Chemistry Test-13

The defects in which stoichiometry of a crystal is not affected, are called stoichimetric defects. Thus, in such defects, the ratio of positive and negative ions are same as indicated by chemical formula of the compound.

Number of hybridized orbitals \(=\frac{1} {2}\) (group number of central atom + number of monovalent atoms)

Number of hybridized orbitals \(=\frac{1}{ 2}(8+6)=7\)

Hence, by VSEPR theory its hybridization is \(\mathrm{sp}^{3} \mathrm{~d}^{3}\).

It has only \(6\) bonded atoms, that result in its shape as a distorted octahedron.

Zr(40) and Hf(72) have similar properties as they belong to the same group and have the same valence shell electronic configuration.Electronic configuration of \(\mathrm{Zr}(40)\) is \([\mathrm{Kr}] 5 \mathrm{~s}^{2} 4 \mathrm{~d}^{2}\) and of \(\mathrm{Hf}(72)\) is \([\mathrm{Xe}] 6 \mathrm{~s}^{2} 4 \mathrm{f}^{14} 5 \mathrm{~d}^{2}\).

As we know that the energy,

\(\mathrm{E}=\mathrm{hv}=\frac{\mathrm{hc}}{\lambda}\)

Also, the energy of absorbed photon = Sum of the energies of emitted photons

So,

\(\mathrm{E}=\mathrm{E}_{1}+\mathrm{E}_{2}\)

\(\frac{h c}{\lambda}=\frac{h c}{\lambda_{1}}+\frac{h c}{\lambda_{2}}\)

Given,

\(\lambda=355 \mathrm{~nm}, \lambda_{1}=680 \mathrm{~nm},\)

Let \(\lambda_{2}=x \mathrm{~nm}\)

So,

\(\frac{h c}{355}=\frac{h c}{680}+\frac{h c}{x}\)

\(\Rightarrow \frac{1}{355}=\frac{1}{680}+\frac{1}{x}\)

\(\Rightarrow \frac{1}{x}=\frac{1}{355}-\frac{1}{680}\)

\(\Rightarrow \frac{1}{x}=\frac{680-355}{355 \times 680}\)

\(\Rightarrow \frac{1}{x}=0.00134\)

\(\Rightarrow x=\frac{1}{0.00134}=746.2 \mathrm{~nm}\)

That is near to \(743 \mathrm{~nm}\).

\(\therefore\) The other wavelength is \(743 \mathrm{~nm}\).

Given,

Mass of solid = \(6 \mathrm{~g}\)

Molar mass of solid \(=72 \mathrm{~g}\)

Moles of solid \(=\frac{6} {72}=0.0834 \mathrm{~mol}\)

Volume of water \(=30 \mathrm{ml}=30 / 1000 \mathrm{~L}=0.03 \mathrm{~L}\)

Molarity = Moles of solute/Volume of solution

Molarity \(=\frac{0.0834}{0.03}=2.78 \mathrm{M}\)

\(\therefore\) The molar concentration of the solution is\(2.78 \mathrm{~mol} / \mathrm{L}\).

Alkali metals exist as hydrated ions in an aqueous medium.

Degree of hydration decrease the ionic size down the group.

This result in maximum hydration for Li⁺ and minimum hydration for Cs⁺.

As mobility is reverse of the size of ions.

Therefore, the correct order of mobility of hydrated ions of alkali metal in aqueous medium is:

Cs⁺ >Rb⁺ >K⁺ >Na⁺ >Li⁺

Metallic or non-stoichiometric hydrides can only be formed by d-block and f-block elements like Pt, Pd, La, Y etc.

Rest of the elements form either ionic or molecular hydride.

As we know,

\(E_{c e l l}=E_{c e l l}^{0}-\frac{0.0591}{n} \log Q\)

Given,

\(\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(0.1 \mathrm{M}) \rightarrow \mathrm{Zn}^{2+}(1 \mathrm{M})+\mathrm{Cu}(\mathrm{s})\)

\(Q=\frac{\left[Z n^{2+}\right]}{\left[C u^{2+}\right]}=\frac{1}{0.1}=10\)

\(E_{c e l l}=1.10-\frac{0.0591}{2} \log 10\)

\(E_{c e l l}=1.10-0.0295\)

\(E_{c e l l}=1.0705 \mathrm{~V}\)

Given,

\(SO_{2}+\frac{1}{2} O_{2} \rightleftharpoons SO_{3} \cdots \cdots (A)\)

Thus, \(k_{1}=\frac{\left[SO_{3}\right]}{\left[SO_{2}\right]\left[O_{2}\right]^{\frac{1}{ 2}}}\)

\(2SO_{3} \rightleftharpoons 2 SO_{2}+O_{2} \ldots \ldots (B)\)

Thus, \(k_{2}=\frac{\left[SO_{2}\right]^{2}\left[O_{2}\right]}{\left[SO_{3}\right]}\)

\(k_{2}=\frac{1}{k_{1}^{2}}\)

\(k_{2}=\left(\frac{1}{k_{1}}\right)^{2}\)

The addition polymers are formed by the repeated addition of monomer molecules possessing double or triple bonds, e.g., the formation of polythene from ethene. The given molecule is ethene.

Peptide bond is planar covalent and has partial double bond character.The shown figure represents a peptide bond. The partial double bond character is due to resonance.

A mixture of 1°, 2° and 3° amines can be separated by Hinsberg’s method.

Mixture of primary, secondary and tertiary amines can be separated by  Hinsberg's method and fractional distillation.

They give different products with Hinsberg's. These products differ in solubility. Hence, they can be separated with Hinsberg's  method. They differ in boiling points. Hence they can be separated by fractional distillation.

Substitution of the hydroxyl hydrogen atom is even more facile with phenols, which are roughly a million times more acidic than equivalent alcohols.This phenolic acidity is further enhanced by electron-withdrawing substituents ortho and para to the hydroxyl group. It is noteworthy that the influence of a nitro substituent is over ten times stronger in the para-location than it is meta, despite the fact that the latter position is closer to the hydroxyl group. Furthermore additional nitro groups have an additive influence if they are positioned in ortho or para locations. In substituted phenols, the presence of electron withdrawing groups enhance the acid strength; while electron releasing groups decrease the acid strength.

In the given complex, the ligand is oxalate ion and the central metal ion is nickel. As we mentioned, coordination number is the number of ligand donor atoms to which the metal is directly attached. Since oxalate ion \(\left(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right)\) is a bidentate ligand, it donate two electron pairs to nickel . Also there are three oxalate ion present in \(\left[\mathrm{Ni}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{4-}\). So the total number of atoms attached is

Coordination number \(=2 \times 3\)

\(=6\)

(A) In this process, carbon anode is oxidised to CO and CO₂.

(B) It is a fact.

(C) At cathode, Al³⁺ from Al₂O₃ is reduced to Al.

(D) Al₂O₃ is the electrolyte, which is undergoing the redox process. So, Al₂O₃ serves as electrolyte and Na₃AlF₆, although an electrolyte, serves as solvent.