Chemistry Test - 2

Bond order \(=\frac{\text { Number of bonds }}{\text { Number of Resonating structures }}=\frac{5}{4}=1.25\) In a given resonance structure, the 0 atom that forms double bond has formal charge of 0 and the remaining 30 atoms have formal charge of -1 each. In the resonance hybrid, a total of -3 charge is distributed over 40 atoms. Thus the formal charge of each 0 atom is \(\frac{-3}{4}=-0.75\)

Note:

To calculate the formal charge, the following formula is used. Formal Charge \(=[\) Number of valence electrons on atom \(]-[\) non-bonded electrons \(+\) number of bonds \(]\) For 0 atom that forms double bond with \(P\) atom, Formal Charge \(=6]-[4+2]=0\)

For 0 atom that forms single bond with \(P\) atom, Formal Charge \(=6]-[6+1]=-1\)

Hence, The correct answer is C.

Reaction mechanism is given below:

So ,(1) is correct option.

\(\mathrm{CaSO}_{4(29)}+\mathrm{BaCl}_{2(\mathrm{aq})} \rightarrow \mathrm{BaSO}_{4(\mathrm{s})}+\mathrm{CaCl}_{2(\mathrm{a})}\)

\(136 \mathrm{g}\) CaSO \(_{4}\) produces \(233 \mathrm{g} \quad \mathrm{BaSO}_{4}\)

Weight of \(\mathrm{BaSO}_{4}=0.617 \mathrm{g}\)

Mass of \(\mathrm{CaSO}_{4}\) reacted \(=\frac{0.617 \times 136}{233}=0.3601 \mathrm{g}\)

Percentage purity of \(\mathrm{CaSO}_{4}=\frac{0.3601 \times 100}{0.400}=90.025\)

Hence, The correct answer is B.

\(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}()\)

\(\mathrm{At} 300 \mathrm{K} \Delta \mathrm{H}=-336.2 \mathrm{kcal}\)

\(\Delta \mathrm{n}_{9}=2-(1+3)=-2\)

\(\mathrm{T}=300 \mathrm{K}, \Delta \mathrm{H}=-336.2 \mathrm{kcal}\)

\(\mathrm{R}=2 \times 10^{-3} \mathrm{kcal}\) degree \(^{-1} \mathrm{mol}^{-1}\)

\(\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_{9} \mathrm{RT}\)

\(-336.2=\Delta \mathrm{U}+(-2) \times 300 \times 2 \times 10^{-3}\)

\(\Delta \mathrm{U}=-336.2+1.2=-335.0 \mathrm{kcal}\)

Hence, The correct answer is B.

\(\frac{r^{+}}{r}=\sqrt{\frac{3}{2}}-1=0.225\) suggests that the cation is occupying the tetrahedral void. Therefore,the coordination number of the cation is 4 and the option (1) is correct.

The anions will form the lattice (ccp or hcp) and in both the cases, they will touch along the face diagonal. Hence, option (2) is also correct.

There are two tetrahedral voids on each body diagonal, and in the given compound, one of them is occupied. Hence, option (3) is correct.

When the concentration of \(\mathrm{B}\) alone was doubled, the rate did not change. Hence the reaction is zero order in B. When the concentration of A alone was doubled, the rate increased by two times. Hence, the reaction is of first-order in A. \(r=k[A]^{1}[B]^{0}\)

The overall order of the reaction is \(1 .\)

For the first-order reaction, the unit of \(k\) is \(\sec ^{-1}\)

Hence, The correct answer is A.

Hence, The correct answer is D.

trans-2-butene is more stable than 1-butene due to the presence of alkyl groups, (CH3^-)^- which have a +ve inductive effect (they are slightly electron releasing, they push electrons toward the double bond, which helps stabilise it). So, the more substituted the C's of the double bond, the more stable the bond. According to given graph, trans-2-butene is more stable than cis-2-butene by:

30.3 kcal - 27.6 kcal = 2.7 kcal,

Hence options B is the correct answer.

