therefore P_{C} \propto \sqrt{P_{A}}\)

Hence, if \(\mathrm{P}_{\mathrm{A}}\) is doubled, the \(\mathrm{P}_{\mathrm{C}}\) increases by a factor of \(\sqrt{2}\)

Note: However, if the total pressure had been increased, the change would then favour the backward reaction and \(P_{C}\) would then decrease. The reaction proceeds with an increase in the number of gaseous moles \(\left(\Delta \eta_{9}>0>\right) K_{p}=K_{c}(R T)^{2 n_{0}}\)

since, \(\Delta \eta_{0}=2-1=1\)

\(\therefore K_{p}=K_{c}(R T)\)

ie. \(K_{p}>K_{c}\)

The forward process being exothermic, increase in temperature will favour the backward reaction, but both \(\mathrm{K}_{\mathrm{p}}\) as well as \(\mathrm{K}_{\mathrm{C}}\) will decrease.

Hence, option (4) is correct.

Hydrogen gas is evolved when ethanol reacts with sodium.

\(2 {Na}+2 {CH}_{3} {CH}_{2} {OH} \rightarrow 2 {CH}_{3} {CH}_{2} {O}{Na}+{H}_{2}({g})\)

Alcohols react with sodium leading to the evolution of hydrogen. With ethanol, the other product is sodium ethoxide \(({CH}_{3} {CH}_{2} {ONa})\).

\(10 \mathrm{ml}\) of \(0.2 \mathrm{NHCl}+30 \mathrm{ml}\) of \(0.1 \mathrm{NHCl}=40 \mathrm{ml}\) of

\(\mathrm{NaOH}(0.61 \mathrm{g}\) organic acid \()\)

meq. of \(H C l=\) meq of \(N a O H=\) meq of organic acid \(10 \times 0.2+30 \times 0.1=\frac{0.61}{E} \times 1000\)

\(5=\frac{0.61 \times 1000}{E}\)

\(E=\frac{160}{5}=122\)

Hence, The correct answer is C.

Cannizzaro reaction is a chemical reaction that involves the base-induced disproportionation of an aldehyde lacking a hydrogen atom in the alpha position.

The oxidation product is a salt of a carboxylic acid and the reduction product is an alcohol.

The Reimer Tiemann reaction is a chemical reaction used for the ortho-formylation of phenols with the simplest example being the conversion of phenol to salicylaldehyde.

From the above examples it is clear that in Cannizzaro reaction C-C bond formation won't takes place.

Hence option D is correct.

The meta product can be controlled by protecting the –NH₂ group by acetylation reaction with acetic anhydride, the nitration reaction can be controlled and the p-nitro derivative can be obtained as the major product. The reaction then becomes:

\(z=1+\frac{0.0169 \times 600}{0.083 \times 500}=1.244\)

\(\therefore \quad V_{m}=\frac{Z R T}{P}=\frac{1.244 \times 0.083 \times 500}{600}\)

or \(V_{m}=8.6 \times 10^{-2} \mathrm{L}\)

Hence, The correct answer is C.

Tautomerism is spontaneous interconversion of two isomeric forms with different functionalgroups. The prerequisite for this is the presence of the \(\mathbf{C}=\mathbf{O}, \mathbf{C}=\mathbf{N}\) or \(\mathbf{N}=\mathbf{O}\) in the usual cases andan alpha \(\mathrm{H}\) atom. In case of keto enol tautomerism, the keto form is more stable. Enols can beformed by acid or base catalysis from the ketone and are extensively used in making \(\mathrm{C}-\mathrm{C}\) singlebonds in organic synthesis, Compound in ( 1 ), ( 3 ) and (4) exhibit tautomerism as shown below.

Hence options A ,B and C are the correct answers.

tautomerism is not possible

Benzoquinone is highly stable due to conjugation and does not exhibit tautomerism.

We know that \(K=a n t i \log \left(\frac{\Delta G^{o}}{2.303 R T}\right) \cdots(i)\)

Given that, \(\Delta G^{\circ}=-81 . k J / m o l, R=8.314 \times 10^{-3} k J K^{-1} \quad m o l^{-1}, T=1000 K\)

Substituting these values in eq. (i) we get \(K=\operatorname{anti} \log \left(\frac{-(-8.1)}{2.303 \times 8.314 \times 10^{-3} \times 1000}\right)=2.65\)

Hence, The correct answer is D.

Option (1) is incorrect as cold alkaline solution of KMnO4 would produce vicinal diols.

Option (2) is correct as KMnO4 is stronger oxidising agent and in the acidic medium, it oxidises the alkene by breaking the carbon-carbon double bond and replacing it with two carbon-oxygen double bonds.

Option (3) is correct as the process is ozonolysis by ozone, which is arrested at the aldehyde stage by adding zinc water.

1) Number of Ba ions per unit cell = 1/8 x 8 = 1

Number of Ti ions per unit cell = 1 x 1 = 1

Number of O ions per unit cell = 1/2 x 6 = 3

Hence, the empirical formula of the substance should be BaTiO₃.

Thus, (1) is correct.

2) Titanium ions can be assumed to occupy a hole in the lattice formed by barium and oxide ions. The titanium ions occupy the body centres of the face centred cubic and this is one of the octahedral holes in the fcc lattice.

So, (2) is also true.

3) The octahedral holes at the centres of unit cell constitute just one-fourth of all the octahedral holes in a face centred cubic lattice. 

Thus, (3) is also true.

In acidic medium \(M n O_{4}^{-}\) acts as strong oxidizing agent and itself is reduced to \(\mathrm{Mn}(II)\) ions. In the acidic medium, \(F e^{2+}\) acts as a reducing agent and itself is oxidized to \(F e^{3+}\). Hence, the colour changes from dark violet to light yellow when dil \(\mathrm{HCl}\) is added to a mixture of \(\mathrm{MnO}_{4}^{-}\) and \(\mathrm{Fe}^{2+}\).

Hence, The correct answer is A.

The equilibrium constant \(K_{p}\) for the reaction \(A(g) \rightleftharpoons B(g)+C(g)\)

\(K_{p 1}=1\) at \(300 k ; K_{p 2}=4\) at \(320 k\)

\(\log _{10} \frac{4}{1}=\frac{\Delta H}{2.303 R}\left[\frac{1}{300}-\frac{1}{320}\right]\)

\(0.6=\frac{\Delta H}{2.303 \times 2} \times \frac{20}{300 \times 320}\)

\(\Delta H=13.31 \mathrm{Kcal} / \mathrm{mole}\)

for \(B(g)+C(g) \rightleftharpoons A(g)\)

\(\Delta H=-13.31 K\) cal \(/\) mole.

Hence, The correct answer is A.

As we move down the group alkaline earth metal carbonates require more heating to decompose, so carbonates become more thermally stable down the group.

\[\mathrm{BaCO}_{3}>\mathrm{SrCO}_{3}>\mathrm{CaCO}_{3}>\mathrm{MgCO}_{3}\]

Hence, The correct answer is C.

Hence, The correct answer is B.