\(\frac{1}{2} N_{2(g)}+\frac{3}{2} H_{2(g)} \rightleftharpoons N H_{3(g)}\)

Partial pressure of \(N H_{3}=\frac{24}{100} \times 100\)

\(=24.0 a t m\)

Pressure of \(N_{2}+H_{2}=100-24=76 a t m\)

Partial pressure of \(H_{2}=\frac{3}{4} \times 76=57 a t m\)

Partial pressure of \(N_{2}\) \(=\frac{1}{4} \times 76=19\) atm

\(K_{p}=\frac{P_{N H_{3}}}{p_{N_{2}}^{1 / 2}+p_{H_{2}}^{3 / 2}}\)

\(=\frac{24}{(19)^{1 / 2} \times(57)^{3 / 2}}\)

\(=\frac{24}{4.358 \times(7.549)^{3}}\)

\(=\frac{24}{4.358 \times 430}=0.0128 a t m^{-1}\)

Hence, The correct answer is A.

In SO₂, the oxidation state of S is +4. It can vary its oxidation state between -2 and +6 in its compounds and act both as an oxidising agent as well as a reducing agent.

In MnO₂, the oxidation state of Mn is +4. It can vary its oxidation state between +2 and +7 in its compounds and act both as an oxidising agent as well as a reducing agent.

In CrO, the oxidation state of Cr is +2. It can vary its oxidation state between +1 and +6 in its compounds and act both as an oxidising agent as well as a reducing agent.

Hence options D is the correct answer.

All ionic carbides, on hydrolysis, yield hydrocarbons, while covalent carbides, on hydrolysis, do not yield hydrocarbons. Out of given, Sic is covalent, while rest are ionic carbides. Thus, \(\mathrm{CaC}_{2}, \mathrm{Be}_{2} \mathrm{C}\) and \(\mathrm{Al}_{4} \mathrm{C}_{3}\) yield hydrocarbons on hydrolysis as shown below:

\(\mathrm{CaC}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Ca}(\mathrm{OH})_{2}+\mathrm{C}_{2} \mathrm{H}_{2}\)

\(\mathrm{Be}_{2} \mathrm{C}+4 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{Be}(\mathrm{OH})_{2}+\mathrm{CH}_{4}\)

\(\mathrm{Al}_{4} \mathrm{C}_{3}+12 \mathrm{H}_{2} \mathrm{O} \rightarrow 4 \mathrm{Al}(\mathrm{OH})_{3}+3 \mathrm{CH}_{4}\)

Hence option D is the correct answer.

Ag⁺¹ ion has 18 valence electrons so it is more covalent due to greater polarizability on anion.

Hence, The correct answer is B.

A) Acidic nature depends on elecronegativity. In the periodic table, the trend in E.N is \(N>P>\) As.Thus the trend in acidic nature is reverse of that of E.N and can be written as; \(A s H_{3}>\) \(P H_{3}>N H_{3}\)

B) lonic radii increases on moving down a group from top to bottom, therefore the trend in ionic size is; \(C s^{+}>K^{+}>N a^{+}>L i^{+}\)

D) lonisation potiential increases on moving from left to right in a period.But first I.E of Be is more than \(\mathrm{B}\); because Be has stable fully filled valence orbital \(\left(2 s^{2}\right)\) Therefore the trend in I.E is ; \(C>B e>B>L i\)

Hence, The correct answer is D.

The ionization energy is given to be \(=2.18 \times 10^{-18} \mathrm{J},\) which equals \(13.6 \mathrm{eV}\). Therefore, the energy of electron in the first level is \(-13.6 \mathrm{eV}\). We can thus use the formula, \(E=-13.6 \times \mathrm{Z}^{2} / \mathrm{n}^{2}\) to

calculate energy of electron in the second orbit. \(E_{2}=\frac{-2.18 \times 10^{-18}}{2^{2}}=-5.45 \times 10^{-19} J\)

Hence, The correct answer is D.

Aromatic aldehydes do not give the Fehling test.

[Fructose, or fruit sugar, is a simple ketonic monosaccharide found in many plants, where it is often bonded to glucose to form the disaccharide sucrose. It is one of the three dietary monosaccharides, along with glucose and galactose, that are absorbed directly into blood during digestion.]

Hence, The correct answer is C.